Analytic Geometry for Colleges, Universities & Schools eBook
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ll
mmill
AfTRONOMY
ANALYTIC GEOMETRY FOR
COLLEGES, UNIVERSITIES, AND TECHNICAL SCHOOLS.
BY E.
VV.
NICHOLS,
PROFESSOR OF MATHEMATICS IN THE VIRGINIA MILITARY INSTITUTE.
LEACH, SHEWELL, & SANBORN, BOSTON. NEW YORK. CHICAGO.
COPYRIGHT, iwi,
BY LEACH, SIIKWELL, & SANBOKN.
C.
J.
PETERS & SON,
TYPOGRAPHERS AND ELECTROTYPERS. PRESS OF BERWICK
&
SMITH.
Nsf PEEFACE. THIS text-book
is
designed
and Technical Schools. to prepare a
make
it
work
for
The aim
for beginners,
Colleges,
of
and
the
Universities,
author has been
at the
same time to
for the requirements
of comprehensive For the methods of develop ment of the various principles he has drawn largely upon his sufficiently
the usual undergraduate course.
experience in the class-room. all
home and
authors,
In the preparation of the work
foreign,
have been freely consulted. In the first few chapters discussion of each principle.
whose works were
available,
elementary examples follow the In the subsequent chapters
sets
of examples appear at intervals throughout each chapter, and are so arranged as to partake both of the nature of a review and an extension of the At the preceding principles.
end of each chapter general examples, involving extended application of the are principles deduced,
the benefit of those
who may
a more
placed for
desire a higher course in the
subject.
The author takes pleasure cussion
of
Surfaces,"
Mathematics
in
appears as the
He
by
in calling attention to a
and Lee University, chapter in this work.
Washington
final
"Dis
A. L. Nelson, M.A., Professor of
which
takes pleasure also in acknowledging his indebtedness
i
PREFACE.
v
Prof. C. S. Venable, LL.D., University of Prof. William Cain, C.E., University
of
to
and
Virginia, to
North
Carolina,
of Pennsylvania, to Prof. E. S. Crawley, B.S., University
for assistance rendered in reading
and revising manuscript,
and for valuable suggestions given. E.
LEXINGTON, YA. January, 1893.
W. NICHOLS.
CONTENTS. PAET
I.
PLANE ANALYTIC GEOMETRY. CHAPTEK
I.
CO-ORDINATES.
PAGES
ARTS. 1-3.
4-6.
The Cartesian or Bilinear System. The Polar System. Examples
CHAPTER
Examples
....
1
4
II.
LOCI. 7.
Locus of an Equation.
8.
Variables.
9.
10-16. 17, 18.
The Equation
Constants. Examples Relationship between a Locus and
of a
...
9 10
its
Equation
Discussion and Construction of Loci. Methods of Procedure. Examples
CHAPTER
Locus
....
11
...
11
Examples
23
III.
THE STRAIGHT LINE.
21.
The Slope Equation. Examples The Symmetrical Equation. Examples The Normal Equation
22.
Perpendicular Distance of a Point from a Line.
19.
20.
25 29 32
Ex
26.
amples Equation of Line, Axes Oblique. Examples General Equation, Ax + B?y + C = Equation of Line passing through a Point. Examples Ex Equation of Line passing through Two Points.
27.
Length
23.
24.
25.
37 .
38 39
amples of
33 35
Line joining
Two
Points.
v
Examples
...
41
CONTENTS.
vi ARTS. 29. 30.
Intersection of
Two
PAGES
....... = ......
.
28.
Examples
Lines.
Ax + By + C + K(A a5 + BV + e ) Examples. Angle between Two Lines.
Ex
General
.................
amples
CHAPTER
IV.
TRANSFORMATION OF CO-ORDINATES. 31. 32. 33*
34 35
Objects
of.
............
Illustration
Examples to a Parallel System. an Oblique System. Rectangular Rectangular System to also Rectangular. Examples. System to Another System Polar a to System. From a Polar Rectangular System Examples. General System to a Rectangular System.
From One System
Examples
....... ......... CHAPTER THE
V.
CIRCLE.
Equation of Circle
38.
Generation of Circle. General Equation of
39.
Polar Equation of Circle
36, 37.
~rv
ou
.
.
Circle.
....
Concentric
Circles.
59
Ex
................. ............ Supplemental Chords ............. Tangent. Sub-tangent ............ Normal. Sub-normal ............. General Equations Tangent and Normal. Examples Tangent .............. Length amples
40.
41. 42. 43. 44.
45, 46.
of
of
Radical Axis.
Radical Centre.
Examples
.....
47.
Condition that a Straight Line touch a Circle.
48.
Equation of Tangent Chord of Contact Pole and Polar
49, 50. 51.
^7
.
............ ............... .....
Conjugate Diameters.
Examples.
CHAPTER
General Examples
.
VI.
THE PARABOLA. 52, 53.
70
Slope
Generation of Parabola. tions
Equation of Parabola.
Defini
................ ........... Parabola
54.
Construction of
55.
Latus-Rectum.
Examples
77
CONTENTS.
Vil
PAGES
ARTS. 57-59.
Polar Equation of Parabola Tangent. Sub-tangent. Construction of Tangent
60, 61.
Normal.
Sub-normal
62.
Tangents
at Extremities of
63.
ici
56.
= +
88 .
.
89
90
Latus-Rectum
91
92
64.
Examples y% Tangent Line makes Equal Angles with Focal Lines and Axis Condition that a Straight Line touch the Parabola.
95
65.
66.
67.
68. 69. 70.
71, 72. 73.
+
c$
Slope
Equation of Tangent Locus of Intersection of Tangent and Perpendicular from Focus Locus of Intersection of Perpendicular Tangents Chord of Contact Pole and Polar Conjugate Diameters Parameter of any Diameter. Equation of a Diameter
...
.
at the Extremities of a Chord. General Examples
Tangents
CHAPTER
95 96 97
97 98 98 100
Examples. 101
VII.
THE ELLIPSE. 74, 75.
76. 77.
Generation of Ellipse. Definition. Equation of Ellipse Focal Radii
78.
Construction of Ellipse
Latus-Rectum.
Polar Equation of Ellipse
81.
Supplemental Chords Tangent. Sub-tangent Tangent and Line through Centre
85.
86,87. 88.
89.
90.
91. 92. 93.
94.
108 Ill
79.
84.
106 109
80.
82,83.
.
Eccentricity.
Examples .
.
.
114 115
Point
of
Tangency and
H8
...
Methods of constructing Tangents Normal. Sub-normal. Examples Normal bisects Angle between the Focal Radii Condition that a Straight Line touch the Ellipse.
...
118 119
....
122
Slope
Equation of Tangent Locus of Intersection of Tangent and Perpendicular from Focus Locus of Intersection of Perpendicular Tangents Equation of Chord of Contact Pole and Polar Conjugate Diameters .
.
.
123
124 125 125
126 126
CONTENTS.
via ARTS. 95, 96.
97. 98. 99.
PAGK8 Equation of a Diameter. Conjugate Diameter 2 a"
+
b
-
=
a*
+
Co-ordinates of Extremities of
129
b*
.
130
.
...
Parallelogram on a pair of Conjugate Diameters Relation between Ordinates of Ellipse and Circles on
Axes 100, 101.
131
133
Construction of Ellipse. General Examples
Area
CHAPTER
of Ellipse.
Examples. 134
VIII.
THE HYPERBOLA. 102,103. 104, 105.
100, 107. 108.
109. 110. 111.
112,113. 114.
Generation of Hyperbola. Definitions. Equation of Hyperbola Focal Radii Eccentricity. Construction of Hyperbola. Latus-Rectum Relation between Ellipse and Hyperbola
....
116, 117. 118.
119.
120.
121.
122. 123. 124. 125.
126, 127.
128. 129.
130. 131.
144
146
Conjugate Hyperbola. Examples Polar Equation of Hyperbola
146
Supplemental Chords Tangent, Sub-tangent Tangent and Line through Point of Tangency and
150
149
Centre 115.
141
143
Method Normal.
151
of constructing
Tangents
151
Sub-normal.
Examples Angle between the Focal Radii Condition that a Straight Line touch the Hyperbola. Slope Equation of Tangent Locus of intersection of Tangent and Perpendicular through Focus Locus of intersection of Perpendicular Tangents Chord of Contact Pole and Polars Tangent
bisects
.
.
.
.
.
Conjugate Diameters Conjugate Diameters lie in the same Quadrant Co-ordinates of Equation of Conjugate Diameter. Extremities of Conjugate Diameter 2 = a12 a2 b2 Parallelogram on a pair of Conjugate Diameters. Ex
...
-
150
b"
amples Asymptotes Asymptotes as Axes. Vertex
152 154 155 155 155
156 156 156 157 157
158 159 160
Rhombus on
Co-ordinates of 162
CONTENTS.
i
ARTS. 132.
PAGES
Tangent Line, Asymptotes being Axes. The Point of Tangency Intercepts of a Tangent on the Asymptotes Triangle formed by a Tangent and the Asymptotes Intercepts of a Chord between Hyperbola and its
.............
133. 134.
135.
x
....
Examples.
General Examples
CHAPTER
165
.
165
...
165
.
Asymptotes.
164
IX.
GENERAL EQUATION OF THE SECOND DEGREE. 136, 137.
The General Equation.
138.
First Transformation.
139.
Second Transformation a = 0. c -
140, 141. 142. 143.
144. 145.
146.
....... .... ..........
Discussion
Signs of Constants
........... Summary ............. .............. = ........... ............ ..... General Summary. Examples .........
..."
6-<4ac
62
62
4 ac
>
.
.
.
4 ac
CHAPTER
170 171
172 173 175
175 176 177 173
X.
HIGHER PLANE CURVES. 147.
148.
149. 150. 151.
152.
Definition
EQUATIONS OF THE THIRD DEGREE. The Semi-cubic Parabola Duplication of Cube by aid of Parabola The Cissoid Duplication of Cube by aid of Cissoid The Witch
........... ...... ............ ................
188 190 1 91
193 194
EQUATIONS OF THE FOURTH DEGREE. 153. 154.
155. 156.
......... ..... The Limacon ............ The Lemniscate ........... The Conchoid
Trisection of an Angle by aid of Conchoid
196 198
JQQ
201
TRANSCENDENTAL EQUATIONS. 158.
The Curve The Curve
159.
The, Cycloid
157.
........... ........... ........
of Sines of
Tangents
203 204 206
CONTENTS.
X
PAGES
ARTS.
SPIRALS, 160.
Definition
161.
The The The The The
162. 163. 164.
165.
PART
208
Spiral of
Archimedes
208 210 212 213
Hyperbolic Spiral Parabolic Spiral Lituus
214
Examples
Logarithmic Spiral.
SOLID ANALYTIC GEOMETRY.
II.
CHAPTER
I.
CO-ORDINATES. 166.
The Tri-planar System.
167.
Projections
168, 169.
Length
of
Line
joining
170.
Angles. Examples The Polar System
171.
Relation
217 219
Examples
Two
Points.
Directional
220 222
between Systems. ordinates. Examples
CHAPTER
Transformation of Co 223
II.
THE PLANE. 226
172.
Equation of Plane
173.
Normal Equation
174. 175.
Symmetrical Equation of Plane General Equation of Plane
176.
Traces.
177.
178.
Perpendicular from Point on Plane Plane through three Points
179.
Any Equation between
227 229
of Plane .
.
Intercepts
229 .
three Variables.
230 231
232
Discussion.
233
Examples
CHAPTER
III.
THE STRAIGHT LINE. 180. 181.
Equations of a Straight Line Symmetrical Equations of a Straight Line
236 237
CONTENTS.
xi
ARTS.
PAGES
182.
To
183.
184.
Planes Line through One Point Line through Two Points.
185.
Intersecting Lines
242
Angle between Two Lines Angle between Line and Plane
243
247
191-193.
Transformation of Co-ordinates The Cone and its Sections
194, 195.
Definitions.
186, 187. 188. 189, 190.
find
where a given Line pierces the Co-ordinate 238 239
Examples
246
Equation of a Conic.
CHAPTER
239
Examples
.
.
.
250 254
IV.
DISCUSSION OF SUHFACES OF THE SECOND OKDER. General Equation of the Second Degree involving three Variables. Transformations and Discussion .
The Ellipsoid and varieties The Hyperboloid of One Sheet and varieties The Hyperboloid of Two Sheets and varieties The Paraboloid and varieties Surfaces of Revolution.
Examples
.
.
259 262
.... .
.
.
265 267
269 273
PLANE ANALYTIC GEOMETRY.
CO-ORDINATES.
PAST
L
CHAPTElt
I.
THE CARTESIAN OR BILINEAR SYSTEM.
of objects are determined by other some objects whose positions are referring assumed as known. Thus we speak of Boston as situated Here the ob west. in latitude north, and longitude is referred are the equator and the Boston which to jects meridian passing through Greenwich. Or, we speak of Bos ton as being so many miles north-east of New York. Here the 1.
THE
relative
them
positions
to
reference are the meridian of longitude through and New York itself. In the first case it will be each observed, Boston is referred to two lines which intersect other at right angles, and the position of the city is located when we know its distance and direction from each of these objects of
New York
lines.
In like manner, if we take any point such as P x (Fig. 1) in the plane of the paper, its position is fully determined when we know its distance and direction from each of the two lines which intersect each other at right angles in and
X
Y
This method of locating points is known by the that plane. name of THE CARTESIAN, or BILINEAR SYSTEM. The lines of 1
PLANE ANALYTIC GEOMETRY. X, Y, are called CO-ORDINATE AXES, and, when read separately, are distinguished as the X-AXIS and the Y-AXIS. The point 0, the intersection of the co-ordinate
reference
axes, is called
the ORIGIN OF CO-ORDINATES, or simply the
ORIGIN.
x and y which measure the distance of the from the Y-axis and the X-axis respectively, are point P!
The
lines
FIG.
1.
the distance (x ) from called the co-ordinates of the point the Y-axis being called the abscissa of the point, and the dis f tance (y ) from the X-axis being called the ordinate of the point.
Keferring to Fig. 1, we see that there is a point in each of the four angles formed by the axes which would satisfy the conditions of being distant x from the Y-axis and distant 2.
y from the X-axis.
This ambiguity vanishes when we com In the case
bine the idea of direction with these distances.
of places on the earth s surface this difficulty is overcome by using the terms north, south, east, and west. In analytic geome are used to serve the same try the algebraic symbols -f- and
purpose.
All distances measured to the right of the Y-axis
CO-ORDINA TES. are
called positive
abscissas
measured
those
;
3 to
the
left,
negative ; all distances measured above the X-axis are called all distances below, negative. With this positive ordinates the co-ordinates of the become understanding, (V, y ) point Pj. ;
of
P
2,
(-
x
f
,
y
)
;
of
P (- x f
3,
s
f
y
)
;
of
P4
-y
r
(x
,
,
;
f
).
3. The four angles which the co-ordinate axes make with each other are numbered 1, 2, 3, 4. The first angle is above
the X-axis, and to the right of the Y-axis the second angle above the X-axis, and to the left of the Y-axis the third ;
is
;
angle is below the X-axis, and to the left of the Y-axis the fourth angle is below the X-axis and to the right of the ;
Y-axis.
EXAMPLES. 1.
Locate the following points
(2.
1,
-
(0, 1),
(-
1,
;
-
2), (3,
What
(-
4, 0), (0,
-
(0, 1),
3).
are the lengths of its sides ?
Ans.
Vl3,
The ordinates of two points are each them situated with reference
the line joining
Ans.
line
0),
Locate the quadrilateral, the co-ordinates of whose ver (2, 0), (0, 3),
5.
2,
- 4).
tices are,
4.
(-
1),
Locate the triangle, the co-ordinates of whose vertices
are,
3.
:
- 1), (1, 2), (2, 3), (3, - 4). (0, 0), (3, 0), (0,
The commoji joining them
abscissa of two points situated ?
is
=
5, 5,
b
;
V 13.
how
is
to the X-axis ?
Parallel, below.
a
;
how
is
the
6. In what angles are the abscissas of points positive ? In what negative ?
In what angles are the ordinates of points negative ? 7. In what angles positive ?
PLANE ANALYTIC GEOMETRY.
4
In what angles do the co-ordinates of points have like In what angles unlike signs ? signs ? 8.
9.
The base
X-axis and the origin
its
of an equilateral triangle coincides with the is on the Y-axis at the distance 3 below
vertex
required the co-ordinates of its vertices ? Ans. (i- V12, 0), (0, 3), (-i V"l2,
;
-
ordinate
is
and how
is it
situated ?
A
Ans.
straight line passing through the an ing angle of 45 with the X-axis. 11.
0).
moves that the ratio of its abscissa to its = 1, what kind of a path will it describe, always
If a point so
10.
The extremities
of a line are the points
origin,
and mak
1,
(2, 1), (
2)
:
construct the line.
=
the ordinate of a point is 0, on which of the If the abscissa is ? co-ordinate axes must it lie ? 12.
If
=
13.
Construct the points
the line joining them 14.
Show
is
2,
(
3), (2, 3),
bisected at
that the point (m, n)
and show that
(0, 0). is
distant
Vm + 2
?i
2
from
the origin. 15. Find from similar triangles the co-ordinates of the middle point of the line joining (2, 4), (1, 1).
Ans.
($,$).
THE POLAR SYSTEM. 4. Instead of locating a point in a plane by referring it to two intersecting lines, we may adopt the second of the two methods indicated in Art. 1. The point P 1? Fig. 2, is fully
determined when we know its distance PI X (= 0) from some given point tion P! to X. If we give all values from line values from
to 360
to
(= in oo
and direc some given to r, and all r)
easily seen that the position be located.
0, it is
of every point in a plane may This method of locating a point
is
called the
POLAR SYSTEM.
CO-ORDINA TES.
5
The point is called the POLE the line X, the POLAR Axis, or INITIAL LINE the distance r, the EADIUS VECTOR ; the angle 0, the DIRECTIONAL or VECTORIAL ;
;
The POLAR CO
ANGLE.
distance r and the angle ORDINATES of a point.
0,
(r, 0),
are called the
In measuring angles in this system, it is agreed (as in trigonometry), to give the positive sign (+) to all angles meas5.
FIG.
ured round to the sign
(-)
to those
2.
left from the polar axis, and the opposite measured to the right. The radius vector
considered as positive (+) when measured from the pole toward the extremity of the arc (0), and negative (-) when measured from the pole away from the extremity of the (r)
is
arc
(0).
A
few examples will make
points clear. If r 2 inches and
=
this
method
of locating
=
45, then (2, 45) locates a point PI 2 inches from the pole, and on a line making an ano-le of -f 45 with the initial line. If r
=-2
=
inches and 45, then (- 2, 45) locates a 2 inches from the pole, and on a line making an angle of 45 with the initial line also but in this case the point is on that portion of the boundary line of the angle which has been produced backward the point
P
3
;
If
r
=2
inches and 9
= - 45,
through then (2,
pole.
- 45)
locates a
PLANE ANALYTIC GEOMETRY.
6
below inches from the pole, and out on a line lying point P 4 2 it. with 45 of the initial line, and making an angle 45 ) 2, 45, then ( 2 inches and 6 If r the to pole), locates a point P 2 directly opposite (with respect the point
P4
,
-
-
=-
=-
- 45).
(2,
in analytic geometry oi express 6. While the usual method it and seconds (, minutes, in is degrees, ing an angle of terms in convenient to express angles frequently becomes radius of the is equal in length to the arc whose the angle UNIT. CIRCULAR is called the measuring circle. This angle ot centre the at that angles from ,
We
same
know
"),
geometry
each other as the arcs included between and ff be two central angles, we hence, if
circle are to
their sides;
have,
arc
_.
7
& Let
ff
= unit
angle
;
~"arc
then arc
=r
(radius
of
measuring
circle).
_ Hence
= 360,
Hence,
r
circular unit .
If
arc
-.
r 6
= arc
common r
X
360
X
circular unit.
measure, then arc
=2
TT
r
X
= 2irr.
circular unit.
Therefore the equation,
= 2 X circular
360
TT
unit,
...
(1)
between the two units of measure. expresses the relationship
j EXAMPLES. an What is the value in circular measure of From (1) Art. 6, we have,
1.
= 30 X 12 = 2 circular unit. 30 = - circular unit. TT
360 ..
angle of 30 ?
6
CO-ORDINA TES. 2.
What
7
are the values in circular measure of the following
angles ?
1, 45, 60, 3.
What 3
90,,
120, 180, 225, 270, 360.
are the values in degrees of the following angles?
^
2
4"
4
8
*
* 4"
4.
What
5.
Locate the following points
is
,
8
8"
6
the unit of circular measure ?
Ans.
(2,
(-
40),
4,
57, 17
,
45.".
:
90), (3,
-
135).,
^3,
(-
-
1,
180),
3
- 1, - *), 6.
2,
-
Locate the triangle whose vertices are, / 3 \ \ 5 1 -7T \, M, -TT 4 -
7. The base of an equilateral triangle (= a) coincides with the initial line, and one of its vertices is at the pole re quired the polar co-ordinates of the other two vertices. ;
(a>
8.
The polar
co-ordinates
of
a point
are ( 2, \
three other ways of locating the
same
point,
~\ 4
()).
Give
/
using polar
co-ordinates.
Ans.
9.
a.,
/
-
i
2
Construct the line the co-ordinates of whose extremities
PLANE ANALYTIC GEOMETRY.
8 10.
How
being ( 3,
is
-\
the line, the co-ordinates of two of f 3,
\
its
points
situated with reference to the initial
Ans.
Parallel.
Find the rectangular co-ordinates of the following points 11-
=
13
4,
\
:
LOCL
CHAPTER
II.
LOCI.
THE Locus OF AN EQUATION
7.
generatrix as
it
moves
is
the
path described by
its
in obedience to the laiv expressed in the
equation.
The EQUATION OF A Locus the
law subject
to
the algebraic expression of
is
which the generatrix moves in describing that
locus.
If we take any point P 3 equally distant from the X-axis and the Y-axis, and impose the condition that it shall so move ,
FIG.
that the ratio of
its
ordinate to
3.
its
abscissa shall always be
equal to 1, it will evidently describe the line algebraic expression of this law is It
x
and
is
The
=
1,
or y
P
3
Pi-
= x,
called the Equation of the Locus. line is called the Locus of the Equation.
P^
The
Again
:
PLANE ANALYTIC GEOMETRY.
10
we take the point P 4 equally make it so move that the ratio of
if
,
distant from the axes, and ordinate to its abscissa
its
at any point of its path shall be equal to 1, it will describe In this case the equation of the locus is the line P 4 P 2 .
y~
=
1,
x
and the 8. first,
line
P4 P 2
or y
=
x,
the locus of this equation.
is
be observed in either of the above cases (the for example), that while the point P 3 moves over the
It will
line P 8 PU its ordinate and abscissa while always equal are yet in a constant state of change, and pass through all values For this reason y and x are from oo, through 0, to -|- GO.
called the
VARIABLE
or
GENERAL CO-ORDINATES
of the line.
we
consider the point at any particular position in its x as at P, its co-ordinates ( ?/) are constant in path, of to this the point, and to and value, position correspond If
,
this position alone.
The
variable co-ordinates are represented
by x and
of the moving y, and the particular co-ordinates point for any definite position of its path by these letters with a dash or subscript or by the first letters of the Thus (x ?/), (x lt yi), (a, ), (2, 2) alphabet, or by numbers. correspond to some particular position of the moving point. ;
,
EXAMPLES. 1.
Express in language the law of which y
algebraic expression. Ans. That a point shall so
move
= 3x
-f-
2
is
the
in a plane that its ordinate
shall always be equal to 3 times its abscissa plus 2. 2.
A
3.
The sum
+
a quantity a is point so moves that its ordinate b to its abscissa a ; required the quantity always equal of the law. expression algebraic b. Ans. y -\- a \x a
of the squares of the ordinate and abscissa of a 2 what is the is always constant, and
=
moving point
equation of
its
path
;
?
Ans.
x2
+y
2
==
a2
.
LOCI.
11
Give in language the laws of which the following are
4.
the algebraic expressions
:
^-2,*= -6.
2y--.|. y
2
=
4
4 z2
z.
2 x 2 -f 3
2 ?/
= 6. if
= 2px.
a^f
5 y2
+
6
2
z2
= 18. =a6 2
2
.
9. As the relationship between a locus and its equation constitutes the fundamental conception of Analytic Geometry,
important that it should be clearly understood before We have entering upon the treatment of the subject proper. it
is
been accustomed in algebra to treat every equation of the x as indeterminate. Here we have found that this form y
=
equation admits of a geometric interpretation ; i.e., that it repre sents a straight line passing through the origin of co-ordi nates and making an angle of 45 with the X-axis. We shall find, as
we
dental,
which does not involve more than three variable quan
tities, is
proceed, that every equation, algebraic or transcen
susceptible of a geometric interpretation.
We
shall
geometric forms can be expressed alge braically, and that all the properties of these forms may be find, conversely, that
deduced from their algebraic equivalents. Let us now assume the equations of several loci, and let us locate and discuss the geometric forms which they represent. 10. Locate the geometric figure whose algebraic &wwdent is
y
We its
know
= 3x+2.
that the point where this locus cuts the Y-axis has in the equa 0. If, therefore, we make x
abscissa x
=
=
we
shall find the ordinate of this point. Making the substitution we find y 2. Similarly, the point where the
tion,
=
locus cuts the X-axis has
for the value of its ordinate.
Mak-
t+T*
PLANE ANALYTIC GEOMETRY.
12
=
=
in the equation, we find x ing y the axes and marking on them the points
we
will have
two points of the required
successively equal to 1, 2, 3,
-
1,
-
2,
- 3,
f.
locus.
Drawing now
Now make
x
etc.
in the equation, and
find the corresponding values of y. the result thus let us tabulate convenience
For
:
Values of x
Corresponding
_
2 3
"
"
-
1
_2
Values of y 5 8 11
1
FIG.
1
4
"
4.
a line through them we Locating these points and tracing have the required locus. This locus appears to be a straight
LOCI.
13
We
line and it is, as we shall see hereafter. shall see also that every equation of the first degree between two variables The distances Oa and Ob represents some straight line.
which the
line cuts off on the co-ordinate axes are called INTERCEPTS. In locating straight lines it is usually sufficient to determine these distances, as the line drawn through their extremities will be the locus of the equation from which their
values were obtained.
EXAMPLES. 1.
Locate the geometric equivalent of
I,-..-l_2* Solving with respect to y
y=-2x + The extremities
we
in order to simplify,
have,
2.
of the intercepts are (0,2),
(1,0).
Locating these points, and drawing a straight line through them, we have the required locus. Construct the loci of the following equations
5.
6.
12.
y
2y Is
=
cx
-cl
= 3x.
10.
i-
11.
X
the point (2, 1) 2x 3 ? Is y
=
-
-y=-3 y
:
_2x.
on the line whose (6, 9) ?
NOTE. If a point is on a line, the equation of the line.
its
Is (5, 4) ?
equation Is (0,
co-ordinates
must
is
- 3) ? satisfy
PLANE ANALYTIC GEOMETRY.
14 13.
Which
of the following points are on the locus of the
2 equation 3 x
(2, 1),
14.
+2y = 6? 2
(V2,
(0,V3),
0),
=
2x
1,
Construct the lines y
show by similar
11.
-2x = 3y
0), (2,
V3)
line
6.
Construct the polygon, the equations of whose sides are
y 16.
(- V2,
1, 3),
Write six points which are on the - y
15.
(-
Discuss
y
=
= sx
a;,
= 5.
y
and y
-f-
=
saj
+ 4,
and
triangles that they are parallel.
and x*
construct the equation
+ y* =
Solving with respect to
y =
=t
:
16.
y,
we
Vie - X
have, 2-
The double sign before the radical shows us that for every we assume for x there will be two values for y, equal
value
and with contrary
This is equivalent to saying that signs. for every point the locus has above the X-axis there is a cor responding point below that axis. Hence the locus is symmet
with respect to the X-axis. Had we solved the equation with respect to x a similar course of reasoning would have shown us that the locus is also symmetrical with respect to the Y-axis. Looking under the radical we see that any value of x less than 4 (positive or negative) will always give two real values for y ; that x 0, and that any J- 4 will give y = value of x greater than -j- 4 will give imaginary values for y. Hence the locus does not extend to the right of the Y-axis x 4. farther than x -f- 4, nor to the left farther than rical
=
=
Making
= 0, = 0, y
x
we have y "
"
x
= =
_j_
4
-j-
4.
LOCI.
15
Drawing the axes and constructing the points, - 4), (4, 0), ( 4, 0), we have four (0, 4), (0, points
of
the locus
i.e.,
;
B,
B
,
A, A!. Y B
FIG.
Values of x
5.
Values of y
Corresponding
+ 3.8 and 3.8 + 3.4 and - 3.4 + 2.6 and - 2.6
1
2 3
4
-1 -2 -3 -4
+ 3.8 and - 3.8 + 3.4 and 3.4 + 2.6 and - 2.6
Constructing these points and tracing the curve,
we
find
it
to be a circle.
This
might readily have been inferred from the form
of the equation, for
we know
that the
sum
of the squares
PLANE ANALYTIC GEOMETRY.
16 the
(OC) and ordinate (CP^ of any point Pj equal to the square of the radius (OPj). We might, therefore, have constructed the locus by taking the origin as centre, and describing a circle with 4 as a radius. NOTE, x for any assumed value of y, or y -JL _j_ 0, of
abscissa
in the circle
is
=
=
any assumed value of x always indicates a tangency. Re ferring to the figure we see that as x increases the values of y decrease and become -j- when x = 4. Drawing the line for
represented by the equation x to the curve.
We
= 4,
we find that it is tangent we proceed that any two
shall see also as
coincident values of either variable arising from an assumed or given value of the other indicates a point of tangency. 12.
Construct and discuss the equation 9 x2 16 y 2 144.
=
+
Solving with respect to
we have
?/,
-
144
9 x2
16
= y =
x
X
"
Drawing the axes and laying four points of the locus
Values of x 1
2 3 4 1
3 4
;
= =
gives y
i.e.,
B,
off
B
,
-j-
;
these distances,
A,
Corresponding "
3
4-4:.
A
.
Values of y
4 2.9 and - 2.9
"
+2.6
((
_i_
"
we have
Fig. 6.
2
+2.9
"
-r-
2
2.6
"
2
"
2.9
"
"
-
2
"
-J-
Locating these points and tracing the curve through them, the required locus. Referring to the value of y we see from the double sign that the curve is symmetrical with
we have
respect to the X-axis.
The form
of the equation (containing
LOCI.
17
of the variables), shows that the locus only the second powers with Looking is also respect to the Y-axis.
symmetrical
FIG.
6.
under the radical we see that any value of x between the 4 will give two real values for y ; and that 4 and limits will give imaginary values for any value beyond these limits Hence the locus is entirely included between these limits. y. with which we shall have more to do hereafter, This
+
is
curve, called the ELLIPSE.
13.
Discuss and construct the equation 2 2/
Solving,
=
we have
is symmetrical with respect to the of the as and equation contains only the first power X-axis, As to the Y-axis. with not it is that respect symmetrical x, will always give real values for y, every positive value of x the locus must extend infinitely in the direction of the posi tive abscissae and as any negative value of x will render y
We
see that the locus
;
PLANE ANALYTIC GEOMETRY.
18
imaginary, the curve can have no point to the left of the hence the curve Making x 0, we find y -j- 0; passes through the origin, and is tangent to the Y-axis.
=
Y-axis.
Making y
= 0,
we
=
find
x
=
;
hence the curve cuts the
X-axis at the origin.
Values of x
Corresponding
1
+
"
2 3 4
From is
Values of y 2 and
+2.8
+ 3.4 +4
these data
we
-2 - 3.4 -4 -2.8
easily see that the locus of the equation
represented by the figure below.
FK;.
This curve 14.
is
called the
7.
PARABOLA.
Discuss and construct the equation 4 z2 9 ?2 36
=
Hence
We see from the form of the equation that the locus must be symmetrical with respect to both axes. Looking under
19
LOCI.
the radical, we see that any value of x numerically less than 3 will render y imaginary. Hence there is no _j_ 3 or We see also that point of the locus within these limits.
+
any value of x greater than values for
The
y.
3 or 3 will always give real locus therefore extends infinitely in the
direction of both the positive 3. limits x
=i Making x = 0,
we
find
y
and negative
=
-J-
2
V
1
abscissae
;
from the
hence, the curve
does not cut the Y-axis.
Making y
= 0,
we
find
the X-axis in two points
Value of 4
x.
x
=
(3, 0), (
-j-
3
;
hence, the
Corresponding.
Values of y
5
+ + 2.6 + 3.4 + 1.7 + 2.6
-6
4-3.4
1.7
5 6
-4
FIG.
This curve
is
called the
curve cuts
3, 0).
8.
HYPERBOLA.
and "
"
"
- 1.7 - 2.6 - 3.4 - 1.7 -2.6
"
- 3.4
PLANE ANALYTIC GEOMETRY.
20 15.
We
have in the preceding examples confined ourselves
RECTANGULAR equations of equations whose loci were referred to rectangular Let us now assume the POLAR equation axes. to the construction of the loci of
;
i.e.,
=
r
and discuss and construct
Assuming values
for
-
6 (1
cos 0)
it.
6,
we
find their cosines
convenient table of Natural Cosines. values,
we
Substituting these
find the corresponding values of
Values of 6
Values of cos 1.
.86
6
.50
6 6
160
- .50 - .94
180
-
200 240
1.
-
.94 .50
270
(1 (1 (1
6 (1
30
90
r.
Values of r
60
120
from some
1
)
.86) .50) )
+ .50) + .94) 6 (1 + 1 6 (1 + .94) 6 (1 + .50) 6 (1 6
(1
6 (1
)
)
300
.50
6 (1 -- 50)
330
.86
6 (1
-
.86)
= = .84 = 3. = 6. = 9. = 11.64 = 12. = 11.64 = 9. = 6. = 3. = .84
as the the initial line OX, and assume any point a series of lines, making the draw this point Through pole. assumed angles with the line OX, and lay off on them the corresponding values of r. Through these points, tra the have we required locus. cing a smooth curve,
Draw
LOCI.
FIG.
This curve, from 16. Discuss
and
its
A
9.
heart-like shape,
is
called the CARDIOID.
construct the transcendental equation
y NOTE.
21
= log x.
transcendental .equation
is one whose decree power of analysis to express. Passing to equivalent numbers we have = x, when 2 is the base of the system of logarithms selected. As the base of a system of logarithms can never be nega tive, we see from the equation that no negative value of x can Hence the locus has none of its satisfy it. to
ti-anscends the
&
the left points the other hand, as every positive value of x will give real values for y, we see that the curve extends the direction of the infinitely of the Y-axis.
m
If
y = 0,
On
positive abscissae
then
PLANE ANALYTIC GEOMETRY.
22 If
x
=
0,
then 2*
The
=
.-.
y
= log
.-.
y
=
co.
at a unit locus, therefore, cuts the X-axis
s
distance on
the positive side, and continually approaches the Y-axis with out ever meeting it. It is further evident that whatever be the base of the system of logarithms, these conditions must x. hold true for all loci whose equations are of the form a?
=
Values of x 1
2
Corresponding
Values of y
"
1
"
4
"
8
"
2
3
-
.5
1
2
"
.25
traced through them will Locating these points, the curve
be the required locus.
FIG.
10.
This curve is called the LOGARITHMIC Curve, being taken from its equation.
its
name
LOCI.
23
The preceding examples explain the method employed
17.
in constructing the locus of any equation. While it is true that this method is at best be made approximate, yet it
may
sufficiently accurate for all practical purposes by assuming for one of the variables values which differ from each other
by very small quantities. It frequently happens (as in the case of the circle) that we may employ other methods which are entirely accurate. In the discussion of an equation the first step, usually, it with respect to one of the variables which enter
18. is
to solve
into
it.
The question
of which variable to select
is
immate
rial in principle,
yet considerations of simplicity and conven ience render it often times of The sole great importance. difficulty, in the discussion of almost all the higher forms of If this difficulty can equations, consists in resolving them. be overcome, there will be no trouble in tracing the locus and discussing it. If, as frequently happens, no trouble arises in the solution of the equation with respect to one of the vari
then that one should be selected as the dependent variable, and its value found in terms of the other. If it is equally convenient to solve the equation with to either ables,
respect
of the variables
selected
which enter into
it,
whose value on inspection
then that one should be will afford the
discussion.
EXAMPLES. Construct the loci of the following equations 1.
2.
+ l-0.
2jr-4 y
-., 2
3.
2y
4.
4aj 2
10.
= 10.
+ 5s -9 =-36. 2
5.
fi+4
6.
**+y-
7.
r
8.
35
2
a
?/
:
= ^ cos = logy. 2
simpler
PLANE ANALYTIC GEOMETRY.
24
Construct the loci of the following 9 10. 11.
si.y^O. x + 2 ax + a = 0. x - a* = 0. 2
2
2
12
yi_9 = 0.
13.
7/
2
-
2 xy
+ x - 0. 2
:
14.
x*-x-6 = 0.
15.
or -j-
2
2
x
6 ==
+4x - 5
0.
-
16.
x
17.
x*-7x + l2 = 0.
18.
x2
0.
+ 7 z + 10 =
Factor the first member: equate each factor NOTE. and construct separately.
0.
to 0,
THE STRAIGHT LINE.
CHAPTER
25
III.
THE STRAIGHT
LINE.
19. To find the equation of a straight line, given the angle which the line makes with the X-axis, and its intercept on the Y-axis.
FIG.
11.
Let C S be the line whose equation we wish to determine. Let SAX and OB b. Take any point P on the line
= PM
=
to OY and BN to OX. Then (OM, MP) = (x, y) are the co-ordinates of P. From the figure PM = PN + OB = BNtan PEN +
and draw
||
||
BN = OM = x, ..
and tan PBN = tan SAX = tan Substituting and letting tan a = s, we have, (I)
.
b,
but
PLANE ANALYTIC GEOMETRY.
26
Since equation (1) is true for any point of the line SC, it true for every point of that line hence it is the equation of the line. Equation (1) is called the SLOPE EQUATION OF THE is
;
STRAIGHT LINE
COROLLARY
If
= tan b = in
=
sx
s
;
1.
y
(
.
(1),
.
.
called the slope.
is
)
we
have,
(2)
a line which passes through the
for the slope equation of origin.
COR.
=
If s
2.
y which
is,
as
it
we have
in (1),
=b
ought to
be, the
equation of a line parallel to
the X-axis.
COR.
3.
If
=
s
oo,
=
then
90, and the
line
becomes
parallel to the Y-axis.
Let the student show by an independent process that the equation of the line will be of the form x = a.
SCHOLIUM. We have represented by the angle which the makes with the X-axis. As this angle may be either acute or obtuse, s, its tangent, may be either positive or nega
line
tive.
The
line
may
also cut- the Y-axis either above or below
the origin; hence, b, its Y-intercept, From these considerations negative.
y
=-
sx
y
-{-
it
appears that
+b
represents a line crossing the == sx
be either positive or
may
first
angle
;
b
represents a line crossing the second angle
y =
sx
b
represents a line crossing the third angle
y
= sx
;
b
represents a line crossing the fourth angle.
;
THE STRAIGHT
LINE.
27
EXAMPLES.
The equation of a and intercepts.
1.
slope
line
Solving with respect to
Comparing with
=
we
y,
22 1
=
y
x H
,
(1)
Art.
Making y
Y-intercept.
is
2y
+x=3
;
required
its
have,
3 .
19,
we
=
1
and
2
=
= 3 = X-intercept.
find s
in
the
equation,
b
=
^
2
we have
=
3. Construct the line 2 y -\- x The points in which the line cuts the axes are
2.
O,
?V
and
(3,
3).
-*/
Laying these points
off
on the axes, and tracing a straight
through them, we have the required
line
thus
~F-+fLay
off
OB =
b
=
|
;
draw
BN OX and make it = 2, also NP OY and make it = + 1. ||
||
The
line
through
P and B
the required locus. T^I^T
-*
For
=
locus.
solving the equation with respect to
:
tan
BAX. .-.tan
BAX
is
y,
we
Or otherwise have,
PLANE ANALYTIC GEOMETRY.
28 3.
=
x
Construct the line 2 y
we
Solving with respect to y,
have,
Lay ||
to
also
A
3.
off
OX NP
||
BO = 5 = |
and make it = 2 to OY and make
straight line
BK
Draw
.
draw
;
it
=
1.
P and B
through
will be the required locus.
= - = tan PBN = tan
For
BAX =
Hence, in general,
s.
Y
to the left of
BN
is
laid off to the right or
according as the coefficient of
x
is fiositive
or negative.
Give the slope and intercepts of each of the following and construct
lines
:
4.
2
Ans. 5.
s
=
,
&
=
*
= - 2, a = - -
1,
^^
x-
6.
.
* ==
- 12, 8.
y
~ 3
+2x = 1 -
.
y.
9.
_
a and 6 in the answers above denote the X-intercept NOTE. and the Y-intercept, respectively.
THE STRAIGHT What
14.
29
angle does each of the following lines cross ? 10.
y
= 3x +
11.
y
= - x + 2.
12.
1.
y
=2
a;
1.
\<^
Construct the figure the equation of whose sides are
2y 15.
LINE.
+ s-l = 0, 3^=2x + 2, y = - x -
I.
Construct the quadriiftfc&rai the equations of whose sides
are
x
= 3,
y
=
x
+
y
l,
= 2,
x
= 0.
20. To find the equation of a straight line in terms of
its
intercepts.
Y
FIG.
Let S C be the
line.
Then OB
OA =
12.
a
= b = Y-intercept, = X-intercept.
The slope equation of Art. 19. equation (1), y
a line
=
sor
and
we have determined
+
/>.
to be
PLANE ANALYTIC GEOMETRY
30
From
the right angled triangle
tan
GAB =
tan
AOB, we
BAX =
-
s
have,
= OB OA
.-.-.--*. a
Substituting in the slope equation,
This
is
called the
we
have,
SYMMETRICAL EQUATION
of the straight
line.
COR.
1.
+ and b +, then we have, -= 1, for a line crossing the
If a
ni
rvt
\-
first
j-
and
If a
b
+, then
---f-^ = lisa line crossing the a
If
and
a
b
,
b
a
-f-
and
--
^
a
b
b
,
=1
is
a line crossing the third angle.
then
----- *-
a
second angle.
then
-
If
angle.
=1
is
b
a line crossing the fourth angle.
EXAMPLES. 1.
Construct
NOTE. Y-axis.
^
=
1.
Lay off 3 units on the X-axis and Join their extremities by a straight
2 units on the line.
Across which angles do the following lines pass
?
THE STRAIGHT
LINE.
31
Give the intercepts of each, and construct.
+ |=1.
2-
|
/v,
q
\
Q o
7/ if
=
4.
i*^ *
1
^ ^
57^ =
n
3
"Q
y
r>
-i -
1
^^
AXV-~)C7^-
z--<r^-
-
K"
Write the slope equations construct
-f~V=l.
the
of
following lines, and
:
=
x
6.
o
r-IS-7. 3
.
s.
y 2
T ?.6 = _ +
i.
y-ix-2. O .
^
^
-J
O J
5 10.
Write y
=
e
+ 6 in a symmetrical ."
7-
;
=
form.
1 1.
/-t
A O*^"
/>
J
Given the following equations of straight their slope and symmetrical forms
lines, to write
:
11.
+ 3z-7 = z + 2. ^^ = ~ 3
2y
23
13.
a?
12.
.
14
2? x
=
43
~
V
_2x-l
PLANE ANALYTIC GEOMETRY.
32
To find the equation of a straight line in terms of the perpendicular to it from the origin and the directional cosines 21.
of the perpendicular.
The Directional cosines of a line are the cosines it makes with the co-ordinate axes.
NOTE.
the angles which
FIG
Let OS Let
be-
the
line.
AOP = triangles AOP and BOP, we OA = -OP- OB = OP
OP = p, BOP
From
the
13.
==
.
7,
have
,
COS 7
COS
that
a
is,
=
P
7.
*
,
b
=
P
^
cos w
cos 7
Substituting these values in the symmetrical equation, Art. 20. (1),
_
a
+ y-b =
x cos
which
is
1,
we
have, after reducing,
+ y cos 7 = p
the required equation.
.
.
.
(1)
of
THE STRAIGHT Since
=
y
90
,
This form
and
COR.
and the COR.
.
.
NORMAL EQUATION
(1)
of the straight line.
becomes parallel to the Y-axis.
line
= 90,
If
2.
.
= 0, then x =p
If
1.
;
more frequently met with than that given in
is
called the
is
= sin hence + y sin = p (2)
cos y
cos
cc
33
LINE.
then
y*=p and the
If x
22.
becomes parallel to the X-axis.
line
+ y sin = p be the equation of a given line, + y sin a = p -- d is the equation of a parallel
cos
then x cos
For the perpendiculars p and p j- d coincide in direction line. hence the lines since they have the same directional cosines to which they are perpendicular are parallel. ;
COR.
Since
1.
d
p it
is
evident that d
therefore, (x
the origin
,
p
is
y
)
-J-
-p = is
the distance between the lines.
_j_
If,
be a point on the line whose distance from d,
we have
+ y sin = p = x cos d y
x cos u .-.
d
"
-\-
-J-
d.
sin
*
p
.
.
.
(1)
a point (x y ) from the line found -\- y by transposing the constant term to the first member, and substituting for x and y the co Let us. for example, find the ordinates x y of the point. distance of the point ( y~3, 9) from the line x cos 30 y sin
Hence the distance sin a
x cos a
=p
of
,
is
,
+
30
=
5.
From
(1)
d
= V3 cos 30
-f 9 sin 30
-
5
PLANE ANALYTIC GEOMETRY.
34
From .*.
Fig.
p
sin= / =
,
b
-2
_L
7
t=
<2
/
(
y
i
|-
a
i
1
^b
+6
Vtt 2
2
the expression for the distance of the point (x
line
-
J/
-
a
=
Hence is
=
13 we have cos
whose equation
is
of the form -
a
+b=
r
?/)
,
from a
1.
Let the student show that the expression for d becomes
VA + B 2
when the equation
of the line
See Art. 24, Equation
2
given in
is
general form.
its
(1).
EXAMPLES. 1.
line
The perpendicular Kit makes an arfgle
= 5 and
equation of the
fall
from the origin on a straight with X-axis required the
of 30
;
line.
V3
Ans.
sc
+ = //
2. The perpendicular from the origin on a straight makes an angle of 45 with the X-axis and its length =
required the equation of the
line.
Ans. 3
What
is
the distance of the point
(2.
Ans.
.
Find the distance of the point from the following cases
x -\-y
4) from the
x
From
(2,
5.
From
(3, 0)
to
6.
From
(0, 1)
to 2
7.
From
(a, c) to
5) to
-
=
-
1.
?/
= - x = 2. 1.
|
y
=
SJT
+
5
= 2. line
line in each of the
| |
line
V2
i.
:
4.
10.
b.
^^
THE STRAIGHT LINE.
35
23. To find the equation of a straight line referred to oblique axes, given the angle between the axes, the angle which the line makes with the X-axis and its Y-intercept.
NOTE.
Oblique axes are those which intersect at oblique
angles.
Y
1
M N>
FIG. H. %
it
Let CS be the line whose equation we wish to determine, being any line in the plane YOX. Let YOX = ft SAX = OB = b. Take any point P on the line and draw PM to OY and ON to SC ,
From
NP = OB = b.
== ,
the figure
y
From
;
||
||
PM = y, OM = a, NOX
then,
triangle
= MN + NP = MN + b
ONM, we
MN
sin
OM
sin
.
.
.
(1)
have,
NOM MNO sin
sin
Substituting the value of (1),
we
drawn from
.M"N"
this equation in
have,
y
Sin
= sin
(/?
"
-
x )
+b
...
(2)
PLANE ANALYTIC GEOMETRY.
36
This equation expresses the relationship between the co Bnt as the point ordinates of at least one point on the line. selected was any point, the above relation holds good for every point, and is, therefore, the algebraic expression of the law which governed the motion of the moving point in de It is therefore the equation of the line.
scribing the line. COR. 1. If b
= ?/
0,
then sin
=
-
sin (p
x
.
.
.
(3)
a)
the general equation of a line referred to oblique axes passing through the origin. is
COR.
2.
If b
= y
=
and
=o
.
.
.
then
0,
(4)
the equation of the X-axis.
COR.
3.
If
=
ft
and
=
x
.
=
/3 .
.
;
y
But tan This 19
is
<t
=s
5.
.
= tan n x y = sx
if
-J-
-\-
the axes are
made
rectangular,
ft.
ft-
See Art.
the slope equation heretofore deduced.
(1).
Cor.
.
then
(5)
the equation of the Y-axis. COR. 4. If (3 = 90 i.e.,
then
,
If
/3
= 90 and = 0, then y = sx. See Art. 19, Cor. ft
1.
EXAMPLES. makes an two units
Find the equation of the straight line which 1. angle of 30 with the X-axis and cuts the Y-axis distant from the origin, the axes
making an angle
with each other. Ans. 2.
had been assumed rectangular what would have been the equation ?
If the axes
ple above,
Ans.
y
of
=x
in the
=
-JL-
60
-f-
2.
exam
+ 2.
THE STRAIGHT LINE.
37
The
co-ordinate axes are inclined to each other at an of angle 30, and a line passing through the origin is inclined to the X-axis at an angle of 120, required the equation of the 3.
line.
Ans.
24. is
y y
= f\
,/q vo
Every equation of the first degree between two variables
the equation of a straight line.
Every equation of the first degree between two variables can be placed under the form
Ax in
y,
+ By + C
-
.
which A, B, and C may be either Suppose A, B, and C are not zero.
we
and whose slope
from which If
(1)
finite or zero.
Solving with respect to
with
(2)
(1) Art.
the equation of a straight line -
.
have,
Comparing equation is
.
A = 0,
it
then
s
=
was derived ?/
=
is
A
;
.19,
we
see that
whose Y-intercept
hence
(1),
the
b
it
=
equation
the equation of a straight
line.
*
B
the equation of a line parallel to the X-axis. If
B
= 0,
then x
=
A
the equation of a line parallel to the Y-axis. If
C
= 0,
then y
=-
A B
X)
the equation of a line passing through the origin. Hence, for all values of A, B, C equation (1) is the equa tion of a straight line. ^
J
PLANE ANALYTIC GEOMETRY.
38 25.
To find the equation of a straight
line
passing through
a given point.
Let (V, y
)
be the given point. is to be straight,
its
Since the line
y = sx + b
which
in
s
and
...
equation must be
(1)
be determined.
b are to
of a line expresses the relationship which of every point on it; hence co-ordinates the exists between when the co-ordinates of any satisfied be must its
Now, the equation
equation for the general co-ordinates x and point on it are substituted the We equation of condition. have, therefore, y.
+
=sx
y
But a straight
...
b
(2)
cannot in general be made to pass cut off a given distance (b) on
line
through a given point (x /), the Y-axis, and make a given angle (tan. = s) with the X-axis. We must therefore eliminate one of these requirements. By ,
subtracting (2) from
y
which
is
COR.
-
(1),
we have,
=s
y
(
-
x
x
(
}
3)
the required equation.
1.
If
x
=
y
0,
-
then
= SX
y
.
.
.
(4)
on the the equation of a line passing through a point Y-axis. COR. 2. If y 0, then
is
=
y is
=s
(x
x
.
.
.
(5)
the X-axis. the equation of a line passing through a point on
COR.
3.
If
x
= y
and y
= 0,
then
= sx
the equation (heretofore through the origin. is
)
determined), of a line passing
THE STRAIGHT
LINE.
39
EXAMPLES. 1.
Write the equation of several lines which pass through
the point 2.
is the equation of the line which passes through and makes an angle whose tangent is 2 with the
2),
(1,
(2, 3).
What
X-axis ?
Ans. y
A
3.
straight line passes through ( with the X-axis. What
angle of 45
1, is its
3),
=
2 x
4.
and makes an
equation ? Ans. y
=x
2.
Required the equations of the two lines which contain and 60, respectively (a, b), and make angles of 30
4.
the point
with X-axis. Ans.
y 26.
y
-b = ^.(x-a).
^
,
(x",
Since the line
which
s
and
we
line passing
through
be the given points.
straight its equation
= sx + b
b are to
Since the line y"),
y")
is
y
(x",
V3
To find the equation of a straight
two given points. Let (x r y ),
in
-b = ?L=JL
y
.
.
.
must be
(1)
be determined.
required to pass through the points (x have the equations of condition.
y y"
is
= =
8X #t"
+ +
b
*
.
.
.
,
y ),
(2)
...
(3)
As a straight line cannot, in general, be made to fulfil more than two conditions, we must eliminate two of the four con ditions expressed in the three equations above. Subtracting (2) from (1), and then (3) from (2), we have, y y
- y = s (x - x - = s (x y"
f
) x")
PLANE ANALYTIC GEOMETRY.
40
member by member, we
Dividing these,
y
y
_
x
tf-y"
Hence
y
-
y
=
-x"
^^
-
(x
x
.
.
)
(4)
.
OC
3C,
is
have,
~ xx
the required equation. r then COR. 1. If y
= -y = y y",
which
is,
COR.
tf>
it
=
2.
is
=
or y
should be, the equation of a line then If x
as
to the X-axis.
||
x",
x
x
which
>
0,
or x
the equation of a line
||
=
x
1
,
to the Y-axis.
EXAMPLES. 2, 8) required both Given the two points ( 1, 6). ( line the passing through the slope and symmetrical equation of 1.
;
them. Ans.
y
vertices of a triangle are required the equations of its sides. 2.
The
=(-
+ 4,
2 x
(-
2, 1),
| 3,
+|=
-4)
1.
(2, 0)
= 5 X + 11 oy = 8 -\4:X (y + x = 2. (
Ans.
y
Write the equations of the lines passing through the points 3.
Ans. 4.
?/
(1,4), (0,0)
Ans. 5.
6.
(-2,3),(-3,-l)
y
(0,2),(3,-1) Ans.
= 4x-f-ll. 7.
7 y
2.
+ 2 x = 24.
(-3,0)
Ans. y 8.
+ =
(2,0),
= 0.
(-l,-3), (-2,4) Ans.
:
(-2,4)
Ans.
= 4 x. x
(5,2),
;
y
+ 7 x + 10 = 0.
THE STRAIGHT LINE. 27.
To find the length of a
line joining
FIG.
Let (x
f
,
i/),
(x",
y"}
41
two given points.
15.
be the co-ordinates of the given points
L = FT" = required length. Draw and P A to OY, and PC to OX. We see from the figure that L is the hypothenuse of a
P,
P".
P"B
||
||
right
angled triangle whose sides are
PC P"C
Hence, PT"
= AB = OB - OA = - x - EC = if - y = x"
P"B
=L=
1.
If
L
~
- xY
fa"
x and y the origin, and we have Con.
2 V*"
for the distance of a point
,
and
.
= 0,
+
2 y"
+
(y
r::~y lY
the point
...
from the
P
.
-
-
(l)
coincides with
(2)
origin.
EXAMPLES. Given the points (2, 0), ( 2, 3) required the distance between them also the equation of the line passing through them. Ans. L = 5, 47/-f3;r =6. 1.
;
;
PLANE ANALYTIC GEOMETRY.
42 2.
what
The
vertices of a triangle are are the lengths of its sides ?
(2, 1)
1,
(
2)
Give the distances between the following points 0)
(2, 3), (1,
(4,
-
5), (6,
(-
(0, 2),
Ans. 11.
What
is
Ans. Area 28.
To
;
(-
2,-l),
9.
(a,
ft),
10.
(-
2, 3),
:
(0, 1)
;
(2,
0)
d)
(c,
V5.
(-
a, 6).
2.
the expression for the area of a triangle whose (V, y
),
(a",
-
= \ \x ./iTicZ
8.
0)_
0)
(0, 0), (2,
vertices are
0)
V20. 1,
Ans. 6.
(-
3,
-_!)
^ws. 5.
3,
VlO.
Ans. 4.
7.
2)
-
V8, VlOJ V2fr
Ans.
3.
(
(y"
y
")
(*
y")>
+
",
y
^Ae intersection
? ")
- /) +
f *"
(y"
xf
1
"
(y
y")]-
of two lines given by their
equations.
Let
y y
= sx =s +6
-f- b,
and
a:
be the equations of the given lines. Since each of these equations is satisfied for the co-ordinates of every point on the locus it represents, they must -ttt-tke samer-tim* be satisfied for the co-ordinates of their point of Hence, for the intersection, as this point is common to both. So the equations are simultaneous. co-ordinates of this
treating them,
we
pointfind b
s b
b
sb
for the co-ordinates of the required point.
EXAMPLES. 1.
Find the intersection of y
= 2x + 1
and 2 y Ans.
= x - 4. (-2, -3).
THE STRAIGHT LINE. 2.
The equations
43
of the sides of a triangle are
required the co-ordinates of
its vertices.
(MH-i, \o oj
Ans.
7
\
Write the equation of the
= ]
which shall pass through and 3 y x 8=0, and make an angle with the X-axis whose tangent is 4. 3.
the intersection of 2 y
3x
-f-
line
-f-
2
=
Ans. 4.
What the
=
x -f
2,
y
= 3 x -f 2, Ans.
5.
x2
The equation 2
-f-
?/
=
= 4 x -|- 10.
are the equations of the diagonals of the quadri equations of whose sides are y x-\-l =Q,
lateral
y
y
10
is
y
+2x+2 =
and y
23 y
9
;c
?
+ 2 = 0, 3y
A*
A
and
a;
=
6.
whose equation is required the length of the chord.
of a chord of the circle
=x
-f-
2
;
Ans. 29. If
30 x
+ By -f C = +By+C =
.
.
.
.
.
L
(1) .
(2)
be the equations of two straight lines, then
Ax -f By + C
+K
(A a;
+ B y +C = )
...
(3)
(K being any constant quantity) is the equation of a straight line which passes through the intersection of the lines repre sented by (1) and (2). It is the equation of a straight line because
it
variables.
an equation of the first degree between two See Art. 24. It is also the equation of a straight
is
which passes through the intersection of (1) and (2). it is obviously satisfied for the values of x and y which simultaneously satisfy (1) and (2). Let us apply this principle to find the equation of the line which contains the point (2, 3) and which passes through the intersection of y = 2 x -\- 1 and 2 y + a; = 2. line
since
PLANE ANALYTIC GEOMETRY.
44
From
(3)
we have y
2a
l
+ K(2y +
sc
2)=0
for the equation of a line which passes through the intersec But by hypotheses the point (2, 3) is tion of the given lines.
on
this line
;
hence 3
4-1 + K K
Substituting this value for
we
(6
+ 2-2) =0
have,
o
x
y
or,
1 ==
Let the student verify this result for the required equation. by finding the intersection of the two lines and then finding the equation of the line passing through the two points. 30.
To find the angle between
given by their
lines
tivo
equations. M,
Yrf
FIG.
Let
y y
16.
= sx +- and = sx+ V />.
be the equations of s
1
SC and MN,
= tan
and
s
respectively
=
tan
.
;
then
THE STRAIGHT LINE. From
45
the figures
= .
,
(f
(f
=
-\-
a
u
a.
From trigonometry, tan
.-.
cp
)
(
= tan
= tan -
1
-
.
.
.
1 -f ss
COR.
1.
If s
= <p
tan a
s
,
(2).
then
= tan -1
= 0.
.. qp
the lines are parallel.
COR.
2.
If 1
+ = ss
(f
/.
= tan
substituting;
Or,
.*,
cp
then
0,
= tan -
1
oo
.-. g>
= 90
the lines are perpendicular.
These results may be obtained geometrically.
SCHOL.
If the lines are parallel, then, Fig. 16,
If they are perpendicular
= 90 .-.
= s = tan
tan a
.-.
1
+ (90
+ ss = 1 tan 1-1 = 0. -{-
a
-f
)
tan
= -
cot
=
.
tan a
=
1 4- tan
=
|
tan
EXAMPLES. 1.
What
and 2y
is
the angle formed by the lines y
x
1
+ 2cc + l=0?
y
=
= 90.
PLANE ANALYTIC GEOMETRY.
46
2. Required the angle formed by the 8 = 0. 6x and 2 y
+
3.
(2,
Ans.
3x y Perpendicular to 2 ?/
5
(b)
= 0.
= 0.
3x
-
5=0. 2 y
= 8,
(5)
3y
-j-
2x
= 1.
Given the equations of the sides of a triangle a -f- 2 and y = 3 required. l, y = 2#
+ y=
(a)
q>
Required the equation of the line which passes through 1) and is
Ans. (a) 3 x 4.
=
2
y -\-ox
+
(a) Parallel to 2 (b)
lines
;
The angles of the triangle. The equations of the perpendiculars from
vertices to
sides. (c) 5.
The lengths
What
relation exists
y y y y 6.
What
of the perpendiculars.
between the following
lines
:
= SX + = SX 3. = SX -\- 6. = SX + m. b.
relation exists between the following:
y y
= sx -[= sx
b.
-j- c.
Find the co-ordinates of the point in which a perpen 1 = 0. 2 a; dicular through ( 2, 3) intersects y 7.
+
,. 8.
Find the length of the perpendicular
origin on the line
2y-\-x
(,r
let fall
from the
L
= - V80.
= 4.
Ans.
o
=
be the equations of three straight lines, and /, m, and n be three constants which render the equation
B ty-j-C"
THE STRAIGHT I
+ By + C) + m
(Ax
C")
+
LINE.
47
+
+ Wy -f
+
By n (Af x CO (Afx an identity, then the three lines meet in a point.
=
Find the equation of the bisector of the angle between Ax + By + C = and Afx -f B y -f C = 0. Ax + By = __ (A x +j^y + CQ
10.
the two lines
+_C "
VA + B 2
VA + B
2
2
2
GENERAL EXAMPLES.
A
1.
straight line makes an angle of 45 with the X-axis off a distance 2 on the Y-axis what is its equation
=
and cuts
when
;
the axes are inclined to each other at an angle of 75 ? Ans. y -Y/2 x -j- 2.
=
2.
Prove 3
y 3.
a;
-f-
that
f
If
the
lines
y
=x (1,
-{- 1,
3 intersect in the point
(x
y
,
)
and
tremities of a line,
ordinates of
its
(x",
y"}
show that
y
=2x+2
and
0).
are the co-ordinates of the ex I
t~,
S-
1_2.
a rf the co j
middle point.
of the sides of a triangle are y = x -f- 1, x 1 required the equations of the sides of the triangle formed by joining the middle points of the sides 4.
a;
The equations
= 4,
y
=
;
of the given triangle.
(y Ans.
lyo ^
5.
=
=x ,
JL
_
x
+4 4
q O.
Prove that the perpendiculars erected at the middle
points of the sides of a triangle meet in a
NOTE.
Take the origin
common
at one of the vertices
point.
and make
the X-axis coincide with one of the sides. Find the equations of the sides and then find the equations of the perpendiculars ;
at the
middle points of the
sides. The point of intersection of any two of these perpendiculars ought to satisfy the equa tion of the third.
PLANE ANALYTIC GEOMETRY.
48 6.
Prove that the perpendiculars from the vertices of a meet in a point.
triangle to the sides opposite 7.
Prove that the line joining the middle points of two of is parallel to the third side and is equal
the sides of a triangle to one-half of it.
8. The co-ordinates of two of the opposite vertices of a square are (2, 1) and (4, 3) required the co-ordinates of the other two vertices and the equations of the sides. Ans. (4, 1), (2, 3) y 4. 2, x 1, y 3, x ;
=
;
9.
=
=
=
Prove that the diagonals of a parallelogram bisect each
other.
Prove that the diagonals of a rhombus bisect each other
10.
at right angles. 11.
Prove that the diagonals of a rectangle are equal.
12.
Prove that the diagonals of a square are equal and
bi
sect each other at right angles.
The distance between the points (x, y) and the algebraic expression of the fact. give 13.
Ans. (x
The points
14.
-
+
1)2
A
15.
3) are equi-distant
points
(x",
at an angle
L
y
2), (
What
16.
(
_
2
2)
= (x -
2
2)
+
(y
-
circle circumscribes the triangle
(3, 4), (1,
Ans.
+
2
I)
=4
- 2) = 4 2
(y
;
2 .
from the point
Express the fact algebraically.
(x, y).
(x
(1, 2), (2,
-
(1, 2) is
=
is
y"),
(3
1,
2)
;
a
3)
;
x
or,
whose
+ y = 4.
vertices are
required the co-ordinates of its centre. Ans. (2, 1).
the expression for the distance between the inclined (x ?/ ), the co-ordinate axes being ,
?
- xj + 2
V<V
(y"
- yj + 2
-x (x"
)
(y"
-y
}
cos
THE STRAIGHT LINE.
49
17. Given the perpendicular distance (p) of a straight line from the origin and the angle () which the perpendicular makes with the X-axis required the polar equation of the line. ;
Ans. cos (0 18.
origin on the line 19.
)
Required the length of the perpendicular from the
What
the point
is
|
+|=
2 4
AnSt
the equation of the line which passes through and makes an angle of 45 with the line
2),
(1,
1.
whose equation
is
y
-\-
2x
=
1 ?
Ans.
y
=
- x
+_
.
of two lines passes through the points (1, 2), the other passes through the point (1, 3), and makes an angle of 45 with the first line required the 20.
One
4,
(
3),
;
equations of the lines. Ans.
= 1. x + 1, and y = 3, or x y in the normal equation of a line, through 21. If p = what point does the line pass, and what does its equation Ans. (0, 0) y = s x. become ? ;
22.
r sin
Required the perpendicular distance of the point (r cos $, Ans. r from the line x cos y sin = p. p.
+
0),
Given the base of a triangle the squares of its sides == 4 c 2 the vertex is a straight line. 23.
of
.
= 2 a, Show
and the difference that the locus of
24. What are the equations of the lines which pass through the origin, and divide the line joining the points (0, 1), (1, 0), into three equal parts. 2 y x. Ans. 2 x ?/,
=
=
25.
If (x
,
y ) and
show that the point
(x",
(
y")
be the co-ordinates of two points,
+ nx m + "/^divides the m+n m n
mx
"
y"
^
ii
ne
-\-
joining
m
:
n.
them
into
two parts which bear to each the
ratio
PLANE ANALYTIC GEOMETRY.
50
CHAPTER
IV.
TRANSFORMATION OF CO-ORDINATES. IT frequently happens that the discussion of an equa and the deduction of the properties of the locus it represents are greatly simplified by changing the position of the axes to which the locus is the referred, thus 31.
tion,
simplifying
equation, or reducing
some desired form. The operation accomplished is termed the TRANSFORMATION it
to
by which this is OF CO-ORDINATES.
FIG.
The equation
of the line PC, Fig. 17,
y
when axes
= sx
and
X
its
y
is
-f b
referred to the axes
Y
17.
Y
and X.
If
we
refer
equation takes the simpler form
= sx
.
it
to the
TRANSFORMATION OF CO-ORDINATES. we
If
refer
it
to
Y"
and
X",
51
the equation assumes the yet
simpler form y"
appears that the position of the axes materially
it
Hence,
affects the
- 0.
form of the equation of a locus referred to them.
The equation
NOTE.
of a locus
which
referred to rec
is
of tangular co-ordinates is called the KECTANGULAR EQUATION the locus when referred to polar co-ordinates, the equation is called the POLAR EQUATION of the locus. ;
To find the equation of transformation from one system
32.
of co-ordinates to a parallel system, the origin being changed.
O
FIG.
18.
X
and Y as axes, be any plane locus referred to be any point on that locus. Draw PB to OY then from the figure, we have, of P when referred (OB, BP) (x, y) for the co-ordinates
CM
Let
and
let
P
;
||
=
to
X
and
Y
;
= (# ,/) for the
(O A, AP) to
X
Y DO
and
(OD, origin.
co-ordinates of
P when
referred
;
)
(a,
b)
for the
co-ordinates
of
,
the
new
PLANE ANALYTIC GEOMETRY.
52
From
OB = OD
the figure
x
hence
=
a
+
-f
DB
x and y
=b
BP = BA
and
;
-\-
+ AP
;
y
are the desired equations. As these equations express the relations between x, a, x y, on the locus they express the relations b, and y for any point f
,
Hence, since the
between the quantities for every point.
CM
Q expresses the relationship between equation of th locus the co-ordinates of every point on it if we substitute for
x and y in that equation their values in terms of x and y the resulting or transformed equation will express the rela tionship between the x and y co-ordinates for every point on it.
EXAMPLES. 1.
What
origin
is
=
does the equation y
removed
+1
3x
become when the
to (2, 3) ?
Ans. y
x
Construct the locus of the equation 2 y fer the origin to (1, 2) and re-construct. 2.
3.
The equation
of a curve
is
2 ?/
2 -f x
(2,
-
origin
2) ?
What
does the equation
become when the origin
is
2 ?/
- 2 x* -
removed
to (
,
9
The equation
of a circle
is
x*
+
2 ?/
4.
8
=
;
taken at
+ y = 16. + 6x - 3 =
2y 1
2
? )
Ans. 2 5.
is
-\-
Trans
x2
Ans. 4.
= 2. 4 x
4 y
-\-
what does the equation become when the
= 3x
2 ?/
=a
2
4 x2
when
=
-
1.
referred
does this rectangular axes through the centre. What left-hand the at equation become when the origin is taken
to
extremity of the horizontal diameter
?
TRANSFORMATION OF CO-ORDINATES. 33.
53
To find the equations of transformation from a rectangu-
lar system to
an
oblique system, the origin being changed.
B
L
D FJG.
19.
Let P be any point on the locus CM. Let O Y O X be the new axes, making the angles y and with the X-axis. Draw PA to the Y -axis also the lines ,
;
||
O D, AL, PB AF, O K to ||
From
the figure,
we
to the Y-axis, the X-axis. ||
and
have,
OB = OD PB = DO
+ O N + AF, + AN + PF.
and
OB = x, OD = a, O N == x cos 6, AF = PB = y, DO = b, AN = x sin 0, PF =
But
hence, substituting,
we
cos g>,
sin <p
;
have,
=a+x = b x y x
+ y cos + y sin
cos sin
-}-
)
,^
q>
<p
)
for the required equations.
COR.
1.
If a
= 0,
have,
x
y
and
b
= x! cos = x sin B
= 0, -f-
-f
y cos y sin
coincides with 0, and
<p
go
)
...
we
(2)
.
for the equations of transformation from a rectangular system to an oblique system, the ori-gin remaining the same.
PLANE ANALYTIC GEOMETRY.
54
= 90 + 6, coincides with 2. If a = 0, b = 0, and and the new axes X and Y are rectangular. Making these substitutions, and recollecting that COR.
we
g>
cos
(p
sin
(f
= cos (90 + 0) = sin = sin (90 + 0) = cos
0,
and
9,
have,
=x y = a
x
y sin 9 \
cos
+ ?/ cos
sin
,^
j
one rectangular system /or &e equations of transformation from the to another rectangular system, origin remaining the same. If we rind the values of x and y in equations (2) NOTE. in terms of x and y we obtain the equations of transforma tion from an oblique system to a rectangular system, the
origin remaining the same.
EXAMPLES.
=
16 become when the does the equation x* if ? 45 of an angle axes are turned through Ans. The equation is unchanged. 2.
tion
+
What
1.
1 required the equa The equation of a line is y = x axes to referred when making angles of line same the of ;
45 and 135
with the old axis of
x.
Ans.
3.
when
y
=
-\A-
does the equation of the line in Example 2 become an referred to the old Y-axis and a new X-axis, making
What
angle of 30
with the old X-axis. Ans.
2 y
= (V3 -
1)
x
-
2.
TRANSFORMATION OF CO-ORDINATES. 34.
To find the equations of transformation from a rectangu
lar system to a polar system, the origin
FIG.
Let angle
55
(a, b) g>
be the pole and
with the X-axis.
point on
it.
From
non-coincident.
20.
O
S the
initial line,
CM
Let
the figure,
and pole
we
making an
be any locus and have,
P
any
OB = OD BP = DO But
OB = x, BP = y,
+ O F, + FP. = OD a, O F = O P cos PO F = r cos (B + DO = FP = O P sin PO F = r sin (0 +
g>)
b,
<p)
hence, substituting,
we
=a = b y x
have,
-f-
-(-
;
r cos
qp))
+?
r sin (B
)
)
for the required equations.
COR.
1.
If the initial line
usual to so take
it)
x y
cp
= 0,
O
S
is
parallel to the X-axis
(it is
and
= a + r cos = b + r sin
^
) )
become the equations of transformation. COE. line
2.
If the pole is taken at the origin 0,
made coincident with
the X-axis a
and the
initial
= 0, b = 0, and = 0. g>
PLANE ANALYTIC GEOMETRY.
56
Hence, in this case,
s-rcosfl) (3)
_
= r sin 6
y
)
will be the required equations of transformation.
35.
system
1.
To find the equations of transformation from a polar to a rectangular system.
When
From
and
the pole
initial line coincides
equations
=
2
adding
r
by division
tan
origin are coincident,
and when the
with the X-axis. 2
cc
we
Art. 34,
(3),
+y
=
2 ;
have, by squaring and
and,
.
x
for the required equations.
We
have, also, from the same
equations,
Vx +
r
2.
When
2
the pole
y
and
Vx + y
r
2
2
2
origin are non-coincident,
and when
the initial line is parallel to the X-axis.
From
a simi equations (2) of the same article, we have, by
lar process,
= (x- ay + tan. 6 = y^~~-
r*
x cos
=x
sin0
=_y
,,
a
=
a
;
(y
-
also
a
x
-
2
b)
b
V(* for the required equations.
-
ay
+
(y
-
W
TRANSFORMATION OF CO-ORDINATES.
57
EXAMPLES.
The rectangular equation of the circle is x + y = a what is its polar equation when the origin and pole are coin 2
2
2
1.
cident and the initial line coincides with the X-axis ? Ans.
The equation
2.
of a curve
is (x
2
+y
2
2
)
=a
2
2
= a.
r
y
(x
;
2
)
re
;
and initial being taken as quired its polar equation, the pole in the previous example. a 2 cos 2 0. Ans. r 2
=
Deduce the rectangular equation of the following curves, assuming the origin at the pole and the initial line coincident with the X-axis.
= a tan sec ^ws. o^ = a? y. sec 4. r = a tan = a y. 2
r
3.
;c
3
r2
=a
6.
2
sin 2 2
Ans.
2
2
2
5.
(x
+
a (cos 2 Ans. x 2 y r ==
+
2
2
2 ?/
)
=2a
scy.
sin 0)
=a
y}
(x
GENERAL EXAMPLES. Construct each of the following the origin to the point indicated, the to the old, and reconstruct
straight lines, transfer
new axes being
parallel
:
1.
2. 3. 4.
= 3 x + 1 to (1, 2). to (- 1, 2). 2 y - x - 2 = _ i = to 4 (- 2, - 1). y y + x _f_i = o to (0, 2).
y
_|_
What do referred to
5.
6. 7.
8.
= sx + b to d). - 2). y + 2 = to (2, = mx to w). y -4o; + c = 0to(d,0).
y
(c,
a;
?/
(Z,
the equations of the following curves become when a parallel (rectangular) system of co-ordinates
passing through the indicated points ? 9.
10
"
11.
3 x2
?
+2y = 2
= 4 *(1
9 y2
6, (
,
0).
>>-
- 4x = 2
V2
12.
36
(3, 0).
/_ 2
*
PLANE ANALYTIC GEOMETRY.
58
=
What
13.
X-axis
is
Y-axis
is
does the equation x 2 -f- y 2 4 become when the turned to the left through an angle of 30 and the turned to the right through the same angle ?
14. What does the equation x axes are turned through an angle of 2
15.
What
2
,y
=a 45
become when the
?
=
the polar equation of the curve y 2
is
pole and origin being coincident, and the ing with the X-axis ? 16.
2
The polar equation
of a curve
r
is
2 px, the
initial line coincid
=a
(1
-f-
2 cos 0)
;
required rectangular equation, the origin and pole being coincident and the X-axis coinciding with the initial line. its
2
Ans.
-f
(x
y
2
2 ax) 2
=
a 2 (x 2
-f
y
2
).
Required the rectangular equation of the following curves, the pole, origin, initial line, and X-axis being related as in
Example 17.
r
2
16..
*
=
20.
r
=
asec 2 ?-. 2
cos 2
Ans. x 2 18.
19.
2
y
= a sin r = aO.
r
0.
=a
2 .
21.
22.
r r
2
= a sin 2 0. -2r (cos + V3 sin
Find the polar equations of the equations are 23. 24.
2
2
whose rectangular
:
= y (2 a - x). 4a x = y (2 a x). xs
loci
2
25. 26.
= a x - x\ x* -f y* = a*. 2
a*y
2
4
= 5. 6>)
THE CIRCLE.
CHAPTEE
59
V.
THE CIRCLE. 36.
THE
generated by a point moving in remain at the same distance from a will be observed that the circle as here de
circle is a curve
the same plane so fixed point.
fined
is
It
as to
the same as the circumference as defined in plane
geometry. 37.
Given the centre of a
circle
and
radius
its
to
deduce
its
equation. Y
Let C
(^ $)
be the centre of the
point on the curve.
OX
;
then
circle,
PM
to ||
and
OY
let
P
and
be any
CN
^^
= (*(, & MP) = (x, y)
(OA, AC)
(OM,
Draw CA and )
are the co-ordinates of the centre C.
are the co-ordinates of the point P.
||
to
PLANE ANALYTIC GEOMETRY.
60
=
Let CF
From
a.
CN ON NP
But
2
have,
+ NP = CP ... (1) = (OM - OA) = -^, = (MP - AC) = (y -^) and =a 2
2
;
2
2
(a;
2
2
2
,
CP 2
2
Substituting these values in 2
(x
we
the figure,
-1^)
+
we
(1),
have,
-$) = a 2
(y
.
.
.
for the required, equation. For equation (2) relation existing between the co-ordinates of
the circle
(2)
expresses the
any point (P) on between the co
hence it expresses the relation ordinates of every point. It is, therefore, the equation of the ;
circle.
If in (2)
we make x a*
+ ,f CC
or,
symmetrically,
= =a
2
.
2
- 4-
= 0,
and y .
.
=1
?/
have,
*
(3)
.
.
.
(4)
L
CL
we
*
for the equation of the circle when referred to rectangular axes passing through the centre.
Let the student discuss and construct equation
(3).
See
Art. 11.
COR. factor,
1.
we
If
we transpose x z
have, 2 ?/
i.e.,
=
(a
+ x)
in (3) to the second
x)
(a
in the circle the ordinate
is
member and
;
a mean proportional between
the segments into which it divides the diameter. COR. 2. If we take L, Fig. 21, as the origin of co-ordinates, and the diameter as the X-axis, we have,
LC
H = x = a and y = 0.
These values of x and y (x
-
2
a)
+
+
in (2) give 2 ?/
=a
2 ,
=
2 2 2 ax reduction, x y (5) for the equation of the circle when referred to rectangular
or, after
.
.
.
THE CIRCLE.
61
axes taken at the left hand extremity of the horizontal diameter. 38.
Every equation of the second degree between two varia which the coefficients of the second powers of the variables are equal and the term in xy is missing, is the equa bles,
in
tion
of a
circle.
The most general equation these conditions obtain
ax 2
of the second degree in
which
is
+ ai/ + cx+dy+f=Q.
.
.
.
(1)
Dividing through by a and re-arranging, we have, a If to both
members we now add c
2
4 a2 the equation
may
+
d*
4 a2
be put under the form
4 a2 this with (2) of the preceding article,
Comparing that
it is
we
see
the equation of a circle in which c ~a
d ~2~o
are the co-ordinates of the centre and
=
COR. 1. If ax 2 -{be the equation -f ex dy -f- m of another circle, it must be concentric with the circle repre sented by (1) for the co-ordinates of the centre are the same. ay"-
+
;
Hence, when the equations of
circles
have the variables in
PLANE ANALYTIC GEOMETRY.
62
their terms affected with equal coefficients, each to each, the
Thus
circles are concentric.
are the equations of concentric circles.
EXAMPLES.
What
is
of the
the equation
circle
when
the origin
is
taken.
What
1.
At D,
Fig. 21 ?
Ans.
x2
2.
At K,
Fig. 21 ?
Ans.
x2
3.
At H,
Fig. 21 ?
Ans.
x2
+ f - 2 ay = 0. + if + 2 ay = 0. + if + 2 ax = 0.
are the co-ordinates of the centres, and the values of
the radii of the following circles ? 4.
Ans. 5.
a; a-t-
2 ,*
7.
+2
2 z2
X2
2
,
?/
=
2, 3),
a
= 4.
a
= 2.
a
= 3.
4) a
= 5.
+ 4a;-6y-3=0. (-
^tns.
6
a
(1, 1),
2
= 0.
8x
?/
Ans.
(2,0),
_ 6 x = o. (3, 0),
8.
xz
+
2
?/
4 x -f 8
5
?/
= 0. .
9.
x 2 H-
10.
x2
11.
x2 2
12.
x
13.
x2
2 ?/
mz
+^ +G= 2/
+ = m. 4* = + =c +^ cz + =/ 2
?/
2
?/
2
2
2
.
2/
2
7/
wy.
(2,
THE CIRCLE. Write the equations of the
circles
63 whose
and whose
radii
centres are 14.
a
= 3,
18.
(0, 1).
x*+y -2y = 8. 15. a -2, (1, - 2). 19. Ans. x + if -2x + 4 y + 1 =0. 20. 16. a = 5, (- 2, - 2). + y -f 4 + 4 = 17. 17. a = 4, (0, 0). 21. = 16.
a
= m,
a
=
a
= 5,
(b, c).
2
Ans.
b, (c,
-
d).
2
^4rcs.
z
2
ic
The radius
of a circle -\-
y
5
is
;
what
is its
\
24.
(2, i).
equation
x 2 -f
2
common
centre the point
Find the equation of a
(2,
circle passing
4
?/
Write the equations of two concentric
have for their
A;,
if it is
4ic=2?
2
Ans.
v
=
?/
concentric with x 2
23.
&).
2
2
cc
22.
(7,
a;
?/
cc
circles
1).
= 21. which
,
through three
given points. 39.
To deduce the polar equation of the
circle.
B FIG.
The equation
of the circle (X
-XY+
22.
when
(y
referred to
_ y Y = a\
OY,
OX
is
PLANE ANALYTIC GEOMETRY,
64
To deduce the curve, then
polar equation let
P
be any point of the
(OA, AP) = (x, y) - (* y ) (OB, BO ) = (r, 0) (OP, POA) ,
= (/, = r cos
(00 O OB) ,
From
A=
the figure,
OB = hence, substituting, (
a;
we
=r
cos
,
)
AP = y = r sin BO = y = r sin
2
)
+
(r sin
- / sin
Squaring and collecting, we have,
+ sin 0) + r (cos + sin =a sin r + r - 2 r/ cos (0 -
+ sin
2
ff
)
2
)
=
a
2 .
- 2 r/ (cos
cos
)
2
&e polar equation of the
is
:
2
2
i.e.,
2
2
2
2
^(cos
0,
0,
have,
r cos
r C os 6
x
)
=
a*
.
.
.
(1)
circle.
This equation might have been obtained directly from the triangle*
COR. tre
OO P.
1.
If
= 0, the initial line OX passes through the cen
and the equation becomes r2
COR.
2.
Tf
+/
= 0,
a
2 rr cos
and /
= a,
=a
2 .
the pole lies on the circum
ference and the equation becomes
= 2 a cos = 0, and / = 0,
r
COR.
3.
If
0.
the pole
is
at the centre
and
the equation becomes r
40.
= a.
To show that the supplemental chords of the
each other. perpendicular The supplemental chords of a
circle are
to
circle are those chords
diameter and pass through the extremities of any other on the circumference.
which
intersect each
65
FIG.
23.
Let PB, PA be a pair of supplemental chords. prove that they are at right angles to each other. The equation of a line through B ( a, o) is
For a
line
A
through y = s
Multiplying these, if
=
(a, o),
We
wish to
we have
a).
(x
member by member, we have a 2) ... (a) ss (x*
for an equation which expresses the relation between the co-ordinates of the point of intersection of the lines.
Since the lines must not only intersect, but intersect on the circle
whose equation
this equation
above
;
is
must subsist
hence, dividing,
same time with equation
at the
(a)
we have
"1
or,
1
+ ss =
...
(1)
Hence the supplemental chords
of a circle are perpendicular
to each other.
Let the student discuss the proposition for a pair of chords passing through the extremities of the vertical diameter.
PLANE ANALYTIC GEOMETRY.
66
To deduce the equation of the tangent
41.
FIG.
to the circle.
24.
Let CS be any line cutting the Its equation is
circle in the points
P
r
(#
,
y
),
P"
(x",
y").
r
x
(x
Since the points (a/, y ), the equations of condition /**
(x",
I
/"I
~~
J
*
*
(Art. 26, (4)
are on the circle,
y")
2
/2
i
).
*
).
we have
\
\
/
These three equations must subsist at the same time hence, subtracting (2) from (1) and factoring, we have, ;
f
(x
+ y x
x")
y
-
(*
x")
+
y,
_|_
X
y
+
y"
^
.x"
+
(y
l
y")
(y
Substituting in the equation of the secant line
If
point
we now revolve the secant
P
when the
will approach
secant
P
;/
CS becomes
line
-
y")
-
;
"
it
becomes
upward about
P"
the
will finally coincide with it tangent to the curve. But when
and
THE CIRCLE. P
coincides with
in (3)
we
P",
x"
and
?/
=
?/";
hence, substituting
have,
-
y or, after
=
x
67
y"
=-
^
(*
-
(*)
*")>
reduction,
+ yy _;..;.. "
a-x"
or,
(5)
symmetrically,
for the equation of the tangent.
SCHOL. The SUB-TANGENT for a given point of a curve is the distance from the foot of the ordinate of the point of tangency to the point in in Fig. 24,
which the tangent intersects the X-axis
AT
its
value
we
have,
P".
make y
OT = But
thus,
;
the sub-tangent for the point To find in the equation of the tangent (5) and
is
x
=
.
AT = OT - OA =
-
x"
x"
= an ~ x = V,//2 -2
sub-tangent
"2
.
To deduce the equation of the normal to the The normal to a curve at a given point is a dicular to the tangent drawn at that point. 42.
The equation 24, is
of
any
-
y
y"
through the point
line
= s (x -
x")
...
circle.
line
perpen
P"
(x",
y")
Fig.
(1)
In order that this line shall be perpendicular to the tangent P"T, we must have 1
But
Art. 41, (4)
-f s
88
=
= 0.
~ y"
;
hence,
we must have
s
=V x"
.
PLANE ANALYTIC GEOMETRY.
68
Therefore, substituting in
y
y"
we
(1),
-*(*-
have, (2)
*")
;
*
or, after
reduction,
yX
-X
"
y"
=
...
(3)
for the equation of the normal. see from the form of this equation that the the circle passes through the centre.
We
normal to
SCHOL. The SUB-NORMAL for a given point on a curve is the distance from the foot of the ordinate of the point to the In the circle, point in which the normal intersects the X-axis.
we
see
from Fig. 24 that the Sub-normal =
By methods
43.
the last two articles, (x
to be
-x
x" .
precisely analogous to those developed in we may prove the equation of the tangent
- x )* + (y- y Y = *
- x ) + (y - y ) ) and that of the normal to be - x ) -(x(y -y") (x
r
(x"
(y"
- y } = a*
-y Let the student deduce these equations. x")
(x"
(y"
)
=
.
.
.
...
(1)
(2)
EXAMPLES. 1.
dy
What is
+f=
0,
+
+
+
the polar equation of the circle ax 2 ex ay* the origin being taken as the pole and the X-axis
as the initial line ?
Ans.
r2
+ (- cos \a
2.
x2
What
is
+ f = 25
equation of (3,
- sin a
\r J
What
+ /a = 0.
the
tangent to the circle of the subAns. 3 x 4y 25 J^.
The value
4) ?
+
tangent ? 3.
the
at the point
-f
=
;
the equation of the normal to the circle 37 at the point (1, 6) ? What is the value of the
x 2 -f if = sub-normal
is
? --
Ans. y
= 6x
;
1.
THE CIRCLE. What
4.
x2
circle
are the equations of the tangent and normal to the 20 at the point whose abscissa is 2 and ordi-
if
-f-
69
=
Give also the values of the sub-tangent and sub-normal for this point. nate negative ?
2
Ans.
cc
- 4 = 20 =8 ?/
Sub-tangent
2
;
;
?/
+4 = = 2. cc
;
sub-normal
Give the equations of the tangents and normals, and the values of the sub-tangents and sub-normals, to the following circles 5. 6.
7.
:
+ ?f = 12, at (2, + V8). = 25, at (3, - 4)! x + = 20, at (2, ordinate +). x + x -f if = 32, at (abscissa +, 4). x + y = a at y = m, at (1, ordinate +). = &, at (2, ordinate x + = 18, at (m. ordinate -f). = 45 and the Given the circle x + x* z
2
?/
2
2
?/
2
8.
9.
2
2
2
2
2
,
10.
a?
-|-
2
11.
(6, c).
2
-j- ?/
12.
cc
2
).
2
7/
2
13.
+ =
2
line 2 y x 2 required the equations of the tangents to the circle which are parallel to the line. ?/
Ans
What
14.
x~
+
2 2/
f3*
{3*
;
+ 6y = 45. + 6y=-45.
are the equations of the tangents to the circle
= 45 which are perpendicular to the line ,
2y
-f-
^
=2?
(3y- 6x^45.
- 3 y = 45. x + y = 9
16 x
The point
1 2 (3, 6) lies outside of the circle the of the required equations tangents to the circle which pass through this point. X 3 Ans. A 4 3x 15. y
16.
;
=
=
PLANE ANALYTIC GEOMETRY.
70
What
17.
_
(x
2
2)
equation of the tangent to the circle the point (4, 4) ? Ans. 2 x -f y 12.
the
= 5 at
2
3)
(y
=
The equation
18.
the
+
is
x2
circle
-\-
y*
of one of
=9
is
y
=
two supplementary chords of x -|- 2, what is the equation
of the other ?
2
?/
-f 3
.x
== 9.
Find the equations of the lines which touch the circle 2 2 = r* and which are parallel to y = sx -f- c. a) -\- (y b)
19.
(x
x 2 + ?/ 2 4 x + 4 ?/ = 9 of the normal at the the point whose required equation abscissa == 3, and whose ordinate is positive. Ans. 4 x y = 10.
The equation
20.
of a circle
is
;
44. To find the length of that portion of the tangent lying between any point on it and the point of tangency.
Let (xu this point
be the point on the tangent. The distance of 1/1) from the centre of the circle whose equation is
x
(x V(a>!
Y
-f-
y
(y
-xY+
Y
<*?
- y y.
(y,
is
evidently
See Art. 27,
(1).
But
this distance is the hypothenuse of a right angled tri whose sides are the radius a and the required distance angle
d along the tangent d*
COR.
1.
If
x
= d2
as
it
;
=
hence (x,
-xy+
and y
= 0,
(2/1
then
= x^ + ^ - a
- yY - ^ (1)
...
(i)
becomes
2 .
.
.
(2)
ought.
45.
To deduce the equation of the radical axis of two given
circles.
The RADICAL AXIS OF TWO CIRCLES is the locus of a point from which tangents drawn to the two circles are equal.
THE CIRCLE.
71
T
FIG.
Let
x
(x \
Y
+ (v + (y
/
x")
y/
\*/
I
2
(x
25.
2
} /
y")
=a =b
2 ,
2
be the given
Let P (XD yi) be any point on the radical axis the preceding article, we have,
d2 = d2 = .:
+
6/1
2
b
y")
(x"
,
x
)
(y"
Calling, for brevity,
w iH
2
(
)
2
(
y")
2
)
1
y)
-f (y t
hence, reducing,
- x x,+2
-f-
(#15 2/i)
+ Vl - y + yi 2
x")
( Xl 2
2
-
(x,
by definition 2
-x )
(x,
=
y ) y, 2 a2 b
we
then from
;
a2 b2
=
a2
2
( Xl
x")
have, 2
x"
2
circles.
-
x
2
+
-y
2 y"
2
.
the second
member m, we
see that
satisfy the equation. 2
(x"
-x )x+2
(y"
y
)
y
=m
.
.
.
(1)
But (xu 1/1) is any point on the radical axis hence every It is, therefore, the re point on that axis will satisfy (1). ;
quired equation.
COR. then, is
1.
If c
=
and
c
c
=
c
=
be the equation of two circles,
the equation of their radical axis.
PLANE ANALYTIC GEOMETRY.
72
2. From the method of deducing (1) it is easily seen the two circles intersect, the co-ordinates of their points of intersection must satisfy (1) hence the radical axis of two
COR.
that
if
;
intersecting circles is the line joining their points of intersection,
PA,
Fig. 25.
Let the student prove that the radical axis of any two circles is perpendicular to the line joining their centres.
To show that the radical axes of three given a common point.
46.
circles in
tersect in
Let
c
= 0,
c
= 0,
and
c"
=
be the equations of the three circles. Taking the circles two and two we have for the equations of their radical axes c c
-c = - = _c =o
.
c"
.
.
(1)
...
(2)
"
c
.
.
.
(3)
evident that the values of x and y which simultaneously will also satisfy (3) hence the proposition. and
It is
satisfy (1)
(2)
;
The intersection of the radical axes of three given circles
THE RADICAL CENTRE
called
is
of the circles.
EXAMPLES. Find the lengths of the tangents drawn to the following circles 1.
:
(x
-
2
2)
+
(y
-
2
3)
=
16 from
(7, 2).
Ans. 2.
x>
+
(y
+ 2)
2
-
10 from
4. 5.
_ a y + if = 12 from x 2_f2/ 2_2^+4?/-2 from x + y = 25 from (6, 3).
VlO.
d=^/3.
(b, c).
(a-
2
=
(3, 0).
Ans.
3
d
(3, 1).
2
Ans.
d
=
V20.
THE CIRCLE. 6.
7>
g.
+y 2
x2
_
(3
X
2
a )2
+y
i
=
2 x
10 from
73
(5, 2).
= 3.
Ans. d
+
_
_
(^
6)2
=
4y
=
rom
c f
10 rom f
(d,
/).
(0, 0).
Give the equations of the radical axis of each of the follow ing pairs of circles
:
9.
10.
13.
+_ y 3)2 | (x (
1
x2
2 |X
_^_
* |x
_|_ (
(
x
2
^2
4y ^2
_|_
_ _
y
2 y
3)2
=
0.
_
16
+- y = 16. = I) +
^s.
o.
-4?w.
3
= 2 y.
a?
a?
=
i-
= 0.
z
2
2
1 (x
= 0. _9 _
2/
a?.
14.
Find the co-ordinates of the radical centres of each of the following systems of circles 15.
:
= 9. - 2 = 25. 2
16.
+ if - 4 x + 6 y - 3 = -4* = 12. = 6 | x = (.T + = 9. J I) + f (^ __ 2 4. 4 y = 10. ( ^2 + yj x + if - k* = -j= d. // (
J
x-
_j_
?/
_|_
y>
7.
2
2
.
7/
2
a,
18.
(
G,
^rcs.
(1,
-
3).
0.
x8 4-3^ .2
17.
-4w.
2
ft
2
2
?/
?H
.
2
-f-
//
I)-
PLANE ANALYTIC GEOMETRY.
74 47.
To find the condition that a straight
= sx + b + =
line
y
2 a2 must fulfil in order that it may touch the circle x y~ In order that the line may touch the circle the perpendicu lar let fall from the centre on the line must be equal to the .
radius of the circle.
From
p
= bcosy = p
r2 (1
hence, is
we have
Art. 21, Fig. 13,
-
= VI
+ tan.
2
=r=
+s =b 2
2
)
.
.
.
(1)
the required condition. COR. 1. If we substitute the value of b
the equation y
-
-==
sec y
= sx + y = sx
b,
AVC
_j_
r
drawn from
(1) in
have
Vl +s
2
...
(2)
for the equation of the tangent in terms of its slope.
48. Two tangents are drawn from a point without the circle ; the points oftangency. required the equation of the chord joining
FIG.
26.
y ) be the given point, and tangents through it to the circle. Let
P
r
f
(x
,
let
PT",
P P,
be the
THE CIRCLE.
75
deduce the equation of The equation of a tangent through It is required to
PP
/X .
is
P"
y")
(#",
\yy^_
xx"
O
P
Since
on this
is (a/, ?/)
the equation
i
line, its co-ordinates
must
satisfy
hence
;
x
The point
_
rt
I
(x",
1/1/ = 1 + ^~ a
x"
*
.
2 aT
2
?/"),
therefore, satisfies the equation
A
it is a point on the locus represented by similar (1). source of reasoning will show that P is also a point of this locus. But (1) is the equation of a straight line hence, since .-.
;
it
is satisfied
for the co-ordinates of both
P"
equation of the straight line joining them. the required equation. 49.
A
chord of a given
and P, It
circle is revolved
is,
it is
the
therefore,
about one of
its
points ; required the equation of the locus generated by the point of intersection of a pair of tangents drawn to the circle at the points in which the chord cuts the circle.
Let
P
f
f
Fig. 27, be the point about which the chord It is required to find the equation of the locus generated by Pj (x l5 yj, the intersection of the tangents revolves about P 1? 1? as the line P
P AB
(x
,
y
),
revolves.
AP BP From
AB
.
the preceding article the equation of the chord
X^X
,
?/!?/ =
-,
^"""^
Since
P
(a/,
y)
on this
is
a2
xx hence Cb
^
we have
line,
a"
.
-\
y y == l 1 % Cf
.
.
/1
,
(1J
AB
is
PLANE ANALYTIC GEOMETRY.
76 is satisfied
P
for the co-ordinates of
the locus represented by
(1).
any pair of tangents drawn
x
But
(x lt
P
x
y-^)
is
;
hence
P
x
lies
on
the intersection of
to the circle at the points in
hence (1) position, cuts the circle will be satisfied for the co-ordinates of the points of intersec
which the chord, in any
;
tion of every pair of tangents so drawn.
the equation of the required (1) is, therefore, observe that equation (1) is identical with (1) of the preceding article hence the chord PP" is the locus whose
Equation
locus.
We
;
equation
we
sought.
The point P
(**
+& =
THE POLAR 1-1-21 (i
=
1
f
(x
Y
,
y)
is
and the
of the point
called line
P
THE POLE
PP"
(x>,
y>)
of the line
(^ + ^f
=
1\
is
PP"
called
with regard to the circle
THE CIRCLE.
77
the principles here developed are perfectly general, the circle. within or be without^ on, pole may Let the student prove that the line joining the pole and the centre is perpendicular to the polar.
As the
The terms pole and polar used no connection with the same terms used NOTE.
I.
co-ordinates, Chapter
If the polar of the point P (x y ), .Fig. 27, passes through XD yi), then the polar of P (x 1? yi) will pass through P 1
50.
PI
(
The equation
of the polar to
xx ~a*
In order that
P
x
+
yu
But f
(x
,
(x
,
y
) is
=
y^ may be on
(aj 1?
this line,
we must
have,
a2
this is also the equation of condition that the point f y ) may lie on the line whose equation is
^ _i_M
1
"
a2
a2
But
P
^~
a2
P
in this article have in treating of polar
this is the equation of the polar of P! (x^ y-^
;
hence
the proposition.
To ascertain the relationship between the conjugate diam
51. eters
of the
A pair
circle.
to be conjugate when they are the curve is referred to them as axes its equation will contain only the second powers of the variables.
of diameters are said
so related that
when x2
Let
+
2 ?/
ascertain
OY OX ,
,
a2
.
.
.
(1)
its centre and axes. becomes when referred to axes making any angle with each other, we must
be the equation of the
To
=
what
circle,
referred to
this equation
substitute in the rectangular equation the values of the old
PLANE ANALYTIC GEOMETRY.
78
co-ordinates in terms of the new.
have
=x =x
x y
+y +y
cos 9 r
sin
From
the
1
y
*
+ 2 x y cos
-
((f
FIG.
1,
we
cos 9 sin cp
equations of transformation. values in (1) and reducing, we have,
for
Art. 33, Cor.
Substituting these
+x =a 2
0)
2 .
.
.
(2)
28.
diameters Now, in order that OY OX may be conjugate the term containing x y in (2) that related so be must they must disappear; hence the equation of condition, ,
- 6) = = 90, 0*9-e 9
cos ...
(cp
;
The conjugate diameters of the As there dicular to each other.
circle are therefore
are an infinite
perpen
number
of
the circle which satisfy the condition of being pairs of lines in the circle there at right angles to each other, it follows that in diameters. are an infinite number of conjugate
THE CIRCLE.
79
EXAMPLES. 1.
-|-
Prove that the line y = V3 x -+- 10 touches the circle 2 = 25, and find the co-ordinates of the point of tangency. ?/
_ y f5 __
2.
y
What must be the value of b in 2 may touch the circle x +
=2x +b
order that the line
=
2
?/
16
?
^7i. 3.
y
What must be the value of 4 may touch the circle x
= sx
s
=
6
4. _|-
The y
2
=
V80.
-L,
in order that the line
=2?
2
-f if
Ans.
#2
5 3,
s
=
-L_
V7.
slope of a pair of parallel tangents to the circle 16 is 2 required their equations. ;
Ans.
^j
= 2x -
V80.
Two tangents are drawn from a point to a circle required the equation of the chord joining the points of tangency in each of the following cases ;
:
5.
From
6.
From
7.
From
8.
From
(4, 2)
(3,
to x
4) to or
(1, 5) to
x
+ if = 9.
2
2
2
+ if = +
4x
+2y =
^4ws.
3x
+ 4 y = 8.
8.
2
=
2
=c
?/
Ans.
16.
Ans. x (a, 6) to
2
cc
+
7/
9.
10.
Of
= 16.
2 .
v4?is.
What
5 y
-\-
9.
ax
by
-{-
=c
2
are the equations of the polars of the following points (2, 5)
Of
(3,
with regard to the
4)
circle
with regard to the
x
circle
2
.
:
+ y- = 16 ?
^
2
cc
.
5 y
-,
16+16 x + if = 9 ? ^715. 3 x + 4 y = 9. 2
PLANE ANALYTIC GEOMETRY.
80 11.
Of
(a, b)
with regard to the
circle
x2
-\-
What 12.
are the poles of the following lines
Of 2 x
-f 3 y
= m?
y-
Ans.
ax
:
= 5 with regard to the circle x
2
-f
y
Ans. 13.
Of ~
+ y = 4 with regard to the Jg
14.
Of y
2 -
= 25 ?
(10, 15).
circle
^
+ fg = 1 ?
= sx + 6
= m.
by
-\-
5
-
(
2 4 )>
with regard to the circle
Ans
-
(-^r>i
Find the equation of a straight line passing through x 2 -\- if 3 x -\- 4 y = 0. 0) and touching the circle Ans. y = - x.
15. (0,
GENERAL EXAMPLES. 1.
Find the equation of that diameter of a drawn parallel to y = sx + b.
circle
which
bisects all chords
Ans. 2.
line
sy
+#=
0.
Required the co-ordinates of the points in which the x-|-l = intersects the circle
2y
3. Find the co-ordinates of the points drawn through (3, 4) touch the circle
in
which two
lines
and the [The points are common to the chord of contact circle.]
THE CIRCLE. 4. (4,
The centre
0)
;
required
which touches the Y-axis
of a circle its
81
equation. 2
Ans.
(x
4)
is
at
+ y = 16. 2
Find the equation of the circle whose centre is at the = x -|- 3 is tangent. origin and to which the line y Ans. 2 x2 + 2 y* = 9. 5.
+y =
=
+
2
6.
Given x 2
2 16 and (x 4 required the if 5) equation of the circle which has their common chord for a diameter.
7.
Required the equation of the circle which has the dis (3, 4) from the origin as its diameter. Ans. x* -j- ?/ 2 4 y = 0. 3x
5
tance of the point
8. Find the equation of the circle which touches the lines x. 0, and y 3, y represented by x
=
9.
=
=
Find the equation of the
points
(-
(1, 2),
(-
2, 3),
circle
- 1).
1,
which passes through the
10. Required the equation of the circle which circumscribes the triangle whose sides are represented by y 4 x, 0, 3 y 4x and 3 y 6. Ans. x 2 -f ?/ 2 0. | x 1| y
=-
+
=
Required the equation of the circle whose intercepts are a and b, and which passes through the origin. 11.
12. is
The
points
in the line
13.
circle 14.
y
The point x2
ax
4
;
(4,
on a
whose centre
its
required Ans. 2 x 2
2) is the
(3.
6) lie
circle
by
equation.
+2
2 ?/
17 x
y
= 30.
middie point of a chord of the
= 16 required the equation of the chord. = 16 and the chord y 4 x = 8. x +
-\- if-
Given
and
(1, 5)
=x
= 0.
x 2 -f y z
Ans.
;
2
2
?/
Show
that a perpendicular from the centre of the circle bisects the chord. 15.
Find the locus of the centres of
pass through
(2, 4), (3,
-
2).
all
the circles which
PLANE ANALYTIC GEOMETRY,
82 16.
Show
that
if
the polars of two points meet in a third is the pole of the line joining the first
point, then that point
two points. 17.
= 8,
Required the equation of the and whose sub- normal = 2.
circle
whose sub-tangent
= 20.
x 1 -f y 2
Ans.
Required the equation of ihe circle whose sub-normal = 2, the distance of the point in which the tangent intersects the X-axis from the origin beinr = 8. 18.
x2
Ans. 19.
ax 2
Required the conditions
in
order
that
2
-f- ?/
the
= 16.
circles
+ aif + cx + dy + e = Q and ax + af + kx + ly + m =
may
z
be concentric. Ans.
20.
c
d
k,
I.
Required the polar co-ordinates of the centre and the
radius of the circle r2
2 r (cos
+
^/3 sin
0)
= 5.
Ans. 21.
A
line of fixed length so
moves that
60)
(2,
its
;
r
= 3.
extremities
remain in the co-ordinate axes required the equation of the circle generated by its middle point. ;
22.
Find the locus of the vertex of a triangle having given = 2 a and the sum of the squares of its sides = 2 b 2
the base
.
Ans.
x2
-j-
y
2
=
&
2
a2
.
Find the locus of the vertex of a triangle having given the base = 2 a and the ratio of its sides 23.
=
.
Ans.
A
circle.
n 24. Find the locus of the middle points of chords drawn from the extremity of any diameter of the circle
THE PARABOLA.
CHAPTER
83
VI.
THE PARABOLA. THE
parabola is the locus generated by a point moving same plane so as to remain always equidistant from a fixed point and a fixed line. The fixed point is called the Focus the fixed line is called the DIRECTRIX the line drawn through the focus perpendic ular to the directrix is called the Axis the point on the axis midway between the focus and directrix is called the VERTEX 52.
in the
;
;
;
of the parabola. 53.
To find the equation of the parabola, given the focus and
directrix.
Let
EC
be the directrix and let
F
the axis of the curve, and the tangent
be the focus.
OY
drawn
Let OX,
at the vertex
PLANE ANALYTIC GEOMETRY.
84
0, be the co-ordinate axes. to OY, PB
and draw
PA
||
(OA, AP)
From
Take any point P on the curve OX and join P and F. Then
to
||
=
y) are the co-ordinates of P.
(x,
FAP, we have 2 ... (1) generating the curve, we have
the right angled triangle
= AP = FP - FA 2
if
But from the mode of
2
FP = BP = (AO 2
and from the
+ OD) =
2
FA = (AO - OF) = 2
2
Substituting these values in
=
or, after
(
X
2
y
reduction,
As equation
true for every point
;
then
2
= 2 px
hence
(x
-
(x
-
OF)
OD = OF =
true for
is
(3)
(x
2 ,
.
OF)
2 .
we have
(1),
+ OD) -
DF =p,
Let
+ OD)
2
we have
figure,
if
;
.
.
2 .
.
.
(2)
hence
(3)
of the parabola it the equation of the curve.
any point
it is
is
COB. 1. If (x y ) and (x are the co-ordinates of any two points on the parabola, we have, 1
,
hence
,
2 ?/
2 :
y"
\:x
\
y")
x"
;
i.e., the squares of the ordinates of any two points on the bola are to each other as their abscissas.
para
SCHOL. By interchanging x and y, or changing the sign of the second member, or both in (3), we have 2 y 2px for the equation of a parabola symmetrical
=~
with respect to X and extending to the left of Y; x 2 = 2 py for the equation of a parabola symmetrical with respect to Y and extending above X. x2 = 2 py for the equation of a parabola symmetrical with respect to Y and extending below X. Let the student discuss each of those equations. See Art. 13.
THE PARABOLA. 54.
85
To construct the parabola, given the focus and
FIG.
First Method.
Let
DR
directrix.
30.
be the directrix and
let
F
be the
focus.
From F
FD on the directrix it Take a triangular ruler ADC and make its base and altitude coincide with the axis and Attach one end of a string, whose directrix, respectively. the perpendicular
let fall
;
will be the axis of the curve.
A
is AD, to the other end to a pin fixed at F. Place the point of a pencil in the loop formed by the string and stretch it, keeping the point of the pencil pressed against the
length
;
base of the triangle. Now, sliding the triangle up a straight edge placed along the directrix, the point of the pencil will describe the arc OP of the parabola for in every position of the pencil point the condition of its being equally distant from the focus and directrix is satisfied. It is easily seen, for ;
instance, that
FP = PD
when
the triangle
is
in the position
ADC
that
.
Second Method.
Take any point C on the
axis
and erect
PLANE ANALYTIC GEOMETRY.
86
the perpendicular P CP. as a centre and DC (=
With F
Measure the distance DC.
as a radius describe the arc of a
FP)
P CP in P and P P and F will be points of the parabola. By taking other points along the axis we may, by this method, locate as many points of the curve as may be cutting
circle,
.
desired.
55. To find the Lotus-rectum, or parameter of the parabola. The L AT US-RECTUM, or PARAMETER of the parabola, is the
double ordinate passing through the focus. The abscissa of the points in which the latus-rectum pierces
the parabola
is
=
x
.
2i
Making
this substitution in the equation
= 2px * = 2^y* = 2 p 2 y = 2 p. 2
?/
we have Hence COR.
1.
Forming a proportion from the equation if
i.e.,
any
= 2px, (
we have
x:y:\y. 2p;
the latus-rectum of the parabola is a third proportional to abscissa and its corresponding ordinate.
EXAMPLES. Find the latus-rectum and write the equation of the parab ola which contains the point 1.
5.
8,
?/-
=
8
Ans. 4.
What is it
is
?/
=
8
x.
(_,
,
y
2
= b- x. a
2).
-,?/=a parabola x* =
Ans.
the latus-rectum of the
defined in this case ?
2
b*
a
2
8,
(a, I).
x.
(-2,4). Ans.
How
3.
(2,4).
Ans. S.
:
-x. a
THE PARABOLA.
87
6. What is the equation of the line which passes through the vertex and the positive extremity of the latus-rectum of any parabola whose equation is of the form if 2px ?
=
Ans. y 7.
The focus
of a parabola
vertex of the curve
;
what
is
is its
when symmetrical with
(a)
"
"
at 2 units
= 2 x.
distance from the
equation respect to the X-axis ?
"
"
"
(b)
Ans.
y
(a)
2
"
Y-axis?
= 8x,
(b)
x*
=
8
y.
Construct each of the following parabolas by three differ ent methods. 8. 9.
12. 2 ?/
10.
x2
11.
x2
= 6 y. = - 10 y.
What
are the co-ordinates of the points on the parabola where the ordinate and abscissa are equal ? Ans. (0, 0), and (6, 6).
=6x 13.
x2
= Sx. - 4 x. if = if
Required the co-ordinates of the point on the parabola whose ordinate and abscissa bear to each other the
=4y
ration 3
:
Ans.
2.
What
(6, 9).
the equation of the parabola when referred to 2 the directrix and X-axis as axes ? Ans. y 2 2px p 14.
is
.
Find the points of intersection of the following 15.
if
16.
x2
= 4 x and 2 y
x
Ans.
17.
18.
:
= 0. (0, 0), (16, 8).
= 6 y and y x 1 = 0. 8 x and y* = + 3 = 0. = = 8. 2x and x + if a:
2
2
?/
Ans. 19.
x*
20.
x2
(2, 2), (2,
= - 4 y and 3 x + 2 if = 6. = 4 y and if = 4 x. 2
- 2).
PLANE ANALYTIC GEOMETRY.
88
To deduce the polar equation of
56.
the parabola, the focus
being taken as the pole.
The equation To
.
.
.
we have
FX
= + r cos = r sin
0.
0.
Substituting these values in
(1),
we have
= p~ + 2pr cos sin = 1 cos 9 = p -\-2pr cos 9 + r cos 9 = (p + r cos 0) r cos r =p 2
2
9.
?-
But
sin
2
r2
.-.
2
;
2
2
.-.
2
2 ,
9,
-\-
solving, 1
cos B
the required equation.
We Let
might have deduced this value directly as follows P (r, 9) Fig. 29 be any point on the curve then ;
= D A = DF -f FA = p + r cos 9 r=p + rvosO. r = cos 9 1 9 = 0, r = oo e = 90, r = p. 9 = 180, r = | 9 = 270, v = p. 9 = 360, r =
FP i.e.,
Hence COR.
*-
.
If If If
If If
An
;
.
.
-,
1.
is
and the pole F
1,
y
is
Fig. 29,
for the equation of transformation, Art. 34,
]
x
or,
OY, OX,
(1)
refer the curve to the initial line
,
Cor.
of the parabola referred to
f = 2px
.
.
oo.
these results. inspection of the figure will verify
:
THE PARABOLA.
89
To deduce the equation of the tangent to the parabola. be the points in which a secant line cuts (V, y ),
57. If
y")
(x",
the parabola, then
y-y =
y
,~
y
will be its equation.
parabola,
(-**)
",
(!)
Ou
OC-
Since (V,
?/),
(#",
y")
are points of the
we have y
2 /2
7/
= 2px = 2X
...
(2)
/
(
3)
These three equations must subsist at the same hence, subtracting (3) from (2) and factoring, we have y x
i.e.,
y"
time
2p
== f
-x"
y
+y"
Substituting this value in (1), the equation of the secant
becomes
When
the secant, revolved about y ) coincides with
(x",
to the parabola (x
=
y
or,
y".
Making
,
y"},
this substitution in (4),
simplifying, recollecting that yy"
=p(x+x")
2 y"
=2
...
becomes tangent hence x r
(x",
;
y")
we
px".
x",
have,
we have
(6)
for the equation of the tangent to the parabola.
To deduce the value of the sub-tangent. Making y = in (6), Art. 57, we have = OT, (Fig. 31) x = -
58.
x"
which the tangent intersects But the sub-tangent CT is the distance of this
for the abscissa of the point in
the X-axis.
point from the foot of the ordinate of the point of tangency; i.e., twice the distance just found; hence
Sub-tangent
=2
x" ;
PLANE ANALYTIC GEOMETRY.
90 i.e.,
the sub-tangent
is
equal
to
double the abscissa of the point
of tangency. 59.
The preceding
principle affords us a simple method of constructing a tangent to a parabola at a given point. Let be any point of the curve. Draw the ordinate P"C, and measure 00. Lay off OT = 00. P"
(a;",
y")
FIG.
A at
line joining
T and
P"
31.
will be tangent to the parabola
P".
60.
To deduce the equation of the normal
The equation
of any line through
= s(x-
y-y"
We have found Art. 57, s
r
-
p
to the
P"
(#",
.
.
x")
.
y"}
Fig. 31,
(1)
-
hence, for the slope of the normal
,P
is
(5) for the slope of the tangent
~7"
.__*:.
parabola.
P"N,
we have
P"T
THE PARABOLA. Substituting this value of
y
-y"
for the equation of the
=-
we have
s in (1),
-
^ (*
normal
91
(2)
*")
to the
parabola.
To deduce the value of the sub-normal.
61.
Making y
=
in (2) Art. 60,
we
have, after reduction,
x=*p+af = ON; Fig. 31, Sub-normal = NC = p .-.
Hence to the
-\-
the sub-normal in the parabola
is
x"
x"
constant
= p.
and equal
semi-parameter FB.
62.
To
shoiv that the tangents
drawn
at the extremities of
the latus rectum are perpendicular to each other. The co-ordinates of the extremities of the latus-recturn are ? ,
p
j
for the upper point,
and
I
,
p
for the lower point. j
Substituting these values successively in the general equa tion of the tangent line, Art. 57 (6), we have
yp
= plx
-yp=plx + or, cancelling,
y
=
x+|
.
.
.
(1)
y=-*-f
(2)
for the equations of the tangents. As the coefficient of x in (2) is minus the reciprocal of the coefficient of x in (1), the lines are perpendicular to each other.
COR. that x
1.
=
Making y = ?.
2
;
in (1)
and
(2),
we
find in each case
hence, the tangents at the extremities of the
PLANE ANALYTIC GEOMETRY.
92
and the same point.
latus-rectum in the
The values
directrix meet the axis of the parabola
of the coefficients of
these tangent lines
make
x in
angles of 45
(1) and (2) show that with the X-axis.
63. To deduce the equation of the parabola when referred to the tangents at the extremities of the latus-rectum as axes.
FIG.
The equation
of the parabola if = 2px
We
.
.
wish to ascertain what
curve
is
Let
P
DY DX
referred to (x
,
,
.
when
referred to
the figure,
OC
CP )
we
OY, OX,
is
(1).
this equation as axes.
becomes when the
,
be any point of the curve
i/)
(OC,
From
32.
= (x,
y),
;
and (DC
then, Fig. 32 ,
CP
)
have,
= DC - DO = DK + C M - DO;
=
(a*
,
tf).
THE PARABOLA.
93
but
DK = x
= r^C, C M = /cos 45 =
cos 45
V2
,
=
x
hence
_
;
*
V2
V2
Substituting the values of x and (2/
^
- *T -
V2
?/
DPF, we
we
in (1),
(C
In order to simplify this expression the triangle
;
?
= MP - C K
..,
| *
DO =|
-\
CP
have, also,
V2
p --^=-- 5r 2 V2
JK
V2
We
-X-,
+
let
-^
)
2/
have, 2
DP = a
.
-
-
;
then from
(2)
have,
DF =p =
=
a cos 45
-^=-
.
V2<
Substituting this value of
by
2,
we
have,
x )2
(/
+
2
?/
or,
Adding 4
a?
2
ic
^
= 2a 2
a:
/
.-.
transposing, x ...
or,
is
a2
4- ?/)
2
^
2
= 4;ry,
2
<X+/-a)
x
(V
and multiplying through ,
/+
a2
= 0.
both members, the equation takes the form
?/ to
or
in (2)
+y -[-
^x
f
a
=
2x *y *
yx
=
-J-
-+- ?/
a
2
a;
?/
=a ...
symmetrically, dropping accents,
the required equation.
^
^ ;
;
(3)
PLANE ANALYTIC GEOMETRY.
94
/EXAMPLES.
What is the polar equation of the parabola, the pole 1. being taken at the vertex of the curve ? Ans. r 2p cot cosec 0.
=
Find the equation of the tangent to each of the following value of the subtangent in each case parabolas, and give the 2 Ans. y = x + 1 2. = 4 x at (1, 2). 2. :
y
3.
;
x2 2
4.
v/
5.
x2
= 4 y at (- 2, 1). = - 6 x at 6, ord = 8 y at (abs +, = 4 ax at (a, 2 a). = at (m, m). = ^y at (abs +, (
2
6.
?/
7.
y
2
2
8.
a;
9.
x*
y + 1 = =3 Ans. 2 y + ^rcs. x + y = 2
Ans.
a;
+
oj
+). 2).
2.
;
12.
;
;
4.
??ia;
= 2py
at
(abs-, \
p).
p8
Write the equation of the normal to each of the following parabolas 10. 11.
12.
13.
:
= 16 x at (1, 4). To x = - 10 y at (abs +, - 2). mx at To if = m, m). To x = 2 m?/ at ofo To
if 2
(
2
,
-
o
\ \
=
K
* of a parabola is x* _L_ y ? curve the of are the co-ordinates of the vertex 14.
The equation
Ans.
5
-a, -- a 4 V4 I
= 4x
and the line y which is, the of the tangent equation required 15.
Given the parabola
if
what
x
=
;
to the line, (a) parallel (b)
perpendicular to the line. x Ans. (a) y 1, (b) y
= +
+ + 1 = 0. a;
THE PARABOLA.
95
16. The point (1, 2) lies outside the parabola if = 6x; what are the equations of the tangents through the point to
the parabola ?
The point (2, 45) is on a parabola which is symmetri with respect to the X-axis required the equation of the parabolk, the pole being at the focus. - 2 V2) x. Ans. if (4 17.
cal
;
=
=
18. The subtangent of a parabola 10 for the point (5, 4) required the equation of the curve and the value of the sub normal.
Ans.
64.
The tangent
focal line
drawn
to the
= ^x;*. 5
>
parabola makes equal angles with the
the point of tangency
to
y*
and
the axis of the
curve.
From
We
Fig. 31
we
have,
FT
= FO + OT =
4-
x".
have, also,
= DC = DO + OC = f + FT =
FP"
.-.
The
triangle
x".
FP".
FP"T is FP"T
=
therefore isosceles and FTP".
/
65.
To find
the condition that the line
in order to touch the parabola y 2
= 2 px.
y
= sx + c must fulfil
Eliminating y from the two equations, and solving the resulting equation with respect to x, we have,
p
;
- SC
V
(C8
- PY ~ C S
2 2
m
for the abscissae of the points oflntersection of the parabola and line, considered as a secant. When the secant becomes
PLANE ANALYTIC GEOMETRY.
96
a tangent, these abscissas become equal but the condition for in the numerator of equality of abscissas is that the radical ;
(1) shall
be zero
hence
;
c s
2
= 0,
= --
c
or, solving,
2
2^
(cs
S
the condition that the line must
is
fulfil in
order to touch the
parabola.
COR.
we
1.
Substituting the value of sx c, y
= + V = sx +
have,
c in
the equation
(2)
^-S
for the equation of the tangent in terms of
its slope.
To find the locus generated by the intersection of a tan to it from the focus as the point of gent, and a perpendicular the curve. around moves tangency 66.
The equation
of a straight line through the focus
I?
is )
In order that this line shall be perpendicular to the tangent
y
we must hence
r
have,
s
= sx + L =
.
.
.
(2)
;
S
y
=
-
+ ^-
-
;
-
(3)
the equation of a line through the focus perpendicular to the tangent. Subtracting (3) from (2), we have is
or,
x
= 0, =
is the But x equation of the required locus. the perpendiculars from the Y-axis the of hence, equation
for the
;
THE PARABOLA. focus
to the
97
on the tangents of a parabola intersect the tangents
Y-axis.
67.
To find
the locus generated by the intersection of two tan to each other as the point of tan-
gents which are perpendicular curve. gencij moves around the
The equation
of a tangent to the parabola
y
The equation
= 8X + f-g
-
-
-
=
-l*-fA
...*
=-
we
(1),
-
-
|.
is
...(2)
S
Subtracting (2) from
Art. 65 (2),
(1)
of a perpendicular tangent
y
is,
have,
(3)
the equation of the required locus. But (3) is the equa hence, the intersection of all perpendicu lar tangents drawn to the parabola are points of the directrix.
is
tion of the directrix
;
68. Two tangents are drawn to the parabola from a point without ; required the equation of the line joining the points of
tangency.
be the given point without the parabola, and let f Since (x y is X be the points of tangency. ( 2, 2/2) (X on both tangents, its co-ordinates must satisfy their equations
Let (V, >
?/)
,
2/")?
~)
;
hence, the equations of condition,
y
y"
2/2/2
The two points
yy r
yy
+X"\
+x
(x
of tangency
satisfy
or
=p(x
=P
=p =p
(* (x
+ +x
z ).
(x",
y"),
(x 2
,
2/ 2 )
must therefore
),
)
.
.
.
(1)
Since (1) is the equation of a straight line, and is satisfied for the co-ordinates of both points of tangency, it is the equation of the line joining those points.
PLANE ANALYTIC GEOMETRY.
98 69.
To find the equation of the polar of the pole
(x,
y
)
with
=
2 px. parabola if The polar of a pole with regard to a given curve is the line generated by the point of intersection of a pair of tangents drawn to the curve at the points in which a secant line through
regard
to the
the pole intersects the curve as the secant line revolves about the pole.
a course of reasoning similar to that of Art. 49, prove the required equation to be
By
As the reasoning by means fectly general, the pole
may
we may
which (1) is deduced is per be without, on, or within the
of
parabola.
COR.
1.
If
we make,
hence, the directrix
is
in (1), (x
r ,
y
)
= | 0\ I \z (
,
we have
the polar of the focus.
70. To ascertain the position and direction of the axes, other than the axis of the parabola and the tangent at the be referred its equation will vertex, to which if the parabola
remain unchanged in form.
FIG.
33.
THE PARABOLA. Since the equation
to retain the
is
99
form
= 2px ... (1) y* y = 2p x ... (2) f
*
let
when referred to the we are now seeking.
be the equation of the parabola
axes,
It is whose position and direction be the whatever that obvious at the outset position of may must be -axis Y new the each to the axes relatively other, on the must be new the and the to origin curve, tangent 0, a curve for, if in (2) we make x = 0, we have y = ;
which we can only account for by assuming the Y -axis and the new origin in the positions indicated. This conclu verified by the analysis which sion, we shall see, is fully result
follows.
Let us refer the curve to a pair of oblique axes, making in the any angle with each other, the origin being anywhere
The equations
plane of the curve. Art. 33
(1),
x
y
= a -j- x cos 6 + y cos = b x sin + y sin r
-f-
q>
cos <p)
y
Now,
+
+2
(1),
(b
sin
-f-
we have, x
2
sin 2
(9
+2
=
2
(a)
sin
(b)
sin 2
(d)
= 0, then sin
and
p
(b sin
sin g>
b
2
= 0. 2 pa = 0.
-
b sin
p
g>
sin g>
.
p
q>
.
satisfied
:
= 0. cos
<p
= 0.
=
and sin 2 = 0; i.e., conditions assumed value of 6. But is -axis makes with the old X-axis
(b) are satisfied for this
X
the angle which the new hence, these axes are parallel. If (a, b) be a point of the parabola ?/ 2 pa is an analytical expression of the fact
;
= 2px,
that the
.
we must have the following conditions
(d)
(a)
(1), q>
cos 0) x -f b 2 pa (3) in order that this equation shall reduce to the same
form as
If
q>
g>.
Substituting these values in 2 2 x y sin sin sin 2
?/
transformation are,
of
new
orisrin lies
on the curve.
;
then
hence
(c)
b
2
=2
shows
PLANE ANALYTIC GEOMETRY.
100 S
1!L^
If
cos
= tan
= P, then
<f
b
gp
p
But
(d) is satisfied.
-
the
is
o
whose ordinate is b, Art. slope of the tangent at the point is the slope of the new Y -axis and tan. hence, the new Y -axis is a tangent to the parabola at the point whose ordinate is b .-. at (a, b) .% at the new origin.
57, (5),
;
<p
;
;
COR.
1.
Substituting
= cos =
that cos
and (d) in (3), recollecting after dropping accents,
(a), (&), (c),
we have,
1,
2
-^S-*sin* y % =p
y
f
letting
or,
,
sin
we have
y
"
cp
= 2p x
2
.
.
.
(4)
for the equation of the parabola when referred to O The form of (4) shows that for every value Fig. 33.
y has two
for x,
OX
Y OX
values, equal but of opposite sign
bisects all chords,
drawn
parallel to
OY
and
line
which
is
,
,
assumed ;
hence,
therefore a
diameter of the parabola.
A DIAMETER of a curve NOTE. tem of parallel chords.
is
a
bisects
a sys
\l
To show that the parameter of any diameter
71.
to
the focus four times the distance from that diameter cuts the curve. Draw the focal line FO and the normal
to
Since the triangle 2
O FN
=
Since 61.
O FT
N
Hence
is
a
normal
at
in the triangle
AO = FO
AO N =
,
<y>
and
sin 2
g>
= FO
NO A, AO = AN cot v = FO
2 sin
cp
AN = p,
FO A 2 sin
In the triangle
hence
cos
equal
O N, Fig. 33. Art. 64, the angle isosceles;
is
<r.
O
is
the point in which
<JP
=p
sin
.
cp
-
cos
cp
;
(jp
cos
<JP.
Art.
THE PARABOLA. .-.
P
FO =
2 sin 2
q>
i* -4 Sin* 2p = 4 FO
But
101
go
f
.-.
72.
.
To find the equation of any diameter in terms of the
and
slope of the tangent
The equation
of
= AO =
y But from the triangle b
the semi-parameter.
any diameter as
AO N,
= AX cot =
*L_ tan
is
y
=
,
Fig. 33,
is
we have,
<p
hence
OX
b.
CP
=
<:
s
-
.
.
.
(1)
the required equation.
To show that the tangents drawn at the extremities of chord meet in the diameter which bisects that chord. any 73.
R
FIG.
Let
Y
f
(x
,
/),
P"
(x",
y")
FP";
then
y
-
y
=
y
34.
be the extremities of the chord
y"
n (_;*>... (1)
PLANE ANALYTIC GEOMETRY.
102 its
is
The equation
equation. are
P"
of the tangents at
P
(V,
y")
0",
^-p(* + y/ -*(* Eliminating x from
(2)
-
+
and
(
2) (
*")
3 >
(3) by subtraction,
we
have,
intersection of the tangents. for the ordinate of the point of ~~
x
But
X
(see (1)
"
is
n
f
the reciprocal of the slope of chord
Hence, since the chord
).
are parallel,
we
P
P"
P
and the tangent
P",
YT
have,
y
-y"
Substituting in (4)
it
becomes
-*.value of y with (1) of the preceding is on the diameter. the see that point of intersection
Comparing article,
we
this
EXAMPLES. 1.
y J
= 2.
y
What must be the value of c 4 x _L_ c may touch the parabola
?/
What
=3x + 3.
in order that the 2
Ans.
is
.
the line the parameter of the parabola which 2 touches ? is
to the parabola y slope of a tangent ? the of tangent the equation
The
What
line
=8x?
2
=
6x
is
=
= 3 ^ _L
3.
i
on a tangent to a parabola required the equation of the parabc the equation of the tangent and ,= the slope of the tangent - 16 *. l; 4a 4
The point
(1,
3) lies
;
4.^
_
^=
THE PARABOLA.
=
8 x
is
y
In the parabola y z diameter whose equation 5.
what
the parameter of the
is
W=
103
?
Ans.
136.
6. Show that if two tangents are drawn to the parabola from any point of the directrix they will meet at right angles.
From
=
2 the point 8 x 5) tangents are drawn to ?/ required the equation of the chord joining the points of Ans. 5 ?/ 4^ 8 0. tangency.
7.
8.
What
(2,
the
are
equations of the tangents to
which pass through the point
2,
(
;
+ = = 6x 2
?/
4) ?
Find the equation of the polar of the pole in each of the following cases 9.
Of (-
:
3)
1,
with regard to y 2
= 4*. Ans.
10.
Of
(2, 2)
with regard to if
=
4
11.
Of
12.
Given the parabola
(ay b)
with regard to y
?/
=x
Q.
2y
+ 2a; + 4 = 0.
= 4 x. Ans.
2
+2=
x.
Ans. 2
2x
3y
2x
by
and the point
(
2 a 4,
= 0.
10)
;
to
find the intercepts of the polar of the point.
a
Ans. 13.
The latus-rectum of a parabola 8x
of the line y
-
4,
=4
= 0.
;
= 4:,b=-.
5 required the pole
Ans.
Given
=
=
(i,i).
10 x and the tangent 2 y x 10 required the equation of the diameter passing through the point of tangency. Ans. y 10. 14.
if-
;
=
GENERAL EXAMPLES. 1.
Assuming the equation
point on the curve directrix.
is
of the parabola, prove that every equally distant from the focus and
PLANE ANALYTIC GEOMETRY.
104
Find the equation of the parabola which contains the
2.
points
(0, 0), (2, 3),
2, 3).
Ans.
What
3.
(-
are the parameters of the parabolas
through the point
3 x2
which pass
(3, 4) ?
Ans.
Find the equation of that tangent 4 = 0. 2 x parallel to the line y 4.
Ans.
The parameter
5.
y
=
of a parabola
is
4
;
?/
8y
Y, and
f.
= 9x
which
is
16 a
9
=
0.
required the equation
perpendicular to the line tangent 2 x -\- 2. Give also the equation of the normal which is
which
line
the
of
to
2
= 4 y.
is
line. parallel to the given
=
A
2 4 x makes an angle of 45 with the tangent to ?/ X-axis required the point of tangency. Ans. (1,2).
6.
;
Show
that tangents
drawn
at the extremities of a focal
chord 7.
Intersect on the directrix.
8.
Meet
9.
That a
focus 10. 11.
is
at right angles.
line joining their point of intersection with the perpendicular to the focal chord.
Find the equation of the normal
in terms of its slope.
Show that from any point within may be drawn to the curve.
the parabola three
normals 12
Given the parabola
r
=
to construct the tan-
+ cos 6 vectorial angle = 1
60, and to find the gent at the point whose line. initial angle which the tangent makes with the 60. Ans. 61
=
13.
Find the co-ordinates of the
extremity of the latus-rectum being
pole, the its polar.
normal
at one
THE PARABOLA.
=
In the parabola y 2 4 x what chord which the point (2, 1) bisects ? 14.
is
105 the equation of the
Ans. 15.
The polar
of
any
point in a diameter
is
=2x
y
3.
parallel to the
ordinates of that diameter. 16.
The equation
of a chord of y*
=
10 x
is
y
=2x
1
;
required the equation of the corresponding diameter. 17.
Show
that a circle described on a focal chord of the
parabola touches the directrix. 18.
The base
of a triangle
gents of the base angles vertex is a parabola.
=
b.
=2a
and the sum of the tan
Show
that the locus of the
Required the equation of the chord of the parabola =2px whose middle point is (m, n}. n = x m A
19.
y-
Ans.
.
p 20.
A
angle
=
n
=
chord of the parabola if 2 px makes an with the X-axis required its length.
focal cp
y
;
Ans.
-?-.
Show
that the focal distance of the point of intersec to a parabola is a mean proportional to the focal radii of the points of tangency. 21.
tion of
two tangents
Show that the angle between two tangents to a parab one-half the angle between the focal radii of the points of tangency. 22.
ola
is
23.
=
The equation
is y = a
;
of a diameter of the parabola y 2 2 px required the equation of the focal chord which this
diameter bisects. 24. The polars of all points on the latus-rectum meet the axis of the parabola ?/ 2 2 px in the same point required the co-ordinates of the point.
=
;
Ans.
I
PLANE ANALYTIC GEOMETRY.
106
CHAPTER
VII.
THE ELLIPSE. 74.
THE ellipse
that the
sum
is
the locus of a point so moving in a plane
of its distances from
constant and equal to a given called the Foci of the ellipse.
two fixed points
line.
If
is
always
The the
fixed points are points are on the
the given given line and equidistant from its extremities, then line is called the TRANSVERSE or MAJOR Axis of the ellipse. 75.
To deduce
the equation of the ellipse, given
the,
foci
and
the transverse axis. Y
FIG.
Let F, F! be the
OY
J_ to
AA
foci
at its
co-ordinate axes.
and
35.
AA
the transverse axis.
Draw
OX
as the
middle point, and take OY,
THE ELLIPSE. Let also
P
PD
=
Let
FF
X
AA =
V?/
From
the
+ (x
2
X
2
c)
mode
y) are the co-ordinates of P.
=
=
20F
201\
hence
and
FPD
Vy +
=2
=2a Vy + (x - c) + V y + /
and FjPD, we
2
have,,
2
(#-{- c)
.
we
.
.
(a)
have,
2
+ c) = 2 a 2
(x
;
.
.
.
(1)
and reducing,
clearing of radicals,
+ x -c 2
(if
and
r
;
2
a2
FP =
c,
2
r
of generation of the curve,
r -f
or,
(x,
the right angled triangles
=
r
2a,
= /.
From
draw
;
OY.
to
Then (OD, DP)
FP
Draw PF, PFi
be ary point of the curve. ||
107
)
2
=
x2
a 2 (a 2
-
2
c )
...
(2)
As
this equation (2) expresses the relationship between the co-ordinates of any point on the curve, it must express the
relationship between the co-ordinates of every point the required equation.
;
hence
it is
Equation
(2)
elegant form.
may be made, however, to assume Make x = in (2), we have,
=
2 ?/
a2
-
c
2
for the square of the ordinate of the point in
curve cuts the Y-axis this distance
by
we
b, 2
.-.
2
2
C
(= OB
which the
/2
).
Eepresenting
2 ?
-b* ...
Substituting this value of
or,
2
have,
=a _ c = a 2
OB
i.e.,
;
a more
c
2
in (2)
(3)
and reducing, we have,
symmetrically,
for the equation axes.
of ellipse
when
Let the student discuss equation
referred to
(4).
its
centre and
See Art. 12.
PLANE ANALYTIC GEOMETRY.
108 COR.
we make
If
1.
2
x -f y
which
is
COR.
=a =a 2
the equation of a
circle.
we interchange a and
If
2.
we have,
in (4),
b
2
for the equation of an ellipse lies along the Y-axis.
COR. 3. If (z y we have from (4) ,
y
f
)
and
=
*L
(z",
(^
-z
we
b in (5),
have,
whose transverse axis
y")
- a/
are
2
+
)
(=2 a)
two points on the curve,
and
= -^ (a -
2
2
y"*
z"
^
-
+
)
;
x ) (a (a ) (a ordinates the of any two points on the of i.e., the squares other as the rectangles of the segments in ellipse are to each
hence, y
2
*
::
:
(a
y"
:
which they divide the transverse COR.
4.
= x a and y and dropping accents, ^x 2
tf ,f
_|_
2 atfx
for the equation of the ellipse, of co-ordinates.
The
line
BB
,
Fig. 35,
x")
;
.
axis.
By making x
after reduction
76.
x"}
=
=
in
y
.
.
.
(4),
we have
(7)
A
being taken as the origin
is
called
the
CONJUGATE
or
and A from the figure that bisects all lines drawn through it and terminating the point the CENTRE of is called For this reason in the curve.
MINOR
axis of the ellipse
VERTICES of the
ellipse.
the points
;
A
are called the
It is evident
the ellipse.
The is
~ ** - -?
ratio
called the
this ratio is
a
ECCENTRICITY always
<
See
.
a
1.
(3)
of the ellipse. of c
The value
ures the distances of the foci F,
F from x
Art. 75
...
(1)
It is evident that
= i Va - b 2
the centre.
2
meas
THE ELLIPSE. If a
=b
then
in (1),
If b
=
=
e
=1
in (1), then e
when
the ellipse becomes a
zero. i.e.,
;
when
the ellipse becomes a
becomes unity.
straight line the eccentricity
77.
i.e.,
;
becomes
circle its eccentricity
109
To find the values of the focal
radii,
/, of a point on
r,
the ellipse in terms of the abscissa of the point.
The FOCAL RADIUS of a point on the ellipse of the point from either focus.
From equations
(a),
Art. 75,
= Vy +
we
2
r
from the equation of the
;
ellipse, Art.
^_ ( a 2 a2
2
have,
cf
(x
the distance
is
x 2\
^2
we
75
(4),
_
&_ X 2
have,
.
a2
hence, substituting *
\
o 2 ex
/ t/V -
+
2 ex
T
= r
Similarly
we
c
a
78.
c-
,.-
a
x
;
See
ex.
(1) Art.
76 ... (1)
find
r
%
=
,
<Z-
a hence
2
-(X
+
= a + ex
.
.
.
(2)
Having given the transverse
axis
and the
foci of
any
ellipse, the principles of Art. 75 enables us to construct the ellipse by three different methods.
First Method.
verse axis
AA
.
Take a cord equal Attach one end of
in length to the trans at F, the other at F
it
.
Place the point of a pencil in the loop formed by the cord and stretch it upward until taut. Wheeling the pencil around, while keeping, the point on the paper and tightly pressed
PLANE ANALYTIC GEOMETRY.
110
against the cord, the path described will be an arc of the After describing the upper half of the ellipse, re ellipse. move the pencil and form the loop below the transverse axis.
By
a similar process the lower half
FIG
may
be described.
It is
36.
evident during the operation that the sum of the distances of the point of the pencil from the foci is constant and equal to the length of the cord i.e., to the transverse axis. Take any point C on the transverse axis Second Method. ;
and measure the distances
CA
A C,
AC.
With F
as a radius describe the arc of a circle
;
as a centre
also with
F
and as a
CA as a radius describe another arc. The points these arcs intersect are points of the ellipse. which in B/ R, the radii two other points P, P may be interchanging By A smooth curve traced through a number of determined. centre and
points thus located will be the required ellipse.
Let the axes AA = 2 on any straight edge = OA = a and DL = OB
Third Method.
Lay given. will do)
off
KD
MX
a,
BB =
2 b be
(a piece of paper == b. Place the
the position indicated in the straight edge on the axes in and L slide along the axes, the point Then as figure.
K
D
THE ELLIPSE. describe the
will
DKE
For from the
ellipse.
are similar triangles
DK = --
Ill
:
DL lji
-a
_-
;
i.e.,
=
x
_ b
-//
DLH
figure
2
(x
and
and y being
the co-ordinates of D).
Hence, squaring, clearing of fractions, and transposing, we have 2 z a?tf _j_ &2X = a b ^
That is the locus described by D is an ellipse. ment based upon this principle is commonly used
An for
instru
drawing
the ellipse. 79.
To find the
The
latus rectum or
parameter of an ellipse. parameter of an ellipse is the double or dinate passing through the focus. The abscissas of the points in which the latus rectum 2 b -JSubstituting either pierces the ellipse are x of these values on the equation of the ellipse latus rectum, or
=
we have Hence
y
2
=
2
a2
(a
vV
-
Latus rectum
Forming a proportion from
2
(a
.
- b )) = 2
=2y=
.
a2
9 A2
-
-
a
.
y
...
=
.
a (1)
this equation there results,
2y:2b::b:a ; hence i.e.,
2y
:
2 b
::
2
b
:
2 a
;
the latus rectum is a third proportional to the two axes.
EXAMPLES. Find the semi-axes, the eccentricity, of each of the following ellipses
and the latus rectum
:
1.
2.
3 x 2 -f 2 y 2
= 6.
^,^ = L
3.
4>
x2 4
+ 3 if = 2. o __ 6 = 2 2
,
3.2
PLANE ANALYTIC GEOMETRY.
112
60y c
i
-\-
9
x-
=
O
7
a.
o.
Write the equation of the 9.
The transverse
=
V -^-
l
-f-
m
?i.
having given the distance between the foci
ellipse
= 10
axis
9
x-
;
:
or
Ans.
25 10.
Sum
= 18
of the axes
;
difference of axes
36
=
Transverse axis
10
-4^
the conjugate axis
;
=
l.
9
= 6.
Ans. 11.
$-
+
l-
= 1.
9
= y2
the
transverse axis.
x*
A 12.
tween
Transverse axis
= 20
conjugate axis
;
=
distance be
foci.
Ans.
13.
=1
4y2 25
25
Conjugate axis
= 10
;
-
distance between foci
Ans.
-j-
y
= 50.
2
= 10.
^+
y*
= 25.
Given 3y 2 + 4 a? 2 = 12 required the co-ordinates of the point whose ordinate is double its abscissa. 14.
15.
;
Given the
ellipse 3
2 ?/
+ 2 x = 12, and the line y 2
x
1
;
to find the co-ordinates of their points of intersection.
X1 (--=3-1, and the abscissa of a 64 15 .-2
16.
Given the
f
ellipse
point on the curve
=
;
required the focal radii of the point. 8T Ans. r 7 T%, /
=
= V
THE ELLIPSE. 80.
113
To deduce the polar equation of the
ellipse, either focus
being taken as the pole.
FIG.
Let us take F as the
=
From
the figure,
x
i.e.,
and
pole,
the co-ordinates of any point r a ex (1) we have, .
37.
P .
.
-\-
or,
reducing,
P FA)
=
From
(r, (9)
be
Art. 77
;
e
(ae
-f-
(1),
we have
r cos 0),
we have ,
*L
_ 1
+ e cos
for the polar equation of being taken as the pole. From Art. 77 (2),
FP
We
,
r cos 0.
Substituting this value of x in
=a
(FP
(1)
OD = OF + FD
= ae r
let
of the ellipse.
...
(2)
the ellipse, the right-hand focus
= / = a + ez
.
readily determine from this value
1
-
e
cos
for the polar equation of the ellipse, the left-hand focus being taken as tlje pole.
PLANE ANALYTIC GEOMETRY.
114 COR. If
= a (1 = FA, = / a (1 + = F A. =a-a r = a (1 - e r
0,
0)
e)
= 90,
If
2
= (1 + = FA = a (1 - = F A FM r = a (1 - e
r
ft
/ If
= 270,
If
= 360,
To deduce
e)
,
e)
.
2
)
r
/ 81.
,
= a (1 - e) = FA, = a (1 + = F A. e)
the equation of condition for the supplemental
chords of an ellipse. Y
FIG. 38.
Let AP,
AP
The equation
be a pair of supplemental chords. of a line through A (a. o) is y
2
= a(l-e )=a-a- -=-^- = - = FN. = 180,
If
)
= FM.
= r
2
=
s (a;
r/).
THE ELLIPSE. The equation
V =
Where
A
of a line through s
(x
?/
= ss
a, 6) is
(
+ a). we must have
these lines intersect 2
115
f
2
a 2)
(x
.
.
.
(1)
In order that the lines shall intersect on the ellipse their equations must subsist at the same time with the equation of the ellipse
Dividing
*- 5 (a ^^ (1)
or
(2),
ss
=-
.
.
for the required condition. COR. If a the ellipse b,
=
ss
f
2)
we have
by
comes
(
= -
.
(3)
becomes
-a circle
and
(3)
be
1,
a relationship heretofore deduced. Art. 40 (1). SCHOL. The preceding discussions have developed a remark able analogy between the ellipse and circle. As we proceed we shall find that the circle is only a particular form of the ellipse and that all of the equations pertaining to it may be deduced directly from the corresponding equations deduced for the ellipse
82.
Let P"S
by simply making a
To deduce the equation of the tangent P"
(x",
y"),
V
cuts the ellipse.
As
= b in those equations.
2
= i!_(V2
2 2/"
the ellipse.
,
we must have
the points are on the ellipse, ,
to
(x ?/ ) be the points in which a secant Its equation is, therefore,
= 4 (a- a-
AA
3/2}
2 *"
)
...
(3)
PLANE ANALYTIC GEOMETRY.
116
FIG.
39.
These three equations must subsist at the same time subtracting (3) from (2) and factoring, we have
-
2/0
(/ r
hence
y x
r
+ 2/0 =
P_
T
^
becomes
v^alue in (1) it /
/,2
Eevolving the secant
_|_
line
"
becomes a tangent and x we have
f
(
"-
=
,
x",
y
=
y"
;
for the equation of the tangent
(x",
.
.
(5)
P"
?/")
and
secant
hence, substituting,
,*;... 1 2
P"
x
r
xx b
-
P (x y ) will approach When this occurs the
i.e.,
a2
X)
upward about the point
the other point of intersection will finally coincide with it.
or
x
-^-7T7
>-
hence
-
- i/\ -x
Substituting this
;
(4)
THE ELLIPSE. COR.
If b
= a,
117
we have _i_
yy n
I
a* for the equation of the tangent to the circle. SCHOL. If we make x and y successively tion of the tangent (5),
we have y
=
-
f
See Art. 41
=
and x
values of the variable intercepts
=
,,
hence
y"
IP
,
OT d2 x
=
..
and x
y
These values
,
(6).
equa
= ^ for x
the
"
y ,
in the
OT, Fig. 39
;
.
in the equation 2
2 *" .
^~"
?/
_
J
give, after reduction,
for the equation of the ellipse, the intercepts of its tangents
on the axes being the variables. 83.
To deduce the value of the sub-tangent.
Making y
=
.-.sub-tangent
in (5), Art. 82,
= DT = ^1 -
we have
x"
COR. If .
.
a*
~x x
x"
b = a,
=
"*
ff
then from Art. 41, Schol. a 2
sub-tangent in the circle
=
2 x"
rr<2
2
.
SCHOL. The value of the sub-tangent being independent of the value of the minor axis (2 b) it follows that this value is the same for every ellipse which is concentric with the given
ellipse,
and whose common transverse axis
is
2
a.
PLANE ANALYTIC GEOMETRY.
118
of condition that a line shall pass through the centre of the ellipse and the point of tangency is, Fig. 39,
84.
The equation
=
tx",
y"
.-.
the slope of this line
t==
The
is
^7
slope of the tangent at
a2
(x",
y")
is,
Art. 82,
y"
Multiplying,
member by member, we have
But
(3)
Art. 81
...
i.e.,
ss
=
the tangent to the ellipse
and
the line joining the centre
and
the point of tangency enjoy the property of being supplemental I)
chords of an ellipse whose semi-axes bear
is
COR. If
s
parallel
to
then s = t a diameter of the
=
f
t,
;
parallel to the tangent diameter.
chord
85.
is
to
each other the ratio
-.
chord if one supplementary ellipse, the other supplementary i.e.,
drawn
at the extremity of that
84 afford us two different principles of Arts. 83, a given of constructing a tangent to the ellipse at
The
methods point.
First Method.
Art. 83, Schol. Let draw the ordinate
P",
Through point. until it meets the circle described P"
of the ellipse Join at F.
(AA )
P"
and
in
T
;
F; draw P"T
Fig. 40, be the given
P"D
and produce
it
upon the transverse axis tangent to the circle
FT
will be the required tangent.
THE ELLIPSE.
FIG.
P"
||
86.
from
RA
40.
Draw
Art. 84 and Cor.
Second Method. centre, and to
A
draw
AR
119
||
to
P"R
;
will be tangent to the ellipse at
To deduce the equation of the normal
The equation
of
any
-
line
through
=
P"R
P"T
P".
to the ellipse.
P"
-
(x",
through the
drawn through
y"),
Fig. 39, is
S (X y (1). In order that this line and the tangent at shall be perpendicular their slopes must satisfy the condition ?,"
.
.
.
X")
P"
1
We
+ S8 = o
.
.
.
(x" ,
y")
(2).
have found Art. 82 for the slope of the tangent
/==
;
"5-^ hence, the slope of the normal
Substituting this value of
for the equation of the
is
s in (1),
normal
to
we have
the ellipse.
PLANE ANALYTIC GEOMETRY.
120 COR.
1.
.
=
a
If
then
b,
-
"
yx
which
xy"
becomes, after reduction, 0,
normal
line to the circle.
To deduce the value of the sub-normal
=
Making y
...
-
is the equation of the
87.
we
(3)
in the equation of the normal, (3), Art. 86,
ON = x
have, Fig. 39,
*x",
Sub-normal COR.
If
1.
=
a
b,
then
Sub-normal for the
circle
=
x".
EXAMPLES. Deduce the polar equation of the ellipse, the pole being with the X-axis. the centre and the initial line coincident
1.
at
A Ans.
r
=
-. 2
sin
2
+b
2
cos 2
the
of following Write the equation of the tangent to each of the sub-tangent in each case. value the and give ellipses,
2z
2.
El
3.
2
+ 4y = 38at 2
+ 2.
(1, 3).
Ans.
!.
+
+ 6 y = 19
;
18.
at (1, ordinate positive).
1,
An,.
4.
x
._l,
a
at (2,0). 0.
5
6
.
.
2
^ + 3^- 11
J/l
a
+
_ o
=
1,
at
(2,-
at (0,
1)
V)
4as
.
_ 3y . 11;i
.
THE ELLIPSE.
8.
9.
+ te - 2, 2
*
y
m +
=
2 ?/
- V2 -
at (1,
at (afo
1,
+,
121
6).
.5).
Write the equation of the normal to each of the following the sub-normal. ellipses, and give the value of 7/
+ 4 x = 39,
at
if
+2
= 44,
at
2
2
10.
3
11.
4
12.
^!_
13.
-+{ o o
14.
^1
_|_
4-
ic
2
=
j
7/2
=
m
2
2 ?/
-f- ?
2
1,
at (1,2).
at
^ ord
=mn 2
1,
(
2 ,
ord negative).
ord
at
1?
2
ic
2,
(
1.
a,
15.
(3, 1).
).
_|_^
at (m, o).
=
+
2x 6 of a chord of an ellipse is y the equation of the supplementary chord, the axes of the ellipse being 6 and 4 ? Ans. y f f x 16.
what
The equation
=
17.
;
is
Given the equation
j-
y
MID
= 1,
and
2
?/
+
=
;
.
re-
quired the equation of the tangents to the ellipse at the points in which the line cuts tne curve. 18.
2
Given the ellipse
= 0; (a)
%
= 1,
and the
line
x
y
required
The equation "
(b)
(-
"
of a tangent to the ellipse
||
"1
to the line. "
"
"
-{-
PLANE ANALYTIC GEOMETRY.
122 19.
The point
(4, 3) is
1G
+ 1P
outside the ellipse
1;
required the equations of the tangents to the ellipse which pass through the point. 88. The angle fanned by the focal lines drawn to any point of an ellipse is bisected by the normal at that point.
FIG.
41.
Let P"N be a normal at any point
We
have found, Art.
From
Art. 76
P" (a;",
?/ ).
= OF = ae; NF = OF OX = ae NF = OF + OX = ae + NF NF r
FP"
:
FP"
.-.
::
:
NF NF
::
P"F
hence
e?x"
But
P"F,
87, that
we have OF
.-.
Draw
(a
:
ex")
:
::
e*x"
(a
:
(
ex")
(a
+
FP"
:
ex")
F
= e (a - ex = e (a + +
)
ex")
ex")
Art. 77, (1)
P".
ff
and
(2)
3
THE ELLIPSE. The normal,
F
P"F
into
cent sides.
therefore,
divides the base of the
triangle
two segments which are proportional to the adja
Hence ==
FP"jST
SCHOL.
123
1.
If
P"T
F
F
P"N.
be a tangent drawn at
=
P"C
FP"T
P",
we must have
;
between a FP"N and the the angle Hence, right angle (= FP"N). tangent to the ellipse makes equal angles with the focal radii for each of these angles is equal to the difference
drawn
to
SCHOL.
method
the point of tangency. 2.
The
principles of this article afford us another
drawing a tangent to the ellipse at a given point. be a point at which we wish to draw a tangent. Let Pro to R, making P"R = FP" join F and R. duce F A line J_ to FR will be tangent to the ellipse P"T, drawn through of
P"
P"
;
P"
at
P".
89. To find the condition that the straight line y must fulfil in order that it may touch the ellipse
4 + y. = a
If
b
= SX O
a2
we obtain the
I
7
-f-
c
i.
2
we consider the line y
= sx
as a secant
and combine the equations
-\- C,
O
?
lr
co-ordinates of the points of intersection.
Eliminating y from these equations, we have
-
sa 2 c
ab
for the abscissas of the points of intersection. Now, when the secant line becomes a tangent, these abscissas become equal.
Looking
at (1)
we
see that the condition for equality of ab-
PLANE ANALYTIC GEOMETRY.
124
numerator shall disappear
scissas is that the radical in the
hence
sV
2
or is
s*a
-f-
+
2
^
2
b
2
-
c
=c
2
=
;
0,
2 .
.
.
(2)
the required condition.
COR. If we substitute the value of equation of the line, we have
y
= SX -.
V*
2
+&
a2
for the equation of the tangent
drawn from
c
...
(2) in the
(3)
the ellipse in terms of
to
its
slope.
To find the
90.
gent
to
a tan a from focus as
locus generated by the intersection of
and a perpendicular
the ellipse
to it
the point of tangency moves around the curve. The equation of a straight line through the focus (ae, o)
is
=
s (x ae). y In order that this line shall be perpendicular to the tangent
=
its
SX y be must equation
y
J-
V* 2
= - i (x -
a
+b
ae)
.
2
(1),
.
.
(2)
S
we now combine (1) and (2) so as to eliminate the slope the resulting equation will express the relationship be (s), tween the co-ordinates of the point of intersection of these hence it will be the lines in every position they may assume If
;
equation of the required locus. Transposing sx to the first member in (1), of fractions
and clearing
(2)
and transposing, we have y sy
= Vs a + b x = ae.
SX -}-
2
-t-
2
2 .
Squaring these equations and adding, remembering that 2 a z e 2 Art. 76, we have, ab
=
or
,
THE ELLIPSE.
125
hence, the circle constructed on the transverse axis of the ellipse
and
the locus of the intersection of the tangents let fall from the focus on them.
is
the perpen
diculars
This circle
known
is
(See Fig. 45.
ellipse.
as the Major-Director circle of the
)
To find the locus generated by the intersection of two tangents which are perpendicular to each other as the points of tangency move around the curve. 91.
The equation
of a tangent to the ellipse
= SX +
y
The equation
s
a2
+b
is
2 .
.
.
(1)
of a tangent perpendicular to (1)
is
hence, by a course of reasoning analogous to that of the pre
ceding
article,
we have xz
+ ?/ = a
2
2 -f b
.
.
.
(3)
The required locus is, therefore, a circle 2 ellipse and having its radius equal to Va 92. out
;
concentric with the -f-
b
2 .
Two tangents are drawn to the ellipse from a point with required the equation of the line joining the points of
tangency.
Let
P
7
?/), Fig. 42, be the given point, and let be the points of tangency. Since 2/2) (x y ) is a point common to both tangents, its co-ordinates must satisfy
?2
(a:
P"
,
V
(*2>
their equations
;
x
(x" ,
f
,
hence, x"
a2
x xz a2
Hence
(x",
y")
and
~W_ y yz
(a; 2
.
1
y z ) will satisfy the equation
y"),
PLANE ANALYTIC GEOMETRY.
126
As (1) is the equation of a straight line, and is satisfied for the co-ordinates of both points of tangency, it must be the equation of the straight line which joins them. 93.
regard
To find the equation of the polar of the pole (V, /), with to
the ellipse 2 ~
_
1--1
FIG.
42.
the aid of Fig. 42, and a course of reasoning similar to PiP", the polar to P may be shown to be
By
that of Art. 49, the equation of
xx a2 COR. PI
P
(^i>
,y_y_
,
1.
b*
If the polar of the point P (x f ?/) passes yi)j then the polar of P l (x^ T/J) will pass
(V, /).
,
through through
(See Art. 50.)
94. To deduce the equation of the ellipse when referred to a pair of conjugate diameters as axes. A pair of conjugate diameters of the ellipse are those diam-
THE ELLIPSE.
127
which if the ellipse be referred its equation will contain the second powers of the variables. only eters to
The equation axes
of the ellipse
when
referred to its centre and
is
we
refer the ellipse to a pair of oblique axes at the centre, we have, Art. 33, Cor. 1, origin If
x y
=x =x
having the
+ i/ cos + y sin y
cos
(p
sin
for the equations of transformation.
Substituting in
(1),
we
have
+
2 2 (a sin 2 2 (a sin
+ W cos 0) x + (a sin + tf cos ij sin + 6 cos cos a y = aW ... (2) 2
2
2
2
2
2
q>
q>)
2
(?)
q>
for the equation of the ellipse referred to oblique axes. But, by definition, the equation of the ellipse when referred to a
pair of conjugate diameters contains only the second powers of the variables hence ;
a- sin
sin
2
cp
-f b cos
=
cos y
the condition that a pair of axes must conjugate diameters of the ellipse. is
Making the
co-efficient of
.
.
.
(3)
fulfil in
x y equal to zero in
order to be
(2),
we have
after dropping accents 2 2 (a sin
+6
2
cos 2 0) x 2 -f (a 2 sin 2
<p
+6
2
cos 2 <p)
y
2
=a6 2
2
... (4)
for the equation of the ellipse when referred to a pair of con jugate diameters. This equation, however, takes a simpler form when we introduce the semi-conjugate diameters. Mak
ing y
=
and x
= 0,
successively, in (4),
a2
sin 2
(p
-M
2
cos 2
+6
2
cos 2
<p
we have
J
PLANE ANALYTIC GEOMETRY.
128 r
which a and have (5). we
in
ft
From
represent the semi-conjugate axes.
a? sin
2
+
|72
2
2
2 ft
cos 2
S
2
= ^L _
^2
;
2 ft
in Substituting these values of the co-efficients
(4),
we
have,
after reduction,
for the required equation.
COR. As equation (6) contains only the second powers of the variables, it follows that each of the two diameters to which the curve is referred will bisect all chords drawn parallel to the other.
SCHOL. The equation of condition for conjugate diameters be put under the forms (3) may
= --7 2
tan 6 tan
.
.
(7)
.
<p
we
see that the
81, Comparing this expression with (3) same result was obtained for the supplementary chords of an
Art.
if ellipse; hence, Fig. 40,
ary chords, then
RP",
A R R A be
PR",
,
a pair of supplement
drawn through the centre
parallel
Again to these chords, will be a pair of conjugate diameters. we see that the same relation Art. with 84, (1) comparing (7) for a diameter and the tangent drawn at ship was obtained diameter and P"T its hence, Fig. 40, if P"R be a :
extremity be a tangent drawn at its extremity, then PR", drawn through to RP". the centre parallel to P"T, is the conjugate diameter a The equation of condition (7) being single equation con ;
unknown quantities (tan. 0, tan. assume any value we please for one of them, and will make known the value of the other hence,
taining two
;
there are
an
infinite
<r),
we may
the equation in the ellipse
number of pairs of conjugate diameters.
THE ELLIPSE.
129
To find the equation of a conjugate diameter.
95.
Let
P"R,
RP
be a pair of conjugate diameters.
to find the equation of
R/P
We
wish
.
Y
FIG.
The equation P" y"}
(*",
of the
wr"
By
tangent line
i
in/ yy
drawn through
#
f
_i
2
P R/
Art. 94, Schol., the diameter
equation must be the same
its
P"T,
is
xx a2
hence
43.
is parallel to P"T as that of the tangent,
;
the constant term being zero.
xx .
=
.
.
.
(1)
or
...
is
the equation of a diameter expressed in terms of the co
(2)
ordinates of the extremity of its conjugate diameter. COR. Let s represent the slope of the diameter P"R
from
(2) _
bV 2
.
.
?/
2
jr
a
2
1 s
;
then,
PLANE ANALYTIC GEOMETRY.
130
^=^=
since
i ;
s
DP" y"
hence we have
-~x a
y=
2
!
.
.
s
(3)
for the equation of a diameter in terms of the slope of its
conjugate diameter.
To find the co-ordinates of either extremity of a diam
96. eter,
the co-ordinates of one extremity of
its
conjugate diameter
being given.
Let Let
P"R
(x",
RT
and
y")
Fig. 43, be a pair of conjugate diameters. be the co-ordinates of We wish to find the ,
P".
f
P
terms of the co-ordinates of The equation of condition that P (a/, y ) shall be on the diameter P R is, Art. 95, (1) co-ordinates (x
,
y
X
Since
P
f
(x
,
)
of
P."
X"
y ) is on the V2 o/2
Eliminating y and
we
in
a;
,
ellipse,
we have
also
successively, from these equations,
find
These expressions, taken with the upper signs, are the co ordinates of P taken with the lower signs, they are the ;
co-ordinates of 97.
R
.
To show that the sum of the squares on any pair of
semi-conjugate diameters
on the semi-axes. Let P"
(x",
y"}
and
is
P
equivalent f
(x
,
y
).
to the
sum of the squares
Fig. 43, be the extremities
THE ELLIPSE.
131
two semi-conjugate diameters. Let OP" = a the triangles ODP", OD P we have, from then, of any
,
OP =
b
r ;
,
...
"
and
V
But, Art. 96,
cc
=x +y
2
(1)
2
2
2
.
.
.
(2)
=~
and
hence
=
2
b
Adding
(1)
and
+
2
b
2
i/
2 x"
_
98.
2
a
hence,
_|_
P"R
The area
(
= 2a
),
is
+ -TT
v
//
i
\
.
;
2
==
-
2
-f
ft
.
.
(4)
.
PR
(
=
2ft
of the parallelogram
From
the figure sin (180
),
to
the rectangle constructed
Fig. 44, be
OP"TR
is
OP"TR
=
0) )
= afb
OP"
=a
sin
-
sin (<p
P"OR
sin (>
(9)
.
<9)
.
.
any two conju
= area BB H H.
P"P.
P"P
(<p
area of
equivalent
To prove that area CTC T
OR X
.-.
2
/
To show that the parallelogram constructed on any two
gate diameters.
= a!
2_ = l
-f 6
conjugate diameters on the axes.
Let
(3}
.
we have
(3),
70X
but
.
a2
;
(1)
PLANE ANALYTIC GEOMETRY.
132
B
c Fm.
From
the triangles
OD P
sn
Hence
sin
,
44.
0DP",
we have .y"
OF
(qp
Mb ab
a fb
0)
=
V
sin
go
cos 6
cos
g>
sin
THE ELLIPSE.
133
Substituting this value in (1) and multiplying through by 4,
we have
area
OP"TB
area
CTC
X 4 = 4 ab = area BB H H. ;
i.e.,
T"
To show that the ordinate of any point on
99.
the ellipse is
the ordinate of the corresponding point on the circumscribing circle as the semi-conjugate axis of the ellipse is to the semitransverse axis. to
Let
P
(*
,
DP
,
DP"
/) and
Since
P
Since
P"
be the ordinates of the corresponding points
P"
(x",
r
(x
,
(x" ,
y
) is
y"}
y"}.
on the
is
ellipse,
on the
circle
we have
whose radius
is a,
we have
Dividing these equations, member by member, we have /
=
7
2 ,
(since
oj
=
a;"
)
;
PLANE ANALYTIC GEOMETRY.
134 Similarly
we may prove x1
x.2
:
::
a
:
that b,
any point on the ellipse, and x 2 the corresponding abscissa of a point on the inscribed circle.
where
Xi is the abscissa of
is
The
100.
method
principles of the preceding article give us a of describing the ellipse by points when the axes are
given.
From
0, Fig. 45, as a centre with radii equal to the semiDraw any OB describe the circles RA, BCB
A
axes OA, radius
draw
N
OB
MN
||
OA
R
,
Since MJST a point of the ellipse. have we the of O, triangle is
BD D N D R OM OR ::
:
i.
y
e.,
.
of the larger circle, cutting the smaller circle in in to cutting the or din ate let fall from
b y"
^
:
is
M
;
N
;
parallel to the base
;
;
hence, the construction.
101.
To show that the area of the
the circumscribing circle as the to its
semi-major
ellipse is to the
semi-minor axis of the
axis.
FIG.
46.
area of
ellipse is
THE ELLIPSE. Inscribe in the ellipse any polygon its vertices draw the ordinates
from
135
AEE E E E A 2
1
ED^D^
3
4
/ ,
and
producing etc. Joining
them upward to meet the circle in P, P 1? P 2; these points we form the inscribed polygon
etc.,
APP^P^A
in
the circle.
Let (x, y ), (x yj, y 2 ) etc., be the co-ordinates of r P, P 15 P 2 etc., and let (a;, y), (x , /), be the co-ordi y"), etc., nates of the corresponding points E, 1} 2 etc., of the ellipse. ,
(x",
,
(aj",
R R
,
Then Area Area RPD.R, Area
hence
But, Art. 99,
a
Hence
AreaRDD^ Area
We
may prove in like manner that every corresponding pair of trapezoids bear to each other this constant ratio; hence, by the Theory of Proportion, the sum of all the trapezoids in the ellipse will bear to the sum of all the trapezoids in the circle the same ratio. these sums 2,t and Representing 2T, respectively, we have 2*
2T As
this
=
by
b
a
relationship holds true for any
zoids, it holds true for the limits to
number
of trape
which the sum of the
trapezoids of the ellipse and the sum of the trapezoids of the circle approach as the number of trapezoids increase.
PLANE ANALYTIC GEOMETRY.
136
But these the circle
limits are the area of the ellipse
area of ellipse area of circle
b
a
Since the area of the circle
COR.
and the area of
hence
;
area of ellipse TT
Since
a2
a2
:
TT
a 2 we have ,
b
a
area of ellipse TT ab TT ab
.. TT
is
:
:
= :
TT
ab.
TT b*,
we
see that the area of the ellipse tween the areas of the circumscribed
a mean proportional be
is
and
inscribed circles.
EXAMPLES. 1.
What must
be the value of
c
in order that the line
y = 2 x + c may touch the ellipse ! + JL == i ? 9 4 2
Ans.
c
= 5.
of an ellipse is 10 what must be the value of the semi-conjugate axis in order that the ellipse 2.
may
The semi-transverse
touch the line 2 y
-\-
;
x
14
= 0?
Ans. 3.
What
~,2 x
_i_
whose inclination
drawn
,,,2 y
i
to X-axis
locus of
-4 a2
ellipse
= 45
?
the intersection of the tangents to the
+
=l b
2
at the extremities of conjugate diameters
required
its
V24.
6
5
The
=
are the equations of the tangents to the ellipse
"*"
4.
b
is
an ellipse
;
equation.
Ans
,
+
=2.
THE ELLIPSE. 5.
Tangents are drawn from the point a
+
frequired the
2 2/
137 (0,
8) to the ellipse
= i; the
equation of
joining the points Ans. 8 y 1
line
of
=
tangency.
Eequired the polar of the point following ellipses 6.
*2
9.
What
with respect to the
:
= 9.
3/ 2
(5, 6)
0.
=1.
7.
are the polars of the foci ?
x
Ans.
=
-e
10.
What
is
the pole of y
= 3x
-f-
1 with respect to
(-
.
11.
The
line 3
y T2
_^L
=
5x 2
_i_
4
7/ iL
is
_i
.
9
required the equation of the conjugate diameter. Ans. 20 y 12.
A
12, 9).
a diameter of
+ 27 # = 0.
pair of conjugate diameters in the ellipse ?/ 2
T 2
_-f JL.^1
16
make
angles whose tangents are
with the X-axis 13.
9
What
is
;
o
o
^ and
^,
respectively,
required their lengths.
the area of the ellipse
?4
+
-
= 1?
10 .
2
TT
VlO.
PLANE ANALYTIC GEOMETRY.
138
The minor
14.
axis of an ellipse
to the area of a circle
and its area is equal what is the length
is 10,
whose diameter
16
is
;
25.
Ans.
of the major axis ? 15. The minor axis of an ellipse is 6, focal radii to a point on the curve is 16
axis, the distance between the
foci,
and the sum of the required the major
;
and the
area.
GENERAL EXAMPLES.
What
1.
(2, 4) 2.
(6.
(
The major
4) is
3.
The
lines
y
;
1
=
=
18, and the point the equation of the ellipse. required
axis of
on the curve
tal chords
x
an
ellipse is
13
+ 6 and y = ~ x -f ^ are supplemen
drawn from the extremities of the transverse
of an ellipse 4.
the equation of the ellipse which passes through the centre being at the origin ?
is
2, 4),
;
The minor
axis of an ellipse
=
is
12,
and the
centre divide the major axis into four equal parts the equation of the ellipse. 5.
sum is
in
Assuming the equation of the distances of
constant and 6.
=
of the ellipse
any point on the for a point
an ellipse whose eccentricity
7.
What
;
foci
and
required
show that the
ellipse
from the
foci
to the transverse axis.
The sub-tangent
of the ellipse.
axis
required the equation of the ellipse.
is
whose abscissa ;
is
2
is
=6
required the equation AnSf
^L
+ ]_ =
l.
are the equations of the tangents to
which form with the X-axis an equilateral triangle
?
that the tangents drawn at the extremities of any chord intersect on the diameter which bisects that chord. 8.
Show
THE ELLIPSE.
139
9. What are the equations of the tangents extremities of the latus-rectum ?
drawn
at the
10. Show that the pair of diameters drawn parallel to the chords joining the extremities of the axes are equal and conjugate.
A chord
11.
of the ellipse H
16
:
"~9~
passes through the point (2, 3) and is bisected by the line x required the equation of the chord.
=
y
;
What
12.
are the equations of the pair of conjugate 2 9x 2 144 which are equal ? ?/
eters of the ellipse 16
+
=
diam
Show
that either focus of an ellipse divides the major two segments whose rectangle is equal (a) to the rectangle of the semi-major axis and semi-parameter to 13.
^
axis in
;
the square of the semi-minor axis. 14.
Show
(&)
that the rectangle of the perpendiculars let fall is constant and equal to the square
from the foci on a tangent of the semi-minor axis. 15.
A system
of parallel chords which make an angle whose 2 with the X-axis are bisected tangent by the diameter of an ellipse whose semi-axes are 4 and 3; required the equation of the diameter.
=
16. Show that the polar of a point parallel to the conjugate diameter. 17.
on any diameter
is
Find the locus of the vertex of a triangle having given = 2 a, and the product of the tangent of the angles
at the base
=
.
^2
Ans.
PLANE ANALYTIC GEOMETRY.
140 18.
Find the locus of the vertex of a triangle having given = 2 a, and the sum of the sides = 2 b.
the base
Ans 19.
-
Find the locus of the intersection of the ordinate of
the ellipse produced with the perpendicular let fall from the centre on the tangent drawn at the point in which the ordi
nate cuts the ellipse. 20. Find the locus generated by the intersection of two tangents drawn at the extremities of two radii vectores (drawn from centre) which are perpendicular to each other. a2 Ans. a^
bW =
21.
A
line of fixed length so
remain in the co-ordinate axes
by any point of the
;
moves that
its
extremities
required the locus generated
line.
=
The angle AOP" y (Fig. 45) is called the eccentric Show that (a; ?/) P the of (# tf) on the ellipse. angle point and from these values of the co-ordinates (a cos cp, b sin 22.
,
,
=
<p)
deduce the equation of the 23.
ellipse.
Express the equation of the tangent at
(a;",
y")
in terms
of the eccentric angle of the point.
Ans.
- cos
a 24.
If
(a/, ?/),
(x",
?/ )
4-
y~
sin
<p
=
1.
b
are the ends of a pair of conjugate and qo show that cp
diameters whose eccentric angles are
90.
q>
,
THE HYPERBOLA.
CHAPTER
141
VIII.
THE HYPERBOLA. THE hyperbola is the locus of a point so moving in a the difference of its distance from two fixed points that plane The fixed is always constant and equal to a given line. 102.
points are called the Foci of the hyperbola. If the points are on the given line produced and equidistant from its extremities, then the given line is called the TRANSVERSE Axis of the hyperbola.
103.
and
To deduce the equation of the hyperbola, given the foci
the transverse axis.
FIG.
Let F,
F
be the
OY 1 to AA
at its
foci,
and
47.
AA
the transverse axis.
Draw
OX
as the
middle point, and take OY,
PLANE ANALYTIC GEOMETRY.
142
PF
;
draw
P
Let
co-ordinate axes.
PF.
also
PD
||
OY.
= (x, y) are = AA 2 a, FF = 2 OF
Then (OD, DP)
the co-ordinates of P.
Let
== 2
OF =
the right angled triangles
FPD
=r
Draw
be any point of the curve. to
2
c,
FP =
r
and F P
r .
From r
= Vy + 2
From
the
and /
2
(x
mode
~c)
= Vy +
and
2
(x
of generating the curve,
/
F PD, we
+ c)
have
2 .
.
.
(a)
we have
= 2 a.
?
Hence, substituting,
Vy +
+ c) 2
2
or,
(x
V?/
2
+
(x
-
clearing of radicals and reducing,
_a (c*
2
)
- a if = a
x2
2
2
2
(c
2
c)
= 2a;
.
.
.
(1)
we have
-
a 2 ) ...
(2)
for the required equation. This equation, like that of the ellipse (see Art. 75), may be put in a simpler form.
Let
c
This value in
or,
2
=b
a2
2 .
.
(3)
.
(2) gives, after
changing signs,
symmetrically,
4a~
b~
= l...(5)
for the equation of the hyperbola and axes.
Let Cor.
1.
the
student
If b
discuss
a in
x2
(5),
if
=
this
when
referred to
its
centre
(See Art. 14)
equation.
we have a2
.
.
.
(6)
The curve represented by this equation is called the ^Equi lateral Hyperbola. Comparing equation (6) with the equation of the circle
we the
see that the equilateral hyperbola bears the
common hyperbola
same
that the circle bears to the
relation to
ellipse.
THE HYPERBOLA. COR.
If (V,
2.
143
are the co-ordinates of two y ) and we have from (4) (x" ,
y")
points on the curve,
y
hence
= -*! (x -
2
2
a
/
2
f
2 : ?/"
a 2) and if 2
::
a) (x
(x
=
+ a)
_
-*i
a2
(35"*
a)
:
(x"
^)
(x"
;
-f a)
;
the squares of the ordinates of any two points on the hyperbola are to each other as the rectangles of the segments in which they divide the transverse axis. i.e.,
COR.
= x a and y and dropping accents,
By making x
3.
after reducing
V-
2
b
x2
+ 2 ab x = 2
for the equation of the hyperbola,
104.
From equation b =
4-
...
A
(3) Art. 103,
=y
f
in
(4)
we have
(7)
being taken as origin.
we have
vr~
this distance off above
and below the origin on the the points B, B Fig. 47, Art. 103. The line BB i called the CONJUGATE Axis of the hyperbola. The points A and A are called the VERTICES of the curve. The Laying
Y-axis,
we have
7
,
bisects all lines drawn through it and point terminating in the curve for this reason it is called the CENTRE of the ;
hyperbola.
The
ratio
Vet 2
-I-
62 -
=
a
1 a
>
=
e.
See
(3) Art.
103 ... (r
called the ECCENTRICITY of the This ratio is hyperbola. 2 1. The value of c b 2 measures the evidently -t Va distance of the foci F, from the centre. is
=
>
F
If b
=a
in (1),
we have
e
= V2
+
for the eccentricity of the
equilateral hyperbola.
105. To find the values of the focal radii, r, r of a point on the hyperbola in terms of the abscissa the of point. From equations (a) Art. 103, we have
PLANE ANALYTIC GEOMETRY.
144
From
the equation of the hyperbola, (4) Art. 103,
we have
Hence, substituting r
=
x2
\/
V a2
b
2
+x
2
a2
= hence
r
Similarly,
we
106.
Tio
and
= ex
a
2 ,
Art. 104 (1),
:
a ...
= ex + a
.
.
(1)
.
(2)
construct the hyperbola having given the transverse
the foci of the curve.
FIG.
First Method. foci.
+c
find
/
axis
x
,
2 ex
48.
Let A A be the transverse axis and F, F the ruler whose length is L and attach
Take a straight-edge
,
THE HYPERBOLA.
145
its ends at F so that the ruler can freely revolve about Cut a piece of cord so that its length shall be that point. L 2 a, and attach one end to the free end of the ruler,
one of
=
and the other end to the focus F.
Place the ruler in the
position indicated by the full lines, Fig. 48, and place the Stretch point of a pencil in the loop formed by the cord. the cord, keeping the point of the pencil against the edge of
the ruler.
If
we now
revolve the ruler upward about
F
,
the
the pencil, kept firmly pressed against the ruler, point will describe the arc AP of the hyperbola. By fixing the end of the ruler at F, we may describe an arc of the other of
It is evident in this process that the difference of the distances of the point of the pencil from the foci F ,F, is always equal to 2 a.
branch.
Take any point D on the transverse axis. Second Method. Measure the distances A D, AD. With F as a centre and A D with F as a centre and as a radius describe the arc of a circle ;
AD
as a radius describe another
arc.
The
intersection
of
these arcs will determine two points, P 1? P 2 of the curve. By interchanging centres and radii we may locate the points E, 1? ,
R
2
In this manner we
n the other branch. >
many
may determine as may require.
points as the accuracy of the construction
107. To find the lotus-rectum or parameter of the hyperbola. The LATUS-RECTUM, or PARAMETER of the hyperbola, is the double ordinate passing through either focus.
Making x
we have
=
-j-
y
V&
2
+
=
.-.
a
b
2
in the
2y
=
Forming a proportion from
2y:2b i.e,
equation of the hyperbola
.
a this equation,
we have
::b:a-,
the latus-rectum of the hyperbola is a third proportional to
the axes.
PLANE ANALYTIC GEOMETRY.
146 108.
The equation of the
and axes
is
a 2y 2
The equation and axes
-+-
b
2
x2
ellipse
=ab 2
of the hyperbola
in the sign of b 2 2
its
centre
its
centre
.
when
referred to
is
expressions
-
referred to
2
Comparing these equations, we is
when
.
If,
see that the only difference
therefore, in the various analytical for the ellipse, we substitute
we have deduced
+ V
for & 2 or,
what is the same thing, I 1 for b, we will obtain the corresponding analytical expressions for the b
,
hyperbola.
To deduce the equation of the conjugate hyperbola. hyperbolas are CONJUGATE when the transverse and con jugate axes of one are respectively the conjugate and trans verse axes of the other. 109.
Two
FIG.
Thus
49.
AA
be the transverse axis of the hyper in Fig. 49, if for its conjugate axis, then the hyperbola
bola which has
which has
BB
BB
for its transverse axis
and
AA
for its conjugate
THE HYPERBOLA. axis
its
is
147
conjugate; and, conversely, the hyperbola whose is BB and conjugate axis is AA has for its
transverse axis
conjugate the hyperbola whose transverse axis
whose conjugate axis is BB We have deduced, Art. 103,
is
AA
and
.
*a1
= b
(5),
1
.
.
2
(1)
for the equation of the hyperbola whose transverse axis lies wish to find the equation of its conju along the X-axis.
We
has
from the figure that the hyperbola which for its conjugate axis for its transverse axis and
It is obvious
gate.
BB
AA
bears the same relation to the Y-axis as the hyperbola whose and conjugate axis is BB bears to the transverse axis is
AA
X-axis
;
hence, changing a to I and b to
a,
x to y and y to x
we have
in (1),
-
=
-...
for the equation of the conjugate hyperbola to the hyperbola
whose equation is (1). Comparing (1) and (2) we see that the equation of any hyperbola and that of its conjugate differ only in the sign of the constant term.
COR.
Since
V^
2
+ u2 = V
any hyperbola and those of
The
2
its
+b
2 ,
the
focal
distances
eccentricities of conjugate
hyperbolas, however, are
For the hyperbola whose semi-transverse axis a and semi-conjugate axis is b, we have
not equal.
Art. 104, (1) e
For
its
=
Vj
of
conjugate are equal.
.
conjugate hyperbola, we have
is
PLANE ANALYTIC GEOMETRY.
148
EXAMPLES. and
Find the semi-axes, the eccentricity of each of the following hyperbolas
the- latus-rectum
:
1.
9
2.
^1
2,2
_
4 X2
-
-^!
4 2
3.
?/
=
36
5.
3 y2
6.
a?/
1.
-
2
2
cc
bx*
2
=
12.
=-
aft.
9
-
16 x 2
= - 16.
7.
4
4 z2
4.
=_
-
16 y z
= - 64.
8.
y
2
- if = m. -
=
ma?
TO.
Write the equation of the hyperbola having given
foci
axis
= 12
axis
= 10
The transverse
9.
:
the distance between the
;
= 16.
10.
The transverse
=
parameter
;
8.
Ans.
25 11.
Semi-conjugate axis
=6
;
The equation
-L 36 64 X2
13. The conjugate axis double the conjugate.
is
10,
x2
The transverse
axis
is
8,
.<
1.
-
3 if
3 yz
=
3? -
Io
and the conjugate axis
=
distance between foci.
16
6.
+ 6 = 0.
and the transverse axis A Ans
14.
2
?/
j
2 of the conjugate hyperbola, x
Ans.
l.
= 10.
the focal distance
Ans.
12.
-= 20
-fID
is
THE HYPERBOLA. 15.
149
Given the hyperbola 2
.2
10
,,2
jr
_i
.
4
required the co-ordinates of the point whose abscissa its
is
double
ordinate.
2
l 16.
Vf Vj
Write the equation of the conjugate hyperbola
to each
of the hyperbolas given in the first eight examples above. 2 4 x* = Given the hyperbola 9 y focal radii of the point whose ordinate
36
17.
is
;
=1
required the
and abscissa
positive. 18.
Determine the points of intersection of
49 = ^~
= + $16
and
1.
16
1.
110. To deduce the polar equation of the hyperbola, either focus being taken as the pole.
F as the pole, Fig. 47. PFD) = (r, 0) be the co-ordinates
Let us take Let (FP,
on the curve.
From
From
r = ex - a = OD OF + FD r cos x = ae
Art. 105, (1),
FP =
i.e.,
Fig. 47,
.
of
any point
P
we have .
.
(1)
;
0.
-f-
and reducing, we have
Substituting this value in (1)
a
(I
1
-^ e
...
cos
(2)
for the polar equation of the hyperbola, the right being taken as the pole.
Similarly from Art. 105,
(2),
a-* / = *1 e cos
hand focus
we have
2
)
for the polar equation, the left
.
(3 )
hand focus being the
pole.
PLANE ANALYTIC GEOMETRY.
150
=
If
COR.
r
0,
/ 90,
r
= =
r
=
a
If
If
a
ae2
-f-
=
= a ae = FA = a + ae = F A.
=
-
=
,
semi-latus rectum.
= semi-latus rectum. = = a a r = - a + ae = FA, T = a -ae= - F A b = semi-latus rectum. a 4- ae r =
ae 2
= ISO
,
.
_
If $
2
270
2
a
=
r
a
ae
2
= semi-latus rectum.
= a
111. To deduce the equation of condition for the supple mentary chords of the hyperbola. By a method similar to that of Art. 81, or by placing b~ 2 for b in (3) of that article, we have =
aa hence, the product of the slopes of any pair of supplementary chords of an hyperbola is the same for every pair. If
COR.
=
a
we have
b,
1, or, s
ss
=
1 ,
S .
.
tan
sum of
= cot
;
the two acute angles which
any pair of sup make ivith the plementary chords of an equilateral hyperbola hence, the
X-axis
is
equal
to
90.
To deduce the equation of the tangent
112.
to
the hyperbola.
adopted in the entirely analogous or substituting 57 Arts. 41, 82, circle, or ellipse, or parabola, 2 2 find we Art. of b for b in (5) 82,
By
a
to that
method
;
-
xx"
~a7"
yy"_ "
b
2
1
a)
to be the equation of the tangent to the hyperbola.
THE HYPERBOLA.
151
To deduce the value of the sub-tangent. operating on (1) of the preceding article (see Art. 83),
113.
By we
find
Sub-tangent
114.
The
hyperbola
(0,
=
x"
-
x
=
*-
"*
~
a* .
slope of a line passing through the centre of an is 0) and the point of tangency (x",
y")
x
The
slope of the tangent t
=
Art. 112, (1)
is,
^L
.*1
a2
y"
member by member, we have
Multiplying these equations,
= Comparing
a)
$...
(1) of this article ss
=
tt
.
.
.
with
(1) of Art. Ill,
we
find
(2)
Hence, the line from the centre of the hyperbola to the point of tangency and the tangent enjoy the property of being the supplemental chords of an hyperbola whose semi-axes bear to each other the ratio a
=
.
COR. If s = t, then s t if one i.e., supplementary chord of an hyperbola is parallel to a line drawn through the centre, then the other supplementary chord is to the ;
parallel
tangent drawn to the curve at the point in which the line through the centre cuts the curve. 115. The preceding principle affords us a simple method of drawing a tangent to the hyperbola at any given point of the curve.
PLANE ANALYTIC GEOMETRY.
152
FIG.
50.
Let P be any point at which we wish to draw a tangent. 7 to P O join C and A. Join P and 0, and from A draw A C will be the required from P to CA The line P T, drawn 7
||
;
||
tangent.
To deduce the equation of the normal to the hyperbola. can do this by operating on the equation of the tangent, 2 in the equa as in previous cases, or by changing b into tion of the normal to the ellipse, Art. 86, (3). By either 116.
We
b"
method, we obtain
for the required equation. 117. To deduce the value of the sub-normal. By a course of reasoning similar to that of Art. 87,
sub-normal
COR. If
b
d
we have
*/ %
$-
= a, sub-normal
i.e.,
=
7/2
in the equilateral
=
x"
;
hyperbola the sub-normal
the abscissa of the point of tangency.
is
equal to
THE HYPERBOLA.
153
EXAMPLES. Deduce the polar equation of the hyperbola, the pole
1.
being at the centre. 7.2
=
^!__ a 2 sin 2
b
-\-
2
cos
2
Write the equation of the tangents to each of the follow ing hyperbolas, and give the value of the sub-tangent in each case.
-
3.
4
-
9 y2
2.
-
"I
._
= - 36, 1,
at
=
at
(
4 ^2
J
= _ 36j = abj
5.
y
6.
a?/
7.
^1-J^ = 1,
2
m
at
~=T7T
-
i
-
4 x2
to 2
(4,
(5,
ord.
ord.
4 Ord >
+).
+).
+)
at
at
at
(aft, ord. +).
(Vm,
0).
7^
Write the equation of the normal to each of the above hyperbolas, and give the value of the sub-normal in each 8.
case. 9.
=
;
x 6 The equation of a chord of an hyperbola is y what is the equation of the supplemental chord, the
axes of the hyperbola being 12 and 8 ? A Ans.
10.
y
= 4- x
-
8
_
.
Given the equations -
Y-
=~
-
-f-
1,
and y
-x=
;
required the equations of the tangents to the hyperbola at the points in which the line pierces the curve.
PLANE ANALYTIC GEOMETRY.
154 11.
9
One
of z
?/2
the supplementary chords of the hyperbola 144 i s parallel to the line y -=x what are
_ if} x = _
;
the equations of the chords ?
16
Ans.
aj
y
Given the
12.
2x
hyperbola
What
is
if
T is
118.
=
6
3 the
required
;
at the positive
end of
the equation of a tangent to
x2
which
3 if
normal
equations of the tangent and the right hand focal ordinate. 13.
2
_16.
9
(
_ :
_,
"T
x
parallel to the line 2?/
+ l = 0?
lines drawn to any point by the tangent at that point. Art. equation of the tangent line,
The angle formed by the focal
of the hyperbola
is bisected
in the
o Making y we have
112, (1),
x
= OT.
=
Fig. 50.
x
From
OF = OF = ae OF - OT = FT = ae - -^ =
Art. 104, (1)
hence
;
9
x
OF
\ x
(ex"
+ OT = F T =ae + -^ = ~-a FT F T + a. JC
.-.
::
:
+ a);
**.
ex"
:
ex"
(ex"
a).
But from Art. 105 we have a FF = = FT +a -a FF FT FT F T FF FT ex"
ex"
.-.
Hence
:
PT
;
::
:
::
ex"
:
:
ex"
+ a.
;
FFF
divides the base of the triangle i.e., the tangent are proportional to the adjacent which into two segments, the angle at the vertex. bisect therefore must it sides ;
THE HYPERBOLA.
155
P N, Fig. 50, is perpendicular to bisects the external angle formed by the focal
Since the normal
COR.
the tangent,
it
radii.
The
SCHOL.
method
of
Let
point.
FP FT ,
.
P
this
article
us
gives
Draw
be the point, Pig. 50.
The
P T drawn
line
between the focal 119.
of
principle
another
drawing a tangent to the hyperbola at a given so as
the focal radii
to bisect
the angle
radii will be tangent to the curve at
To find the condition that the
fulfil in order that
it
line
y
P
= sx + c
.
must
touch the hyperbola
may
^ K=l a2
By
a
b
2
method similar s
2
a2
to that b
2
=c
employed in Art.
89,
we
find
2 .
.
.
(1)
for the required condition.
COR. 1. Substituting the value of equation of the line, we have y
= sx
sV -
4-
b
c
2
drawn from
...
(1) in
the
(2)
for the equation of the tangent to the hyperbola in terms of its slope.
120.
To find the to the
hyperbola and a perpendicular
to it
tangent as the point of tangency moves around the curve.
x 2 -f if is
of a a from focus
locus generated by the intersection
=
a2
.
.
.
(1)
the equation of the required locus.
(See Art. 90.)
121. To find the locus generated by the intersection of tivo tangents which are perpendicular to each other as the points of
tangency move around the curve. x* is
+if
=a
2
b*
.
.
the equation of the required locus.
.
(1)
(See Art. 91.)
PLANE ANALYTIC GEOMETRY.
156
Two
122.
without
tangents are draivn
;
tangency.
^_ a2
is
yy_ b
=i
.
.
.
2
(i)
To find the equation of the polar of
with regard
to the
the pole (x
1
,
/
),
hyperbola
_-!... is
hyperbola from a point
(See Art. 92.)
the required equation.
123.
to the
required the equation of the line joining the points of
the required equation.
(1,
(See Arts. 49 and 93.)
To deduce the equation of the hyperbola when referred a pair of conjugate diameters.
124. to
A pair
of diameters are said
to be
conjugate
when they
are so
related that the equation of the hyperbola, when the curve is referred to them as axes, contains only the second powers of the variables. ~2
a is
2
,,,2
%b
=1
.
:
.
2
(1)
the required equation, and
a 2 sin
sin
tan 6 tan
or
2 b cos 9 cos
<p
q>
=~
.
.
.
a2 -
is
y
= 0,
(2)
the condition for conjugate diameters. (See Art. 94.) From the form of (1) we see that all chords
COR.
parallel to
SCHOL.
drawn
one of two conjugate diameters are bisected by the other.
From
hence
Art. Ill, (1)
ss
= tan = tan
we have
tan
<p.
we have
=
tan cp i.e., if one of s s 0, a to is diameters tivo conjugate chord, the other conju parallel to the is diameter supplement of that chord. parallel gate If,
therefore,
;
THE HYPERBOLA. From
157
we have
Art. 114
= il; a2
hence If,
tt
therefore,
t
= tan = tan
we have
0,
t
= tan
<p
;
i.e.,
if one of
a tangent of the hyper the other conjugate diameter coincides with the line joining
two conjugate diameters bola,
tan
the point of tangency
125.
From
is
and
to
parallel
the centre.
the condition for conjugate diameters, tan
tan
9?
=
b
2 .
a*
we
see that the products of the slopes of any pair of conju is positive; hence, the slopes are both positive
gate diameters
or both negative.
gate diameters must
126.
It appears, therefore, that lie in the same quadrant.
any two conju
To find the equation of a conjugate diameter.
FIG.
51.
PLANE ANALYTIC GEOMETRY.
158 Let
P"R"
be any diameter
P
then
;
B/,
drawn through the /A
parallel to the tangent at Art. 124, Schol. jugate diameter. The equation of the tangent at
centre
(P
P"
N)
P"
a2
xx"
or
y
_
=
y")
is
2
b
hence, the equation of
(#",
will be its con
PB
is
yy"
_
*
"
7)2
^.^x a
.
.
(2)
2
y"
^V
But
is
= cot
=1
P"OX
;
s
the equation of a diameter in terms of the slope of
its
conju
gate diameter.
127.
To find the
diameter, the
of either extremity of a
co-ordinates
of one
co-ordinates
extremity of
its
conjugate
diameter being given.
Let the co-ordinates of
By
P"
y"),
(x",
Fig. 51, be given.
a course of reasoning similar to that of Art. 96,
The upper
the point
signs correspond to
lower signs to the point B/
(
x
y
,
P
we
f
(x
,
y)
;
find
the
}.
128. To show that the difference of the squares of any pair the of semi-conjugate diameters is equal to the difference of squares of the semi-axes. By a course of reasoning similar to that of Art. 97, or, by 2 2 2 for b in (4) of that article, we b for b\ - b substituting
-
find
a
*
-
b
2
=
a2
-
b
2 .
.
.
(1)
THE HYPERBOLA.
=
=
b COR. If a b, then a has equal conjugate diameters.
129.
;
i.e.,
the equilateral hyperbola
To show that the parallelogram constructed on any two
conjugate diameters on the axes.
By
f
159
is
to the
equivalent
rectangle constructed
a method similar to that of Art. 98,
4 a
Area
i.e.,
-
r
b sin (g>
0)
= 4 ab
we can show
that
;
MNM N = Area CDC D
Fig. 51.
.
EXAMPLES.
The
1.
line
y
= 2 x + c touches _ _ ""
-i
=
the hyperbola
.
4
9
what
,y
is
the value of
A
tangent to the hyperbola
c ?
Ans. 2.
10 its
Y-intercept
=
j i 12 2 required ;
its
V32.
-L,
slope and equation.
VI76
Ans.
A
2
tangent to the hyperbola 4?/ angle of 45 with the X-axis required 3.
=
=
_*!
has
c
;
y
2x
2
= =
VlTO x 6
+ 2.
makes an
its equation. 2 are drawn to the 9 x2 tangents hyperbola 4 y - 36 from the point (1, 2) required the of the chord equation of contact. ;
4.
=
Two
;
Ans. 9 x 5.
What
focus ? 6.
the equation of the polar of the Of the left-hand focus ?
What
is
is
the polar of 4 if x2
(1,
=
7.
)
=
8 y
36.
right-hand
with regard to the hyperbola
^5.
4?
Find the diameter conjugate
to
y
=x
2
cc
?/
= 4.
in the hyperbola
.2
J/_
16
Ans.
y
=
PLANE ANALYTIC GEOMETRY.
160 8.
Given the chord y
^ _ t.
= 2x =1
6 of the hyperbola
;
4
9
-f-
required the equations of the supplementary chord. Ans. y
=
x
.
9. In the last example find the equation of the pair" of conjugate diameters which are parallel to the chords. 2 x. Ans. y 2 x, 9 y
=
10.
144
The point
(5,
^)
lies
on the hyperbola 9
=
if
16 x 2
=
required the equation of the diameter passing through also the co-ordinates of the extremities of its conjugate it diameter. ;
;
130.
To deduce the equations of the
rectilinear asymptotes
of the hyperbola.
An ASYMPTOTE
of a curve
is
a
line
passing within a finite
distance of the origin which the curve continually approaches, and to which it becomes tangent at an infinite distance.
FIG.
52.
THE HYPERBOLA.
161
The equation of the hyperbola whose transverse axis along the X-axis may be put under the form y*
=
~ (^ -
2
)
...
lies
(i)
(ju
DD CO
The equations of the diagonals, constructed on the axes BB are ,
,
AA
=
y or,
squaring,
,
of the rectangle
-x, a
2y"
=
7,2
x2
.
.
.
(2)
a/ r
where y represents the ordinates of points on the diagonals. Let P (x, y] be any point on the X-hyperbola and let Sub (x, y } be the corresponding point on the diagonal DD tracting (1) from (2) and factoring, we have
D"
;
.
<y-z/)0/+2/)
y =
r
hence
y
D"P
=
=
2 ;
-
+
.
.
(3)
y y As the points P recede from the centre, 0, their ordi nates D"]ST, P N increase and become infinite in value when and P are at an infinite distance. But as the ordinates D",
D"
increase the value of the fraction
(3), which represents their becomes zero when y and y are and P are continually approach ing each other as they recede from the centre until at infinity they coincide. But the locus of during this motion is the
difference, decreases and infinite hence, the points ;
D"
D"
DD
infinite
diagonal hence, the diagonals of the rectangle constructed on the axes of the hyperbola are the asymptotes of ;
the curve.
Therefore
y
= + * x and y = - - x a a
are the required equations. COB. 1. If a b, then
= y =
i.e.,
45
-4-
x and
?/
=
x
;
the asymptotes of the equilateral hyperbola with the X-axis.
make
angles of
PLANE ANALYTIC GEOMETRY.
162
COR. 2. The equation of the hyperbola COD jugate to may be put under the form
Subtracting (1) from
_
y
y
hence, aw hyperbola and totes of each other.
COR.
3.
ti"
if/ie
p
i
its
p/ == _
conjugate are curvilinear
Subtracting (2) from
y hence,
we have
(4),
=
i,
y
-=
(1)
"P
asymp
we have
(4),
Y\"
"\v
y
+y
-jr~>
rectilinear asymptotes of
/
>
an hyperbola and of
its
con
jugate are the same. 131.
To deduce the equation of the hyperbola when referred
to its rectilinear
asymptotes as axes.
FIG.
The equation is
53.
of the hyperbola
when
referred to
OY, OX,
THE HYPERBOLA.
163
We wish to ascertain what this equation becomes when OY OX the rectilinear asymptotes are taken as axes. Let P be any point of the curve let Y OX = XOX = 0. Then (OC, CF) = (x, y) (OD, DP = (x y From the figure, OC = OK + DR CP = RP - DK ,
;
)
;
}.
,
;
;
y = But from the triangle OAB, we have x
i.e.,
=
hence, x
=
+y
f
(x
r
)
cos
;
(ij
-x
)
sin
B.
r
(x
-+-
+
a
Va^+ft
Substituting these values in f
(x or,
(9
+ 7/) 2
(y
we have
(1),
-
a
reducing and dropping accents, 2 a2 b Xy ...
= -+
J__
34
2
)
= a* -f ^
2 ,
(2)
for the equation of the hyperbola referred to its asymptotes. In a similar manner we may show that
is
the equation of the hyperbola conjugate to
ferred to
COR.
asymptotes as axes. Multiplying (2) by sin 2 6
(1),
when
re
its
we may
place the result in
the form
yx sin 2 that
is
therefore
=
^1A
+*
2 "
.
DP P H = ON. AH area ODP P = area OMAN
8jn
2
;
.
;
;
hence, the area of the parallelogram constructed upon the co ordinates of any point of the hyperbola, the asymptotes being axes, is constant and equal to the area of the structed upon the co-ordinates of the vertex.
rhombus con
PLANE ANALYTIC GEOMETRY.
164 132.
when
To deduce
the equation of the tangent to the hyperbola
the curve is referred to its rectilinear asymptotes as axes. Y,
FIG.
54.
a course of reasoning similar to that employed in Arts, 41, 57, 82, we find the required equation to be
By
y-Y = -^(*-*") X or,
...
(i)
symmetrically,
1^=2... COR. If we make y
=2 OM = x
But
=
x"
in (2),
=
x",
OT.
.-.
-
(2)
we have Fig. 54.
OM = MT
.-.
TD
= TD
;
hence, the point of tangency in the hyperbola bisects that por tion of the tangent included between the asymptotes.
THE HYPERBOLA.
D
Since
133.
=
+ b\ = a +b OT OT = a + b
4 2
or
x"y"
x"
.
a2
2
2 if
2
2 ;
2
.
i.e.,
we have
a point of the hyperbola,
is ?/")
(x",
(see Fig. 54)
165
.
.
.
(1)
hence, the rectangle of the intercepts of a tangent on the totes is constant
and equal
to the
sum of
asymp
the squares on the
semi-axes.
From
134.
(1) of the last article
sm
^
ing through by
we
have, after multiply
,
OT-^LfYl v sin 20 =
n- J_ A 2 "f-^-
sin 2
=
2
(a
+b
2
)
sin
cos
0.
But, Art. 131,
=
sin
-J 2
;
2
sin 2
=
arm OTT
i.e.,
(t
Va + b
2
OT or
hence
.-.
6
=
cos
,
Va +
= a&
is
equivalent
to
;
arm OAD B.
Ae triangle formed by a tangent
asymptotes
2
to
the hyperbola
Draw the chord RB/, Fig. 54, parallel Draw also the diameter OL through D. Since TD = T D, we have B/L = BL.
135.
T
and
its
the rectangle on the semi-axes. to the tangent
T.
Since
i.e.,
OL
is
LK = LH LK = BL LH
a diameter,
we have
BL B K = BH
;
hence
;
;
hence, the intercepts of a chord between the hyperbola asymptotes are equal.
and
its
PLANE ANALYTIC GEOMETRY.
166
EXAMPLES.
What
1.
are the equations of the asymptotes of the hyper~2
bola
,.2
f-gAns.
What
y
=
| x.
-J-
are the equations of the asymptotes of the follow
ing hyperbolas
:
_. 3.
3 ^2
_
2 ^2
Ans.
=_ y = -^ Vf x.
mx
5
2
_
Uy 2
=
c
6. What do the equations given in the four preceding ex amples become when the hyperbolas which they represent are referred to their asymptotes as axes ?
=
The semi-conjugate axis of the hyperbola xy what is the value of the semi-transverse axis ?
7.
6
;
25
Ans.
What
is
8.
are the equations of the tangents to the following
hyperbolas
:
8.
To xy
=
9.
To xy
=+
10. at (1, 10).
Ans.
Ans. 10.
To xy
11.
To xy
m, at
= - p,
1,
(
at
(
\
+ 10 x = 20. = 3 x 12.
y
12, at (2, 6).
y
-\-
m). 2, /
Required the point of the hyperbola xy 4. the sub-tangent 12.
=
12 for which
=
Ans.
(4, 3).
an hyperbola
13. The equations of the asymptotes of whose transverse axis = 16 are 3 y = 2 x and 3 y
+
2x
=
;
required the equation of the hyperbola. *
"
64
256
THE HYPERBOLA.
167
14. Prove that the product of the perpendiculars from any point of the hyperbola on the asymptotes stant and
let fall is
con
GENERAL EXAMPLES. 1.
10
;
The point
(6, 4) is on the hyperbola whose transverse required the equation of the hyperbola.
is
-
Am. 25
400
Assume
the equation of the hyperbola, and show that the difference of the distances of any point on it from the foci is constant and 2 a. 2.
=
Required the equation of the hyperbola, transverse axis 2 x and 3 y 6, which has 5 y 13 x for the equa tions of a pair of conjugate diameters. 3.
=
=
=
2
^_^/_ = 1
An*.
9 78 that the ratio of the sum of the focal radii of any point on the hyperbola to the abscissa of the point is con stant and 2 e. 4.
Show
=
5.
fulfil
What
are the conditions that the line
y
= sx
-f c
must
a
= 0.
in order to touch
-^ a 2
-
-2 b
=
1 at infinity J ?
Ans.
s
=
_L_
,
c
6. Show that the conjugate diameters of an hyperbola are also the conjugate diameters of the conjugate hyperbola. 7. Show that the portions of the chord of an hyperbola
included between the hyperbola and its conjugate are equal. 8. What is the equation of the line which passes through the focus of an hyperbola and the focus of its conjugate hyperbola ?
Ans.
x
PLANE ANALYTIC GEOMETRY.
168
Show
9.
that
^
=
e
when
e
and
e
b
_ a
are the eccentricities of
two conjugate hyper
bolas.
Find the angle between any pair of conjugate diame
10.
ters of the hyperbola.
Show that in the hyperbola the curve can be cut by of two conjugate diameters. one only 11.
Find whether the line y = f x intersects the hyperbola 9 x2 = 144, or its conjugate.
12.
16 y~
Show that the conjugate diameters of the hyperbola make equal angles with the asymptotes. 13.
Show
14.
eral
drawn from any point
that lines
equilateral
of the equilat
hyperbola to the extremities of a diameter
make equal
angles with the asymptotes.
In the equilateral hyperbola focal chords drawn parallel to conjugate diameters are equal. 15.
A
from the focus of an hyper perpendicular is drawn show the bola to asymptote that the foot of the perpendicular is at the distance a 16.
:
(a)
from the centre, and (b)
that the foot of the perpendicular
from the 17.
is
at the distance b
focus.
For what point of an hyperbola
is
the sub-tangent
=
the sub-normal ? that in the equilateral hyperbola the length of the normal is equal to the distance of the point of contact 18.
Show
from the 19.
centre.
Show
chord of
that the tangents drawn at the extremities of any diameter which hyperbola intersect on the
the
bisects the chord.
THE HYPERBOLA. 20.
Find the equation of the chord of the hyperbola J?L -
J^!
is
=1
12
9
which
169
bisected at the point
(4, 2).
Required the equations of the tangents to
21.
J*L.
I*-
16
10
=
which make angles of 60
1
with the X-axis.
that the rectangle of the distances intercepted on drawn at the vertices of an hyperbola by a drawn at any point is constant and equal to the
Show
22.
the tangents
tangent
square of the semi-conjugate axis.
Given the base of a triangle and the difference of the
23.
tangents of the base angles
Show
24.
;
required the locus of the vertex.
that the polars of
(m, n) with respect to
the
hyperbolas
^1 2
_ J/l =
a
If
25.
1,
b*
ll
- ^1 =
V2
a2
1 are parallel.
from the foot of the ordinate of a point
hyperbola a tangent be drawn to the
circle
(x,
y) of the
constructed on
the transverse axis, and from the point of tangency a line be drawn to the centre, the angle which this line forms with the Show transverse axis is called the eccentric angle of (x, y). values deduce from these and b tan sec that go, go), y) (a
=
(x,
the equation of the hyperbola. 26.
If
r
(x
,
y
),
(x",
y")
are the extremities of a pair of
conjugate diameters whose eccentric angles are that
<p
+
<p
=
90.
go
and
cp,
show
PLANE ANALYTIC GEOMETRY.
170
CHAPTER
IX.
THE GENERAL EQUATION OF THE SECOND DEGREE. 136. The most general equation of the second degree be tween two variables is 2 2 (1) ay + bxij + ex + dy + ex
+f=
.
.
.
whatever. d, e, /are any constant quantities this which loci the of the To investigate equation properties of the constants as to represents under all possible values
in
which
a, b,
c,
sign and magnitude
is
the object of this chapter.
The equations of the lines in have had to do in preceding chapters, 137.
By + C = 0. 2 = (Ax + B# + C) _ _ Two o. yS. 2/2
Ax
-f
_j_
x2
7/2 _|_
xt
7/2
2 7/
=a _ o.
2
= 2px.
.
0.
a plane, with which are
we
Straight line. Two coincident straight lines.
straight lines.
Circle.
Two
imaginary straight
lines.
Parabola.
= a 2^ Ellipse. _ = crb tfxay Hyperbola. = a 2j^ Hyperbola, a 2^2 _
0,2^2 _|_
2^,2
2
.
#83.2
we Comparing these equations with the general equation, them may be deduced from it by making the constants fulfil certain conditions as to sign and magnitude. the lines which We therefore, prepared to expect that see that all of
are,
these equations represent will appear among the loci repre sented by the general equation of the second degree between
two
variables.
find that these
equation.
In the discussion which lines
is
to ensue
are the only loci represented
we
by
shall
this
EQUATION OF THE SECOND DEGREE.
171
DISCUSSION. 138. To show that the
locus represented by
a complete equa
tion of the second degree between two variables is also represented by an equation of the second degree between two variables, in which the term containing xy is wanting.
Let us assume the equation ex* +dy ex aif bxy (1) and refer the locus it represents to rectangular axes, making the angle with the old axes, the origin remaining the same. Prom Art. 33, Cor. 2, we have
+
+
+f=
+
x
=x
cos 6
y
x
sin 6
in
+bx
cos
Substituting these values
+ V + d y + e x +/= 2
C
y>
.
y sin
+y
for the equations of transformation. in (1), we have,
a y*
.
.
.
.
.
(2)
which
= a cos = 2 (a c = a sin d! = d cos e = d sin a!
V
+ c sin
2
2
6 sin B cos 2 -\- c cos e sin -J- e cos
c)
2
cos
b sin
}
+ b (cos 6 + 6 sin cos 2
sin 2 0) I
.
.
.
(3)
0, the angle through which the axes have been turned, entirely arbitrary, we are at liberty to give it such a value as will render the value of b equal to zero. Supposing it to
Since
is
have that value, we have or
2 (a
c)
(a
sin 2
c)
or
cos 6
sin
tan
20
=
2
-
2
(cos
=
.
^ .
c
Since any real
+b
-f b cos
a
number between
-f-
=
sin 2 0)
.
.
0,
(4)
(5) GO
and
o>
is
the tan
gent of some angle, equation (5) will always give real value for 20; hence the above transformation is always possible.
Making
b
=
in (2),
a lf for the
we
have, dropping accents,
+ eV + d y + e x +f =
equation of the locus represented by equation, then, we shall confine our attention.
.
.
(1).
.
(6)
To
this
PLANE ANALYTIC GEOMETRY.
172
To find the value of a and c in terms of a, b, Adding and then subtracting the first and third of
COR.
and
1.
c.
the equations in
we have
(3),
+a =c+a (7) = (c - a) cos 2 + b sin
c
-
c
Squaring
a
(4)
and
...
c
.
...
2
-
a
=
V(c
-
(7)
+ b*
8
a)
and
.
.
.
(8)
we have
adding to the square of (8),
Subtracting and then adding
a
.
.
(9)
we have
(9),
= i\ G + a- V(c - ay +Tj = $ \c + a + V(c-a) + ^
.
.
(10)
.
2
c
COB.
and
To
2.
find the signs of a!
and
c
y= ...
Hence, the 2
a
c
i |( C
+
=-
i
of a
si#?is
(V
and
-(( -- 4 ac)
a
2
= 0,
.-.
2
c
.
) .
-
(12)
depend upon the sign of the
ac.
<
b
+ * )h
2 <0
4 quantity b The following cases present themselves 2 4 ac. The sign of the second 6 1.
2.
Multiplying (10)
we have
(11),
positive,
(11)
.
a and
c
= 4 ac. The or c = 0.
:
member of (12) are both positive, or both negative. second
member
of (12)
becomes
zero,
is
.-.
cannot be equal to zero would reduce (6) to at the same time, for such a supposition an equation of the first degree.] 2 4 ac. The sign of the second member of (12) is 3. b and c negative, or a must be a must be be observed that [It will
a and
c
>
negative,
positive
.-.
negative and
c positive. 7
To transform the equation a if + c x -f a ?/ + e a -f the vari f=Qinto an equation in which the first powers of 2
139.
ables are missing.
EQUATION OF THE SECOND DEGREE.
173
Let us refer the locus to a parallel system of rectangular From Art. 32, we axes, the origin being at the point (m, n). have a y = n x = m y Substituting these values in the given equation, we have 2 * = c x + +f" (2)
+
y +
in
+
f
+
,
.
e"x
.
.
.
d"ij
which d"
c"
/"
Since
m
=2an +d = 2 c w -f e = a + cm + ?i
2
+
rf w-
e
??t
+/
n
j
make
as to
+d =
and 2
c
w
-f e
=
m=
and
= .
2 a see
3)
;
we may make
in general,
We
(
f
and w are entirely arbitrary, we may, in general, 2 a n
i.e.,
*
2
them such values
give
]
2
.
f
(4)
c
from these values that when a! and c are not zero, is possible and equation (2) becomes,
this transformation also
;
after dropping accents,
a if +cV+/"
Equation
;
...
(5)
we
observe, contains only the second power of hence it is satisfied for the points (x, y) and
(5),
the variables
=
But only the equation of curves with centres can satisfy this condition hence, equation (5) is the equa tion of central loci. When either a f or is zero, then n or in (
y).
a?,
;
</
and the transformation becomes impossible. two cases which require special consideration.
is infinite
arise
CASE 1. a! == o. Under this supposition equation
Hence
140.
(6),
Art. 138,
cV-MV + ^+/=0
...
becomes
(1)
Referring the locus of this equation to parallel axes, the origin being changed, we have for the equations of trans formation x m x y n -f- y
=
+
,
=
.
PLANE ANALYTIC GEOMETRY.
174
Substituting in C
V + dy + (2 2
Now, make i.e.,
we may
in general,
2
c
m+
we have
(1),
m + e ) x + c m* + d n + e m
c
= 0,
e
c
+f=Q
.
.
.
(2)
and w such values as to
m + d w + e m +/= 2
;
we may make
m= m If d mation
is is
-|- e^/^i, -f-
/_
e
x
7
2
4/c 4
V
rZ
c
=
not zero (since a! 0, c is not zero), this transfor possible and (2) becomes, after dropping accents,
x2
c
Or 2
COR.
and
f ,
g
^
If
d
=
-\-
= 0,
dy
jr a
_
,
-^^...
0, (1)
c x* or,
and
m
give
/Q\ (3)
becomes
+ e x+f=0
solving with respect to
...
(4)
x,
CASE 2. c = o. Under this supposition equation
141.
(6),
Art. 138, becomes
Transforming this equation so as to eliminate ?/ and the constant term, by a method exactly similar to that of the preceding
article,
we
find r
n =
d
J* d*
nyi
- 4 a!f J_
A
i
r
4fcV and,
if e
is
not zero,
we have
f=--,*
(of is
not zero since
(2)
c
= 0)
EQUATION OF THE SECOND DEGREE. COB. If
142.
e
= 0,
equation
Summarizing the
175
becomes
(1)
results of the preceding articles,
we
find that the discussion of the general equation
aif
+
bxy
+ ex + dy + ex +/ = 2
has been reduced to the discussion of the three simple forms 1.
a! if
+ cV +
x2
/"
= -
.
c
\
=
--
y
.
x.
f
= 0.
:
Art. 139, (5)
Art. 140, (3)
Art. 141, (2)
^-.
Art. 140, (5)
3.
^-.
Art. 141, (3)
The discussion now involves merely a consideration sign and magnitude of the constants
of the
which enter into these
equations.
143.
tf
<
4
ac.
Under this supposition, since of and d are both positive or both negative, Art. 138, Cor. 2, neither a! nor c can be zero hence, forms 2 and 3 of the preceding article are excluded ;
from consideration. The first form becomes either
ay
or
which
c a
f
of and c may have any real value and and any sign any value. Hence arise four cases
in
f"
:
may have
PLANE ANALYTIC GEOMETRY.
176
CASE
1.
If
has a sign different from that of a and c whose semi-axes are ,
f"
equations (1) are equations of ellipses
GASP:
2.
If
f"
has the same sign as that of a and
c
,
equa
curves.
tions (1) represent imaginary has a different sign from that of c and CASE 3. If a
=
f"
has If circles. equations (1) are equations of the same sign as of and c then the equations represent imagi
a and
e
/"
,
,
nary
curves.
CASE nary
4.
If/"
= 0, equations (1)
are equations of taw imagi
the origin. straight lines passing through
Hence, when
b
z <
4
ac, every
equation of the second degree
an imaginary curve, between two variables represents an ellipse, at the a circle, or two imaginary straight lines intersecting origin. b
Under
=
;
2
= 4 ac.
either a this supposition, Art. 138, Cor. 2, form (1) of Art. 142 is excluded.
- 0,
or
hence,
Resuming the forms
(3)
y
we have
=
~~
(>
Z
a
four cases depending upon the sign and magnitude
of the constants. are not zero, CASE 1. If d and c in the first form of (2) not are of zero, then form second the in (2) and if e and a of parabolas. equations (2) are equations
EQUATION OF THE SECOND DEGREE.
177
Since the first form of (3) is independent of y, it lines parallel to each other and to the Y-axis. two represents The second form of (3) represents, similarly, two lines which
CASE
2.
are parallel to the X-axis.
CASE
3.
If
2 e" <
4/b the
form of
first
(3)
represents two
lines.
imaginary two imagi If d 2 4/o/, the second form of (3) represents lines. nary CASE 4. If e 2 = 4/c the first form of (3) represents one <
,
straight line parallel to the Y-axis. 4 fa the second form of (3) represents one straight If d 2 line parallel to the X-axis.
=
,
=
2 4 ac, every equation of the second degree Hence, when b between two variables represents a parabola, two parallel straight lines, two imaginary lines, or one straight line.
2
145.
l>
>
4
ac.
Under this supposition, Art. 138, Cor. 2, since a! and f must have opposite signs, neither a nor c can be zero hence forms (2) and (3) of Art. 142 are excluded from con The first form becomes either sideration under this head. c
;
ay
-<fc
-y +
or
We
+/"
cV+/"
1.
If
=OJ
f has a different sign from that of a
(1) are equations of hyperbolas
/r
0=v If
1
have here three cases.
CASE
still
-
,
equations
whose semi-axes are
andJ=v/r,
f has a different sign from that of
c
,
equations (1) are
equations of hyperbolas.
CASE
2.
If a
eral hyperbolas. CASE 3. If f"
=
c
,
= 0,
secting straight lines.
equations (1) are equations of equilat
equations (1) are equations of two inter
PLANE ANALYTIC GEOMETRY.
178
2
4 ac, every equation of the second degree between two variables represents an hyperbola, an equilateral hyperbola, or two intersecting straight lines.
Hence, ivhen
b
>
The preceding discussion has
SUMMARY.
146.
elicited the
following facts 1. That the general equation of the second degree between two variables represents, under every conceivable value of the :
constants which enter into bola, or one 2. 3.
it,
of their limiting
When When
b
2
b
2
When
b
2
4&c
<
=
it
4ac
an
ellipse,
an
represents it
a parabola, an hyper
cases. ellipse, or
a limiting
case.
represents a parabola, or a limiting
case. 4.
4&c
>
if.
represents
an hyperbola,
or
a limiting
case.
EXAMPLES. 1.
Given the equation 3y 2
-\-
2 xy
+3x
2
Sx
Sy
=
;
to classify the locus, transform and construct the equation. Write the general equation and just below (a) To classify. it the given equation, thus :
+ bxy + ex + dy -f ex +/ = 8x = 8y 3 if + 2 xy + 3 x ay
2
2
2
Substituting #>
_ 4 aCj
the
co-efficients
we have
4 ac
tf
the
in
=4-
2 ac. b hence and the locus belongs to the ellipse
36
class
= -
.
.
.
(1)
characteristic
32
;
<4
(b)
To
class, Art. 146.
refer the locus to axes such that the term containing
xy shall disappear.
From
Art. 138, (5),
we have
=
tan 2
hence
tan
20
=
-
o
o .-.
2(9
= 90
.-.
6>
=+ = 45
oc,
.
.
.
(2)
EQUATION^OF THE SECOND DEGREE. new X-axis makes an angle of X-axis. Taking now (10), (11), (3), Art. ing values, we have i.e.,
the
of
179
45 with the old
-j-
138,
and substitut
= i \c + a - V(c - a)- + = 2. = i \c + a + V(c- a) + J=4. = dcosO e sin = V2 (d e) = = d sin + e cos = V^ (^ + e = l>*\
f
c
d e
2
these
being zero), 2 y2
8 V2.
)
in
values
0.
J-
(9
Substituting
2
Art.
(6),
+ 4 x - 8 ViT x =0 2
138,
...
we have (/
(3)
To re/er ^^^ ^oci^s to its centre and axes. (c) Substituting the values found above in (4), Art. 139,
n
have
d
=
=
U.
2c
= an + cm 2
Hence/"
(3).
2
8
-f
cf
found above in 2 if
f"
8,
Art. 139,
+4x
2
for the reduced equation.
= V2 (d)
Draw X-axis.
curve
and
To
b
=
together with the values of a
we have
(5), Art. 139,
8
+ %- = 1
or
a
+ em +/=
dn
.
Substituting this value of
and
we
=
...
0,
(4)
The semi-axes
of the ellipse are
2.
construct.
the axis
OX
See
Draw OY
when
(b).
,
making an angle of 45 with the old The equation of the -Lto OX
referred to these axes
.
is
given in
(3).
Constructing
PLANE ANALYTIC GEOMETRY.
180 the point (c).
(V2,
Draw O
when
Y"
referred to
i-
See 0) we have the centre of the ellipse. The equation of the curve to OX" at .
O
Y",
X
O
as axes is given in (4).
FIG.
55.
V2 and 2, we can construct the by either of the methods given in Art. 78.
Having the semi-axes, ellipse
DISCUSSION. If
in (1),
y
x If
x
If
x
to the If
y
= 0,
x
for the X-intercepts 0,
=-
=
in (1),
we have
=
in (3),
we have y
OD,
.
for the Y-intercepts 0,
=
_|_0;
i.e.,
OC,
the ellipse
is
tangent
Y -axis.
=
in (3),
x If
we have
x = o in
(4),
we have
=
0,
x
we have
y=
=1=2.
for the
X -intercepts
0, OB,
= 2 V2. for the
Y"-intercepts
O
A,
OA
.
EQUATION OF THE SECOND DEGREE. If
y
=
in (4),
we have
=
x
X -intercepts O B, O O,
for the
V2.
-j-
Given the equation ?/ 2 2 xy -f # 2 2y 1 and construct the equation. To classify.
2.
181
=
0, class
ify the locus, transform (a)
+
aif
y hence
b
2
2
2
bxy
+ ex
ajy 4-
4 ac
a?
2
dy
-f
2
2y
=4
+/ =
-f ex 1
=
.
.
.
= 0,
4
(1)
hence the locus belongs to the parabola class, Art. 146. To refer the locus to axes such that the term (b) containing
xy shall disappear.
From
Art. 138, (5),
we have b
20=
tan
a
c
;
hence, substituting
tan 2
=
?
-1
1
= -45
...
...
5
(2)
Substituting the values of the coefficients in (10) of Art. 138,
(11)
(3)
we have
= \ c + a + V(e a) = 2. b*\ = d cos = 2 ( V2) = V2. e sin e = d sin + e cos = - 2 (- 1 V2) = + V2. Substituting these values in (1), Art. 140 (since a = 0), we have 2 x - V2 y + V2^ - 1 = (3) 2
c
-f-
\
d!
2
.
(c)
To
refer the
parabola
to
.
.
a tangent at the vertex and the
axis.
Substituting the values of the constants in (a), Art. 140, r
we have
m=-
n
=
2
e
^-7
~
= - V2 = - -35
e 4fc = ~~T^~~ 4 d c
5
T~7^ 4 V2
=-
-
nearly.
90 nearly. J
PLANE ANALYTIC GEOMETRY.
182
is
Substituting the values of d and not zero), we have
= iV2.y
*
.
.
.
c>
in (3), Art. 140 (since
d
1
(4)
for the reduced equation. (d) To construct.
V
Draw
OX
making an angle
OY 1 to OX
of -- 45
with the X-axis
;
See (). The equation of the parabola when referred to these axes is given in (3). vertex .35, Constructing the point ( .90), we have the
draw
.
and O of the parabola See (c). Draw O parallel to the axes OY respectively. The equation of the parab The curve can now ola referred to these axes is given in (4). be constructed by either of the methods given in Art. 54. .
OX
,
X"
Y"
EQUATION OF THE SECOND DEGREE.
183
DISCUSSION. If
If
If
x
=
y
=
x
=
y
we have
in (3),
we have
_
x
==
OD
,
Y -intercept OK,
.707.
we have
for the
- V2 + yio
Jj
.
X -intercepts OL, OL - V2 - Vio Z=
4 If
OD,
,
V2
in (3),
*/
for the
~=-
=
00
.4.
for the Y-intercept
-l.
=
=-
y
in (1),
00,
for the Y-intercepts
y = 2.4
y If
we have
in (1),
in (4),
4
=
?/
if
;
=
y
x
in (4),
=
_j_
2
-\-
bxy
+
ex 2
-f-
dy
+
2y + 6x
ex
3
-\-
=
f= .
.
0.
= 0, classify
Given the equation ?/ -^ 2 cc 2 2 ?/ + 6 x 3 the locus, transform and construct the equation. (a) To classify. 3.
ay
,
0. .
(1)
hence, the locus belongs to the hyperbola class, Art. 146. (b) To ascertain the direction of the rectangular axes
(xy
being wanting). tan 6
.-.
i.e.,
the
new X-axis
26
=
= 0; is
c-
= -5_ = Q -3
:
parallel to the old X-axis.
To refer the hyperbola
(c)
a
to
its
centre
and
axes,
we
have,
Art. 139, (4),
hence
n
=
1,
m=
-. t
Substituting in the value of
/ = hence
a n2
+cm + 2
f"
dn
=1
+em
/",
Art. 139, (3),
+/ = 1
2
we have
+9
3
;
PLANE ANALYTIC GEOMETRY.
184
This value, together with the values of a? and 2 if 4 x* = - 1 139, gives (3) for the required equation.
-
To
(d)
.
in (5), Art.
c
.
construct.
FIG.
Construct the point
OX, and O
Y" ||
(%, 1),
is
and through
The equation
OY.
to
referred to these axes
57.
given in
We
(3).
tion that the semi-transverse axis
is
-.
u
it
draw O
of
the
X" ||
hyperbola
see from this equa
Laying
tance to the right and then to the left of vertices of the curve A, A
,
off this dis-
we
locate the
.
DISCUSSION. If
x
=
y If
y
we have
in (1),
= o in
(1),
x
= 3, y = we have
=
3
for the Y-intercepts
-
OC,
00
OD,
OD
,
1.
for the X-intercepts
+ V3
to
3- V3
,
EQUATION OF THE SECOND DEGREE.
=
x
If
in (3),
=
y
we have
= iVH^
2/
If
185
in (3),
we have
for the X-intercepts
O
A,
OA
,
.-4 From
this data
the student
readily determine the focal distances of the
may
and the
eccentricity, the parameter,
hyperbola.
= 0,
+
Given the equation y 2 -\- x z 4 ?/ 4# 1 the ify locus, transform and construct the equation. 4.
b2
(a)
<4
=
(b)
ac .-.
the locus belongs to the ellipse class.
.-.
new X-axis
(m,w)=(-2,
(c)
x2
hence
class
-f-
is
to old X-axis.
||
2)and/"
= -9
=9 y-
the transformed equation of the locus, which from the form of the equation is evidently a circle. is
Locate the point
(d)
and with 3 as a
(2,
With
2).
this point as a centre, it will be the re ;
radius, describe a circle
quired locus. 5.
y
2
2 xy
+x
b = 4 ac = 45 2
(a) (b)
.-.
2
new X-axis
We
to the old X-axis.
...
= 0.
parabola class.
.-.
a!
2
= 0, 2x
i.e.,
2
x
c = 2, d = 0, - 2^0; = 1 and x =
are the equations of the locus (c)
inclined at an angle of
45
have also
when
The construction gives the
e
=
1
...
(1)
new
referred to the
lines
OX OY ,
axes.
as the
new
axes of reference.
Equations
drawn
||
(1) are
to the
the equations of the two lines and at a unit s distance from
Y -axis
CM, C it.
M
PLANE ANALYTIC GEOMETRY.
186
FIG.
58.
We may
construct the locus of the given equation without going through the various steps required by the general method. Factoring the given equation, we have (y
hence
-
x
+ V2)
=x
y
V2
(y
-x- V2) =
and y
=x
-f-
;
V2
the equations of the locus. Constructing these lines (OY, being the axes of reference), we get the two
are
OX
CM, C
parallel lines
M
Classify, transform,
equations 6.
2
?/
.
and construct each of the following
:
2 xy
+x +2y
2
+
2
2 x -f 1
= 0. y
7.
8.
if
+
ajy
2
^
x2
- 1 = 0.
4. 5
a-s
_
12 x
- 12 y = 0.
=x
1.
EQUATION OF THE SECOND DEGREE. 9.
10.
2?/
2
4x
+ 2x -4?/2
+1
=0. x2
7/2 _j_ jpa
_ 2 + 1 = 0.
+
x2
+ 2*+2 = 0.
-
2
187
+
2 7/
=
f
.
aj
(1, 0).
11.
2
2/
Imaginary 12.
y
2
ellipse.
+ x - 8 x + 16 = 0. 2
XT/
Parabola. 13.
2 xy -f x
if
2
y
+2x
1
= 0. Parabola.
14.
4 ay
15.
?/
-
2x
+
2
=
0.
Hyperbola.
16.
2
2 or
+ 2 + 1 = 0. ?/
Two
intersecting lines.
- x + 2 y + 2 x - 4 = 0. 2
if
Equilateral hyperbola. 17. 18. 19.
20. 21.
-
+ x + 2 y + 1 = 0. / + 4 xy + 4 - 4 = 0. 2 XT/ + 2 x 2 y + 2 x = 0. 4 + 4 x = 0. if x + 2 = 0. 2 jcy 2
?/
2
2
jcy
2
ic
2
2
2/
2
a;//
2
?/
2
PLANE ANALYTIC GEOMETRY.
188
CHAPTER
X.
HIGHER PLANE CURVES. 147. Loci lying in a single plain and represented by equa tions other than those of the first and second degrees are
We
HIGHER PLANE CURVES. shall confine our atten tion in this chapter to the consideration of a few of those curves which have become celebrated by reason of the labor called
expended upon them by the ancient mathematicians, or which have become important by reason of their practical value in the arts and sciences.
EQUATIONS OF THE THIRD DEGREE. 148.
THE SEMI-CUBIC PARABOLA.
This curve ordinate
OP
TT
let fall
is
the locus generated by the intersection of the common parabola with the perpendicular
of the
from
its
vertex upon the tangent drawn at
T
as
the point of tangency moves around the curve. 1.
To deduce the rectangular equation.
T be the point of tangency, and let P (x y ) be a point of the curve. Let y 2 == px be the equation of the common parabola. to the parabola Since the equation of the tangent line T Let
(x" ,
,
y")
M
is
Art. 57,
(6),
the equation of the perpendicular vertex is
{r
(OM)
let
fall.
from the
HIGHER PLANE CURVES.
189
FIG
Since
TT
is
*=*",
Combining
OY, we have
parallel to
(1)
and (2)>
.
.
for its equation
(2)
we have
But
_ V4j^_
hence
2p Squaring and dropping accents, we have
for the equation of the semi-cubic parabola. This curve is remarkable as being the first curve rectified,
that
is,
the length of a portion of
it
which was was shown to
PLANE ANALYTIC GEOMETRY.
190
be equal to a certain number of rectilinear units. It derives its name from the fact that its equation (3) may be written
xl
= p% y
,
To deduce the polar equation. r Making x = r cos 6 and y
2.
reduction, r
=p
tan
2
sec
sin 9 in (3),
.
.
.
we
have, after
(4)
for the polar equation of the curve.
SCHOL.
Solving
(3)
with respect to
y,
we have
An
inspection of this value shows (a) That the curve is symmetrical with respect to the
X-axis; (b)
That the curve extends
infinitely
from the Y-axis in
the direction of the positive abscissas.
To duplicate the cube by the aid of the parabola. Let a be the edge of the given cube. We wish to con struct the edge of a cube such that the cube constructed on it shall be double the volume of the given cube i.e., that the 3 8 = satisfied. shall be a 2 x condition 149.
;
FIG.
60.
HIGHER PLANE CURVES.
191
Construct the parabola whose equation is 2 2 ax i/ (1) be the curve. Construct also the parabola whose Let
=
.
.
.
MPO
equation
is
x*~y... NPO be this Then OA (= x) Let
}
is
the abscissa of their point of intersection
the required edge.
we have x8
(2)
curve.
For eliminating y between
(1)
and
(2),
= 2 a*.
This problem attained to great celebrity among the ancient We shall point out as we proceed one of the geometricians.
methods employed by them in solving 150.
The
THE
it.
CISSOID.
by the intersection (P) of (OMM T) with the ordinate DN
cissoid is the locus generated
the chord
(OM )
of the circle
s FIG.
61.
PLANE ANALYTIC GEOMETRY.
192
MN (equal to the M on the diameter
MN
ordinate
let fall
from the point
as the chord revolves about
through 0)
the origin 0. It may also be defined as the locus generated by the inter z 8 ax with the section of a tangent to the parabola y let fall on it from the origin as the point of perpendicular moves around the curve.
=
tangency
To deduce the rectangular equation. Let OT = 2 a, and let P (x, y) be any point From the method of generation in this case of the curve. 1.
First Method.
ON = N T. MN = M ON M we have lST
From
.-.
the similar triangles
ONP,
,
NP ON M N ON NP = y, ON = M N = VON N T = ON = 2 a - x :
:
But
.
:
:
.
oj,
V (2 -),
;
.
Hence is
.y x :
y.
=
::
V(2 a
^
x)
...
x 2 a :
x.
(1)
the required equation.
The equation
Second Method. parabola if
=
8 ax
y
=
is
of the tangent line to the
Art. 65, (2)
+ .
SX
2a s
of a line passing through the origin line is this to pendicular
The equation
Combining these equations
so as to eliminate
s.
and per
we have
for the equation of the locus.
This curve was invented by Diocles, a Greek mathematician of the second century, B.C.,
and called by him the cissoid from
HIGHER PLANE CURVES. word meaning
a Greek
It
"ivy."
193
was employed by him
in
solving the celebrated problem of inserting two mean propor tionals between given extremes, of which the duplication of the cube
is
PO^) = 0) = OP = M K = OK - OM/. OK = 2 a sec and OM = 2 a cos r = 2 a (sec cos 0), r = 2 a tan 6 sin
From
the figure (OP,
we have But
(r,
also r
hence
;
or is
a particular case.
To deduce the polar equation.
2.
the polar equation of the curve.
SCHOL. Solving
y
(1)
with respect to
y,
we have
= x
2 a
An
inspection of this value shows (a) That the cissoid is symmetrical with respect to the X-axis.
That x = and x = 2 a are the equations of its limits. That x = 2 a is the equation of a rectilinear asymp
(b) (c)
tote (SS
).
151. To duplicate the cube, by the aid of the cissoid. Let OL, Fig. 61, be the edge of the cube which we wish to Construct the arc BO of the cissoid, CO = a duplicate.
CD = 2 CA = B draw BO LR intersecting BO in R.
being the radius of the base circle. Lay off 2 a and draw DT intersecting the cissoid in
and
at
Then
L
LR
erect the perpendicular is the edge of the required cube
;
;
for the equation
of the cissoid gives
hence
HB = 2 a
The
HB = y, OH =
2
ig PL _L
(since
- x).
CDT and HBT CD CT HB HT.
similar triangles :
::
:
give
x,
and
HT =
PLANE ANALYTIC GEOMETRY.
194 But
CD =
2
CT by .-.
construction
;
hence
HT in the value of HB above gives HB = 2QH3 hence HB = 2 OH HB triangles OHB and OLE are similar hence HB OH LE OL HB OH ::LK ::OL HB = 2 OH hence LB, = 2 OL whence the 2
This value of
2
8
3
.
;
The
;
::
:
3
:
8
8
.-.
But
8
3
:
3
3
,
8
;
struction.
152.
HB = 2 HT
HT = HB
THE WITCH.
FIG.
62.
con
HIGHER PLANE CURVES. The witch nate
the locus of a point
is
P
011
DP
195
the produced ordi-
of a circle, so that the produced ordinate the diameter of the circle as the ordinate
DM
OA
DA
outer segment It
may
to
is
to the
is
of the diameter. of a point P its foot
also be defined as the locus
linear sine
DP
DM of an angle
from
at a distance
011
D
the
equal
to twice the linear tangent of one-half the angle.
To deduce the rectangular equation.
1.
From
First Method.
DP =
y,
we have
x
(2
a
- x),
;
y:2a:: V(2 a
x)
x 2a :
x.
the required equation.
is
Second Method.
Let
But a (1
(1
+ cos
hence
or,
MCO =
;
then by definition
= 2 a \ //a ^ ~ cos ^ V a (1 + cos 0) - cos 0) = a - a cos = OC DC = OD == x, and = a + a cos = OC + DC = OD = 2 a - x 0) u = 2a == 2
?/
a
of generation,
:
.
DA = 2 a hence
mode
::
:
But
the
DP OA DM DA OA = 2 a, DM = VOD DA = Vx
a tan 2
,
;
squaring
y
2
= ~2 a
.
x
This curve was invented by Donna Maria Agnesi, an Italian mathematician of the eighteenth century. SCHOL. Solving (1) with respect to y, we have
y
Hence
(a)
=
=t 2 a
the witch
is
x J-~ V 2a
.
x
symmetrical with respect to the
X-axis. (b) (6-)
x x
= and x = 2 a are the equations of its limits. = 2 a is the equation of the rectilinear asymptote
SS
.
PLANE ANALYTIC GEOMETRY.
196
EQUATIONS OF THE FOURTH DEGREE. THE CONCHOID.
153.
The conchoid is the locus generated by the intersection of a circle with a secant line passing through its centre and a fixed point A as the centre of the circle moves along a fixed
OX. As the
line
intersection of the circle and secant will give two points P, P, one above and the other below the fixed line, it is evident that during the motion of the circle these points will generate a curve with two branches. is called the SUPERIOR BRANCH
MBM
;
FERIOR BRANCH.
(= OB)
is
called the
the DIRECTRIX
1.
;
The upper branch the lower, the IN
The radius of the moving circle O P MODULUS. The fixed line OX is called
the point A, the POLE.
To deduce the rectangular equation.
Let
P
(x, y),
the intersection of the circle
PP P
and the
HIGHER PLANE CURVES. secant
and
let
AO P, be any OA = a.
point of the curve.
The equation
of the circle
The equation
of the line
y
Making y
=
= sx
a
.
.
.
OP
Let
whose centre
AO P
197 b,
f
at
is
= OB = (x
,
0) is
is
(1)
we have
in (1),
s
00
for the distance
But
00 =
x
;
.
hence a
-
^
= b-
.
.
.
(2)
the equation of the circle. If we now combine (1) and (2) the s, resulting equation will express the relationship between the co-ordinates of the locus generated by the intersection of the loci they represent. Substituting the value of s drawn from in we have is
so as to eliminate
(1)
(2),
*V-(P-tf) is
We
might have deduced
triangles
Draw AT ATP and O SP
||
PS:SO e
+
2
2/)
.
.
.
(3)
the required equation.
simple way:
-
<>
this equation in the following very to OY. OX, and Since the
PT
to
are similar,
::PT:
TA
||
we have
;
-
Hence _
This curve was invented by Nicomedes, a Greek mathema who flourished in the second century of our era.
tician
was employed by him in solving the problems of the duplication of a cube and the trisection of an angle. It
PLANE ANALYTIC GEOMETRY.
198 2.
To deduce
From
the polar equation.
the figure
we have (AY being the
the pole)
(AP,
PAB)
But
AP =
hence
r
is
ACT
=
= a sec
and
A
0)
(>-,
OP
-t
initial line,
;
b
the polar equation of the curve. SCHOL. Solving (3) with respect to
x,
we have
An
inspection of this value shows is symmetrical with respect to the (a) That the conchoid
Y-axis. (b) (c)
That y That y
= b and y = = gives x =
b are the -J-
tote.
If
(d)
comes a (e)
a
then
cc
=
-t
equations of its limits. the X-axis is an asymp
_
V&
.
2
a
2/
;
i.e.,
the conchoid be
circle.
If b
(/)
0,
oo,
.
>
If b
a,
=
a,
the inferior branch has a loop as in the figure. coincide and the loop and the points
A
A
disappears. (g)
If b
<
a,
the inferior branch
A
is
similar in form to the
a) is isolated] branch, and the point (o, its co-ordinates though entirely separated from the curve, satisfy the equation.
superior
i.e.,
still
To trisect an angle by the aid of the conchoid. Let PCX be the angle which we wish to trisect. From C with any radius as CD describe the semi-circle DAH. From the point A draw AB J_ to CX and make OB = CD. With A as a pole and OB as a modulus construct a conchoid on CX as a directrix. Join H, the intersection of the inferior bnuich and the circle, with A and produce it to meet the 154.
directrix in
K
;
then
CKA =
PCX.
HIGHER PLANE CURVES.
199
3
FIG.
64.
H and C then from the nature HK = HC = OB. From the figure PCX == CAR + CKA CAK = CH A = 2 CKA but PCX = 2 CKA + CKA. hence Therefore CKA = 1 PCX. For join
;
of the conchoid
;
;
We
might have used the superior branch for the same pur
pose.
155.
THE LIMACON.
FIG.
65.
PLANE ANALYTIC GEOMETRY.
200
The Iima9on is the locus generated by the intersection of lines OP, CP which are so related that during their revo lution about the points and C the angle PCX is always two
equal to | POX. 1. To deduce the polar equation. Let be the pole, and OX the initial point of the curve, and let
(OP,
From
the triangle
POX)
:
r
:
a
::
Let
line.
P
be any
then
;
(r, 0).
r
OCP
sin
sin f
::
a sin
TT Hence
=
a
POC, we have
OP OC i.e.,
OC
<9
sin
:
sin 1
:
OPC
;
0.
*
:
t
From Trigonometry
= 3 sin \ r =
sin
hence
= is
4 sin 3 \ = (3 a (3 4 sin 2 ^ 0),
+ 2 cos 0)
a (1
.
.
4 sin 2 \
.
0)
sin^ 0;
(1)
the polar equation of the limayon. 2.
To deduce
From
the rectangular equation.
we have
Art. 35, r
=
#
+
cos
/
for the equations of transformation from polar to rectangular co-ordinates. Substituting these values in (1), we have
ax
Vx x
or
- ^) =
2
4-
?/
2
2
2
y
2
(*
+y
for the required equation.
SCHOL.
1.
From
the triangle
ODA, we have
OD = OA cos = From hence
2 a cos
OP = a + 2 cos 6 OP - OD = DP = a .
(1)
;
;
0.
a
)
...
(2)
HIGHER PLANE CURVES.
201
ODA
and the limayon i.e., the intercept between the circle of the secant through is constant and equal to the radius of the circle. SCHOL.
If 6
2.
If B If 6 If
156.
= 0, r = 3 a = OB. = 90, r = a = OM. = 180, r = - a = OC = 270, r = a = OM
THE LEMXISCATA.
The lemniscata
is
the locus generated by the intersection of
a tangent line to the equilateral hyperbola with a perpen dicular let fall on it from the origin as the point of tangency
moves around the curve.
FIG. 1.
66.
To deduce the rectangular equation.
Since
T
(a",
?/ ) is a
have, Art. 103, Cor
x
The equation
-
point of the equilateral hyperbola,
1,
y
-
==
a2
.
.
.
of the tangent line xx" yy"
=
(1)
TP
is,
a 2 ... (2)
Art. 112,
we
PLANE ANALYTIC GEOMETRY.
202
rr
Since the slope of this line
OP
pendicular
is
the equation of the per
,
is
,*.
0)
and solving for Treating (2) and (3) as simultaneous y",
we
x"
and
find .r
Substituting these values in -
-+ ff
we have
(1),
2
(x 2
or
(x
+
2
T/
2
)
=a
2
2
(x
-
2 .
7/ )
.
.
(4)
for the required equation.
This curve was invented by James Bernouilli. It is quadto the square constructed on the rable, its area being equal semi-transverse axis
OA.
To deduce the polar equation. We have Art. 34, (3), for the equations of transformation x = r cos 0, y = r sin 9.
2.
These values in 2
2
(4) give
+ sin =a
6 \r (cos r4 therefore r2
or is
=
=^
2
2
z
0)J 2 r cos 2
a 2 cos 2
(9
-
2
\r* (cos
sin
2 <9)};
0,
...
(5)
the required equation. a. 1 .-. r cos SCHOL. If 0, cos 20 values with .-. r has two equal 90 2 cos B cos If 45, <
<
opposite signs. If 45, cos 2
=
If 6
45
and
If $
If
= cos 90 =
(9
<135
.-.
=
0.
.-.
r
=
.-.
r
r
r is imaginary.
= 135, cos 2 = cos 270 = - 1 e = 180, cos 2 6 = cos 360 >
=i
=
=
=
6
.
0.
a.
An examination of these values of r shows that the curve formed by the asymptotes of the occupies the opposite angles hyperbola. The curve
is
to both axes. symmetrical with respect
HIGHER PLANE CURVES.
203
TRANSCENDENTAL EQUATIONS. 157.
THE CURVE OF
This curve takes
y
= sin
and may be defined of the
SINES.
name from
its
its
equation
x,
as a curve
whose ordinates are the sines
corresponding abscissas, the latter being considered as
rectified arcs of a circle.
FIG.
To construct the
6?.
Give values to x which differ from each other by 30, and find from a "TABLE OF NATURAL SIXES the values of the corresponding ordinates. Tabulating the result, we have, curve.
-
Value of x
30
=5=
Corresponding .52
6
60
90
=
^= =^=
1.04
1.56
o
120
= i5 = 2.08 6
Value of y
.50
.87
1.00
.87
PLANE ANALYTIC GEOMETRY. Value of x 150 180
210
Corresponding
Value of y
=~- = 2.60 6 = = 3.14 = 3.66 =
.50
TT
.50
6
240
=
~=
4.18
6
270
-
= 4.70
=
1.00
6
300
= i^ =
-.87
5.22
6
330 360
-
= 5.75 = = 2 = 6.28
.50
TT
a smooth curve points and tracing As x may have locus. the have required through them, we of the the and GO equation to yet satisfy any value from
Constructing
these
the curve itself extends infinitely in curve, it follows that abscissas. the both of positive and negative the direction
158.
THE CURVE OF TANGENTS.
FIG.
68.
HIGHER PLANE CURVES. This curve also takes
y
= tan
its
name from
its
205
equation.
x.
To construct the curve. Give x values differing from each other by 30 and find from a Table of Natural Tangents the corresponding values of
Value of x
y.
Tabulating,
Corresponding
we
have,
Value of y
"
30
=
=
-
.57
"
.52
6
60
=
=
1.04
=
1.56
1.73
"
6
90
=
oo
6
120
= 4-^ = 2.08
"
= 5 - = 2.60
"
1-73
6
150
.57
6
180
210
= = 8.14 = 3.66 =
"
TT
.57
"
6
240
= 4.18
==
1.73
"
6
270
= 4.70
=
oo
"
6
300
=
330
=
360
=
= 5.22 ^= 2
TT
=
"
-
-
5.75
6.28
1.73
.57
"
Constructing these points and tracing a smooth curve through them, we have the locus of the equation. This curve, together with that of the preceding article, belong to the class of Repeating Curves, so called because
they repeat themselves infinitely along the X-axis.
PLANE ANALYTIC GEOMETRY.
206 159.
THE CYCLOID.
This curve is the locus generated by a point on the circum ference of a circle as the circle rolls along a straight line. The line is called the BASE of the cycloid the
OM
;
point P, the GENERATING POINT the circle BPL, the GEN ERATING CIRCLE the line HB perpendicular to OM at its middle point, the Axis. The points and M are the VERTICES ;
;
,
of the cycloid. 1. To deduce the rectangular equation, the origin being taken at the left-hand vertex of tJie curve. Let P be any point on the curve, and the angle
which the
circle has rolled,
of the circle,
=2
ButOA = ayOB
CK =
a cos 6
;
=
AB
and
through Let LB, the diameter
AP
= CB
CK.
=a AB = PK = a sin AP = y, CB = a, 0,
0,
hence, substituting,
=a9 y = a x
0.
a.
OA = OB
Then
PCB
a sin
a cos
we have ...
(1)
HIGHER PLANE CURVES. between these equations, we have
Eliminating
=a
x
207
-
1
V2 ay
-
cos"
- ^2 ay -if
.
y .
.
=
2
1
a
-
vers"
(2)
for the required equation. SCHOL. An inspection of (2) shows values of y render x imaginary. (a) that negative
When
(b)
or 4
TT
=
y
a, or 6
TT
a,
or etc.
When
y
;
= 0;
1
vers"
=2
but a vers" 1
TT
a,
hence there are an infinite number
and M.
of points such as (c)
=a
x
0,
=2
x
a,
=a
1
vers"
=
2
=
a
TT
=
OB
but
;
3 TT a, or 5 ?r a, or 7 TT a, or etc. hence, there are a vers" 2 an infinite number of points such as H. 2 a are equations of the limits. and y (d) y v and 2 a there of y between the limits (e) For e very value are an infinite number of values for x. 2. To deduce the rectangular equation, the origin being at the highest point H, 1
;
=
=
We
have for the equations of transformation x TT a x OA OB - PK 2 a AP B / y
= =
These values in x y
But H,
=
,
(1)
= HK =
+ +
above give
= a (0 = a
a sin
TT)
,\
1
a cos
the angle through which the circle has rolled from TT
Hence
= = H
;
hence x = a 6
-\-
a sin
y
= a (COS ff -
x
=
a
1
vers"
,.-.
]
1) }
-+V
2 ay
2 ?/
.
.
.
(5)
The usually attributed to Galileo. With the exception of the conic sections no known curve The fol possesses so many useful and beautiful properties. invention of this curve
is
lowing are some of the more important
:
PLANE ANALYTIC GEOMETRY.
208 1.
2.
3.
=
HDB = a = 3 HDB = 3 OPHM = 4 HB = 8 a.
Area OPHDB O area Area of cycloid OHMO Perimeter
TT
2
.
TT
a2
.
If two bodies start from any two points of the curve curve being inverted and friction neglected), they will (the reach the lowest point at the same time. 4.
H
5.
point
A
H
body rolling
down
this curve will reach the lowest
in a shorter time than if
it
were to pursue any other
path whatever.
SPIRALS. 160. The SPIRAL is a transcendental curve generated by a point revolving about some fixed point, and receding from it in obedience to some fixed law.
The portion of the locus generated during one revolution of the point is called a SPIRE. The circle whose radius is equal to the radius-vector of the generating point at the end of the first revolution the MEASURING CIRCLE of the spiral. 161.
THE SPIRAL OF ARCHIMEDES.
FIG.
70.
is
called
HIGHER PLANE CURVES. This spiral
209
the locus generated by a point so moving its radius-vector to its vectorial angle is
is
that the ratio of
always constant.
From
the definition,
we have
r
=c r = c
hence is
.
.
.
(1)
the equation of the spiral.
To construct the
spiral.
Assuming values for and finding from ing value for r, we have Values of
45
(1)
the correspond
Values of r
Corresponding
=^
^c
4
4
90=^ 4 135
= ^L
180
=<*
270
^ =^
360
=2
225
2
_^ c 4
3*
=
-
..
4
T"
Constructing these points and tracing a smooth curve through them, we have a portion of the spiral. Since Since
number
= =
gives r
=
0, the spiral passes through the pole. gives r makes an infinite oo, the spiral of revolutions about the pole. (9
=
oo
Since 2 TT gives r the measuring circle.
=
=2
IT c,
OA (=
2
IT
c)
is
the radius of
PLANE ANALYTIC GEOMETRY.
210 162.
THE HYPERBOLIC
This curve
SPIRAL.
the locus generated by a point so moving that the product of its radius-vector and vectorial angle is always constant.
From
is
the definition r
=
we have c,
or for the equation of the spiral.
FIG.
71.
To
construct the spiral. Giving values to 0, finding the corresponding values of
we have Values of
Corresponding
Values of r oc
45
=Z
"
r,
HIGHER PLANE CURVES. Values of
135= 180
211
Values of r
Corresponding
7T
O
4
=
7T
-
"
7T
7T
225
5
=
TT
4 270
OTT
=-TT
~
"
4
315
GTT
=-TT
"
4
360
=
2
TTT "
TT
oo
Constructing the points we readily find the locus to be a curve such as we have represented in the figure. oo there is no point of the spiral Since 6 gives r
=
=
corresponding to a zero-vectorial angle. Since oo gives r 0, the spiral makes
=
=
an
of revolutions about the pole before reaching Since 6 2-n- gives
infinite
number
it.
=
c is
the circumference of the measuring circle. Let P be any point on the spiral then
SCHOL.
;
(OP,
POA) =
(r, 0).
and OP as a radius describe the arc PA. circular r 6, and from (1) c r Arc PA measure, By hence Arc PA c
With
as a centre
=
=
i.e.,
=
;
;
the arc of any circle between the initial line and the
spiral is equal to the circumference of the
measuring
circle.
PLANE ANALYTIC GEOMETRY.
212
THE PARABOLIC SPIRAL. This spiral is the locus generated by a point so moving that the ratio of the square of its radius-vector to its vectorial angle is always constant. From the definition we have 163.
r*
or,
=
c
.
.
.
(1)
for the equation of the spiral.
FIG.
To construct the Values of
45
=
72.
spiral.
Corresponding
Values of r
!T
4 90
=
2* 4
135
=
^ 4
180
=
7T
((
V2 C7T V3c
HIGHER PLANE CURVES. Values of
Corresponding
^
225
=
270
= ^T
3i5
= IJL
360
=2
213
Values of r
-JV5C
4 7C7T
oo
00
Constructing these points and tracing a smooth curve through them we have the required locus. Since gives r 0, the spiral passes through the pole. Since oo gives r oo, the spiral has an infinite num ber of spires.
= =
= =
THE LITUUS
or
This curve has for
its
164.
r2
=
TRUMPET. equation
c,
FIG.
73.
PLANE ANALYTIC GEOMETRY.
214 If
=
0,
=
r
;
6
if
The Logarithmic
165.
This spiral the ratio of vector
oo
is
=
is
its
r
oo,
an asymptote to
initial line as
=
0.
its infinite
This curve has the branch.
/SJjiraZ.
the locus generated by a point so moving that vectorial angle to the logarithm of its radius
Hence
equal to unity.
or passing to equivalent r
a*
.
numbers (a being the .
.
(1)
for the equation of the spiral.
To construct
the spiral. r
is
Let a
=
2,
then
= 29
the particular spiral
we wish
to construct.
base),
we have
HIGHER PLANE CURVES. Values of 6
215 Values of r
Corresponding
1
= 57.3 2 = 114. 6 3 = 171.9 4 = 229. 2 1
2 4
"
8
16 "
CO
<
- 1
= - 57.3
.5
-2=-114.6 - 3 = - 171.9 _ 4 = - 229. 2
.25
.125 .062
oo
A
smooth curve traced through these points
will
be the
required locus.
=
Since a, it
=
1 whatever be the assumed value of gives r follows that all logarithmic spirals must intersect the
initial line at a unit s
Since
number
=
distance from the pole.
=
oo the spiral makes an infinite gives r of revolutions without the circle whose radius 1.
GO
,
OA =
Since 6
r =
=
GO gives 0, the spiral of revolutions within the circle
number
makes an
OA
infinite
before reaching
its pole.
EXAMPLES. 1.
Discuss and construct the cubical parabola B
y
X = ~rp*
2.
What
the polar equation of the Iima9on, Fig, 65, the
is
pole being at
C
?
Ans.
r
= 2 a cos -
0.
3 3.
cata
Let is
OF
= OF = a Vf
Fig. 66.
Show
that the lemnis-
the locus generated by a point so moving that the
PLANE ANALYTIC GEOMETRY.
216
product of its distances from the two fixed points F, constant and
v
F
2
Discuss and construct the loci of the following equations 4. 5.
6.
7.
x
= tan
y
= cos
2/
a;
8.
is
12.
y.
a3
=x
axy.
=
x.
13.
jcl -j-
= sec*.
14.
4 + |r =
y?
a2
= sin y.
15.
?
=
= cot x.
16.
r
=
:
s
1.
l.
&3
sin 2
0.
<?/
sin 2
9.
10.
y
= cosec
y
= 3x ~ l
20.
a?
=a
n
sin 3 -
.
o
Q 2
2
?/
,
-4- a-?/
18.
r 2 sin 2
20
=
1.
o 2
=
1 rv
-|
19.
1.
r
= -1 1
~T~ Sill
C .
sin
Discuss and construct the locus of the equation
y
_
96 a 2 ;/2
+ 100 ciV
2
J- V(ic 21.
r
x3
no .
17.
ic.
Show
asymptotes Ex. 20.
that y of the
6 a) (a
^ x are locus
x^
=
+ 6 a)
or (x
8 a) (x
+ 8 a).
the equations of the rectilinear represented by the equation of
SOLID ANALYTIC GEOMETRY,
PART
II.
CHAPTER CO-ORDINATES.
I.
THE TRI-PLANAR SYSTEM..
determined when we know its distance and direction from three planes which intersect each other, these distances being measured on 166. The position of a point in space
lines it
is
is
Although drawn from the point parallel to the planes. immaterial in principle what angle these planes make
with each other, yet, in practice, considerations and simplicity have made it usual to take They are so taken in what follows. angles. Let XOZ, ZOY, YOX be the CO-ORDINATE Let OX, secting each other at right angles.
CO-ORDINATE AXES and 0, their
intersection,
of convenience
them
at right
PLANES inter OY, OZ be the the ORIGIN of
CO-ORDINATES. - XYZ. Let P be any point in the right triedral angle the we know lengths Then P is completely determined when
and directions of the three lines PA, PB, PC let fall from this point on the planes. As the planes form with each other eight right triedral which satisfy angles, there are evidently seven other points the condition of being at these distances from the co-ordi nate planes.
The ambiguity
is
217
avoided here (as in the case
SOLID ANALYTIC GEOMETRY.
218
of the point in a plane) by considering the directions in which these lines are measured.
Assuming distances
to the right of
YOZ as positive,
distances
to the left will be negative.
4-Z
A/
-Y
-X
4-X
Assuming distances
XOY
6o*;e
as positive, distances
will be negative.
in front
Assuming distances
o/XOZ
as positive, distances ^o
^Ae rear will be negative. r Calling x 7/ z (= BP, ,
AP, CP, respectively) the co-orc?ithe FIRST ANGLE, we have the follow
,
wa^es of the point P in in the ing for the co-ordinates of the corresponding points
other seven
:
SECOND ANGLE, above
XY
plane, to left
of XZ plane, (- x y z ) P 2 THIRD ANGLE, above XY plane, to r of XZ plane, (- x - y z ) P 8
YZ
plane, in front
.
,
,
.
,
,
left
YZ
plane, in rear
CO-ORDINATES.
219
XY plane, to right YZ P y (x below XY plane, to right YZ
plane, in
FOURTH ANGLE, above
XZ
rear of
f
plane,
front of
XZ
,
,
FIFTH ANGLE,
f
plane, (x
y
,
SIXTH ANGLE, below
,
XY
.
)
4
s
PS-
)
plane, to left
YZ
plane,
in
plane, in front
P6 ) y plane, ( SEVENTH ANGLE, below plane, to left YZ plane, in f z ) P7 x rear of XZ plane, ( y EIGHTH ANGLE, below plane, to right YZ plane, in rear
of
XZ
x
.
,
,
XY
.
}
,
of
XZ
-
f
plane, (x
,
XY - P
i/,
)
8.
EXAMPLES. 1.
In what angles are the following points
(1, 2,
-
3),
(-
-
1, 3,
2),
(-
1,
-
2,
-
:
4), (3,
-
2, 1).
2. State the exact position with reference to the co-ordi nate axes (or planes) of the following points :
(0, 0, 2),
- 1, - 1),
0), (0, (5, 1,
3.
(0),
2, 1, 2), (3, 1, 0), (3, (3,
(1, 1,
0,
-
1),
(1,
-
2,
-
3),
1, 2), (2, 0, 3),
(0,
0,
-
2),
((4,
1,
1,
2,
2),
1).
In which of the angles are the X-co-ordinates positive ? In which of the angles are the Y-co?
In which negative
ordinates positive ?
In which are the Z-co-ordinates negative?
167. Projections. The projection of a point on a plane is the foot of the perpendicular let fall from the point on the Thus A, B, and C, Fig. 75, are the projections of the plane. point P on the planes XZ, YZ, XY, respectively.
The
projection of a line of definite length on a plane is the on that plane. OC, Fig. 75, is the projection of OP on the plane.
line joining the projections of its extremities
Thus The projection
XY
of a line of definite length on another line is that portion of the second line included between the feet of the perpendiculars drawn from the extremities of the line of definite length to that line.
SOLID ANALYTIC GEOMETRY.
220
Thus OM, Fig. 75, is the projection of OP on the X-axis. The projections of points and lines as above de ;NOTE. Unless otherwise stated, all projections fined are orthogonal. will be so understood in what is to follow. 168.
To find the length of a
line joining two points in space.
P"
FIG.
76.
be the given points. P (a? tf, ) and and the be L (=P required length. Draw N Join OX. to NB to OY and CD NA to OZ to NO. P M draw and and C We observe from the figure that L is the hypothenuse of a are P M and right angled triangle whose sides
Let Let
V"
(z",
,
y" ,
*"}
P"C
P")
||
;
||
;
||
||
P"M.
Hence
= NO =
but
= ...
L
=
(P"C
CO- OR DINA TE S.
=
= 0,
COR. If x 0, if with the origin and ...
L
=
= 0,
s
+
2 V*"
221
then the point
+
2 y"
2 s"
...
P
coincides
(3)
expresses the distance of a point from the origin.
Given the length and the directional angles of a line joining any point with the origin to find the co-ordinates of the 169.
point.
The Directional angles of a line are the angles which makes with the co-ordinate axes. Let its
P
(x, y, z),
Fig. 75, be
distance from the origin.
the line
OP = L will be POX, POY, POZ = a, fa y,
point, then
any Let
respectively.
OM, ON, OR
Since
on X, Y,
(== x, y, z) are the projections of
Z, respectively,
x
y z
OP
we have
= L cos u = L cos = L cos y
}
I
ft
...
(1)
for the required co-ordinates.
COR.
Squaring and adding equations
+y + =L x* + y + * = L
x* but
hence
That
2
2
2
2
2
2
2
cos u -f cos is,
the
2
/? -f-
(1),
we have
2
2 -f- cos (cos Art. 168 (3)
cos 2 /
sum of the squares of
=
;
1
.
.
ft
-f cos
2
y)
;
.
.
(2)
the directional cosines of a
space line
is equal to unity. SCHOL. The directional angles of any line, as P Fig. 76, are the same as those which the line makes with three lines drawn through 1J/ to X, Y, Z. The on projections of P - x f if tf, z three such lines are Art. 168; hence P",
||
x"
,
x"
y" "
P"
-
x = L COS a - y = L cos z = L cos
ft
"
_^
"1
[
...
(3)
SOLID ANALYTIC GEOMETRY.
222
EXAMPLES. Required the length of the points 1.
(-
(1, 2, 3),
(3,
- 2,
following
-
(0, 3, 0), (3,
(0, 0, 0), (2, 0, 1).
Ans. 5.
0), (2, 3, 1).
(0, 4, 1),
V27.
(-
2,
-
1,
^ws. 6.
1, 0).
Ans. 7.
joining the
V14.
^tws. 3.
4.
2, 1, 1),
Ans. 2.
lines
:
(1,
-
-/5.
-2). V38.
2, 3), (3, 4, 6).
Ans.
5.
Find the distance of the point
(2, 4,
3)
7.
from the origin
;
also the directional cosines of the line. 8.
A
line makes equal angles with the co-ordinate axes. are its directional cosines ?
What 9.
Two
What
is
of the directional cosines of a line are
If (x
10.
Vf
and
the value of the other ? f ,
y
,
z
)
and show that (a;",
y",
z")
are the co-ordinates of the
extremities of a line
x
+
x"
y
+ 2
y"
+
z "z"\ z"
2-J
are the co-ordinates of its middle point.
THE POLAR SYSTEM. 170. The position of a space point is completely determined when we know its distance and direction from some fixed point.
For a complete expression of the direction of the point it is The angles necessary that two angles should be given. usually taken are
The angle which the line joining the point and the makes with a plane passing through the fixed the line join and 2d, The angle which the projection of point line in the fixed a with makes that on the plane points ing 1st,
fixed point ;
plane.
CO-ORDINA TES.
FIG.
223
77.
Let be the fixed point and P the point whose position we wish to determine. Join O and P, and let XOY be any plane passing through 0. Let OX be a given line of the plane XOY. Draw PB 1 to XOY and pass the plane PBO through PB and OP. The intersection OB of this plane with XOY will be the projection of
BOX
and the distance
(go)
OP on XOY. The angles POB (0), OP (r), when given completely de
termine the position of P. For the angle cp determines the determines the line OP in that plane, plane POB, the angle and the distance r determines the point P on that line. This method of locating a point is called the POLAR SYS TEM. The angles and qo are called VECTORIAL and
ANGLES,
the distance r
The point terms of
called the
RADIUS VECTOR
when written (r, 9, POLAR CO-ORDINATES.
P,
its
is
<p),
is
of the point.
said to be expressed in
It is evident <P,
and
may
all
to 360 to and by giving all values from values from to oo to r that every point in space
be located.
171. Given the polar co-ordinates of a point to find its rec tangular co-ordinates. Draw OY 1 to OX and in the plane BOX draw OZ 1 to ;
SOLID ANALYTIC GEOMETRY.
224
OY
and OX, and
Draw
BA
||
to
(OA, AB, BP)
From
=
From
OB =
From
(x,
BOP, we have sin
6.
ABO, we have
= OB
r cos
cos
x
.:
cp.
= r cos
the same triangle
y y x
cos
cp.
we have
= OB sin = r cos 6 sin qo,
= r cos 6 cos = r cos 9 sin y z = r sin
Henee
co-ordinate axes.
then, Fig. 77, y, z) are the rectangular co-ordinates of P.
=r
the triangle
x
But
;
the triangle z
OX, OY, OZ be the
let
OY
(p.
cp
cp
1
V
...
(1)
express the required relationship. Con. If P (x, y, ) be the co-ordinates of any point on a locus whose rectangular equation is given then equations (1) are evidently the equations of transformation from a rectangu lar system to a polar system, the pole being coincident with the origin.
y,
Finding the values of we have
r,
6
and
cp
from
(1) in
terms of x and
r= Vx + if + = tan+ = tan2
1
(9
1
for the equations of transformation from a polar system rectangular system, the origin and 2)ole being coincident.
to
a
225
CO-ORDINA TES. EXAMPLES. Find the polar co-ordinates of the following points 1.
2.
3.
(2,1,1).
(V3,
1,
(10,2,8).
4.
2 V3).
(3,
-
1, 4).
Find the rectangular co-ordinates of the following 5.
(5,
:
:
7.
30, 60).
(6,=,
6
Find the polar equations of the following and origin being coincident
surfaces, the pole
:
9.
10.
+ ^2 + = a z + sx + ty - c = 0.
X2
Ans. r
z
,2
m
sin
+ 5 cos
Find the directional cosines of the ing pairs of points 11.
(1, 2,
12.
(4,
-
-
(-
1, 3, 2).
and 15. If (V, /, ) the point space points, show that (z",
/
\
mx"
g>
cos
-f
sin
a.
qo
lines joining the follow
:
1), (3, 2, 1).
1, 2),
cos
=
+ nx
r
divides the line joining n. each other the ratio :
14.
(0, 2, 0), (3, 0, 1).
1,
5), (4, 5, 6).
be the co-ordinates of two
")
+ ny m+w
them
-
(2,
my"
7wT+ %
m
y",
-
13.
into
+ nz m+n
f
mz"
two parts which bear to
SOLID ANALYTIC GEOMETRY.
226
CHAPTER
II.
THE PLANE. To deduce the equation of the plane. Let us assume as the basis of the operation the following 172.
property If on a perpendicular to a plane two points equidistant from the plane be taken, then every point in the plane is equidistant :
these two points, equally distant.
from
and any point not
in the plane
is
un
FIG.
Let it
ABC
in E.
Draw
be any plane.
Produce
OR
until
the plane is equally distant be any point of the plane let ;
OR 1
to
ABC, and meeting
RR = OR = p. and R from
Every point Let
.
ON, MN,
MR
,
P
in
(x, y,*,)
the co-ordinates
THE PLANE. of
R =
d,
e,
From
Then from
/, respectively.
PR
==
- xY + (e-
(d
the same article, equation
OF = 2
xs
227
(3),
+f+*
we have
Art. 168, (2),
yY + (/- zY we have
2 5
hence, by the assumed property, (d
- xY + (e-
yY
+ (/- *) = x + y + * 2
2
2
2 -
Simplifying this expression, we have 6 + f + ey +fz = d*+ *
dx
^
z
-
/1X (1)
...
for the required equation.
To find the equation of a plane in terms of
173.
pendicular
to it
from
the origin
and
the
per
the directional cosines of
the perpendicular.
be the directional angles of the perpendicu Since ON, MN, lar OR (=2p), Fig. 78. (= d, e, f) the projections of OR on the co-ordinate axes, we have (Art.
Let
/8,
,
and
7
169, (1)
=
MR
7
)
= 2p cos e = 2^ cos
d
f
1
(3
Substituting these values in cos2 ft H~ cos2 Y that cos 2
+
cc
cos
+
?/
cos
for the required equation. EQUATION of the plane.
Since
2
d
(1), 1?
(1)
Art. 172,
+ z cos 7 = p
^8
e
and remembering
we h ave
Equation
OR = 2p = V^ +
cos
...
[
cos 7
2p
2
+/
(2) is
.
.
.
called the
(2)
NORMAL
2 ,
equations (1) give e
ANALYTIC GEOMETRY.
.SOLID
228
Substituting these values in
we have
(2),
d
e ~
~
y
for the equation of the plane expressed in terms of the co-or dinates of a point on the perpendicular to it from the origin
and the perpendicular. COR.
1.
If
p
=
we have
in (2),
+ y cos + z cos y =
x cos
.
ft
.
.
(4)
for the equation of a plane through the origin.
COR.
2.
= 90,
If
y cos is
= 0,
cos
(3
+ z cos 7
hence
=p
.
(5)
.
.
the equation of a plane 1 to the YZ-plane. 90, we obtain similarly ft
If
=
x cos
+ z cos 7 = p
for the equation of a plane If Y 90, then
.
.
.
(6)
to the XZ-plane.
J_
=
x cos is
+
the equation of a plane
COR.
3.
= 90
If
cos
?/
and
1L .--cos
=p
/?
.
.
.
(7)
1
to the XY-plane.
(3
= 90,
...
then
(8)
7
is
1
the equation of a plane
to
YZ
and XZ, and hence
XY. Similarly,
we
find
y
= --* cos
(10)
COS for the equations of planes
to ||
XZ
and
YZ
respectively.
||
to
THE PLANE. COR.
If
4.
=
p
y
and
in (8), (9),
=
...
I
229
(10),
then
(11)
J
are the equations of
XY, XZ, and YZ,
respectively.
To find the equation of a plane in terms of
174.
in
its
tercepts.
OB =
OA
Let, Fig. 78, =a, perpendicular to the plane
=
OC
b,
OE (=p)
Since
c.
ABC, we have from
the right
is
tri
OEA, OEB, and OEC
angles
=
COS
...
0080=| COS 7
(a)
f
= J
Substituting these reducing,
we have * a
values
+ -|b + *c = 1
normal
the
in
.
.
.
and
(1)
for the required equation. Equation (1) METRICAL EQUATION of the plane.
175.
equation
is
called the
Every equation of the first degree between
SYM
three vari
ables represents a plane.
The most general equation of three variables is of the form
the
-f
By
-f
C*
=D
.
.
.
(1)
Dividing both members of this equation by we have
VA + 2
degree between
ABC Ax
first
B2
+C
-2 2
A2
x
VA ~+ B + ni C ^/A22
~"~
+B +C 2
i
T^22
i
2
* y*
i
VA
2
-|-
B2
+C
VA + B + C 2
2
2
2 ,
SOLID ANALYTIC GEOMETRY.
230
of Art. 173, we see that the co the directional cosines of some are efficients of the variables co-ordinates of one of its the of terms in line
Comparing
(2)
with
(3)
expressed
and that the second member measures the distance of the a plane from the origin hence (2) and therefore (1) is
points,
;
equation of a plane.
To find the equations of the traces and the values of the intercepts of a plane given by its equation. 176.
7
FIG.
ABC
Let
be the plane and
Ax
79.
let its
+ By + C2 = D.
To find the equations of the
1.
equation be
traces
AB, BC, AC.
the given plane with traces are the intersections of their equations, we the co-ordinate planes hence, combining
The
;
have
Ax
+ By + Cz = D | z
Ax
Ax
=
/<
Ax
+ B^ =
D<
Trace
011
XY (AB)
.
.
.(1)
)
+ By + C = D | Cz
=D
")
.
Ax + Cs = D I
. j>
r ,
Q,,
__
_
j)^
Trace on
X Z (AC)
Trace on YZ (BC)
.
.
,
(2)
... (3)
THE PLANE.
231
intercepts OA, OB, OC. The points A, B, C are the intersections of the given plane with the co-ordinate planes taken in pairs ; hence, combining
To find the
2.
we have
their equations,
A*
+ By + C* = D (4) -nL
y-Q
-D .
._ 00 COR.
If
the
= cept = OB
oo
.
.
.
.
.
(5)
(6)
plane is perpendicular to XZ its Y-inter0. Making B hence, equation (5), B
Ace
=
=
;
in the general equation,
we have
+O=D
.
.
.
(7)
a perpendic (7) and (2) are the same equations; hence, ular plane and its trace on the plane to ivhich it is perpendic ular have the same equation.
But
=
cos 7 177. If x cos a. -j- y cos ft -|p be the normal equa a plane, then x cos u -\- y cos (3 -j- z cos 7 p -^ d is the
=
tion of
equation of a parallel plane at the distance d from it. For the directional cosines of the perpendiculars are the
same hence, the perpendiculars are coincident hence, the The distance of the planes apart is equal planes are parallel. to the difference of the perpendiculars drawn to them from the origin but this difference is p -j- d p i.e., J- d. Hence, ;
;
;
;
the proposition. COR. If (V, ?/,
from the origin is p X COS _J_ d
=
)
be a point in the plane whose distance d then
-]-
f
-|-
;
if COS
(3
+
COS 7
p
.
.
.
(1)
SOLID ANALYTIC GEOMETRY.
232 is its
distance from the parallel plane whose distance from From equations (a), Art. 174, we have is p.
the origin
=-
cos
a hence
+ cos
cos 2
These values in
(1)
,
cos
+ cos
2
=
j8
cos 7
,
=
P;
6
=
2
/
c
^& + % v
-~ -
+
C
=
1-
2
give
for the expression of the distance of a point which is given in its symmetrical form.
from a plane
Let the student show that the expression for d becomes
_A^Vy
d
*
+ Cz:_-^ +B +C
m
(3)
2
2
of the plane is given in its general form. the significance of the double sign in (1), (2), and
when the equation
What
is
(3)?
178.
the equation of a plane which passes through
To find
three given points.
Let (V, y
,
z
),
(x",
y" ,
*"),
we seek
Since the equation
(*"
is
,
y",
which A, B,
C,
D
")
^ the given points.
that of a plane,
Ax + By + Cz = D
in
*
f
.
it
must be
(1)
are to be determined by the conditions
imposed. the co Since the plane is to contain the three given points, its equation hence, ordinates of each of these must satisfy the following equations of condition ;.
:
Ax Ax"
Ax
+ B// + C.~ = D =D + B/ + = D. C + By + C,~"
"
"
"
These three equations contain tlie/ow unknown quantities from the equations the values of A, A, B, C, D. If we find
THE PLANE. B,
C
D
in terms of
233
and the known quantities, and substitute
term of the resulting equation will (1), each Let as a factor.
these values in
D A=
contain
AT>,
B
Substituting in
= B D,
(1),
C
= CD be
the values found.
we have
A Daj + B D?/ + C Vz = D. A a + B y + C s = 1 ...
...
is
(2)
the required equation.
179.
The preceding discussion has
every equation of
the
first
elicited the
fact that
degree between three variables It remains to be shown that every
represents a plane surface. equation between three variables represents a surface of some kind.
Let
z
=/(,
y)
C
1)
Since be any equation between the three variables (x, y, z). x and y are independent, we may give them an infinite number For every pair of values thus assumed there is a of values. These values in (1) give the corre point on the XY plane. sponding value or values of z, which, laid off on the perpen dicular erected at the point in the XY plane, will locate one
But the number or more points on the locus of the equation. of values of z for any assumed pair of values of x and y are necessarily finite, Avhile the number of pairs of values which be given x and y are infinite a surface of some kind.
may
;
hence
(1)
must represent
If
-/(*,
z
=
q>
y)
(x, y)
.
.
.
(2)
j
be the equations of two surfaces, then they will represent their For these equa line of intersection if taken simultaneously. tions can only be satisfied at the
same time by the co-ordinates
of points common to both. Hence, in general, two equations between three variables determine the position of a line in space.
SOLID ANALYTIC GEOMETRY.
234
z
If
=f(x,y)
= =
<P
V
(^ y) (x, y}
be the equations of three surfaces, then they will represent simul their point or points of intersection when considered as Hence, in general, three equations variables determine the positions of space points.
taneous.
letiveen
three
EXAMPLES. Find
4.
11.
the traces
and intercepts
of the following planes:
2
The
from directional cosines of a perpendicular let fall
of the required the equation the origin on a plane are | o o o |; == 4. plane, the length of the perpendicular ,
\
,
An,
5+^ + f-l.
the plane whose intercepts are Required the equations of as follows
:
12.
1,2,3.
13.
2,
-
14-
1,
15.
3.
i,, -
1,
-
-2. "
,
4.
of the equation of the plane, the equations ? 4 z x 4 and 3 y -jwhose traces are x 4. z 3 ?/ a 16.
What
is
=
=
+ =
THE PLANE. The
17.
co-ordinates of the projection of a point in the 2ov. the plane are (2, 1) required 3y-}-2z
=
plane x
235
XY XY plane.
;
the distance of the point from the
Ans.
f.
Write the equations of the planes which contain the follow ing points
:
(-
-
18.
(1, 2, 3),
19.
(4, 1, 0), (2, 0, 0), (0, 1, 2).
20.
(0, 2, 0), (3, 2, 1),
21.
(2, 2, 2), (3, 3, 3),
1, 2,
1), (3, 2, 0).
((-
1, 0, 2).
!,-!,-
1).
Find the point of intersection of the planes 23. 2 x - y + z = 10. 22. a- + ^ _ z = 4. x + ?/-2 = 3. 2^-33 + ?/ = 10. 2 x - 4 y + 5 = 6. x -f T/ - 2 = 2. 24.
Find the distance of the point
(2, 1, 3),
from each of the
planes 25. .x cos
26.
x
27.
+ y cos
60
=
+ x
+ ^2 + 3 z = 4.
z
+z
60
3y
= 9.
cos 45
8.
--- + - =
325
28.
1.
Find the equation of the plane which contains the and is parallel to the plane x 2 ij -\- z 6. Reduce the following equations to their normal and sym metrical forms 29.
point
=
(3, 2, 2)
:
30.
2x
3y
+z= 2
qo
o<6.
4
31.
.
x
1 ,
~\-
y
-
#
=
cc
+2y
z
= ~.
o.
33. If s, s represent the sides of the triangle formed by the traces of a plane, and a, b. c represent the intercepts, 2 2 2 s 2 (a 2 -f b 2 s c ). show that s 2 s"
,
+
+
"
=
+
236
SOLID ANALYTIC GEOMETRY.
CHAPTEE
III.
THE STRAIGHT 180.
To deduce
LINE.
the equations of the straight line.
straight line in space is determined when two planes which intersect in that line are given. (See Art. 179.) The equations of any two planes, therefore, may be considered as
The
when taken simultaneously. Of the of pairs of planes which intersect in and de termine a space line, two of its projecting planes that is, two planes which pass through the line and are perpendicular representing a space line infinite
number
two of the co-ordinate planes give the simplest equations. For this reason two of these planes are usually selected.
to
THE STRAIGHT LINE. Let then
237
PBM
its
be the plane which projects a space line on XZ, equation will be of the form
x in which s = tan
= sz
a ( see Art. 176, Cor.) and a OA.
-j-
ZBP
=
M
Let P B be the plane which projects the line on YZ, then its equation will be
in
=
y=t*+b,
=
which t tan ZB P and b OA. But the two planes determine the line x
= y
;
hence
= sz + a
are the required equations. COR. 1. If a and b
x
.
=
= sz = tz
0,
then
)
j
which pass through the 0, we have
are the equations of a line and t COR. 2. If s
=
=
for the equation of a line
origin.
to the Z-axis. Since equations (1) express the relation existing between the co-ordinates of every point on the space line, if we eliminate Z from these equations we obtain the immediate
COR.
||
3.
relation existing between x and y for points of the line. But this relation is evidently the same for all points in the pro and therefore for jecting plane of the line which is 1 to
XY
trace on
its
on
XY
;
XY.
But the trace
hence, eliminating, tx
is
the projection of the line
we have
= bs
at (4) for the equation of the projection of the line
sy
.
.
on XY. have found, Art. 169, Schol., for the length of a joining two points the expression
181. line
.
We
T.-^-y _ COS a
y"
-j/
COS
(3
^
*"-:. COS 7
SOLID ANALYTIC GEOMETRY.
238
L and
Eliminating ordinates of
letting
cos
cos for the
182.
z"
x",
y",
(=
x, y, z)
any point on the line, we have x _ y x y _ z
To find where a
x-jx
cos 7
(3
SYMMETRICAL EQUATION
be the co
of a straight line.
line given by the equations
of
its
projections pierces the co-ordinate planes.
Let 1.
X
== __
f
? I be the equations of the fz-\-0 ) y To find where the line pierces the XY-plane.
The equation
of the
line.
"f
XY-plane
is
Since the point of intersection is common to both the line its co-ordinates must satisfy their equations.
and the plane,
Hence
= sz y =
x
t-Z
-j-
a
-\-
b
are simultaneous equations.
So treating them we find
(a, b, 0)
to be the required point. To find where the line pierces the XZ-plane. 2.
The equation
of the
XZ-plane
is
y-o. Combining
this with the equations of the line,
for the required point. To find where the line pierces the XZ-plane. 3.
x y
= sz = tz
-{-
-f-
hence <),
s is
the required point.
b y are
simultaneous
;
we have
THE STRAIGHT LINE. To find the equations of a
183.
pont. Let
r
(x
j
y
,
z
f
239 a given
line passing through
be the given point.
)
Since the line
is
straight its equations are
a)
= te + b
y in
ft
>
which the constants are unknown. Since
to pass through the point (# satisfied for the co-ordinates of
f
it is
must be
the equations of condition
,
y ,z
) its
equations
this point
;
hence
:
= sz + a = tz + b \ f
x
,n\
)
1
y
f
As the
three conditions imposed by these four equations in cannot, general, be fulfilled by a straight line, we must eliminate one of them. Subtracting the first equation in
group (2) from the first in group (1) and the second in group (2) from the second in group (1), we have x = s (z z - y = (Z - Z
x
)
t
y
,,
| }
)
for the general equations of a straight line passing through
a
point.
To find the equations of a
184.
line
passing through two
given points.
Let
f
(x
,
As the
y
,
z
fl
),
(x
,
y",
z")
be the given points.
line is straight its equations are
... in
which the constants are
As
it is
to be determined.
to pass through (x
x
f
= sz = tz
f
r
y
(i)
,
y
,
z
),
we must have
SOLID ANALYTIC GEOMETRY.
240
As
it is
to pass x"
y"
through
= =
sz"
tz"
(#",
+ +b
a
y",
z"),
I
we must have
also
(3)
}
As these six equations impose four conditions on the line, we must eliminate two of them. The conditions of the the line to pass through the hence we must eliminate the other two. Elimiting a and b from groups (1) and (2), by subtraction, we hav6 z ) x x = s (z proposition, however, require
two points
;
x
r
y
-y =
t
Now, eliminating a and X -X =
(z
b
-*
Eliminating
s
-y"
and
=
s (Z t
(z
and
(2)
_ -
4)
(3),
we have
-
z")
z")
between
t
(
from
"
,j
|
) j
t j
and
(4)
(5),
we have
,
for the required equations.
EXAMPLES. 1.
Given the
line
the projection on
x ^
= z 1 =4^ _ 3 *?
-4-
) r
required the equation of
XY. Ans.
2.
How
2x
y
=
5.
are the following lines situated with reference to
the axes ?
x=2\ y=0\
?/
= 0)
a;
=
)
Find the co-ordinates of the points in which the following lines pierce the co-ordinate planes
:
x=-z-l)
,
2x
=
THE STRAIGHT LINE. 6.
Given
(2, 1,
2), (3, 0, 2)
241
required
;
The length of the line joining the points. The equation of the line. The points in which the line pierces the
(a) (7>)
(c)
co-ordinate
planes.
Find the equations of the points
7. (2, 1, 3), (3, 8. (
-
1,
- 1).
9.
- 1, 2, 3), ( - 1,0, 2).
11.
of 45
which pass through the
lines
:
The
(2,
-
-
10. (1,
1,
1, 0), (3, 0, 0).
-
(-
2),
projections of a line on XZ and YZ make angles respectively with the Z-axis, and the line in (1, 2, 3)
;
required the equations of ~ ^^ w. -/
*v
t*/ /y>
y 12.
The
2, 4, 3) 13.
(-
15.
= -4- - V 3 + V3
vertices of a triangle are (2, 1, 3), required the equations of its sides.
Is the point (2,
1,
- 2) ?
3)
Assume two
(3,
0,
1),
on the line which passes through
Write the equations of a
NOTE.
them
is
;
1, 3, 2), (3, 2,
14.
- 3).
and 30
space contains the point the line.
(
-2,
1,
line
which
points in the plane
the plane
lies in
;
the line joining
will be a line of the plane.
Find the equation of a line through y + z = 4.
(1,
2,
2)
which
parallel to the plane x 16.
Find the point
pierces the plane 3 x 17.
-\-
in
2 y
which the z
= 4.
line
^ y
+2^ = 3 z + 2 = 0)
)
Required the equation of the plane which contains the
SOLID ANALYTIC GEOMETRY.
242
Find the point of intersection of the planes z = 4, x x + 3y y + z = 2, 2 x
18.
+ y = 3.
Find the equations of the projecting planes of the =4 a._2v
19.
line
+
20.
planes cross ? z 2y 2, x
Which angles do the following x -y. + z = 4:,2x +y -3z =
185.
To find the
equations.
Let
x
- =
-
1.
intersection of two lines given by their
= sz + a )
y-. + ft|
,
Mld
x
=
+
s z
y-^ +
a \ f
ft ; intersection is com of the Since be the given equations. point mon to both lines, its co-ordinates must satisfy their equations. Hence these equations are simultaneous. But we observe that there are four equations and only three unknown quantities the lines hence, in order that these equations may consist (and a certain relationship must exist between the con ;
intersect),
which enter into them. To find this relationship, we eliminate x between the first and third, y between the second and fourth, and z between the two equations which result. stants
We
thus obtain (s
-
f
s ) (b
-
&
)
-
-
(t
(a
-a = )
for the required equation of condition that the two lines shall If this condition is satisfied for any pair of as intersect. lines the lines will intersect, and we obtain the co-ordinates of this point by treating any three of the four which represent them as simultaneous. So treating
sumed
equations the first, second, and third
we
obtain a
a
s
s
/
,
;
i
^>
a s
for the co-ordinates of the required point. were prepared to expect NOTE.
We
a s
that our analysis
some conditional equation, for in assuming the it would be an exceptional case equations of two space lines
would lead
to
THE STRAIGHT LINE. if
we
243
assumed them that the
so
lines which they represent Lines may cross each other under any angle in space without intersecting. In a plane, however, all lines except parallel lines must intersect. Hence, no conditional
intersected.
equation arose in their discussion. 186.
To find the angle between
tiuo
lines,
equations, in terms of functions of the angles make with the axes.
x
Let
= sz
-f-
given by their
which the
lines
a
be the equations of the two lines. The angle under which two space lines cross each other is measured by the angle formed by two lines drawn through some point parallel to their directions.
FIG. .
Let
OB
and
OC
will be their equations.
angle sought.
.
be two lines drawn through the origin
parallel to the given lines.
Let
81.
cp
Then
The angle between these
(= BOG)
be this angle.
lines
is
the
SOLID ANALYTIC GEOMETRY.
244
which the line BO makes Let a y represent the angles the angle which and with X, Y, Z, respectively P (x y z } Take axes. same the with makes point CO any on CO and join them by on OB and any point ,
,
;
0",
",
y"
f
,
,
P"
Let
OF - I/,
From
OP"==
P
the triangle
But Art.
+ TL
IfO j
P P
OP".
I
T
ff>
168, equation (3)
and
(x"
2
2
2
,
//2
x"*+
=
cos
=L
=
L"
Substituting in cos
<f
(1),
(*"
- *r
we have ~
-
<p
Ov
T~^T"
169, (1)
x x"
,
y"*
Substituting these values in
But Art.
/-i\
(2)
=x +y +* = +^ - xj + (if - yj + L"
=
2
-Lr
2
2
L.
2L07-
L2
L
=
P"
we have
OP",
^
COS
and
L",
z")
y",
(x",
a right line forming the triangle
= cos
cos cos
cos
/?
,
=L = "
L"
(2),
cos
y"
+ cos
/3
cos
/3"
+ cos y
for the required relation. 90 COR. If v
= + cos
r
"
cos
cos
187.
To ymd ^e
^3
cos y cos
L"
y"
)3",
we have
"
cos
=L =
?/
,
",
cos
)8"
+ cos
awr/Ze ^//-ic/i
^<?o
y
cos
= y"
cos
.
.
.
y"
.
.
.
(3)
(4)
each space Zmes maA;e with
the projections other in terms of functions of the angles which axes. the lines make with the co-ordinate
of
be, as
of the lines in the preceding article, the equations
THE STRAIGHT LINE. drawn through the origin
P
(x, y
,
z
),
= sz =
x
y
*
I/ 2
and, Art. 168,
Eliminating,
we
=
parallel to the given lines.
a point on the
is
Fig. 81,
first line,
2 x"
+y +z\ 2
find
*
2
and since have
^
2
vi
r/ P"
(x
,
x"
y"
is
y",
")
=s = M, =x +
L"
+s + 2
1
_~
2
vi
+ *+T
a
a point on the second line,
;
we
z"
2
and, Art. 168,
Since
we have
,
"
vi + s +
245
1
2
"
y"
-f-
^ //2
.
Hence,
Vl +
s*
+
But, Art. 169,
L/
Substituting these values reducing,
we have _
Vl +
2 .9
for the required expression.
in
equation
Vl + s (3),
/2
+
Art. 186,
t
12
and
SOLID ANALYTIC GEOMETRY.
246 COB.
1.
If
<f>
= 0,
are parallel
the lines
and equation
(1)
becomes i
=
T= =4J^===.
Clearing of fractions and squaring,
we have
and Performing the operations indicated, transposing have we lecting,
of these quantities cannot be hence is equal to zero
But the sum of the squares
each separately equal to zero unless
j
=
<
s?
t
=
f
t,
col
st
;
=s
t
.
.
.
(2)
The first lines. are the conditions for parallelism of space in lines two space are two of these conditions show that if are their projections on the co-ordinate planes parallel, then conse mere a is s third condition t) parallel
also.
The
(sf
and quence of the other two,
may
=
be omitted in stating the
conditions for parallelism. to each 90, the lines are perpendicular COB. 2. If 9
=
other,
and equation
(1)
becomes 1
hence is
1
+ ss + =
+ ** +
it
(
3 >
the condition for perpendicularity in space. with any one ot Since the angle which a line makes
188. of the angle which the the co-ordinate axes is the complement to which that axis is line makes with the co-ordinate plane the complements of p, y if we let ft y be perpendicular ,
,
,
respectively,
we have
for the sines of the angles
co-ordinate planes.
which a space
line
makes with the
THE STRAIGHT LINE.
247
TRANSFORMATION OF CO-ORDINATES. 189. To find the equations of transformation from one system of co-ordinates to a parallel system, the origin being changed.
FIG.
82.
Let X, Y, Z be the old axes and X Y Z the new. Let P be any point on the locus CM. Draw PB, ,
OL
||
to
OZ and
meeting
XOY
in B,
,
A K, B and L. Draw BE Y draw LX to BE
and produce it to A BE will be to and LE to OX. Then (OA, AB, BP) ;
||
||
;
=
||
(
x
,
y,
z)
are the
old co-ordinates of the point P.
(O
A A B B P) = ,
,
(a;
,
y
,
z
)
are the
new
co-ordinates of
the point P.
(OX, XL, LO ) new origin From the figure .
=
(a, b, c)
are the old co-ordinates of the
SOLID ANALYTIC GEOMETRY.
248
OA = ON + O A AB ,
hence
x
=
a
-f-
x
==
NL + A B BP = LO + B P ,
,
y
=b
-\- ?/,
z
=c+z
;
are the required equations.
To find the equations of transformation from a rec tangular system in space to an oblique system, the origin being 190.
the same.
FIG.
Let OX, OY,
OZ
83.
be the old axes, and
OX OY OZ ,
,
the
new.
Let u
OZ
Let
OZ
f
be the angles which
OX
makes with OX, OY,
u",
fi",
f be
the angles which
OY
makes with OX, OY,
respectively.
Let
OZ
ft,
,
respectively.
"
,
$
",
f"
be the angles which
OZ makes
with OX, OY,
respectively.
Let
P
be any point on the locus
CM.
Draw PB and PB
THE STRAIGHT LINE. ||
249
OZ and OZ respectively, and let B and B be the points which these lines pierce the planes XOY and X OY
to
in
,
.
A
Draw B
OY
to
||
=
(OA, AB, BP) point P.
(OA
,
then
;
A B B P) =
f
(x
,
are the old co-ordinates of the
y, z)
(x,
!
y
,
)
,
new
are the
co-ordinates of the
point P.
From
B
P,
on the X-axis
A
and
the perpendiculars PA,
let fall
then from the
;
figure,
BT>,
AL
we have
OA = OL + LD + DA and DA are the projections of OA A B and PB respectively on the X-axis, and each, therefore, is equal to the line whose projection it is into the cosine of the angle which that line makes with the X-axis. (See Art. 169 (1) .)
LD
But OL,
,
,
,
OL = OA =x i.e., OL
cos
.-.
cos
LD = A B cos DA = B P cos LD = y cos DA = z cos ,
"
,
",
hence, substituting, we have r X x COS -f- y COS
Similarly
= = x y z = x
cos
"
</
,
p+y
n"
-f z COS a
p + -f if cos f + z
cos y
;
cos
cos f
cos
"
p"
1
...
I
(1)
Y"
Of the nine angles involved in these equations, six only are independent, for since the old axes are rectangular, we must have (See Art. 169, equation cos 2
cos 2
"
Con.
1.
we must have cos
cos
cos
cos
"
cos
"
"
2
I
(
2)
the
new axes
j
to be rectangular also
in addition to equation (2) the following condi See Art. 186, Cor. :
-f cos /? cos cos cos
"
y"
1
2
"
we suppose
tional equations
cos
2
"
If
).
2 -f cos / cos 2
$"
cos 2
(2)
=1 + p cos =1 + + cos + cos + f=1 cos 2
+ +
p cos p
/3"
+ cos f cos = + cos f cos = + cos f cos y = 7"
p"
cos
/"
"
/?
I
(3)
"
j
Hence, in this case, only three of the nine angles involved in equation (1) are independent.
SOLID ANALYTIC GEOMETRY.
250
THE CONIC SECTIONS. 191..
The CONIC SECTIONS, or, more simply, THE CONICS, from the surface of a right circular cone by
are the curves cut
a plane. We wish to show that every such section is an ellipse, a Art. or one of their limiting cases. parabola, an hyperbola, 146.
To deduce the equation of the
192.
FIG.
Let
any element
Draw DP
PK
to ||
84.
CAEA C be the conic surface, CA about OZ as an axis.
the element
OZ,
as ||
KB
meet the base
generated by revolving Let P be any point on
= c and OEC = 6. and intersecting OZ in D; draw XY-plane and K producing it to to OY, and join
CE
to
conic surface.
;
let
OC
||
circle in E.
Then (OB, BK, KP) = (a?, y, z) are the co-ordinates From the similar triangles COE, CDP, we have
50.-00_tan.. DP
OE
.(1)
of P.
THE STRAIGHT LINE. DC = 00 - PK = c -
But
z,
and
251
DP = OK = Vz + 2
y
z
hence, z
c
i.e.,
is
(c
-
2
z)
= tan =
2
(x
;
+ if)
tan
2 .
.
.
(2)
the required equation.
193.
To find the equation of the and a plane.
intersection of
a right
cir
cular cone
GALA
X
be the cone and OY the cutting plane. Let angle which the cutting plane makes with the plane of the cone s base, Let P (x, y, z) be any point on the curve of intersection
Let
X OX, BPB
.
the
=
We
<p.
wish to find the equation of this curve when
OX as axes. Draw PD to OY PL and DK
referred to
r
OY, ||
(OK, KL, LP)
=
;
(x, y, z)
||
to
OZ
;
then
are the space co-ordinates of P,
\
SOLID ANALYTIC GEOMETRY.
252
= (x
and (OD, DP) to
OX
f
y) are the co-ordinates of
,
P when referred
OY.
,
From
OD cos
KL =
the figure
PL = KD = OD sin
DP,
qp,
OK =
go.
=
V
i.e.,
= x s m Vi
z V>
x
=x
cos V-
But these values of x, y, & must subsist together with the curve of equation of the conic surface for every point on the and intersection; hence substituting in (2), Art. 192, reducing 2 2 2 we tan. cos sin that have, cp ?, dropping y remembering
=
accents, 2 ?/
2
tan
l9
+x
2
2
cos 2
+2 casing
tan
<p(tan
c
2
=
.
.
.
(1)
for the equation of the intersection. By giving every value to y from
value from
to
equation
oo,
(1)
to 90 and to c every can be made to represent every
section cut from a cone by a plane except sections planes that are parallel to the co-ordinate planes.
COR.
1.
Comparing
(1)
a
= tan.
c
= cos
with
(1),
Art. 138.
we
made by
find
2
2
2 (JP
(tan
9
tan
2 g>)
Hence, equation (1) represents an ellipse, a parabola, an Art. hyperbola or one of their limiting cases according as, 146. b
2
4 ac
<
IP =.
b
Case
and
c
1.
>
;
>
If 6
b
hence,
and
>
We
c
=
<p
this (0, 0)
4
>
<p.
ellipse.
4 ac
2
hence
it is
find this supposition in (2) gives a 4 ac, i.e., the intersection is
>
an
<
the equation resulting from introducing the point (1) can only be satisfied by
0,
supposition in ;
2
ac.
the equation of two imaginary lines inter
secting at the origin.
THE STRAIGHT LINE. ]f
= 0,
cp
equation
that
CASE c
=
2.
.-.
qp
and
2 ?/
which
is
2
tan
6>
=c
2 ,
a circle.
is
Hence the
=
c
2
This supposition in
g>.
= 4c.
Z>
=
=
(9
2
If
+x
2
the intersection
is,
becomes
(1)
if tan
From
0.
tan
2 <9
253
-
is
a parabola,
=
y
i.e.,
the equation of the X-axis a straight line. 90and<?= oo,then the cone becomes a cp cylinder,
= =
If
and the cutting plane is perpendicular to its base. section is therefore two parallel lines. CASE 3. 9 This supposition makes a cp.
The
<
b
2
4
>
If 6
Hence the
ac.
<
q>
and
=
c if-
intersection
then
tan
(1)
=x
2
2
is
c
0,
>
.-.
and
>
we have
(1),
;
gives a
(2)
intersection
inter
<
an hyperbola.
becomes
cos 2
2
(tan
9)
tan
<p
2
0)
which is the equation of two intersecting lines. CASE 4. Planes to the co-ordinate planes. to XY-plane. Let z = m be the (a) Plane \\
\\
such a plane. Combining surface, we have
it
equation of with the equation of the conic
tan
which (b)
is
the equation of a circle for
Plane
||
such a plane.
2
or
7/
tan
2 which, since b values of n.
(c)
Plane
a plane. tion a;
\\
all values of m. Let x = n be the equation of Combining with (2), Art. 192, we have to
YZ-plane.
2
>
to
z?
4
+ 2 cz + n
ac, is
tan
2
XZ-plane. Let y
z2
*
tan
2
c
2
=
.
.
.
(4)
the equation of an hyperbola for
Combining with 2
...
2
= p be
(2), Art. 192,
+ 2 c +^
2
tan
2
all
the equation of such after reduc
we have c
2
=
.
.
.
(5)
SOLID ANALYTIC GEOMETRY.
254 which, since values of p 2
b
2 >
4
ac, is
the equation of an hyperbola for
all
.
Hence, in
of the cutting plane, the possible positions an ellipse, a parabola, an hyperbola, or one of
all
intersection is
their limiting cases.
NOTE.
Equations
(3),
(4),
(5)
of
case
4 are the equa
on the tions of the projections of the curves of intersection the But are projection of any parallel. planes to which they to the given curve a is a on equal parallel plane plane curve the conclusions of case 4 are true for the curve
;
hence
curves themselves.
We have defined the conies, Art. 191, as the curves the surface of a right circular cone by a plane, and from cut found and discussed their assuming this definition we have 194.
general equation, Art. 193. A conic, however, may be otherwise defined as the locus in a plane that the ratio of its generated by a point so moving is always constant. distance from a fixed point and a fixed line
195.
To deduce the general equation of a
conic.
Y
D
FIG.
the basis of the Let us assume the definition of Art. 194 as the fixed line. Let F be the fixed point and operation. of its path. Let P be the generating point in any position
OY
THE STRAIGHT Draw FO Draw PL
OY, and take OY and OX as co-ordinate axes. OY, PD J_ to OY, and join P and F. Let OF
to
||
e
2
FP =
definition
From but
FL
-
=
e
DP = 2
=
e*x
2
2
=
2 .
These values
in (1) give e
or, after
a constant.
FPL, FP == FL + PL 2 ;...(!) 2 (OL - OF) = (x - p)\ LP 2 = f- and FP 2
triangle 2
255
J_ to
= p. By
LINE.
=
x2
2
reduction,
+
y-
-p)
(x
-e
(1
2
x2
)
+ if
2
- 2px + p* =
.
.
.
(2)
for the required equation.
COR.
Comparing a
hence
6
CASE
1.
2
(2)
=
4 ac
with
1, b
=
(1),
= 0, 4
c 2
e )
(1
we
Art. 138,
and
The fixed point not on
= (1 = 4 (e~
find
2
e ),
1)
.
the fixed line
.
;
.
(3)
i.e.,
p
not
zero.
If e
an
<
1, b
2 <
ellipse.
= 1,
If e
6
2
4
ac.;
= 4 ac
hence equation
(2) is the
hence equation
;
equation of
the equation of a
(2) is
parabola. If e
>
1, 6
2 >
4 ac
;
hence equation
(2) is the
equation of
an hyperbola.
CASE
The fixed point (2) becomes
2.
is
on the fixed
2
^2
line, i.e.,
p
= 0.
In this case
f+ If e
<
1,
(1
-
e )
=
.
.
.
(4)
equation (4) represents two imaginary lines inter
secting at origin. If
e
= l,
equation
(4)
represents
one
straight line (the
X-axis). If e
>
1,
equation
(4) represents
two straight
lines inter
secting at the origin.
Hence, equation iting cases.
(2) represents the conies or
one of their lim
SOLID ANALYTIC GEOMETRY.
256
GENERAL EXAMPLES. Find the point of intersection of the 2 x = 1 x = 2
1.
lines
+
+
and the cosine of the angle between them. ^%s.
Required the equation of the to 2, 3) and is parallel
2.
(1
_92
}
(3, 5, 1)
line
y~l-V}What
3.
g>
^|
x
=2z
?/=
the angle between the lines
is
What
5
-.+
Ans. 4.
5
=
which passes through
^
_]_ 1 )
cos
;
is
the distance of the point
(
3, 2,
V
-
=
90.
from
1)
the line
y 5.
A
= 4z
-\-
3
i
makes equal angles with the co-ordinate axes the angles which it makes with the co-ordinate
line
required
;
planes.
4y 2x The equation of a surface is x -f y* + _ 2 what does the equation become when the surface 2
-
6.
Qz
is
;
at
referred to a parallel system of axes, the origin being 2 16. z Ans. x 2 if ?
+
(1, 2, 3)
7.
Given the
the line on
XY
line
x ?
+ ^
^
2 ,
j
8.
Ans.
Required the distance cut I
the projections of the line
j~_
9
=
of required the projection
and the point on which the
co-ordinate planes.
+
off
__
line pierces the
in part, 2 y
on the Z and
+ = 4. a-
Y
axes by
A
^ x == 9
on Y^s
Ans. ?/
"
= -
_3o
6 .
THE STRAIGHT LINE. How
9.
257
are the following pair of lines related ?
= 2z
x
x
2
= 2z-
10. What are the equations of the line which passes through the origin and the point of intersection of the lines
What is origin ? What 11.
on
the distance of the point (3. 2, angle does this line make with
= 3z\
x
A
from the
4) its
projection
XY ? 12. A
straight line makes an angle of 60 with the X-axis and an angle of 45 with the Y-axis what angle does it make Ans. 60. with the Z-axis ? ;
What
13.
Q
"1
are the cosines of the angles
make with 14.
A
which the
lines
^
through the point
line passes
angles with X, Y,
the co-ordinate axes ?
Z whose cosines
V2
are
and makes
(1, 2, 3)
1
2
1 ?
res P ect
2
"
ively; required (a) the
equation of the line, the equation of the plane J_ to the line at the point, to the (V) to show that the projections of the line are J_ traces of the plane. (b)
15.
The
directional cosines of
two
lines are
212 and
o
Xfrj L
1,1. What
is
,
o
,
o
the cosine of the angle which they
2i
make with each other
?
Ans.
Cos
y>
=
3 -
+2 V2 b
16.
x
The projecting planes
=2y
-\-
2.
jects the line
What on
YZ ?
is
of a line are x
=3z
1
and
the equation of the plane which pro 3. 2y Ans. 3 z
=
SOLID ANALYTIC GEOMETRY.
258
and YZ each form with The projections of a line on the 45 of an equation of the line Z-axis the required angle which passes through (2, 1, 4) parallel to the line.
XZ
17.
;
Find the equation of the
18.
(3,.
2, 1)
and meets the
/
Given the
19.
line
x
lines
quired of (a) the value
*
line
which contains the point
* Z
^ _~3
=
l
at right angles. [
and
**+_l\ }
x y
=
+
*
\ J
|
;
re-
order that the lines may be parallel order that the lines may be perpen in of s (b) the value dicular the value of s in order that the lines may intersect. s in
;
;
(c)
212 - -
directional cosines of a line are
required o o the co-ordi the sines of the angles which the line makes with nate planes. 20.
The
21.
Find the equations of the
o
the origin and -
and y 22.
x
=
2,
i
Z
Find
is
1 )
line
,
;
,
which passes through
_
two lines perpendicular to the ^ Ans. x
r
11 &
)
the
angle
+ By + C* = D
and
included
between
Ax + By +
C
If
=
z
_^_
3z
3
j-
)
2 z\
the two planes
=D
.
AA +
+B + a
2
23.
j-
two planes are parallel show that the
C"
coefficients of
the variables in their equations are proportional. 24.
Find the condition for perpendicularity of the two
planes given in
Example
22.
Ans.
AA + BB + CO - 0.
SURFACES OF THE SECOND ORDER.
CHAPTER
259
IV.
A DISCUSSION OF THE SURFACES OF THE SECOND ORDER. By
A. L. Nelson, M.A., Professor of Mathematics in ton and Lee University, Va.
EVERY equation
Washing
involving three variables represents a sur
If the equation be of the first degree the surface will be a plane. If the equation be of a higher degree the surface will be curved. It is proposed in this to determine face.
chapter the nature of the surfaces represented by equations of the second degree involving three variables. The most general 2 2 form of the equation of the second degree is Ax 2 By/ -f- Cz
+ Day + Vxz + -Fyz +
Gx
+ Hy +
Iz
+
+K=
.
.
.
(1)
where
the coefficients A, B, C, etc., may be of either sign and of any magnitude. Let us suppose the co-ordinate axes to be rectan
The form of equation (1) may be simplified gular. by a transformation of axes. Let us turn the axes without chan ging the origin. The formulae
of transformation are (Art. 190)
=X y = x z =x X
f
f
z COS a + y COS cos p + / cos $ + z cos cos f + if cos / + cos f
COS
"
a"
-f-
"
ft
Substituting these values, equation (1) becomes
AV* + By + C V + DVv/ + E + TV + K = (2). 2
2
.
.
.
W+
F yV
+ GV + Hy
SOLID ANALYTIC GEOMETRY.
260
Since the original axes were supposed rectangular the nine are connected by the three relations angles ft f etc.,
+ cos
cos 2 cos 2 COS 2 a If
2
ft
2 4- cos 2 4- COS
"
"
ft"
=
f
2 4- cos
/J"
we take the new axes
able, the
2 4- cos
y"
2 4- COS
I.
= 1. =
1.
/"
also rectangular,
which
desir
is
nine angles will be connected by the three additional
relations
cos a cos
"
-j-
cos a
cos cos
"
cos
u"
cos
cos
ft
+ cos
"
/?
4- cos
cos
ft"
4- cos
ft"
"
ft
cos
f cos
f=
4- cos y cos
"
/?
-4-
f"
f cos
cos
y"
0.
= 0. = 0.
This will leave three of the nine angles to be assumed arbi Let us give to them such values as to render the co trarily. efficients D E and F each equal to zero in equation (2). The general equation will thus be reduced to the form ,
AV
or,
2
4-
,
By
2
4-
CV
2
4-
GV + Hy 4- IV 4- K = 0,
omitting accents,
Ax 2 In order
By 4 Cs to make a 2
4-
+ GJB + Hy + I
2
Equation
A
H
a 4- #
(3) will
(b
+ //)
.
.
(3)
>
=b +y
y
r ,
z
=
f-
4- %
become
Y +B
(a 4- x
.
move the origin without changing The formulae of transformation
will be (Art. 189)
=
K=
further reduction in the form of the
equation let us endeavor to the direction of the axes.
x
4-
(b
+ y Y + C (c + * )* + G +K=
4- I (c 4- *
(a
+x 4 )
0.
)
O
2 2 2 4- B& Developing, omitting accents, and placing A HZ Ic 4- Ga L, the equation takes the form Ax 2 4- B// 2 C,? 2 (2 Att 4- G) x 4- (2 B^ +H) y 4- (2 Cc +1 )z
+
+
+ + L = 0.
In order
and
c,
now
+
+K= +
to give definite values to the quantities a,
which were entirely
2A
,
arbitrary, let us
= --1L_ 2B
c
assume
= _ _JL or
b,
SURFACES OF THE SECOND ORDER.
= If these values of a, reduces to the form
b,
and
c
Cc
,
+I=
.
.
(4)
be Unite, the general equation
+ By + C* + L = 2
.
261
2
.
.
[A], a form which will be set aside for further examination. It may be remarked that equations (4) are of the first degree, and will give only one value to each of the quantities a, b, and c, and there is therefore only one position for the
new
.
origin.
If,
C
however, either A, B, or
become distance.
be zero, then
a, b,
and the origin will be removed This, must be avoided.
infinite,
Let us suppose then assume 2 B6 assume 2 Aa G
+
A=
0,
B and C are finite. We may and 2 Cc + I = 0, but we cannot
while
+ H = 0,
= 0.
or c will
to an infinite
Having assumed the values of assume the entire constant term
b
and
c as
indicated, let us This will give
equal to zero.
+ Cc + Ga + Hb + I C + K = 0, a = - g + Co + Kb + Ic + K
B6 2
2
2
or
2
G and the general equation will be reduced to the form 2 By + C* a + G* = .
.
.
(B),
a second form set aside for examination. must observe that this last proposed transformation will also fail when G 0, that is, when the first power of x as well as the second power of x is wanting in the general equation.
We
=
,
And
without making the second transformation we have a examination, viz.
third form for
:
IV*
+ C* + Hy + I* + K = a
.
.
(C) Lastly, two of the terms involving the second powers of the variables may be wanting, and the equation then .
(1)
becomes
Cs 2
+
Gx
+ Hy 4. Iz + K =
.
.
.
(D)
262
SOLID ANALYTIC GEOMETRY.
It is apparent, therefore, that
every equation of the second degree involving three variables can be reduced to one or another of the four forms
A* 2 + By 2 2 By + C* B?/
C
We
will
with the
+ C.? + Gx
2
2
2
+ Cz* + L = + Gx = (B) + H// + I* + K + Hy + Iz + K =
2
.
.
examine each of these forms form
first
.
(A)
.
.
.
.
.
.
.
.
.
in order,
(C)
(D)
beginning
:
Ax 2
+ B?/ + 2
C* 2
+
L
=
.
.
.
(A)
This equation admits of several varieties of form according to the signs of the coefficients.
C
L
1.
A, B, and
2.
A, B, C, and L positive. Two of the coefficients as
3.
positive,
and
negative in the
A
B
and
first
member.
positive,
C and L
C
negative,
negative.
Two
4.
and L
Ko
of the coefficients as
A
and
B
positive,
positive.
other cases will occur.
CASE 1. Ax 2 + B?/ 2 + Cs 2 = L, in which form all of the coefficients are
positive.
In order to determine the nature of the surface represented by this equation, let it be intersected by systems of planes
The equations parallel respectively to the co-ordinate planes. of these intersecting planes will be x c. Com a, y b, z bining the equations of these planes with that of the surface,
=
=
=
we
find the equations of the projections on the co-ordinate planes of the curves of intersection.
When "
"
= y = z = x
Thus we
2
a, b, c,
By Ax Ax
2 2
+ C.? = L - Aa + Cz = L - IW + By = L - Cc 2
2
2
2
2
an an an
ellipse.
ellipse.
ellipse.
see that the sections parallel to each of the co
ordinate planes are ellipses.
SURFACES OF THE SECOND ORDER. The
L
made
section
Aa
2
or a
>
<
by the
-j-
y
plane
=a
x
-~ and imaginary ,
is
real
263
when
in the contrary
case.
The ft <
=i=
<
Thus we
>
and imaginary when
see that the surface
lar parallelepiped
When sections
When
a
ft
>
is
c
-J-
y
=b
is
y =c
-[-
s
>
real
when
real
when
.
is
i/
.
enclosed within a rectangu
whose dimensions are
=
become a
plane
made by the plane
section
i y ~7T
by the
an d imaginary when
V/~fT~>
The c
made
section
or 4 points.
= 0, = 0,
and by the co-ordinate planes ft
c
By + 2
Ax Ax
2 2
=
or
= 0, C,-
2
we
=
L.
+ C,v = L. By = L. 2
2
-f~
E
c
=
!
Vc
find the sections
the
made
SOLID ANALYTIC GEOMETRY.
264
These are called the principal sections of the surface. The principal sections are larger ellipses than the sections parallel to them, as is indicated by the magnitude of the absolute term.
The surface
called the Ellipsoid.
is
may be generated by the motion of an ellipse of variable dimensions whose centre remains on a fixed line, and whose It
plane remains always perpendicular to that line, and whose semi-axes are the ordinates of two ellipses which have the
same transverse
but unequal conjugate axes placed at
axis,
The axes of the principal sections right angles to each other. are called the axes of the ellipsoid. c,
If
we
we
shall
represent the semi-axes of the ellipsoid by
a, b,
and
have
and the equation of the surface Ax 2 + B?/ 2 -j- Cz 2 a2
b
2
b c
2
x
2
-f-
2
c
a
2 2
c
y
=L
becomes
2
2 -f-
a2b 2z 2
=
a 2b 2 c 2
.
These are the forms in which the equation of the ellipsoid is
usually given. If
we suppose B
= A, then b = a, and the z x + y _ 2
2
equation becomes
2
1
2~
2
and the surface
is
the Ellipsoid of Revolution about the axis
of Z. If
x
A=B 2
2
-j_ ?/
If
L
-(-
z
2
= 0,
then a
== C,
=a
2 ,
=b=
c,
and the surface
the axes 2
and the equation becomes is
a sphere.
2 i/-^-,
reduce to zero, and the ellipsoid becomes a point.
SURFACES OF THE SECOND ORDER.
265
L
be negative in the second member, the equa 2 L will represent an imaginary B?/ -j- Cz will be no there and geometrical locus. surface, Hence the varieties of the ellipsoid are
CASE
tion
If
2.
Ax 2
+
=
2
The ellipsoid proper with three unequal axes. of revolution with two equal axes. (2) The ellipsoid The sphere. (3) (4) The point. (5) The imaginary surface. CASE 3. In this case the equation takes the form (1)
Ax 2 + By 2 - Cz 2 = L, C, and L are essentially
which A, B, positive. Cutting the surface by planes as before, the sections will be, 2 L Aa 2 a hyperbola, having its when x Cz 2 a, ~By in
=
=
,
when a
transverse axis parallel to the Y-axis parallel to the Z-axis
v/
,
when a
-j-
t/ V
but
,
A.
And when
y
>
<
=
a
-J-
the intersection becomes two straight lines whose pro-
.A.
jections on the plane of
When
y
=
b,
Ax 2
YZ pass through
Cz 2
=L
the origin.
B& 2 a hyperbola, with simi ,
lar conditions as above.
When values of
z
=
c,
Ax 2
+ B?/ = 2
L
+ Cc
2 ,
an ellipse real for
all
c.
Since the elliptical sections are all real, the surface it consists of a single sheet.
is
con
tinuous, or
The
principal sections are found by
= 0, b = 0, c = 0, a
which gives By 2
Ax 2 Ax 2
"
-
Cz 2 Cz 2
+%
2
making successively
= L, a hyperbola. = L, = L, an ellipse.
The surface is called the elliptical hyperboloid of one The equation may be reduced to the form v?_ 2
.y*_ 2
^L *
=
i
sheet.
SOLID ANALYTIC GEOMETRY.
266
may be generated by an ellipse centre remains constantly on whose dimensions is whose and perpendicular to that axis, plane semi-axes are the ordinates of two hyperbolas same conjugate axis coinciding with the Z-axis, This surface
of variable
the Z-axis,
and whose
having the but having different transverse axes placed at right angles to each other.
FIG. B.
If
we suppose
A = B,
then will a
=
b,
and the equation of
the surface becomes
the hyperboloid of revolution of one sheet. If
A=B=
C,
we have x 2
+ if - ^ = a
z ,
the equilateral
one sheet. hyperboloid of revolution of a right cone having an If L 0, the equation represents B this base becomes a circle. base and if
=
elliptical
;
A=
SURFACES OF THE SECOND ORDER. Hence we have the following
267
varieties of the hyperboloid
of one sheet.
The liyperboloid proper, with three unequal axes. The hyperboloid of revolution. 3. The equilateral hyperboloid of revolution. 4. The cone. 2 = - L, where A, B, C, and CASE 4. Ax 2 + B?/ 2 1.
2.
O
are essentially positive.
L
=
a, Intersecting the surface as before we have, when x Aa 2 a hyperbola having its transverse L QgZ axis parallel to the axis of Z.
__
_
_
,
j>^2
When its
ing
When c
-j-
>
y
=
b,
Ax 2
Cz"
=
z
=
I/V C
,
c,
Ax 2
+ B// =
L
2
B6 2 -f
and imaginary when
=
sections between the limits real
L
A
.
hyperbola hav
transverse axis parallel to the axis of Z.
beyond those
c
-j-
Cc 2 an ellipse real when ,
i/ V
/
-J- t
<
Since the
.
V C
G
are imaginary, but
two
limits, it follows that there are
distinct
sheets entirely separated from each other. The surface is called the hyperboloid of two sheets. The principal sections are found by making successively 2 2 a 0, which gives By L, a hyperbola with its transverse axis parallel to the Z-axis.
O O
=
b
= 0,
which gives Az 2
=
2
=
L, a hyperbola with its
transverse axis parallel to the Z-axis. c
= 0,
which gives Ax 2
The semi-axes
+ By = 2
of the
L, an imaginary ellipse.
section
first
Those of the second section are I/
V A
of
the imaginary section
The distances 2
i/-^_X I
,
are
i/ V
A
and
(
,
Jj
V/ B i
and
t
And !/_. V
V
1)
A
and
i /
V
B
/
V
.
C
those
C
2 4/1*1. and 2 \/ T
are
(
1).
are called the axes
SOLID ANALYTIC GEOMETRY.
268
FIG. C.
and Eepreseiiting the semi-axes by a, b, form the to reduced the equation of the surface may be of the surface.
L
5 + -? If
we suppose
A=B
?
then a
c,
=
&,
and the equation
duces to
the hyperboloid of revolution of two sheets.
re
SURFACES OF THE SECOND ORDER. If
A=B=
C, the equation
becomes
+f_^= _
x*
269
a*
9
which represents the surface generated by the revolution of an equilateral hyperbola about
its transverse axis. the surface becomes a cone having an Finally, elliptical base, and the base becomes a circle when B. We have, therefore, the following varieties of the if
L
= 0,
A=
hyper
boloid of two sheets: 1.
2.
3. 4.
The hyperboloid proper having three unequal The hyperboloid of revolution. The equilateral hyperboloid of revolution. The cone.
We
will
now examine the second 2 2 By + C* + G x =
axes.
form, .
.
.
(B)
Three cases apparently different present themselves for examination.
B and C
positive
(2).
B, C, and
G
(3).
B
(1).
G
and
negative in the
first
mem
ber.
CASE
positive.
C and G negative. The equation may be written
positive and
1.
B?/
in
which B,
2
+ O. = Gx 2
G
and
are essentially positive. C, Let the surface be intersected as usual by planes parallel respectively to the co-ordinate planes.
When
x
=
,
By
+O =
2
2
Ga, an ellipse real when a
and imaginary when a 0. 2 When y = b, = Gx
>
0,
<
O
B6 2
,
a
parabola with
its
axis
its
axis
parallel to the axis of X.
When
z
=
c,
By
2
= Gx -
O
2
a parabola
with
parallel to the axis of X.
The and
principal sections are found by
c
When When origin.
making a
= 0, b = 0,
0.
a b
= 0, By + C,? = = 0. C.s = Ga;, a 2
2
2
0,
a point, the origin. its vertex at the
parabola with
SOLID ANALYTIC GEOMETRY.
270
When c, =
=
2 G#, a parabola with vertex at the origin. 0, B?/ Since every positive value of x gives a real section, and every negative value of x an imaginary section, the surface consists of a single sheet extending indefinitely and contin
uously in the direction of positive abscissas, but having no from the origin. points in the opposite direction The surface is called the elliptical paraboloid. It may be
dimensions generated by the motion of an ellipse of variable the same on remains whose centre straight line, constantly
and whose plane continues perpendicular to that line, and whose semi-axes are the ordinates of two parabolas having a common transverse axis and the same vertex, but different to each parameters placed with their planes perpendicular other.
FIG. D.
in the 2. If we suppose G to be positive form the take will the so that equation
CASE
By
2
+C = 2
Go;,
first
member
SURFACES OF THE SECOND ORDER.
271
the sections perpendicular to the X-axis will become imaginary
when x
>
0,
and
real
when x
<
0.
In other respects the results are similar to those deduced case
in
1.
will represent a surface of the same form but turned in the opposite direction from the
Thus the equation as in case
1,
co-ordinate plane of If B = C, the
CASE
surface becomes the paraboloid of revolution.
By
3.
YZ.
2
-C = a
Gx.
Intersect the surface
by planes as before. Gz 2 = Ga, a hyperbola with transverse and in the axis in the direction of the Y-axis when a direction of the Z-axis when a 0.
When
x
= a,
2
By
>
;
<
When
y
=
b.
in the direction of
=
=
Gx
B6 2 a parabola having its the X-axis and extending to the left.
Cz 2
-\-
=
,
axis
Gx -j- Cc a parabola having its axis in When z c B?/ the direction of the X-axis and extending to the right. 2
}
2
,
Since every value of x, either positive or negative, gives a real section, the surface consists of a single sheet extending This indefinitely to the right and left of the plane of YZ.
surface
is
called the Hyperbolic Paraboloid. To find its princi make x, y, and z alternately equal to zero.
pal sections
When x = 0, B^/ = Cz two straight lines. When y = 0, Cz = Gx, a parabola with axis to the left. When = 0, B?/ = Gx, a parabola with axis to the right. 2
2
,
2
2
The hyperbolic paraboloid admits of no variety. Now taking up form (C), By 2 + Cs a + Hy + Iz + K = 0, we see that it is the equation of a cylinder whose elements are perpendicular to the plane of YZ, and whose base in the plane of YZ will be an ellipse or hyperbola according to the signs of B and C.
+
+
+K=
The fourth form (D), Cz Ga; Hy -f Is sents a cylinder having its bases in the planes 2
XZ
repre
and
YZ
SOLID ANALYTIC GEOMETRY.
272
\
VA
A
/
\
/
\ /
FIG. E.
parabolas, and having its right-lined elements parallel to the and to each other, but oblique to the axes of and Y.
plane
X
XY
The preceding discussion shows that every equation
of the
second degree between three variables represents one or an other of the following surfaces the ellipsoid 1. The ellipsoid with its varieties, viz. :
;
the
of
proper, ellipsoid and the imaginary surface.
revolution, the
sphere, the
point,
SURFACES OF THE SECOND ORDER. The hyperboloid
2.
of
one
or
two
sheets,
273
with
their
the hyperboloid proper of one or two sheets, the hyperboloid of revolution of one or two sheets, the varieties, viz.
:
equilateral hyperboloid of revolution of one or two sheets, the cone with an elliptical or circular base.
The
3.
paraboloid, either elliptical or hyperbolic, with the
variety, the paraboloid of revolution.
The
4.
cylinder, having its base either an ellipse,
hyper
bola, or parabola.
The general equation of surfaces be deduced by a direct method, as follows
Surfaces of Revolution. of revolution
may
:
Ri
FIG. F.
Let the Z-axis be the axis of revolution, and tion
x2
of
= fz. Let
P
AB,
the
let
the equa
generating curve in the plane of XZ, be
be the point in this curve which generates the circle
SOLID ANALYTIC GEOMETRY.
274
PQR, and
be the radius of the
let r
The value
circle.
We
will
have
of r2
may also be expressed in terms of z from the equation of the generatrix in the plane of XZas follows :
= CP = 2
r2
2
(JD =fz.
Equating these two values of r
we have
as the general equation of surfaces of revolution. It will be observed that the second value of r 2
is the value of x 2 in the equation of the generatrix. Hence, to find the equation of the surface of revolution we have only to substi 2 tute x 2 if of the surface for x in the generatrix.
+
2 Surface of a Sphere. Equation of generatrix x Hence the equation of the surface of the sphere is
+
z2
=
K,.
2
Ellipsoid of Revolution.
Generatrix
Surface
JL_
x*
-f-
~^~
J!L
^
= 1, 1
-4-
a2
c
2
Similarly, the equation of the hyperboloid of revolution
***-*-* Paraboloid of Revolution. x2 4:pz, the generatrix.
= + IT = 4=pz, the Cone of revolution, z = mx x2
surface of revolution.
-+- j3
~T Hence or
x
x
the generatrix,
~
~~i^~
2 -
m
2
2
(x
+
if)
=
(z
-
2
ft)
.
is
SURFACES OF THE SECOND ORDER.
275
EXAMPLES. 1.
9z 7/2
(^ 3.
2
Of ^ 2
+ if = r- ?
Qf
2
if
Determine the nature of the surfaces x 2
+^
_
= 36 ? + =r ?
the locus in space of 4 x 2 -f 9 y-
is
- 16 y = 144 ? + 8x = o?
2.
7
What
2
4
.2
=
-\-
2
,-
y"-
-(-
4 z2
Of Of
= 25,
79>
Find the equation of the surface of revolution about the Z whose generatrix is z = 3 x -\- 5.
axis of 4.
Find the equation of the cone of revolution whose inter 2 = 9, and whose vertex -f- y
section with the plane of XYisx 2 is
(0, 0, 5.)
5.
Determine the surfaces represented by
+ 4 y + 9 s = 36. + 4 y _ 9 # = 36 - 36. x + 4y = 9 4 y + 9 2 = 36 x. 9z* = 36x. 2
2
x*
2
a;
2
a
2
2
2
2
14
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