Ellipse

ELLIPSE

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ELLIPSE The ellipse is a locus of a point which moves such t hat r at io of it s dist ances fr om a fixed point (called focus) and fr om a fixed line (called dir ect r ix) is always const ant . This const ant (denot ed by e) is called eccent r icit y of t he ellipse which must be less t han 1. Alt er nat ively t he locus of a point which moves such t hat su m of i t s d i st a n ces f r om t w o f i xed p oi n t s is always const ant is called ellipse. We shall lat er see t hat t he fixed point s ar e t wo foci of t he ellipse and t he const ant is equal t o lengt h of t he major axis of t he ellipse.

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For ever y ellipse t her e is a point called cent r e of t he ellipse t hr ough which ever y chor d of t he ellipse is bisect ed. The line joining cent r e and focus of t he ellipse is called major axis of t he ellipse. A line per pendicular t o t he major axis and passing t hr ough t he cent r e of t he ellipse is called minor axis of t he ellipse. The dist ance bet ween t he point s on t he ellipse which lie on major axis is called lengt h of t he major axis. The lengt h of t he minor axis of an ellipse is dist ance bet ween t wo point s on it s which lie on minor axis. I f t he major and minor axis ar e along x and y-axis r espect ively wit h lengt hs 2a and 2b and cent r e of t he ellipse be or igin t he equat ion of t he ellipse can be pr oved t o be

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x 2 y2 + =1 a 2 b2

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D er i v i n g t h e eq u at i on of t h e el l i p se i n t h e st an d ar d f or m : L et us choose a ver t ical line V´V as dir ect r ix and S as focus. Dr aw SZ V´V. L et e < 1 eccent r icit y of t he ellipse. Ther e must be a point A on SZ such t hat SA = e AZ

Z

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N

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...(i )

Ther e must be anot her point A´ (since e < 1) On AS pr oduced such t hat SA´ = e A´Z

...(i i )

L et A´A = 2a and let O be t he mid point of A´A. On adding (i ) and (i i ) we get SA + SA´ = e (AZ + A´Z)  2a = e(OZ – a + a + OZ)

 2a = 2eOZ  OZ =

a . e

...(i i i )

Again, subt r act ing (i ) fr om (i i ) SA´ – SA = e (A´Z – AZ)  (SO + a) – (a – SO) = e(A´A) EL L I PSE

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 2SO = e(2a)  OS = ae

...(i v)

L et us t ake O as or igin, A´A as x-axis and a line  t o A´A and passing t hr ough O as y axis t hen

FG a , 0IJ . I f P(x, y) be any point on t he ellipse t hen PS = e PM wher e He K

point (ae, 0) and Z is t he point

PM is t he per pendicular fr om P on dir ect r ix V´V.

 PS2 = e2 PM 2  (x – ae)2 + y2 = e2

 x2 (1 – e2) + y2 = a2 (1 – e2) 

FG x a IJ H eK

2

.

...(v)

x2 y2 + = 1. a 2 a 2 (1  e2 )

On put t ing b2 = a2 (1 – e2) we get t he r equir ed st andar d equat ion (a) Cent r e is or igin (b) Focus is at (ae, 0) (c) Dir ect r ix is t he line x =

a . e

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...(vi )

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x 2 y2 + = 1 which is an ellipse a2 b2

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The equat ion t o t he ellipse, whose focus and dir ect r ix ar e any given point and line, and whose eccent r icit y is known, is easily wr it t en down. For example, if t he focus be t he point (– 2, 3), t he dir ect r ix be t he line 2x + 3y + 4 = 0, and t he eccent r icit y be

4 , t he r equir ed equat ion is 5

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N 2

(x + 2) + (y – 3)

2

F 4I = G J H 5K

2

(2 x+ 3 y+ 4) 2 22 + 32

i .e., 261x2 + 181y2 – 192xy + 1044x – 2334y + 3969 = 0

Z

Gener ally, t he equat ion t o t he ellipse, whose focus is t he point (f, g), whose dir ect r ix is Ax + By + C = 0, and whose eccent r icit y is c, is ( Ax+ By+ C) 2 (x – f) + (y – g) = e A2 + B2 2

2

2

1 . SECOND FOCUS AND SECOND DIRECTRIX OF THE ELLIPSE On t he negat ive side of or igin t ake a point S´ which is such t hat CS = CS´ = ae and anot her point Z´ t hen CZ = CZ´ = EL L I PSE

a e

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 Co-or dinat e of S´ ar e (– ae, 0) and equat ion of second dir ect r ix (i .e., Z´ M ´) is x = –

a e

L et P(x, y) be any point on t he ellipse t hen S´P = ePM ´ (S´ P)2 = e2 (PM ´)2

or

(x + ae)2 + (y – 0)2 = e2

or

FG x+ a IJ H eK

2

or

(x + ae)2 + y2 = (ex + a)2

or

x2 (1 – e2) + y2 = a2 (1 – e2)

or

x2 y2 + = 1 a2 a2 (1  e2 )

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x 2 y2 + = 1, wher e b2 = a2 (1 – e) a2 b2

or

The equat ion being t he same as t hat of t he ellipse when S(ae, 0) is focus and M Z i .e., x = a/e is dir ect r ix.

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H ence co-or dinat es of foci ar e (± ae, 0) and equat ions of dir ect r ices ar e x = ± a/e. N ot e : 1. Dist ance bet ween foci SS´ = 2ae and dist ance bet ween dir ect r ices ZZ´ = 2a/ e 2. I f e = 0 t hen



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t hen equat ion of ellipse i .e.,

Z

b 2 = a2 (1 – 0) b 2 = a2

x 2 y2 = 1 changes in cir cle + a2 b2

x2 + y2 = a 2

1 .1 Geometrical Construction of an ellipse As t he definit ion of ellipse says t hat it is a locus of a point which moves such t hat t he sum of it s dist ance fr om 2 fixed point s (ae, 0) & (– ae, 0) i .e., focus is always const ant and equal t o 2a. To dr aw an ellipse, fix two point S and S´ and attach an inextensible thread to these point. Note if you draw a cur ve, using a pencil, moving the thread around you will an ellipse as the curve and hence the sum of focal distance of any point on an ellipse is constant and equal to the length of major axis of an ellipse.

UV S´P = a – ex W ´SP = a + ex

These t wo ar e f ocal d i st an ce of a point P r elat ive t o 2 focii S & S´

 SP + S´P = a + ex + a – ex = 2a EL L I PSE

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1 .2 Tracing the curve x 2 y2 + = 1 a2 b2

...(1)

The equat ion may be wr it t en in eit her of t he for ms

or

y = ± b 1

x2 a2

x = ± a 1

y2 b2

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...(2)

...(3)

Fr om equat ion (2), it follows t hat if x2 > a2, i .e., if x > a or < – a, t hen y is impossible. Ther e is t her efor e, no par t of t he cur ve t o t he r ight of A´ or t o t he left of A.

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Fr om equat ion (3), it follows, similar ly, t hat , if y > b or < – b, x is impossible and hence t hat t her e is no par t of t he cur ve above B or below B´.

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I f x lie bet ween – a and + a, t he equat ion (2), gives t wo equal and opposit e values for y, so t hat t he cur ve is symmet r ical wit h r espect t o t he axis of x. I f y lie bet ween – b and + b, t he equat ion (3) gives t wo equal and opposit e values for x, so t hat t he cur ve is symmet r ical wit h r espect t o t he axis of y.

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2 . SOME TERMS RELATED TO AN ELLIPSE L et t he equat ion of t he ellipse

x 2 y2 + = 1, (a > b) a2 b2

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1. Cen t r e : All chor ds passing t hr ough C ar e bisect ed at C H er e C  (0, 0)

Z

2. F oci : S and S´ ar e t wo foci of t he ellipse and t heir co-or dinat es ar e (ae, 0) and (– ae, 0) respectively.

EL L I PSE

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3. D i r ect r i ces : ZM and Z´ M ´ ar e t wo dir ect r ices of t he ellipse and t heir equat ion ar e x = and x = –

a e

a r espect ively. e

4. Ax es : The lines AA´ and BB´ ar e called t he m aj or and m i n or axes of t he ellipse

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·.·

0< e< 1

or

0 < e2 < 1

( 0 > – e2 > – 1)

or

0 < 1 – e2 < 1

(or 1 > 1 – e2 > 1 – 1)

or

a2 (1 – e2) < a2

or

b 2 < a2

(or 0 < 1 – e2 < 1)

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b < a

i .e.,

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5. D ou b l e or d i n at es : I f P be a point on t he ellipse dr aw PN per pendicular t o t he axis of t he ellipse and pr oduced t o meet t he cur ve again at P´. Then PP´ is called a double or dinat e. I f abscissa of P is h t hen or dinat e of P,

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y2 h2 = 1 – b2 a2

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and or dinat e of P´ is

y = –

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H ence, co-or dinat es of P and P´ ar e

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b (a 2  h 2 ) a

FG h, H

(for fir st quadr ant )

b (a 2  h 2 ) a

b a2  h 2 a

IJ K

(for four t h quadr ant )

and

FG h, H



b 2 a  h2 a

IJ K

r espect ively.

6. L ast u s r ect u m : The double or dinat es L L ´ and L 1L 1´ ar e lat us-r ect ums of t he ellipse. These line ar e per pendicular t o major axis A´ A and t hr ough t he foci S and S´ r espect ively. L en gt h of t h e l at u s r ect u m Now let t hen

Z

LL ´ = 2k LS = L ´ S = k

Co-or dinat es of L and L ´ ar e (ae, k) and (ae, – k) lies on t he ellipse x 2 y2 + = 1 a2 b2

 or

a2 e2 k 2 + 2 = 1 a2 b

k 2 = b2 (1 – e2)

Fb I = b G J Ha K 2

EL L I PSE

2

2

[·.· b2 = a2 (1 – e2)]

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

k =



2k =

b2 a

(·.· k > 0)

2 b2 = LL´ a

 L engt h of lat us r ect um L L ´ = L 1 L 1´ =

F b I ; L   F ae,  b I GH a JK GH a JK F b I; L   F  ae,  b I  G  ae, GH a JK H a JK 2

L  ae,

2

A I

2

L1

S L

2 b2 and end of point s of lat us-r ect um ar e a

2

1

respectively.

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7. F ocal ch or d : A chor d of t he ellipse passing t hr ough it s focus is called a focal chor d. 8. Ver t i ces : The ver t ices of t he ellipse ar e t he point s wher e t he ellipse meet s it s major axis. H ence A and A´ ar e t he ver t ices

Eccentricity of the Ellipse

For t he ellipse

2

Z 2

and

A´  (– a, 0)

x 2 y2 + = 1, a > b a 2 b2

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x 2 y2 + = 1, we have a2 b2

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b2 = a2 (1 – e2)  e2 = 1 –

2

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A  (a, 0)



b = a (1 – e )  e =

FG IJ H K

b2 4 b2 2b = 1  =1 2a a2 4 a2

F M i nor axi sIJ 1 G H M ajor axi s K

2

2

This for mula gives t he eccent r icit y of t he ellipse.

3 . EQUATION OF ELLIPSE IN OTHER FORMS I n t he equat ion of t he ellipse

x 2 y2 + 2 = 1, if a > b or a2 > b2 (denominat or of x2 is gr eat er t han 2 a b

t hat of y2), t hen t he major and minor axes lie along x-axis and y-axis r espect ively But if a < b or a2 < b2 (denominat or of x2 is less t han t hat of y2), t hen t he major axis of t he ellipse lies along t he y-axis and is of lengt h 2b and t he minor axis along t he x-axis and is of lengt h 2a. EL L I PSE

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The coor dinat es of foci S and S´ ar e (0, be) and (0, – be) r espect ively. The equat ions of t he dir ect r ices ZK and Z´ K´ ar e y = ± b/ e and eccent r icit y e is given by t he for mula a2 = b2 (1 – e2) or e = 1 

a2 b2

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4 . DIFFERENCE BETWEEN BOTH ELLIPSES

N

x 2 y2 + = 1, a > b a 2 b2

x 2 y2 + = 1, b > a a 2 b2

(0, 0)

(0, 0)

(± a, 0)

(0, ± b)

L engt h of M ajor axis

2a

2b

L engt h of M inor axis

2b

2a

Foci

(± ae, 0)

(0, ± be)

Equat ion of dir ect r ices

x = ± a/e

y = ± b/e

Relat ion in a, b and e

b2 = a2 (1 – e2)

a2 = b2 (1 – e2)

L engt h of lat us r ect um

2 b2 a

2 a2 b

B asi c f u n d am en t al s Centr e Vertices

Z

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Ends of lat us r ect ums

F  ae, GH EL L I PSE



b2 a

I JK

F a , GH b 2

I JK

 be

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Parametr ic co-ordinat es

(a cos , b sin )

(a cos , b sin ) (0   < 2)

Focal r adii

SP = a – ex1 and

SP = b – ey1 and

S´ P = a + ex1

S´ P = b + ey1

Sum of focal r adii SP + SP´ =

2a

2b

Dist ance bet ween foci

2ae

2be

Dist ance bet ween dir ect r ices

2a/ e

2b/ e

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SPECI AL F ORM : I f t he cent r e of t he ellipse is at point (h, k) and t he dir ect ions of t he axes ar e par allel t o t he coor dinat e axes, t hen it s equat ion is

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( x  h ) 2 ( y  k) 2 + = 1 a2 b2

R

I f we shift t he or igin at (h, k) wit hout r ot at ing t he coor dinat e axes, t hen

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x = X + h and y = Y + k

So, t he equat ion of t he ellipse wit h r espect t o new or igin, becomes X2 Y2 + 2 = 1 a2 b

Illustration 1

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F i n d t h e eq u at i on of t h e el l i p se w h ose f ocu s i s at (1, –1), d i r ect r i x i s t h e l i n e x + y = 3 an d eccen t r i ci t y i s

1 . 3

Sol u t i on :

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N

I f (x, y) is any point on t he ellipse t hen will

Or Or

Z

( x 1) 2 + ( y+ 1) 2 =

FG H

1 x+ y 3 3 2

IJ K

18[(x – 1)2 + (y + 1)2] = (x + y – 3)2

17x2 + 17y2 – 2xy – 30x + 42y + 27 = 0

Illustration 2 F i n d t h e eq u at i on of t h e el l i p se w h ose t w o f oci ar e (4, 0) an d (– 4, 0) an d w h ose eccen t r i ci t y is

1 . 3

Sol u t i on : Since t he dist ance bet ween foci = 8 we must have EL L I PSE

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FG H

Also b2 = a2 (1 – e2) = 144 1 

IJ K

1 = 128 9

Now since t he midpoint of t he line joining t he foci is cent r e of t he ellipse and t he mid point of li ne joining (4, 0) and (– 4, 0) is or igin.

 Or igin is cent r e of t he ellipse and x-axis is t he major axis.

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We also conclude t hat y-axis is t he minor axis. Thus t he equat ion of t he ellipse is x 2 y2 x2 y2 + 2 =1  + =1 2 144 128 a b

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Illustration 3

F i n d t h e eccen t r i ci t y of t h e el l i p se i f i t s l at u s r ect u m b e eq u al t o h al f of t h e m i n or ax i s. Sol u t i on : We must have 2

b2 1 = b  a = 4b a 2

As b2 = a2 (1 – e2), we have b2 = 16b2 (1 – e2)

 e =

15 4

Illustration 4

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R

F i n d f ocal d i st an ce of t h e p oi n t (4 3 , 5) on t h e el l i p se 25x 2 + 16y2 = 1600. Sol u t i on : The ellipse is

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x2 y2 + = 1  a2 = 64, b2 = 100 64 100

Z

a2 = b2 (1 – e2) (since a < b)

 64 = 100(1 – e2)  e =

3 5

The focal dist ances of any point (x1, y1) on t he ellipse ey1 which, in t his quest ion, ar e 10 –

x 2 y2 + = 1 (a < b) must be b – ey1, b + a2 b2

3 3 × 5, 10 + × 5 or 7 and 13. 5 5

EL L I PSE

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Illustration 5 Sh ow t h at t h e l ocu s of a p oi n t w h i ch m ov es su ch t h at t h e su m of i t s d i st an ces f r om t w o f i x ed p oi n t s (a e, 0) an d (– a e, 0) i s al w ay s a con st an t eq u al t o 2a i s t h e el l i p se Sol u t i on : L et P be (x, y) t hen fr om t he condit ion given we have



( x ae) 2 + y2 + ( x+ ae) 2 + y2 = 2 a



x 2 + y2 + a2 e2  2 aex + x2 + y2 + a2 e2 + aex = 2 a (wher e  = x2 + y2 + a2 e2)

   2 aex +  + 2 aex = 2 a On squar ing t he last r elat ion we get

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x 2 y2 = 1. + a 2 b2

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 – 2aex +  + 2aex +  2  4 a 2 e2 x 2 = 4a2  2  2  4 a 2 e2 x 2 = 4a2 – 2 On squar ing again we get

2 – 4a2e2x2 = 4a4 + 2 – 4a2 – 4a2e2x2 = 4a4 – 2a2 (x2 + y2 + a2e2)  x2 + y2 + a2e2 – e2x2 = a2 

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x2 y2 + = 1 a 2 a 2 (1  e2 )

Thus locus of a point P t he sum of whose dist ances fr om t wo fixed point s S(ae, 0), S´(– ae, 0) is a const ant (= 2a) is an ellipse. Not e t hat t his const ant is equal t o lengt h of t he major axis of t he ellipse.

Illustration 6

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Red u ce t h e eq u at i on of t h e el l i p se 4x 2 + 3y2 + 8x + 12y = 1 t o t h e st an d ar d f or m an d h en ce d et er m i n e/l ocat e cen t r e, f ocu s, d i r ect r i x of t h e el l i p se. Sol u t i on :

Z

Gr ouping t er ms we get 4x2 + 8x + 3y2 + 12y = 1

 4(x2 + 2x) + 3(y2 + 4y) = 1  4(x2 + 2x + 1) + 3(y2 + 4y + 4) = 1 + 4 + 12  4(x + 1)2 + 3(y + 2)2 = 17 

( x+ 1) 2 ( y+ 2) 2 + = 1 17 / 4 17 / 3

Put X = x + 1, Y = y + 2 we get

X2 Y2 + = 1 17 / 4 17 / 3

EL L I PSE

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 a2 =

17 17 , b2 = since b > a t he major axis of t he ellipse must be along y-axis. 4 3

17 17 1 = (1 – e2)  e = 4 3 2

On which foci will lie. Fr om a2 = b2 (1 – e2) we get Thus t he eccent r icit y of t he given ellipse is 17 2a or 2 , 3

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1 , t he lengt hs of major and minor axes and 2b and 2

17 r espect ively.

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Now obser ve t he following t able for var ious locat ions : L ocati on

New axes

Ol d axes

Centr e

(0, 0)

(1, 2)

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R

(On put t ing X = 0, Y = 0 in X = x + 1, Y = y + 1)

F0, GH

Foci

F0, GH Directrix

17 1 . 3 2



X = ±

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I JK

17 1 . 3 2

F 1, GH

I JK

N 17 / 3 1/2

U T F 1, GH

I JK

17 1 . 1 3 2



I JK

17 1 . 1 3 2

x+ 1= ±

17 / 3 1/2

T h e st u d en t m u st n ot e t h at t h e eccen t r i ci t y , l en gt h of m aj or , m i n or ax es, l en gt h of l at u s r ect u m ar e i n v ar i an t s. T h ey r em ai n sam e i n al l f r am e of r ef er en ce. I n ad d i t i on y ou m u st n ot e t h at i f t h e gen er al eq u at i on of secon d d egr ee r ep r esen t s an el l i p se t h en

Z

(1) eccent r icit y = 1 

eccent r icit y = 1 

coeff. of y2 if coeff. of y2 < coeff. of x2. coeff. of x2

Or coeff. of y2 2 2 2 if coeff. of y > coeff. of x . coeff. of x

(2) I f t he gener al equat ion of second degr ee f(x, y) = 0 r epr esent s on ellipse and xy t er m is absent t hen cent r e of t he ellipse is given by

wher e

f f = 0, =0 x y

f denot es t he der ivat ive of f(x, y) wit h r espect t o x, keeping y const ant . x EL L I PSE

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Illustration 7 F i n d t h e l en gt h s an d eq u at i on s of t h e f ocal r ad i i d r aw n f r om t h e p oi n t (4 3 , 5) on t h e el l i p se 25x 2 + 16y2 = 1600. Sol u t i on : The equat ion of t he ellipse is 25x2 + 16y2 = 1600 x2 y2 + = 1 64 100

or

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H er e b > a a 2 = 64, b2 = 100 a 2 = b2 (1 – e2) 64 = 100(1 – e2)



e = 3/5

 Let

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S L

R

P(x1, y1)  (4 3 , 5)

be a point on t he ellipse t hen SP and S´ P ar e t he focal r adii

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

SP = b – ey1



SP = 10 –

 i .e.,

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N

and

3 × 5 and 5

SP = 7

FG0, 10  3 IJ H 5K

and

S´ P = b + ey1 S´ P = 10 +

3 × 5 5

S´ P = 13

i .e., (0, 6)

and S´ is (0, – be)

5.

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POSITION OF A POINT WITH RESPECT TO AN ELLIPSE The point P(x1, y1) lies out side, on or inside t he ellipse

x 2 y2 x12 y12 + + = 1 accor ding as – 1 > 0, a2 b2 a2 b2

= or < 0.

6.

PARAMETRIC EQUATION OF ELLIPSE I f we draw a cir cle wit h major axis of any ellipse as diameter t hen this circle is known as a u xi l i a r y ci r cl e for that ellipse. Now take any point P on t he ellipse and draw a line t hr ough it par allel t o minor axis. The point where t his line cut s the auxiliary circle such t hat P and Q lies on t he same side of t he major axis, is known as cor responding poi nt. I f the line joining t o t he center of the ellipse, makes an angle of  wit h the major axis,  is known as eccen t r i c a n gl e of point P. EL L I PSE

14

ZION TUTORIALS

I f el l i pse be

x 2 y2 + = 1 (a > b), t hen a2 b2

a u xi l i a r y ci r cl e w i l l be x 2 + y2 = a 2 and point Q will be (a cos , a sin ), hence point P will be (a cos , b sin ). H er e x = a cos , y = b si n  is called as par amet r ic equat ion of t he ellipse

x 2 y2 + = 1 a2 b2

A I

Illustration 8

R

F i n d t h e eccen t r i c an gl es of t h e ex t r em i t i es of l at u s-r ect a of t h e el l i p se Sol u t i on :

U T

O T

The coor dinat es of t he end-point s of lat us r ect a of t he given ellipse ar e

L et  be t he eccent r ic angle of an end of a lat us-r ect um of t he ellipse

N

coor dinat es ar e (a cos , b sin ) of t he ellipse





Z

O I

a cos  = a e and b sin  =

F a GH

S L

x 2 y2 + = 1. a 2 b2

e, 

b2 a

I JK

x 2 y2 + = 1. Then, it s a2 b2

x 2 y2 + = 1. a2 b2

b2 a

b b sin  b2 = 2   t an  =   = t an –1 ae a cos  a e

FG b IJ H aeK

Similar ly,t he eccent r ic angles of ot her ends of lat usr ect a ar e given by  = t an –1 (– b/ae). H ence, t he eccentr ic angles of t he extr emit ies of the lat us-r ect a of t he ellipse by  = t an –1 (± b/ae).

EL L I PSE

x 2 y2 + = 1 ar e given a2 b2

15

ZION TUTORIALS

Illustration 9 Sh ow t h at t h e l i n e l x + m y + n = 0 w i l l cu t t h e el l i p se

x 2 y2 + = 1 i n p oi n t s w h ose eccen t r i c a 2 b2

an gl es d i f f er b y /2, i f a 2 l 2 + b 2 m 2 = 2n 2. Sol u t i on :

S L

Suppose t he line l x + my + n = 0 cut s t he ellipse at P (a cos , b sin ) and Q (a cos (/2 + ), b sin (/2 + )). Then t hese t wo point s lie on t he line. l a cos  + m b sin  + n = 0

 and

l a cos  + mb sin  = – n

 and,

– l a sin  + mb cos  = – n

R

  (l a cos  + mb sin )2 + (– l a sin  + mb cos )2 = n 2 + n 2 l 2 a2 + m 2 b2 = 2n 2.



7.

A I

– l a sin  + m b cos  + n = 0

O T

INTERSECTION OF A LINE AND AN ELLIPSE

U T

The abscissae of point s of t he int er sect ion of t he line y = mx + c and t he ellipse given by t he quadr at ic equat ion 0 (* ). I f t he r oot s be x1, x2 t hen

O I

x 2 y2 = 1, ar e + a2 b2

x 2 (mx+ c) 2 = 1 Or (a2m 2 + b2) x2 + 2a2 mcx + a2(c2 – b2) = + a2 b2

N

x1 + x2 = –

2 a2 mc a2 (c2  b2 ) ; x x = 1 2 a2 m 2 + b2 a 2 m 2 + b2

I f or dinat es of t he point s be y1 and y2 t hen y1 = mx1 + c, y2 = mx2 + c  y1 + y2 = m(x1 – x2)

Z

Now t he dist ance bet ween t hese point s is equal t o lengt h int er cept ed by t he line y = mx + c on t he ellipse

x 2 y2 + = 1 which is equal t o a2 b2 ( x1  x2 ) 2 + ( y1  y2 ) 2 = ( x1  x2 ) 2 + m 2 ( x1  x2 ) 2 =| x1  x2|

1 + m2

= ( x1  x2 ) 2  4 x1 x2 1 + m 2

=

2 ab a 2 m 2 + b2  c2 a 2 m 2 + b2

(on put t ing t he values of x1 + x2 and x1x2) EL L I PSE

16

ZION TUTORIALS

I f the line happens to be tangent to the ellipse the length intercepted by it on the ellipse

x 2 y2 + =1 a2 b2

must be zer o.

 a2m 2 + b2 – c2 = 0  c2 = a2m 2 + b2  c = ± a 2 m 2 + b2 I n case c2 = a2m 2 + b2 t he equal r oot of t he quadr at ic equat ion (* ) is x = –

S L

a2 mc a2m = 2 2 c a m +b

2

2

(I f t he equat ion Ax + Bx + C = 0 has equal r oot t hen it is equal t o – B/2A)

F GH

Also y = mx + c = m 

A I

I JK

a2m  a 2 m 2 + c2 b2 + c= = c c c

Thus we have t he following impor t ant r esult .

O T

The l i ne y = mx + c wi l l be a tangent t o t he ellipse

F GH

t he ellipse at 

I JK

R

x 2 y2 + = 1 if c 2 = a 2m 2 + b 2 and it will t ouch a2 b2

x 2 y2 a m b . The line y = mx ± a 2 m 2 + b2 will be a t angent t o t he ellipse 2 + 2 = , a b c c 2

2

1 for all values of m.

U T

N OT E : I f m = 0 t hen equat ion (3) gives

N

b2x2 + c2a2 – a2b2 = 0 or

b2x2 + (a2m 2 + b2) a2 – b2 = 0



x =±

Z

O I

a2 m b

which gives t wo values of x.

Illustration 10

Sh ow t h at t h e l i n e x cos  + y si n  = p t ou ch es t h e el l i p se

 = p 2 an d t h at p oi n t of con t act i s

Fa GH

2

I JK

cos  b 2 si n  . , p p

Sol u t i on : The given line is

x cos  + y sin  = p

EL L I PSE

x 2 y2 + 2 = 1 i f a 2 cos2  + b 2 si n 2 2 a b

17

ZION TUTORIALS

y = – x cot  + p cosec  Compar ing t his line wit h

y = mx + c m = – cot  and



c = p cosec 

H ence t he given line t ouches t he ellipse t hen c2 = a2m 2 + b2 p2 cosec2  = a2 cot 2  + b2



p 2 = a2 cos2  + b2 sin 2 



F a m , b I GH c c JK 2

and point of cont act is

2

F  a (cot ) , GH p cosec  2

i .e.,

Fa GH

i .e.,

2

I b p cosec  JK

O T

cos  b2 sin  , p p

U T

2

I JK

A I

S L

R

8 .1 EQUATION OF TANGENT IN DIFFERENT FORMS 1 . Point Form

N

The equat ion of t he t angent t o t he ellipse

Z

O I

x 2 y2 + (a > b) at a point (x1 y1) is a2 b2

xx 1 yy1 + 2 = 1 a2 b

T I P : Replace x2 by xx1

T I P : Replace y2 by yy1, in t he equat ion of ellipse

N OT E : The equat ion of t angent at (x1, y1) can be obt ained by r eplacing x2 by xx1, y2 by yy1, x by

x+ x1 y+ y1 xy1 + x1 y , y by and xy by . 2 2 2

This is met hod is apply only when t he equat ion of ellipse is a polynomial of second degr ee in x and y.

2 . Slope Form (a)

y = mx + c

EL L I PSE

...(1)

18

ZION TUTORIALS

meet s t he ellipse in point s whose abscissae ar e given by x2 (b2 + a2m 2) + 2mca2x + a2(c2 – b2) = 0 t he r oot s of t his equat ion ar e coincident if c = ± a 2 m 2 + b2 I n t his case t he st r aight line (1) is a t angent , and it becomes y = m x ± a 2m 2 + b2 This is t he r equir ed equat ion.

A I

S L

...(2)

Since, t he r adical sign on t he r ight -hand of (2) may have eit her + or – pr efixed t o it , we see t hat t her e ar e t wo t angent s t o t he ellipse having t he same m, i .e., t her e ar e t wo t angent s par allel t o any given dir ect ion. I t will be found t hat t he point of cont act is t he point

F GH

 a2 m 2

2

2

a m +b

,

b2

I JK

O T 2

2

a m +b

2

R

(b) By a pr oof similar t ot hat of t he last ar t icle, it may be shewn t hat t he st r aight line x cos  + y sin  = p t ouches t he ellipse, if

U T

p 2 = a2 cos2  + b2 sin 2 

(c) Similar ly, it may be shewn t hat t he st r aight line l x + my = n

t ouches t he ellipse, if

3 . Parametric Form

O I

N

a2l 2 + b2m 2 = n 2.

The coor dinat es of t he point ar e (a cos , b sin ).

Z

Subst it ut ing x´ = a cos  and y´ equat ion,

= b sin  in t he equat ion, we have, as t he r equir ed

x y cos  + si n  = 1 a b

8 .2 THE INTERSECTION OF THE TANGENTS AT THE POINTS  AND ´. The int er sect ion of t he t angent s at t he point s  and ´. The equat ions t o be t he t angent s ar e x y cos  + sin  – 1 = 0 a b

EL L I PSE

...(1)

19

ZION TUTORIALS

x y cos ´ + sin ´ – 1 = 0 a b

and

The r equir ed point is found by solving t hese equat ions. We obt ain x y 1 1 a b = = = sin   sin   cos   cos  sin   cos   cos   sin  sin (   )

i .e.,

x y 1 = =  +      +           2 a cos sin 2 b sin sin 2 sin cos 2 2 2 2 2 2

O T

1 1 ( +  ) sin ( + ) 2 2 H ence, x = a and y = b 1 1 cos (   ) cos (   ) 2 2 cos

R| |STip t o r emember : || T Illustration 11

U T

A I

S L

R

U| |V FG IJ | H K |W

(sum) (sum) sin 2 , y=b 2 x= a (diff.) diff cos cos 2 2 cos

O I

N

F i n d t h e eq u at i on s of t h e t an gen t s t o t h e el l i p se 3x 2 + 4y2 = 12 w h i ch ar e p er p en d i cu l ar t ot h e l i n e y + 2x = 4. Sol u t i on :

Z

L et m be t he slope of t he t angent , since t he t angent is per pendicular t o t he line y + 2x = 4.



m × (– 2) = – 1  m =

Since

3x2 + 4y2 = 12

or

x 2 y2 + = 1 4 3

Compar ing t his wit h

x 2 y2 = 1 + a2 b2

EL L I PSE

1 2

20

ZION TUTORIALS 2

a = 4 and b2 = 3



So t he equat ions of t he t angent s ar e

 or

y =

1 1 x ± 4 +3 2 4

y =

1 x± 2 2

x – 2y ± 4 = 0

A I

Illustration 12

R

F i n d t h e con d i t i on t h at t h e l i n e l x + m y = n m ay t ou ch t h e el l i p se p oi n t of con t act . Sol u t i on : The equat ions of t he line and t he ellipse ar e l x + my = n

U T

x 2 y2 + = 1 a2 b2

and respectively.

O T

S L

x 2 y2 + = 1. Al so, f i n d t h e a 2 b2

...(i ) ...(i i )

L et t he line (i ) t ouch t he ellipse (i i ) at (x1, y1). Then t he equat ion of t he t angent at (x1, y1) is

O I

N

xx1 yy1 + 2 = 1 a2 b

Now, (i ) and (i i ) will r epr esent t he same line if

Z Thus, (x1, y1) =

x1 / a 2 y1 / b2 1 a2 l b2 m = =  x1 = , y1 = l m n n n

Fa l , GH n 2

b2 m n

I JK

is t he point of cont act .

Since (x1, y1) lies on (i i ), t her efor e x12 y12 a2 l 2 b2 m 2 + = 1  + = 1 a2 b2 n2 n2



a2l 2 + b2m 2 = n 2

This is t he r equir ed condit ion of t angency.

EL L I PSE

...(i i i )

21

ZION TUTORIALS

Illustration 13 F i n d t h e l ocu s of t h e f oot of t h e p er p en d i cu l ar d r aw n f r om cen t r e u p on an y t an gen t t o t h e el l i p se

x 2 y2 + = 1. a 2 b2

Sol u t i on : Any t angent of

x 2 y2 + = 1 is a2 b2

y = mx + (a 2 m 2 + b2 )

A I

...(1)

S L

Equat ion of t he line per pendicular t o(1) and passing t hr ough (0, 0) is y = –

or

m = –

1 x m

x y

O T

R

...(2)

Subst it ut ing t he value of m fr om (2) in (1) t hen

U T y = –

x2 + y

Fa GH

2

x2 + b2 2 y

I JK

(x2 + y2)2 = a2x2 + b2y2



N

or changing t o polar s by put t ing x = r cos , y = r sin  it becomes

Illustration 14

Z

O I

r 2 = a2 cos2  + b2 sin 2 

Pr ov e t h at t h e t an gen t s at t h e ex t r em i t i es of l at u s r ect u m of an el l i p se i n t er sect at t h e cor r esp on d i n g d i r ect r i x . Sol u t i on :

The L SL ´ be a lat us r ect um of t he ellipse

x 2 y2 = 1. + a2 b2

F bI The co-or dinat es of L and L ´ ar e G ae, a J H K

F and G ae, H

2

F ae, b I GH a JK 2

 Equat ion of t angent at L

is

EL L I PSE



b2 a

I JK

r espect ively

22

ZION TUTORIALS

x (ae) + a2



y

FG b IJ H aK 2

b2

= 1

xe + y = a



The equat ion of t he t angent at L ´

x ( ae) + a2

FG H

y 

b2 a

b2

IJ K

...(1)

F ae, GH



b2 a

is

A I

= 1

ex – y = a



I JK

...(2)

Solving (1) and (2) we get x =

a e

and

y = 0

U T

O T

S L

R

Thus t he t angent s at L and L ´ int er sect at (a/ e, 0) which is a point lying ont he cor r esponding dir ect r ix i .e., x =

a . e

Illustration 15

N

Pr ov e t h at t h e p r od u ct of t h e p er p en d i cu l ar s f r om t h e f oci u p on an y t an gen t t o t h e el l i p se x 2 y2 + 2 = 1 i s b 2. 2 a b

Sol u t i on :

Z

O I

The equat ion of any t angent t o t he ellipse



x 2 y2 + = 1 is y = mx + a 2 m 2 + b2 a2 b2

mx – y + a 2 m 2 + b2 = 0

...(i )

The t wo foci of t he given ellipse ar e S(ae, 0) and S´(– ae, 0). L et p1 and p2 be t he lengt hs of per pendicular fr om S and S´ r espect ively on (i ). Then, p1 = L engt h of  fr om S(ae, 0) on (i ) =

mae 0 + a 2 m 2 + b2 m 2 + (1) 2

EL L I PSE

23

ZION TUTORIALS

=

mae+ a2 m 2 + b2 m2+ 1

and p2 = L engt h of  fr om S´ (– ae, 0) on (i ) =

=

Now,

p1p 2

Illustration 16

Z

O I

m 2 + (1) 2  mae+ a 2 m 2 + b2

A I

m2+ 1

S L

F mae+ a m + b I F  mae+ a m + b I =G JK GH JK m +1 m +1 H 2

2

2

2

=

N

 mae 0 + a2 m 2 + b2

O T

 m 2 a 2 e2 + a2 m 2 + b2 m2 + 1

U T

R

2

2

2

2

=

a 2 m 2 (1  e) 2 + b2 m 2 { a 2 (1  e2 )} + b2 = 1+ m2 1+ m2

=

m 2 b2 + b2 1 + m2

=

b2 (m 2 + 1) = b2 (m 2 + 1)

[·.· b2 = a2 (1 – e2)]

Sh ow t h at t h e p oi n t of i n t er sect i on of t h e t an gen t s at t w o p oi n t s on t h e el l i p se 1, w h ose eccen t r i c an gl es d i f f er b y a r i gh t an gl e l i es on t h e el l i p se

x 2 y2 + = a 2 b2

x 2 y2 + = 2. a 2 b2

Sol u t i on : Let P (a cos , b sin ) and Q (a cos , b sin ) be two points on the ellipse such t hat  –  = /2. The equat ions of t angent s at P and Q ar e x y cos  + sin  = 1 a b

EL L I PSE

...(i )

24

ZION TUTORIALS

x y cos  + sin  = 1 a b

and,

...(i i )

respectively. Since  –  = /2, so (i ) can be wr it t en as –

x y sin  + cos  = 1 a b

x 2 y2 + = 2. a2 b2

Squar ing (i i ) and (i i i ) and t hen adding, we get

A I

H ence, t he point s of int er sect ion of t angent s at P and Q lie on t he ellipse

Illustration 17

O T

R

S L

...(i i i )

x 2 y2 + = 2. a2 b2

Pr ov e t h at t h e l ocu s of t h e m i d -p oi n t s of t h e p or t i on of t h e t an gen t s t o t h e el l i p se

x 2 y2 + = a 2 b2

1 i n t er cept ed bet w een t h e ax es i s a 2y2 + b 2x 2 = 4x 2y2. Sol u t i on :

U T

L et (x1, y1) be any point on t he ellipse

x 2 y2 + = 1 a2 b2

N

The equat ion of t he t angent at (x1, y1) t o (i ) is

O I

This meet s t he coor dinat e axes at Q

Z

F a , 0I GH x JK 2

...(i ) xx1 yy1 + 2 = 1 a2 b

and R

F 0, b I . GH y JK 1

1

L et L (h, k) be t he mid-point of QR. Then,

a2 b2 b2 a2 h = , k =  x1 = , y1 = 2 x1 2 y1 2k 2h

...(i i )

Since (x1, y1) lies on t he ellipse (i ),



x12 y12 a4 b4 a2 b2 + 2 =1  + =1  + =1 2 2 2 2 2 2 a b 4h a 4k b 4h 4 k2

H ence, t he locus of (h, k) is



2

a2 b2 = 1 + 4 x 2 4 y2

a2y2 + b2x2 = 4x2y2. EL L I PSE

25

ZION TUTORIALS

9 .1 NUMBER OF TANGENTS DRAWN FROM A POINT TO AN ELLIPSE Two tangents can be dr awn fr om a poi nt to an el l i pse. The t wo t angent s ar e r eal and dist inct or coincident or imaginar y accor ding as t he given point lies out side, on or inside t he ellipse. Pr oof : L et t he ellipse be

x 2 y2 + = 1, and let P (h, k) be a point . The equat ion of any t angent a2 b2

t o t he given ellipse is y = mx ±

a 2 m 2 + b2

A I

I f it passes t hr ough (h, k), t hen

k = mh ± a 2 m 2 + b2  (k – mh)2 = a2m 2 + b2

S L

 m 2 (h 2 – a2) – 2 mkh + (k 2 – b2) = 0

R

This equat ion, being quadr at ic in m, gives t wo values of m. Cor r esponding t o each value of m t her e is a t angent t o t he ellipse.

O T

Thus, t wo t angent s can be dr awn fr om a point P(h, k) t o an ellipse. The t angent s ar e r eal and distinct , coincident or imaginary according as t he roots of quadratic in m ar e r eal and dist inct , equal or imaginar y

U T

4k 2h 2 – 4 (h 2 – a2) (k 2 – b2) > or < 0

i .e.,

a2k 2 + h 2b2 – a2 b2 > or = or < 0



2

2

h k + 2 – 1 > or = or < 0 2 a b



N

 (h, k) lies out side or on or inside t he ellipse

O I

h2 k2 + = 1. a 2 b2

9 .2 DIRECTOR CIRCLE

Z

The dir ect or cir cle is t he locus of point s fr om which per pendicular t angent s ar e dr awn t o t he ellipse. Eq u at i on of t h e D i r ect or Ci r cl e : L et t he ellipse be

x 2 y2 + = 1. The equat ion of any t angent a2 b2

t o t he ellipse is y = mx ± a 2 m 2 + b2 L et (h, k) be a point on t he dir ect or cir cle if (i ) passes t hr ough (h, k), t hen k = mh ± a 2 m 2 + b2  (k – mh)2 = a2m 2 + b2

 m 2 (h 2 – a2) – 2mhk + (k 2 – b2) = 0 EL L I PSE

...(i )

26

ZION TUTORIALS

This gives t wo values of m, say, m 1 and m 2. Cor r esponding t o each of t hese values t her e is t angent passing t hr ough (h, k) t o t he ellipse. I f t he t wo t angent s ar e per pendicular . Then, m 1 m 2 = – 1 

k 2  b2 = – 1  h 2 + k 2 = a2 + b2 h 2  a2

S L

So, t he locus of (h, k) i .e. t he equat ion of t he dir ect or cir cle is x 2 + y2 = a 2 + b 2.

Clear ly, it is a cir cle whose cent r e coincides wit h t he cent r e of t he ellipse and r adius equal t o a 2 + b2 .

1 0 . EQUATION OF NORMAL IN DIFFERENT FORM 1 0 .1 . Point Form

O T

A I

R

The r equir ed normal is the st r aight line which passes thr ough the point (x´, y´) and is per pendicular t o t he t angent , i .e., t o t he st r aight line b2 x  b2 y = – 2 x+ y a y

I t s equat ion is t her efor e,

F GH

wher e

m 

i .e.,

Z

O I

U T

y – y´ = m (x – x´)

N

b2 x  a2 y 

I JK

= – 1

m =

a2 y  b2 x 

The equat ion t o t he nor mal is t her efor e, y – y´ =

i .e.,

a2 y  (x – x´) b2 x 

x x y y  = y x 2 b2 a Or

a 2 x b2 y  = a 2 – b2 x1 y1

1 0 .2 . Parametric Form Equat ion t o t he nor mal at t he point whose eccent r ic angle is . The coor dinat es of t he point ar e a cos  and b sin . EL L I PSE

27

ZION TUTORIALS

H ence, in t he r esult of t he last ar t icle put t ing x´ = a cos  and y´ = b sin 

x  a cos  y b sin  = cos  sin  a b

it becomes

ax by – a2 = – b2. cos  sin 

i .e.,

The r equir ed nor mal is t her efor e a x sec  – by cosec  = a 2 – b 2

A I

1 0 .3 . Slope Form The equat ions of t he nor mal of slope m t o t he ellipse

at t he point s

F GH

a2 ( a2 + b2 m 2 )

Illustration 18

, 

N

I. J (a + b m ) K

U T mb2

2

2

2

R

x 2 y2 + = 1 ar e given by a2 b2

O T 2

y = mx 

S L

2

m(a  b )

(a 2 + b2 m 2 )

Pr ov e t h at t h e st r ai gh t l i n e l x + m y + n = 0 i s a n or m al t o t h e el l i p se a 2 b2 (a 2  b 2 ) 2 + = . l2 m2 n2

Sol u t i on :

Z

O I

The equat ion of any nor mal t o

x 2 y2 + = 1 if a 2 b2

x 2 y2 + = 1 is a2 b2

ax sec  – by cosec  = a2 – b2 The st r aight line l x + my + n = 0 will be a nor mal t o t he ellipse ax sec  – by cosec  = a2– b2 Since l x + my + n = 0 and (i ) r epr esent t he same line.

EL L I PSE

x 2 y2 + = 1, if a2 b2

28

ZION TUTORIALS

a sec  b cosec  a2  b2 = = 1 m n



cos  = –



cos2  +sin 2  =



an bn and sin  = l (a2  b2 ) m (a2  b2 )

l2

a2 n 2 b2 n 2 + (a 2  b2 ) 2 m ( a 2  b2 ) 2

A I

( a 2  b2 ) 2 a2 b2 = 2 + 2 2 n l m



Illustration 19

S L

R

I f t h e n or m al at an en d of a l at u sr ect u m of an el l i p se p asses t h r ou gh on e ex t r em i t y of t h e m i n or ax i s, sh ow t h at t h e eccen t r i ci t y of t h e el l i p se i s gi v en b y e4 + e2 – 1 = 0. Sol u t i on : L et

O T

x 2 y2 + = 1 be t he ellipse. The coor dinat es of an end of t he lat usr ect um ar e (a e, b2/a). a2 b2

U T

The equat ion of t he nor mal at (ae, b2/a) is

ax a 2 x b2 y  2 = a2 – b2 or – ay = a2 – b2 e ae b / a

N

I t passes t hr ough one ext r emit y of t he minor axis whose coor dinat es ar e (0, – b).



O I

ab = a2 – b2  a2b2 = (a2 – b2)2

a2 × a2 (1 – e2) = (a2 e2)2  1 – e2 = e4  e4 + e2 – 1 = 0



Z

Illustration 20

An y or d i n at e M P of an el l i p se m eet s t h e au x i l i ar y ci r cl e i n Q. Pr ov e t h at t h e l ocu s of t h e p oi n t of i n t er sect i on of t h e n or m al s at P an d Q i s t h e ci r cl e x 2 + y2 = (a + b) 2. Sol u t i on : L et P (a cos , b sin ) be any point on t he ellipse

x 2 y2 + = 1, and let Q (a cos , a sin ) be t he a2 b2

cor r esponding point on t he auxiliar y cir cle x2 + y2 = a2. The equat ion of t he nor mal at P (a cos , b sin ) t o t he ellipse is ax sec  – by cosec  = a2 – b2 The equat ion of t he nor mal at Q (a cos , a sin ) t o t he cir cle x2 + y2 = a2 is y = x t an 

EL L I PSE

...(i )

29

ZION TUTORIALS

L et (h, k) be t he point of int er sect ion of (i ) and (i i ). Then, ah sec  – by cosec  = a2 – b2

...(i i i )

k = h t an 

...(i v)

and,

Eliminat ing  fr om (i i i ) and (i v), we get ah 1 +

k2 h2  bk 1 + = a2 – b2 h2 k2

(a – b) h 2 + k 2 = a2 – b2  h 2 + k 2 = (a + b)2



A I

H ence, t he locus of (h, k) is x2 + y2 = (a + b)2

1 1 . NORMALS FROM A GIVEN POINT Equat ion of nor mal at (a cos , b sin ) t o t he ellipse

FG H



 L et t an

 2

FG1  t an  IJ H 2K

IJ byFG1 + t an  IJ K H 2K

U T 2

 2 t an 2

2

 = t 2

O I

N

R

x 2 y2 + = 1 is a2 b2

O T

ax sec  – by cosec  = a2 – b2 ax 1 + t an 2

S L

= a2 – b2

 byt 4 + 2(ax + a2 – b2) t 3 + 2(ax – a2 + b2)t – by = 0 I f t his nor mal passes t hr ough a given point (h, k), t hen bkt 4 + 2(ah + a2 – b2)t 3 + 2(ah – a2 + b2)t – bk = 0

Z

This is a four degr ee equat ion in t which means maximum four nor mals can be dr awn fr om a given point t o t he ellipse. L et t 1, t 2, t 3 and t 4 be t he r oot s of t his equat ion, so t hat , t1 + t2 + t3 + t4 = – 2

ah + a 2 e2 bk

t 1t 2 = 0 t 1t 2t 3 = – 2 and

...(i ) ...(i i )

ah  a 2 e2 bk

t1 t2 t4 = – 1

EL L I PSE

...(i i i ) ...(i v)

30

ZION TUTORIALS

We know t an

FG  +  +  +  IJ H2 2 2 2K 1

2

3

4

=

s1  s3 s s = 1 3 =  {fr om t r igo} 1  s2 + s4 0

 1 +  2 +  3 +  4 = n  + 2 2



S L

and hence 1 + 2 + 3 + 4 = (2n + 1)  = an odd mult iple of t wo r ight angles.

1 2 . CONORMAL POINTS LIE ON A FIXED CURVE

A I

L et P(x1, y1), Q(x2, y2), R(x3, y3) and S(x4, y4) be t he conor mal point s. L et nor mals at P, Q, R, S meet in T(h, k). Then equat ion of nor mal at P(x1, y1) is

O T

a2 x b2 y  = a2 – b2 x1 y1

The point T(h, k) lies on it

U T

(a2 – b2) x1y1 + b2kx1 – a2hy1 = 0

R

...(i )

Similar ly, for point s Q, R and S in equat ion (i ), x1, y1 ar e just r eplaced by (x2, y2), (x3, y3), (x4, y4) r espect ively.

 P, Q, R, S all lie on t he cur ve

N

(a2 – b2) xy + b2kx – a2hy = 0, which is called Apollonius a r ect angular hyper bola.

Illustration 21

O I

I f    ar e t h e eccen t r i c an gl es of t h r ee p oi n t s on t h e el l i p se, t h e n or m al s at w h i ch ar e con cu r r en t , t h en si n (  +  ) + si n (  + ) + si n (  + ) = 0.

Z

Sol u t i on : We know t he equat ion by . t 4 + 2t 3 (ax + a2e2) + 2t (ax – a2e2) – by = 0 and Now,







t 1t 2 = 0

...(i )

t 1, t 2 t 3 t 4 = – 1

...(i i )

t 1 t 2 = 0

t 1 t 2 + t 2 t 3 + t 3 t 1 = – t 4 (t 1 + t 2 + t 3) t +t +t t1 t2 + t2 t3 + t3 t1 = 1 2 3 t1 t 2 t 3

t1 t2 + t2 t3 + t3 t1 =

1 1 1 + + t2 t 3 t1 t3 t1 t2

EL L I PSE

LM t t t t = 1OP MM  t = t t1 t PP N Q 1 2 3 4 4

1 2 3

31

ZION TUTORIALS

 t an

      t an + t an t an + t an t an 2 2 2 2 2 2

= cot



 FGH t an 2 t an 2  cot 2 cot 2 IJK = 0



F sin  / 2 sin  / 2  cos  / 2 cos  / 2 I  GH sin  / 2 sin  / 2 cos  / 2 cos  / 2 JK = 0





2



      cot + cot cot + cot cot 2 2 2 2 2 2



2

2

2

FG cos  cos  + sin  sin  IJ FG cos  cos   sin  sin  IJ H 2 2 2 2K H 2 2 2 2 K = 0   4 sin  sin 

( + b) (  ) cos 2 2 =0 sin  sin 

 4 cos 



 2 (cos  + cos ) = 0 sin  sin 



U T

O T

A I

S L

R

 2 sin  (cos  + cos ) = 0 sin  sin 

N

 sin  (cos  + cos ) + sin  (cos  + cos ) + sin  (cos  + cos ) = 0  sin ( + ) + sin ( + ) + sin ( + ) = 0.

O I

N OT E : You can r emember t his as a pr oper t y.

1 3 .1 Equation of the chord of the ellipse whose eccentric angles are  and  :

Z

L et t he ellipse be

x 2 y2 + = 1 a2 b2

...(1)

L et P(a cos , b sin ) and Q(a cos , b sin ) be t wo point s on ellipse (1). Equat ion of chor d PQ is y – b sin  =

b(sin   sin ) (x – a cos ) a(cos   cos )

+   2 cos sin b 2 2 (x – a cos ) = a 2 sin  +  sin    2 2 EL L I PSE

32

ZION TUTORIALS

FG y b si n  IJ si n  +  H b K 2

or

=

FG x a cos  IJ cos  +  H a K 2

y +  + x + +  si n  si n  sin =  cos + cos  cos b 2 2 a 2 2

or

FG H

or

x +  y +  +  cos + sin = cos   a 2 b 2 2

or

x + y +  cos + si n = cos a 2 b 2 2

IJ K

A I

N ot e : When   , chor d PQ becomes t angent at P

S L

R

x y   Equat ion of t angent at P(a cos , b sin ) is cos  + sin  = 1 a b

1 3 .2 Equation of the chord of the ellipse x 2 y2 + = 1 a2 b2

Given ellipse is

U T ...(1)

O T

x 2 y2 + = 1 whose middle point is ( x 1 , y1 ). a 2 b2

Let P(x1, y1) be t he middle point of a chor d AB of t he ellipse (1), t hen t he r equir ed equat ion is

N

xx 1 yy1 x 2 y2 + 2  1 = 12 + 12  1 2 a b a b

or

O I

or T = S1 wher e T and S1 have usual meanings.

Z

1 3 .3 Equation of pair of tangents to an ellipse from an external point ( x 1 , y1 ). L et t he given ellipse be

x 2 y2 = 1 + a2 b2

L et P(x1, y1) be an ext er nal point .

Fx GH a

2 2

+

I Fx JK GH a

y2 1 b2

2 1 2

+

I FG JK H

IJ K

y12 xx1 yy1 1 = + 2 1 b2 a2 b

2

This can be wr it t en as SS1 = T 2 wher e S, S1 and T have usual meanings

EL L I PSE

33

ZION TUTORIALS

1 3 .4 Equation to the chord of contact of tangents drawn from a point ( x 1 , y1 ) is xx 1 yy1 + 2 = 1 a2 b

Illustration 22 F i n d t h e l ocu s of t h e p oi n t s of t h e i n t er sect i on of t an gen t s t o el l i p se an an gl e .

A I

Sol u t i on : Given ellipse is x 2 y2 + = 1 a2 b2

O T

S L

x 2 y2 + = 1 w h i ch m ak e a 2 b2

R

...(1)

Equat ion of any t angent t o ellipse (1) in t er ms of slope (m) is y = mx + (a 2 m 2 + b2 )

U T

Since it s passes t hr ough P ( ) t hen



 = m  + (a 2 m 2 + b 2 )

( – m ) = (a2 m 2 + b2 )





N

( – m )2 = a2m 2 + b2



m 2 (a2 – 2) + 2m + (b2 – 2) = 0

O I

...(2)

(2) being a quadr at ic equat ion in m. L et r oot s of equat ion (2) be m 1 and m 2 t hen





Z

m1 + m2 = –

2 b2   2 , m m = 1 2 (a2   2 ) a2   2

(m 1 – m 2) = (m1 + m2 ) 2  4 m1 m2

=

4 2 2 4(b2   2 )  2 2 2 2 (a   ) (a   2 )

422  4(b2   2 )(a2   2 ) = ( a2   2 ) 2

EL L I PSE

34

ZION TUTORIALS

4{ 22  a 2 b2 + b2  2 + a 2  2   22 } = (a 2   2 ) 2

2 = 2 | (a   2 )|

(a 2  2 + b2  2  a 2 b2 )

·.·  be t he angle bet ween t hese t wo t angent s, t hen t an  =

m1  m2 1 + m1 m 2

A I

S L

R

2 ( a2 2 + b2  2  a2 b2 ) (a   2 ) = b2   2 1+ 2 a  2 2

F GH

U T

t an  =

O T

I JK

2 ( a2 2 + b2  2  a2 b2 ) a 2 + b2   2  2

or

(a2 + b2 – 2 – 2)2 t an 2  = 4(a22 + b22 – a2b2)



(2 + 2 – a2 – b2)2 t an 2  = 4(b2 2 + a22 – a2b2)

 L ocus of P( ) is

O I

N

(x2 + y2 – a2 – b2)2 t an 2  = 4(b2x2 + a2y2 – a2b2)

Illustration 23

Z

Pr ov e t h at t h e ch or d of con t act of t an gen t s d r aw n f r om t h e p oi n t (h , k ) t o t h e el l i p se h2 k2 1 1 x 2 y2 + 4 = 2 + 2 . Al so, f i n d t h e l ocu s + = 1 w i l l su b t en d a r i gh t an gl e at t h e cen t r e, i f 4 2 2 a b a b a b

of (h , k ). Sol u t i on : The equat ion of chor d of cont act of t angent s dr awn fr om P(h, k) t o t he ellipse hx ky + = 1 a 2 b2

EL L I PSE

x 2 y2 + = 1 is a2 b2

...(1)

35

ZION TUTORIALS

The equat ion of t he st r aight lines CA and CB is obt ained by making homogeneous ellipse

x 2 y2 + = a2 b2

1 wit h t he help of (1)

FG H

hx ky x 2 y2 + 2 = 2+ 2 2 a b a b





Fk GH a

2 4



I JK

F GH

IJ K

2

I JK

1 k2 1 2 hk x 2 + 4  2 y2 + 2 2 xy = 0 2 a b b a b

But given ACB = 90°  co-efficient of x2 + co-efficient of y2 = 0

A I

h2 1 k2 1  +  = 0 a 4 a2 b4 b2



1 1 h2 k2 + 4 = 2+ 2 4 a b a b

 H ence locus of (h, k) is

1 1 x 2 y2 + 4 = 2+ 2 4 a b a b

Illustration 24

U T

O T

S L

...(2)

R

Sh ow t h at t h e l ocu s of t h e m i d d l e p oi n t s of ch or d s of an el l i p se, w h i ch p ass t h r ou gh a f i x ed p oi n t , i s an ot h er el l i p se. Sol u t i on :

O I

N

L et P(x1, y1) be t he middle point of any chor d AB of t he ellipse

Z

AB is T = S1.





x 2 y2 + = 1 t hen equat ion of chor d a2 b2

xx1 yy1 x 2 y2 + 2 – 1 = 12 + 12  1 2 a b a b xx1 yy1 x 2 y2 + 2 = 12 + 12 2 a b a b

But it passes t hr ough a fixed point Q(h, k), it s coor dinat es must sat isfy equat ion (i ),



hx1 ky1 x12 y12 + + = a2 b2 a2 b2

t his can be r e-wr it t en as

EL L I PSE

...(i )

36

ZION TUTORIALS

FG x  h IJ FG y  k IJ H 2 K + H 2K 2

1

2

1

a

2

2

b

F GH

I JK

F GH

I JK FG h , k IJ H 2 2K

=

1 h2 k2 + 4 a 2 b2

=

1 h2 k2 + 4 a 2 b2

H ence locus of P(x1, y1) is

FG x h IJ FG y k IJ H 2K + H 2K 2

a

2

2

2

b

which is obviously an ellipse wit h cent r e at

A I

S L

and axes par allel t o coor dinat e axes.

Illustration 25

R

I f t h r ee of t h e si d es of a q u ad r i l at er al i n scr i b ed i n an el l i p se ar e p ar al l el r esp ect i v el y t o t h r ee gi v en st r ai gh t l i n es. Sh ow t h at f ou r t h si d e w i l l al so b e p ar al l el t o a f i x ed st r ai gh t l i n e. Sol u t i on :

U T

O T

L et PQRS be a quadr ilat er al inscr ibed in t he ellipse

Z

O I

N

x 2 y2 + = 1. a2 b2

L et PQ, QR and RS be t he t hr ee sides par allel t o t he given lines. Equat ion of PQ is

FG H

IJ K

FG H

x + y + cos + si n a 2 b 2

I t s slope is –

b cot a

FG  +  IJ H 2 K

IJ K

= cos

FG    IJ H 2 K

which is const ant by hypot hesis



 +  = const ant = 21 (say)

Similarly

 +  = const ant = 22 (say)

and

 +  = const ant = 23 (say) EL L I PSE

37

ZION TUTORIALS

Now t he equat ion of SP is

FG H

IJ K

FG H

x + y + cos + si n a 2 b 2

I t s slope

IJ K

= cos

m = –

But

FG    IJ H 2 K FG H

b + cot a 2

IJ K

 +  = ( + ) – ( + ) + ( + ) = 21 – 22 + 23 = const ant

S L

H ence, t he slope of t he four t h side PS is const ant . H ence t he four t h side is also par allel t o a fixed st r aight line.

A I

Illustration 26

R

F i n d t h e l ocu s of t h e m i d -p oi n t of n or m al ch or d s of t h e el l i p se Sol u t i on :

O T

x 2 y2 + = 1. a 2 b2

L et (h, k) be t he mid-point of a nor mal chor d of t he given ellipse. Then, it s equat ion is

U T

hx ky h2 k2 + – 1 = + 1 a 2 b2 a 2 b2

h2 k2 hx ky + + = a 2 b2 a 2 b2

or

N

[Using T = S´]

...(i )

I f (i ) is a nor mal chor d, t hen it must be of t he for m

O I

ax sec  – by cosec  = a2 – b2

Z





cos  =

h2 k2 + h k a 2 b2 = = a3 sec   b3 cosec  a2  b2

F GH

I JK

F GH

a3 h2 k2  b3 h2 k2 + , sin  = + h (a 2  b2 ) a2 b2 k ( a2  b2 ) a 2 b2

Eliminat ing  fr om t he above r elat ions, we get

F GH

a6 h2 k2 + h 2 ( a2  b2 ) a2 b2



...(i i )

I JK

Fa GH h

2

6 2

F GH

I JK

+

b6 h2 k2 + =1 k 2 (a 2  b2 ) 2 a2 b2

+

b6 k2

IF h JK GH a

2 2

+

k2 b2

I JK

2

= a2 – b2

EL L I PSE

I JK

38

ZION TUTORIALS

H ence, t he locus of (h, k) is

Fa GH x

6 2

+

b6 y2

IF x JK GH a

2 2

+

y2 b2

I JK

2

= (a2 – b2)2

1 3 . DIAMETER

S L

D EF I N I T I ON : The l ocus of the mi d-poi nt of a system of par al l el chor ds of an el l i pse i s cal l ed a di ameter and the chor ds ar e cal l ed i ts doubl e or di nates. The point wher e t he diamet er int er sect s t he ellipse is called t he ver t ex of t he diamet er .

A I

The equat ion of t he diamet er bisecting t he chor ds of slope m of t he ellipse

R

x 2 y2 b2 + = 1 is y = x a2 b2 a 2m

CON J U GAT E D I AM T ERS : Two diamet er s of an ellipse ar e said t o be conjugat e diamet er s if each bisect s t he chor ds par allel t o t he ot her .

O T

L et y =m 1 x and y = m 2 x be conjugat e diamet er s wit h r espect t ot he ellipse

x 2 y2 + = 1. Then, y a2 b2

= m 2 x bisect s t he syst em of chor ds par allel t o y = m 1x. So, it s equat ion is

U T y = –

b2 x a 2 m1

...(i )

Clear ly, (i ) and y = m 2 x r epr esent t he same t ime.



O I

N

 b2  b2  m m = 1 2 a2 m1 a2

Thus, t wo st r aight lines y = m 1 x and y = m 2 x ar e conjugat e diamet er s of t he ellipse

Z

1, if m 1 m 2 =

 b2 a2

x 2 y2 + = a2 b2

R E M A R K : I n an el l i pse, t he major axi s bi sect s al l chor ds par al l el t o t he mi n or axi s and vi ce-ver sa, t h er efor e major an d mi nor axes of an el l i pse ar e conjugat e di amet er s of t he el l i pse but t h ey do not sat i sf y t he con di t i on m 1 m 2 = – b2/a 2 ar e t he onl y per pendi cul ar conju gat e di amet er s.

1 3 .1 . Properties of Conjugate Diameters PROPERT Y 1. The eccent r ic angles of t he ends of a pair of conjugat e diamet er s of an ellipse differ by a r ight angle.

EL L I PSE

39

ZION TUTORIALS

PROOF : L et PC P´ and D CD´ be a pair of conjugat e diamet er s of an ellipse

x 2 y2 + = 1, and a2 b2

let t he eccent r ic angles of t he ext r emit ies P and D be  and  r espect ively. Then t he coor dinat es of P and D ar e (a cos , b sin ) and (a cos , b sin ) r espect ively. b t an  a

Now,

m 1 = Slope of CP =

and,

m 2 = Slope of CD = –

b t an  a

A I

Since t he diamet er s PC P´ and DCD´ ar e conjugat e diamet er s. m 1m 2 = – b2/a2



  

Z

O I

N

U T

O T

S L

R

b b  b2 t an  × t an  = 2 a a a

t an  t an  = – 1  t an  = – cot  t an  = t an

FG  + IJ H2 K

  =

  +    –  = 2 2

PROPERTY I I : The sum of the squares of any two conjugate semi-diameters of an ellipse is contant and equal to the sum of the squares of the semi-axes of the ellipse i.e. CP2 + CD2 = a2 + b2. PROOF : L et CP and CD be t wo conjugat e semi-diamet er s of an ellipse

x 2 y2 + = 1, and let t he a2 b2

eccent r ic angle of P be . The eccent r ic angle of D is /2 + . So, t he coor dinat es of P and D ar e (a cos , b sin ) and (a cos (/2 + ), b sin (/2 + )) i .e., (– a sin , b cos ) r espect ively. EL L I PSE

40

ZION TUTORIALS 2

CP + CD 2 = (a2 cos2  + b2 sin 2 ) + (a2 sin 2  + b2 cos2 )



= a2 + b2 PROPERT Y I I I : The pr oduct of t he focal dist ances of a point on an ellipse is equal t o t he squar e of t he semi-diamet er which is conjugat e t o t he diamet er t hr ough t he point . POOF : L et PCP´ and DCD´ be t he conjugat e diamet er s of an ellipse and t he given point be P whose eccent r ic angle is . Then t he coor dinat es of P and D ar e (a cos , b sin ) and (– a sin , b cos ) r espect ively. L et S and S´ be t wo foci of t he ellipse. Then,

S L

SP . S´ P = (a – ae cos ) (a + ae cos ) = a2 – a2 e2 cos2  2

2

2

2

= a – (a – b ) cos 

2

[·.· b = a2 (1 – e2)]

A I

= a2 sin 2  + b2 cos2  = CD 2

PROPERT Y I V : The t angent s at t he ends of a pair of conjugat e diamet er s of an ellipse fr om a par allelogr am.

R

PROOF : L et PCP´ and DCD´ be a pair of conjugat e diamet er s of t he ellipse

O T

eccent r ic angle of P be . Then t he eccent r ic angle of D is

x 2 y2 + = 1. L et t he a2 b2

 + . So t he coor dinat es of P and D 2

ar e (a cos , b sin ) and (a cos (/2 + ), b sin (/2 + )) = (– a sin , b cos ) r espect ively. The coor dinat es of P´ and D´ ar e (– a cos , – b sin ) and (a sin , – b cos ) r espect ively.

U T

Equat ion of t angent s at P, D, P´ and D´ ar e r espect ively. x y x y cos  + sin  = 1, sin  – cos  = – 1 a b a b

N

x y x y cos  + sin  = – 1 and sin  – cos  = 1 a b a b

O I

Clear ly, t he t angent s at P ar e P´ ar e par allel. Also, t he t angent s at D and D´ ar e par allel. H ence, t he t angent s at P, D, P´, D´ for m a par allelogr am. PROPERT Y V : The ar ea of t he par allelogr am for med by t he t angent s at t he ends of conjugat e diamet er s of an ellipse is const ant and is equal t o t he pr oduct of t he axes.

Z

PROOF : Ar ea of par allelogr am T 1 T 2 T 3 T 4 = 4 (Ar ea of par allelogr am CPT 2D) = 4 (2 Ar ea of  CDP) = 8 (Ar ea of CPD) 1 = 8× 2

0 0 1 a cos  b sin  1 = 4 (ab cos2  + ab sin 2 )  a sin  b cos  1

= 4 ab = 2a × 2b = Pr oduct of t he axes of t he ellipse.

EL L I PSE

41

ZION TUTORIALS

1 3 .2 . The length of the subtangent and subnormal L et t he t angent and nor mal at P, t he point (x´, y´), meet t he axis in T and G r espect ively, and let PN be t he or dinat e of P.

P

The equat ion t o t he t angent at P is xx  yy  + 2 = 1 a2 b

...(1)

To find wher e t he st r aight line meet s t he axis, we

A I

put y = 0 and have x =

a2 a2 , i .e., CT = x CN

CT . CN = a2 = CA 2

i .e.,

O T

a2 a 2  x 2  x = H ence, t he subt angent NT = CT – CN = x x

The equat ion t o t he nor mal is

U T

R

S L ...(2)

x x y y  = x y 2 a b2

N

To find wher e it meet s t he axis, we put y = 0, and have

i .e.,

Z

O I

x  x   y = = – b2 x y a2 b2

CG = x = x´ –

b2 a2  b2 x´ = x´ = e2 . x´ = e2 . CN a2 a2

...(3)

H ence,t he subnor mal NG = CN – CG = (1 – e2) CN NG : NC :: 1 – e2 : 1

i .e.,

:: b2 : a2 Som e p r op er t i es of El l i p se 1.

SG = e.SP, and t he t angent and nor mal at P bisect t he ext er nal and int er nal angles bet ween t he focal dist ances of P.

2.

I f SY and S´Y´ be t he per pendicular s fr om t he foci upon t he t angent at any point P of t he ellipse, t hen Y and Y´ lie on t he auxiliar y cir cle, and SY . S´Y´ = b2. Also CY and S´P ar e par allel. EL L I PSE

42

ZION TUTORIALS

3.

I f t he nor mal at any point P meet t he major and minor axes in G and g, and if CF be t he per pendicular upon t his nor mal, t hen PF . PG = b2 and PF . Pg = a2.

4.

The locus of t he feet of t he per pendicular s fr om t he foci on any t angent t o an ellipse is t he auxiliar y cir cle.

1 4 . REFLECTION PROPERTY OF AN ELLIPSE

S L

I f an incoming light r ay passes t hr ough one focus (S) st r ike t he concave side of t he ellipse t hen it will get r eflect ed t owar ds ot her focus (S´). and SPS´ = SQS´

A I

Illustration 27

U T

O T

R

A r ay em an at i n g f r om t h e p oi n t (–3, 0) i s i n ci d en t on t h e el l i p se 16x 2 + 25y2 = 400 at t h e p oi n t P w i t h or d i n at e 4. F i n d t h e eq u at i on of t h e r ef l ect ed r ay af t er f i r st r ef l ect i on . Sol u t i on :

N

For point P, y-co-or dinat e = 4

O I

·.· given ellipse is

16x2 + 25y2 = 400

Z

16x2 + 25(4)2 = 400

co-or dinat e of P is (0, 4)

e2 = 1 –

e=

16 9 = 25 25

3 5

foci (± ae, 0) i .e., (± 3, 0) Equat ion of r eflect ed r ay (i .e., PS) is x y + = 1 or 4x + 3y = 12. 3 4 EL L I PSE

44

ZION TUTORIALS

1 5 . CONCYCLIC POINTS Any cir cle int er sect s an ellipse in t wo or four r eal point s. They are called concyclic points and the sum of t heir eccentr ic angles is an even mult iple of . I f     be t he eccent r ic angles of t he four concyclic point s on an ellipse, t hen  +  +  +  = 2n  wher e n is any int eger .

A I

Illustration 28

S L

L et P b e a p oi n t on t h e el l i p se w h ose f oci ar e S´ an d S. I f n or m al at P m eet s t h e ax es at G an d G´ t h en i i (i ) S´G = e S´P i (i i ) S´G : SG = S´P : SP (i i i ) PG b i sect s t h e an gl e S´PS Sol u t i on :

U T

O T

R

I f P be (x´, y´) t hen equat ion of PGG´ is ax sec  – by cosec  = a2 – b2 a 2  b2 a2  b2 On put t ing y = 0 we get OG = = a cos  =  e2 x1 a sec  a2

N

Now S´G = S´O + OG = ae + e2x´ = e(a + ex´) = e(S´P)

Z

O I

Also SG = OS – OG = ae – e2x´ = e(a – ex´) = e . SP H ence S´G : SG = S´P : SP (* ) The last r elat ion clear ly implies S´PG = GPS. Thus nor mal at any point P bisect s t he angle bet ween S´P and SP wher e S´ and S ar e t wo foci.

EL L I PSE

45

ZION TUTORIALS

Illustration 29 I f t h e n or m al at an y p oi n t P m eet s t h e m aj or an d m i n or ax es i n G an d G´ an d i f OF b e t h e p er p en d i cu l ar u p on t h i s n or m al t h en p r ov e t h at (i ) PF . PG = b 2

(i i ) PF . PG´ = a 2

A I

Sol u t i on :

O T

L et P be (a cos , b sin ) t hen equat ion of t angent at P is

U T

S L

R

x y cos  + sin  = 1. a b

Now PF = OL = dist ance of or igin fr om t he t angent =

1 cos2  sin 2 + a2 b2

=

ab

a 2 sin 2  + b2 cos2

O I

N

1 2 = e 1 Also t he nor mal at P is ax sec 2 e+ 1 t an 2

2

 PG

Z

Fa b GH a 2

 G is

2

FG IJ H K

2

– by cosec  = a2 – b2  OG =

a2  b2 cos  a

I JK

cos , 0

F a b = G a cos   a H

 PG =

t an

2

2

I cos J K

2

+ b2 sin 2  =

b2 (a2 sin 2  + b2 cos2 ) a2

a 2 b cos2  + a 2 sin 2 . Whence it follows t hat PF . PG = b2 b

F GH

Again on put t ing x = 0 in t he equat ion of t he nor mal we get G´ as 0,

EL L I PSE

a 2  b2 sin  a

I JK

46

ZION TUTORIALS 2

2

2

 PG´ = a cos

 PG´ =

F a b  + G b sin  + a H 2

2

I sin J K

2

=

a2 2 b cos2 + a 2 sin 2  b2

d

i

a 2 b cos2  + a2 si n 2  whence it follows t hat PF . PG´ = a2. b

Illustration 30

S L

I f SY an d S´ Y´ b e t h e p er p en d i cu l ar f r om t h e f oci S an d S´ u p on t h e t an gen t at an y p oi n t P of t h e el l i p se t h en sh ow t h at

A I

(a ) Y an d Y´ l i e on au x i l i ar y ci r cl e (b) SY . S´ Y´ = b 2 (c) OY an d S´P ar e p ar al l el . Sol u t i on : L et P be (a cos , b sin ) t hen equat ion of t angent at P is

Z

O I

O T

R

x y cos  + sin  = 1 a b

N

U T

...(i )

cos  Slope of t angent = – a (x – ae) sin  b

sin  Equat ion of SY must be y – 0 = b (x – ae) cos  a

Rewr it ing equat ion (i ) and (i i ) as bx cos  + ay sin  – ab = 0 and ax sin  – by cos  – a2e sin  = 0

EL L I PSE

...(i i )

47

ZION TUTORIALS

On solving for x and y we get t he co-or dinat es of Y as x =

a(a2 e sin 2 + b2 cos ) a2 b sin  + (1  e cos ) , y = a2 sin 2 + b2 cos2 a 2 sin 2 + b2 cos2

The x-coor dinat e can be simplified t o

F GH

x =

F GH

a 2 sin 2 +

I JK I cos J K

b2 cos  a2

a3 e sin 2  +

2

b a2

2

A I

d i = a d1  cos  + cos   e cos  i 2

R

2

a e e cos  + (1  e )cos 

2

=

2

U T

a2 b sin  (1  e cos )

F b a G sin  + a H 2

N

O T

2

2

F GH

a(1  e cos )( e+ cos ) a( e+ cos ) = 1 + e cos  (1  e2 cos2 )

They y-coor dinat e can be simplified t o

y =

2

S L

2

2 2

I cos J K

=

2

b sin  1  e cos 

b2 = 1  e2 a2

I JK

(* )

(* * )

Not e car efully t hat co-or dinat es of Y´ will be obt ained by simply r eplacing e by – e.i .e. Y´ must be

O I

FG a( e+ cos ) , b sin  IJ H 1  e cos  1  e cos  K

Z

To pr ove t hat Y and Y´ lie on auxiliar y cir cle it is sufficient t o show t hat t heir co-or dinat es sat i sfy x2 + y2 = a2. I ndeed (for Y) x2 + y2 =

a2 (e+ cos ) 2 + b2 sin 2 (1 + e cos ) 2

=

a2 e2 + a 2 cos2 + 2 a2 e cos  + a 2 (1  e2 )sin 2  (1 + e cos ) 2

=

a2 [ e2 cos2  + 1 + 2 e cos  ] = a2 (1 + e cos ) 2

EL L I PSE

48

ZION TUTORIALS

As similar pr oof can be given on show t hat Y´ also sat isfies x2 + y2 = a2. (b) Now SY =  dist ance of S fr om t angent =

S´Y´ =

SY S´Y´ =



e cos   1 cos2 sin 2  + a2 b2

 e cos   1

A I

1  e2 cos2  a2 b2 (1  e2 cos2 ) = cos2  sin 2 b2 cos2  + a2 sin 2  + 2 a2 b

O T

b2 (1  e2 cos2 ) = b2 = 2 b cos2  + sin 2 a2

(c)

U T

We have slope OY =

Slope S´P =

Illustration 31

O I

N

S L

(r eplacing e by – e)

cos2 sin 2  + a2 b2

R

F GH

b2 = 1  e2 2 a

I JK

b si n  a (e+ cos )

b sin   0 b sin  = a cos  + ae a( e+ cos )

= slope OY  OY  S´P.

Pr ov e t h at t h e su m of sq u ar es of t h e p er p en d i cu l ar on an y t an gen t f r om t w o p oi n t s on t h e

Z

m i n or ax i s each d i st an t Sol u t i on :

Any t angent t he ellipse

2 a 2  b 2 f r om cen t r e i s 2a .

x y x 2 y2 + 2 = 1 is cos  + sin  = 1 2 a b a b

I f p1, p2 be t he lengt hs of per pendicular s fr om

p1 =

FH0,

a2  b2

IK

and

FH0,

 a2  b2

a 2  b2 a 2  b2 sin   1   sin   1 b b , p2 = cos2  sin 2  cos2  sin 2  + + 2 a2 b2 a2 b EL L I PSE

IK

t hen

49

ZION TUTORIALS

F a  b sin  + 1I GH b JK F cos  + sin  I GH a b JK 2

2

p12 + p22 =



2

2

2

2

2

2

Fa GH b

2

=

2

2

2

sin 2   sin 2  + cos2  + sin 2  cos2  sin 2  + 2 a2 b

S L

I 2 a F sin  + cos  I JK GH b a JK = = 2a 2

2

2

2

2

cos2  sin 2  + 2 a2 b

A I

Illustration 32

R

2

I f PSQ an d PS´R ar e t w o f ocal ch or d s of an el l i p se an d t h e eccen t r i c an gl es of t h e p oi n t s

O T

1 2 = e1 Q an d R ar e 1 an d 2. Sh ow t h at f or al l su ch P  e+ 1 t an 2 2

FG H

t an

U T

IJ K

2

.

Sol u t i on : L et P be (a cos , b sin ). Then t he equat ion of PQ is x  + 1 y  + 1   2 cos + sin = cos 1 a 2 b 2 2

N

 + 1   2 = cos 1 2 2

Since t his passes t hr ough S (ae, 0), we have e cos







FG H

e cos

O I

IJ = cos  cos  + sin  sin  K 2 2 2 2   I   F eG 1  t an t an J = FG 1 + tan t an IJ H K H 2 2 2 2K

    cos 1  sin sin 1 2 2 2 2

Z

t an

1

1

1

1

  e 1 t an 1 = 2 2 e+ 1

(* )

On r eplacing e by – e and 1 by 2 we get t an

   e 1 t an 2 = 2 2  e+ 1

1 2 = e 1 On dividing (* ) by (* * ) we get 2 e+ 1 t an 2 t an

FG H

(* * )

IJ K

2

EL L I PSE

50

ZION TUTORIALS

Illustration 33 I f PSQ an d PH R b e t w o f ocal ch or d s of t h e el l i p se w h ose t w o f oci ar e S an d H . I f P i s ‘’sh ow t h at eq u at i on of ch or d QR i s

x y 1+ e2 . cos  + si n  + 1 = 0 a b 1  e2

Sol u t i on :

S L

L et Q be (a cos , b sin ) and R be (a cos , b sin ) t hen t he equat ion of t he chor d QR must x + y +  + sin = cos be cos a 2 b 2 2

Expanding and dividing by cos

FG H

  cos we get 2 2

IJ FG K H

x   y   1  t an t an + t an + t an a 2 2 b 2 2

Now fr om pr evious example, t an

t an

U T

O I

N

 = 2

F I F I GG1  1 JJ x + y GG e 1 + e+ 1 JJ GH t an 2 JK a b GH (e+ 1) t an 2 (e 1) t an 2 JK or

Z

FG t an H

FG e 1 IJ 1 , t an  = FG e+ 1 IJ 1 H e+ 1 K t an  2 H e 1 K t an  2

2

  , t an in t he equat ion of t he chor d QR we get 2 2

On subst it ut ing t an

2

O T

e+ 1   t an = e 1 2 2 t an



R

IJ = FG1 + tan  t an  IJ K H 2 2K

  e 1 t an = 2 2 e+ 1

A I

IJ K

 1 x y 2( e2 + 1) 2 . +  a b t an  e2  1 tan 2 2 2 2

F GH

I JK

F I 1 J G = G1 + GH t an 2 JJK 2

 +1 2 =  t an 2 2 tan 2

On dividing by t he expr ession on RH S and mult iplying by – 1 we get t he equat ion of t he chor d QR as

F GH

y 1 + e2 x cos  + b 1  e2 a

I JK

sin  + 1 = 0 EL L I PSE

51

ZION TUTORIALS

Illustration 34 PSQ an d PH R b e t w o f ocal ch od s of an el l i p se w h ose t w o f oci ar e S an d H . Sh ow t h at t h e t an gen t s at Q an d R an d n or m al at P ar e con cu r r en t . Sol u t i on : Following t he last t wo solved examples and it s r esult t an

 = 2

e 1 ( e+ 1)t an

Tangent s at Q and R will int er sect at T

 2

, t an

 e+ 1 = 2 ( e 1)t an  2

F a cos  +  GG 2 ,  GH cos 2

+ 2  cos 2

b sin

O T

I JJ JK

A I

S L

R

I t is sufficient t o show t hat T lies on t he nor mal at P i.e. T sat isfies t he equat ion ax sec  – by cosec  = a2 – b2 Now for t he coor dinat es of T,

U T

+ F a cos  +  I b sin G J 2 2  b cosec  L .H .S. = a sec  G   J  GH cos 2 JK cos 2

O I

N

expanding and dividing by cos

Z

put t ing values of t an

  cos 2 2

= a2

F 1  t an  t an  I G 2 2J b sec  G  J GH 1 + t an 2 t an 2 JK

2

F t an  + t an  I G 2 2 J cosec  G  GH 1 + tan 2 t an 2 JJK

  and t an as calculat ed in pr evious illust r at ion 2 2

1 2

= a sec  .

1+

1 t an 2 1

 2  b2 (cosec 

 t an 2 2

EL L I PSE

FG e 1 + e+ 1 IJ  H e+ 1 e 1 K t an 1

2

1+

1 t an 2

 2

52

ZION TUTORIALS

2

2

= – a – b

 F GG sin  = 2 t an 2 , cos  = 1  t an  GH 1 + t an 1 + t an 2

e2 + 1 e2  1

2

a 2  b2 +1 2 = – a2 – b2 . 2a 2 a b 1 a2

A I

 2 2  2 2

I JJ JK

S L

2 a 2  b2 = – a2 + (2a2 – b2) = a2 – b2 = R.H .S.  b2 Thus t he t angent s at Q and R and nor mal at P must be concur r ent .

= – a2 – b2 .

Illustration 35

O T

R

Sh ow t h at t h e ar ea of a t r i an gl e i n scr i b ed i n a el l i p se b ear s a con st an t r at i o t o t h e ar ea of t r i an gl e f or m ed b y j oi n i n g p oi n t s on t h e au x i l i ar y ci r cl e cor r esp on d i n g t o t h e v er t i ces of t h e f i r st t r i an gl e. Sol u t i on :

U T

L et P(a cos 1, b sin 1), Q(a cos 2, b sin 2) and R(a cos 3, b sin 3) be t he ver t ices of t he t r iangle inscr ibed in t he ellipse

x 2 y2 + = 1. The points on t he auxiliar y circle cor responding to t hese point s a2 b2

ar e P´(a cos 1, a sin 1) Q´(a cos 2, a sin 2) and R´(a cos 3, a sin 3).



Z and,

N

1 1 = Ar ea PQR = 2

O I

1 = ab 2

2 = Ar ea of P´Q´R´ =

1 2

1 = a2 2

Clear ly,

a cos  1 a cos  2 a cos  3 cos 1 cos  2 cos  3

a cos  1 a cos  2 a cos  3

b sin 1 1 b sin  2 1 b sin  3 1 sin 1 1 sin  2 1 sin  3 1 a sin  1 1 a sin  2 1 a sin  3 1

cos 1 cos 2

sin 1 1 sin 2 1

cos 3

sin 3 1

1 b = = const ant . 2 a EL L I PSE

53

ZION TUTORIALS

Illustration 36 I f t h e ch or d j oi n i n g t w o p oi n t s w h ose eccen t r i c an gl es ar e  an d  , cu t t h e m aj or ax i s of an el l i p se at a d i st an ce c f r om t h e cen t r e, sh ow t h at   ca = t an 2 2 c+a

t an Sol u t i on :

S L

The equat ion of t he chor d joining point s whose eccent r ic angle ar e  and  on t he ell ipse

A I

x 2 y2 + = 1, is a2 b2

FG H

IJ K

FG H

x + y + cos + si n a 2  2

IJ K

F    IJ = cos G H 2 K

O T

This will cut t he major axis at t he point (c, 0), if

FG H

c + cos a 2

IJ K

U T

FG  +  IJ + cos FG    IJ H 2 K H 2 K F  +  IJ  cos FG    IJ cos G H 2 K H 2 K cos





Z

O I

N

= cos

=

FG    IJ H 2 K

R



FG  +  IJ H 2 K=a F    IJ c cos G H 2 K cos

a+ c (by componendo & dividendo) a c

a+ c   c a 2 cos  / 2 cos  / 2  t an t an = = a c 2 2 c+ a 2 sin  / 2 sin  / 2

EL L I PSE

54

ZION TUTORIALS

SOME MORE SOLVED EXAMPLES Illustration 1 As el l i p se w h ose sem i -ax es ar e a an d b t ou ch es t h e ax i s of x at t h e or i gi n p r ov e t h at l ocu s of i t s cen t r e i s x 2y2 + (y2 – a 2) (y2 – b 2) = 0. Sol u t i on : L et C() be cent r e of t he ellipse whose semi-axis ar e a and b which t ouches x-axis at or igin. Then y-axis will be nor mal at O. Suppose t his nor mal is met by major and minor axes at G and G´ r espect ively. Dr aw CF  nor mal. L et t he equat ion of t he major and minor axes of t he ellipse be y –  = m (x )

...(i )

O T

1 y –  = – (x – ) ...(i i ) m

wher e m is a par amet er , t hen OG =  – m ,

U T

OG´ =  +

 m

A I

S L

R

OG . OF = b 2, OG´ . OF = a 2

We know

N

( – m ) = b2,



FG +  IJ  = a H mK

2

On eliminat ing m we get 2 2 + (2 – a2) (2 – b2) = 0

O I 2

 locus of cent r e C is x y + (y2 – a2) (y2 – b2) = 0.

Illustration 2

Z

2

L et d b e t h e p er p en d i cu l ar d i st an ce f r om t h e cen t r e of t h e el l i p se

x 2 y2 + = 1 t o t h e t an gen t a 2 b2

d r aw n at a p oi n t P on t h e el l i p se. I f F 1 an d F 2 ar e t w o f oci on t h e el l i p se t h en sh ow t h at (PF 1 – PF 2) 2 = 4a 2

F1  b I GH d JK 2

[I I T 1995]

2

Sol u t i on : I f P be (x1, y1) t hen equat ion of t angent at P must be

EL L I PSE

xx1 yy1 + 2 = 1 a2 b

55

ZION TUTORIALS

| 1|

d =



x12 y12 + a 4 b4



x12 y12 1 = + d 2 a 4 b4

x12 y12 + = 1 a2 b2

Also

On eliminat ing y12 bet ween (* ) and (* * ), we get

F GH

x12 1 x12 1 = + 1  d 2 a 4 b2 a2



F GH

I JK

=

FG H

IJ K

x12 1 1 1  2 + 2 4 2 a a b b

I JK

F GH

I JK

A I

x12 e2 x12 b2 x12 b2 b2 b2 =  1 + 1  1  = 1  = d 2 a2 a2 d2 a2 d2 a2

O T

Now (PF 1 – PF 2)2 = ((a + ex1) – (a – ex1)) = 4e2 x12 = 4a2

Illustration 3 A t an gen t t o t h e el l i p se

U T

R

F1  b I GH d JK 2

2

(* )

S L

(* * )

(* * * )

fr om (* * * )

x 2 y2 x 2 y2 + = 1 m eet s t h e el l i p se + = a + b i n t h e p oi n t s P an d Q. a b a 2 b2

N

Pr ov e t h at t h e t an gen t s at P an d Q ar e at r i gh t an gl es. Sol u t i on :

O I

x2 y2 The second ellipse is 2 + 2 = 1 a b

Z

(* )

wher e a´2= a(a + b) and b´2 = b(a + b)

To pr ove t he asser t ion it is sufficient t o show t hat point of int er sect ion of t angent s at P and Q (say (h, k)) of t he fir st ellipse lies on dir ect or cir cle of (* ) i .e., we must show t hat h 2 + k 2 = a´2 + b´2= (a + b)2

EL L I PSE

(* * )

56

ZION TUTORIALS

Now PQ is a chor d of cont act which has ar isen when t angent s ar e dr awn fr om (h, k) t o t he ellipse x2 y2 + = 1 a  2 b 2

 Equat ion of PQ is

xh yk b2 h b 2 + 2 = 1 or y = – 2 + 2 a b k a k

But t he last line is a t angent t o t he ellipse

x 2 y2 + = 1 a2 b2

Ther efor e c2 = a2m 2 + b2 (wit h usual meanings)



A I

4 2 b2 ( a+ b) 2 h 2 b 4 b2 (a+ b) 2 2 b h 2 a2 . 2 + b2 = a . + b  = 2 2 2 4 2 2 a ( a+ b) k k a k k

 h 2 + k 2 = a2 + b2 which pr oves (* * ).

Illustration 4

O T

S L

R

Con si d er t h e f am i l y of ci r cl es x 2 + y2 = r 2 (2 < r < 5). I f i n t h e f i r st q u ad r an t t h e com m on t an gen t t o a ci r cl e of t h i s f am i l y an d t h e el l i p se 4x 2 + 25y2 = 100 m eet s t h e co-or d i n at e ax es at A an d B t h en f i n d t h e l ocu s of t h e m i d -p oi n t of AB . [I I T 1999] Sol u t i on : x 2 y2 The elipse is + = 1 25 4

N

U T

Any t angent t o t his ellipse may be t aken as y = mx ± 25 m 2 + 4

O I

(* )

I n or der t hat (* ) lies in t he fir st quadr ant we must t ake m < 0 and + sign in t he r adical must be chosen. Since (* ) is a t angent t o x2 + y2 = r 2 also, we must have c2 = r 2 (1 + m 2)



Z

25m 2 + 4 = r 2 (1 + m 2)  m 2 =

r2 4 25  r 2

The given dat a is possible only whe m 2 > 0 i .e., 4 < r 2 < 25 which is infact given t o us. Now, let

FG H

(* ) meet x-axis at A and y-axis at B t hen A and B ar e 

Thus if (h, k) be t he mid point of AB t hen h = –

Now 4k 2 = 25m 2 + 4 =

IJ K

 ,0 m

and (0, ) wher e  = 25 m 2 + 4 .

  k , k = on dividing we get m = – 2 2m h

25 k 2 + 4. L ocus of (h, k) must be 4x2y2 – 4x2 = 25y2. h2

EL L I PSE

57

ZION TUTORIALS

Illustration 5 x 2 y2 + = 1 i n f ou r p oi n t s w h ose eccen t r i c an gl es ar e 1, 2, 3, 4, a 2 b2

I f a ci r cl e cu t s an el l i p se t h en p r ov e t h at

i (i ) 1 + 2 + 3 + 4 = 2n 

Fa b GG 4 a GG H 2

(i i ) cen t r e of t h e ci r cl e i s

2

(cos  1 + cos  2 + cos( 1 +  2 +  3 )),

A I

b2  a 2 (sin  1 + sin  2 + sin  3  sin( 1 +  2 4b

Sol u t i on : (i )

L et t he cir cle be x2 + y2 + 2gx + 2fy + c = 0. I f it cut s at for int er sect ion point s

U T

I JJ J +  ))J K 3

R

x 2 y2 + = 1 at (a cos , b sin ) t hen a2 b2

O T

a2 cos2  + b2 sin 2  + 2ag cos  + 2bf sin  + c = 0

(* )

  2 On put t ing sin  = and cos  = , t = t an 2  2  2 1 + t an 1 + t an 2 2 2 t an

 2

S L

The equat ion (* ) becomes

F1  t I a G H 1 + t JK 2

2 2

2

O I

F 2 t IJ +b G H1+ t K 2

2

2

Z

1  t an 2

N

+ 2 ag

1  t2 2t + 2 fb + c= 0 2 1+ t 1+ t2

Which on, simplificat ion, becomes (a2 – 2ag + c)t 4 + 4bft 3 + (4b2 – 2a2 + 2c)t 2 + 4fbt + a2 + 2ag + c = 0 The equat ion (i i ) must have four r oot s t an

1    , t an 2 , t an 3 , t an 4 whose symmet r ic funct ions 2 2 2 2

(in usual not at ions) ar e given by S1 = –

4 bf 4 b2  2 a 2 + 2 c 4 bf a 2 + 2 ag+ 2 c , S = , S =  , S = 2 3 4 a2  2 ag+ c a 2  2 ag+ c a 2  2 ag+ c a 2  2 ag+ c

Now t an

FG  +  +  +  IJ = S  S H 2 2 2 2 K 1S +S 1

2

3

4

1

3

2

(* * )

=0 4

(·.· 1 – S2 + S4  0 and S1 = S3)  1 + 2 + 3 + 4 = 2n  EL L I PSE

58

(i i )

ZION TUTORIALS

L et us now conver t t he equat ion (* ) in a four t h degr ee equat ion in sin  or cos . I ndeed t he equat ion (* ) is (a2 – b2) cos2  + 2ag cos  + c + b2 = – bf sin  I f t his equat ion is squar ed t he coeff of cos4  = (a2 – b2)2 and t he coeff of cos3 = 4ag (a2 – b2)

 cos 1 + cos 2 + cos 3 + cos 4 = –

4 ag( a2  b2 ) 4 ag = 2 2 2 2 2 (a  b ) a b

But fr om par t (i ) 4 = 2n  – (1 + 2 + 3) cos 1 + cos 2 + cos 3 + cos(1 + 2 + 3) = –

A I

4 ag a 2  b2

 x co-or dinat e of t he cent r e

S L

R

a2  b2 [cos 1 + cos 2 + cos 3 + cos(1 + 2 + 3)] 4a We can similar ly show t hat

– g=

– f =

O T

b2  a2 [sin 1 + sin 2 + sin 3 – sin(1 + 2 + 3)] 4b

Illustration 6

U T

An eq u i l at er al t r i an gl e i s d escr i b ed ab ou t an el l i p se. Pr ov e t h at i f (x, y) ar e t h e co-or d i n at es of on e v er t ex of t h e t r i an gl e r ef er r ed t o t h e p r i n ci p al d i am et er s as ax es of co-or d i n at es, t h e

N

op p osi t e si d e m ak es an an gl e  w i t h t h e m aj or ax i s, gi v en b y Sol u t i on :

Z

O I

1 x 2 + y2  a 2  b2 = 2 2 cos 2 x  y 2  a 2 + b 2

L et t he sides AB, BC, AC of t he equilat er al t r iangle ABC cir cumscr ibing t he ellipse

x 2 y2 + =1 a2 b2

meet at ‘’, ‘’ and ‘’ r espect ively. Equat ion of BC is

x y cos  + sin  = 1 a b

cos  a =  b cot   Slope = sin  a b I f BC makes angle  wit h x-axis (major axis) 

b2 cot 2  2 b a t an  = – cot   cos 2 = a b2 1 + 2 cot 2  a 1

EL L I PSE

...(i )

59

ZION TUTORIALS

+ +b a sin 2 , y = 2 x=     cos sin 2 2 a cos

Now,

Fr om t hese we easily get t an

 b a x   2 ay .t an = , t an + t an = 2 2 a+ x 2 2 b(a+ x)

S L

...(i i )

    2 t an 2 t an 4 tan t an   2 2 2 2 Now, t an t an = . =     2 2 1  tan 2  1  t an 2  1  t an 2 + t an 2 + t an 2 + t an 2 2 2 2 2 2 2

FG H

IJ K

b2 (a2  x2 ) = 2 2 a (b  y2 )

O T

A I

R

...(i i i )

b b  cot  + cot  a a t an 60° = b2 1 + 2 cot  cot  a

Now if  be  or  t han

U T

On squar ing and ar r anging, we for m a quadr at ic in cot  t o get cot  cot  =

Fr om (i i i ) and (i v)

We easily get

Z

O I

N

3 a 2  b2 cot 2  a2 . 3 b2 cot 2   a 2 b2

...(i v)

3 a 2  b2 cot 2  a2 a 2 (b2  y2 ) = . 3 b2 cot 2   a 2 b2 b2 ( a2  x2 )

cot 2  =

On put t ing cot 2 in (i ), we easily get –

a2 (3 a 2 + b2  3 x 2  y2 ) b2 (3 b2  3 y2 + a 2  x2 )

1 x 2 + y2  a 2  b 2 = 2 2 2 2 2 cos 2 x  y  a + b

Illustration 7 x 2 y2 + = 1 m eet on t h e cu r v e, p r ov e t h at t h e a 2 b2

I f t h e n or m al s at t w o p oi n t s on t h e el l i p se

t an gen t s at t h ese p oi n t s m eet on

Fx GH a

2 2

+

y2 b2

I JK

2

=

RS T

(a 2  b2 )2 y2 x2 ( a 2  x 2 ) 2 2 + (b 2  y 2 ) 2 2 4 4 a b b a

EL L I PSE

UV W

60

ZION TUTORIALS

Sol u t i on : L et t he nor mals at ‘’ and ‘’ of t he cur ve

x 2 y2 + = 1 meet at ‘’ on t he cur ve, t hen a2 b2

(a+ ) cos a2  b2 2 cos  cos  a cos  = (  ) a cos 2 ( + ) a b 2 sin  sin  b sin  =  (  ) a sin 2 2

sin

2

I f (h, k) be t he point of int er sect ion of t angent s at ( ), t hen

R

( + ) ( + ) b sin 2 , k= 2 h = (  ) (  ) cos cos 2 2 a cos

O T

A I

S L

I f we eliminat e    fr om (* ) and (* * ), we get t he r equir ed locus. On squar ing and adding t he r elat ion in (* * ), we get cos2

1 + cos (  ) 1 = 2 



and

O I

N

cos ( – ) =



Also

U T

1  = 2 2 h k2 + 2 2 a b

Z

cos

wher e  =

h2 k2 + a 2 b2

2  1 + cos(  ) 1  cos = =  2 2 

+ h   h = cos = . 2 a 2 a

cos ( + ) = 2 cos2

1 h =  a 

+ 2 h2 1 = 2 1 2 a 

Now at (* ), we get ( + ) cos a2  b2 2 a cos  = [cos ( + ) + cos( – )] (  ) 2a cos 2

=

LM N

OP Q

a2  b2 2 h 2 2 h 1+ . 2 2a  a a 

EL L I PSE

(* )

(* * )

91

ZION TUTORIALS

EXERCISE–I • SINGLE CHOICE Each of t h ese q u est i on s h as 4 ch oi ces (a ), (b), (c) an d (d ) f or i t s an sw er , ou t of w h i ch ON L Y ON E i s cor r ect . 1.

2.

S L

An ar c of a br idge is semi-ellipt ical wit h major axis hor izont al. I f t he lengt h of t he base is 9 met er and t he highest par t of t he br idge is 3 met er fr om t he hor izont al; t he best appr oximat ion of t he height of t he ar c 2 met er fr om t he cent er of t he base is (a)

11 m 4

(b)

8 m 3

(c)

7 m 2

(d) 2 m

A I

R

Coor dinat es of t he ver t ices B and C of a t r iangle ABC ar e (2, 0) and (8, 0) r espect ively. The ver t ex

O T

B C A is var ying such a way t hat 4 t an t an = 1. Then locus of A is 2 2

3.

(a)

( x  5) 2 y2 + =1 25 16

(c)

( x  5) 2 y2 + =1 25 9

(d)

N

I f t he line l x + my + n = 0 cut s t he ellipse

O I

 a 2 l 2 + b2 m 2 , t hen = 2 n2

(a) 1 (c) 4

4.

U T (b)

Z

A point on t he ellipse

( x  5) 2 y2 + =1 16 25 ( x  5) 2 y2 + =1 9 25

x 2 y2 + = 1 in point s whose eccent r ic angles differ by a2 b2

(b) 2 (d)

3 2

x 2 y2 = 1 at a dist ance equal t o t he ar it hmet ic mean of t he lengt hs of t he + 16 9

semi-major axis and semi-minor axis fr om t he cent er is

F  2 91 ,  3 105 I GH 7 14 JK F 2 105 3 91 I (c) G  7 ,  14 J H K (a)

F  2 91 ,  3 105 I GH 7 7 JK F 2 105 3 91 I (d) G  14 ,  14 J H K (b)

EL L IPSE

92

5.

ZION TUTORIALS

An ellipse has eccentricity

FG H

IJ K

1 1 , 1 . I ts one directrix is the common tangent, (nearer and one focus at S 2 2

to S) to the circle x2 + y2 = 1 and x2 – y2 = 1. The equation of the ellipse in standard form is

FG H

1 3

(a) 9 x

7.

8.

2

FG x 1 IJ H 2K

(c) 6.

IJ K

12

FG H

+ 12(y – 1)2 = 1

(b) 12 x

1 3

IJ K

2

+ 9(y – 1)2 = 1

2

+

FG H

( y 1) 2 9

(d) 3 x+

1 2

IJ K

2

A I

S L

+ 4(y – 1)2 = 1

The t angent at any point on t he ellipse 16x2 + 25y2 = 400 meet s t he t angent s at t he ends of t he major axis at T 1 and T 2. The cir cle on T 1 T 2 as diamet er passes t hr ough (a) (3, 0)

(b) (0, 0)

(c) (0, 3)

(d) (4, 0)

O T

R

The t wo concent r ic r ect angular hyper bolas, whose axes meet at angles of 45°, cut as (a) 

(b) 90°

(c) not hing can be said

(d) None of t hese

U T

I f x cos  + y sin  = p, a var iable chor d of t he hyper bola

x2 y2  = 1 subt ends a r ight angle 2 a 2 a2

at t he cent r e of t he hyper bola, t hen t he chor ds t ouch a fixed cir cle whose r adius is equal t o (a)

2a

(c) 2a 9.

Z

O I

N

I f t he nor mal at ‘’on t he hyper bola

(b)

3a

(d)

5a

x 2 y2  = 1 meet s t he t r ansver se axis at G, t hen AG × A´G a2 b2

is (Wher e A and A´ ar e t he ver t ices of t he hyper bola). 2

(a) a sec 

(c) a2(e4 sec2  – 1) 10.

Fr om any point on t he hyper bola

(b) a2 (e4 sec2  + 1) (d) None of t hese x 2 y2 x 2 y2   = 1, t angent s ar e dr awn t o t he hyper bola = 2 a2 b2 a2 b2

The ar ea cut off by t he chor d of cont act on t he r egion bet ween t he asympt ot es is equal t o (a)

ab 2

(c) 2ab

(b) ab (d) 4ab EL L IPSE

93

ZION TUTORIALS

11.

Wit h a given point and line as focus and dir ect r ix, a ser ies of ellipses ar e descr ibed, t he locus of t he ext r emit ies of t heir minor axis is (a) ellipse

(b) par abola

(c) hyper bola 12.

13.

(d) pair of lines 2

2

The cur ve xy = c(c > 0) and t he cir cle x + y = 1 t ouch at t wo point s, t hen t he dist ance bet ween t he point s of cont act s is (a) 1

(b) 2

(c) 2 2

(d)

2

I f a r ay of light incident along t he line 3x + (5 – 4 2 )y = 15, get s r eflect ed fr om t he hyper bola x 2 y2  = 1, t hen it s r eflect ed r ay goes along t he line 16 9

14.

R

(a) x 2 – y + 5 = 0

(b)

(c)

(d) 3x – y(4 2 + 5) + 15 = 0

2y – x – 5 = 0

2y – x + 5 = 0

O T

The locus of t he point of int er section of tangent s t o an ellipse at t wo point s, sum of whose eccent r ic angles is const ant is a/an (a) par abola

U T

(b) cir cle

(c) ellipse 15.

A I

S L

(d) st r aight line

I f t he equat ion t he family of t he ellipse is

N

FG H

x2 y2  + =1 0<  < 2 2 4 cos  sin 

IJ K

t hen t he locus of t he

ext r emit ies of t he lat us r ect um is

O I

(a) 2y(1 – x2) = 1 + x2 (c) y2 = (1 – x2)2 16.

Z

(b) 2y2 (1 + x2) = (1 – x2)2 (d) 2y2 (1 – x2) = (1 + x2)2

L et ( ) be a point fr om which t wo per pendicular t angent s can be dr awn t o t he ellipse 4x2 + 5y2 = 20. I f F = 4 + 3, t hen

17.

(a) – 15  F  15

(b) F  0

(c) – 5  F  20

(d) F  – 5 5 or F  5 5

Thr ee point s A; B, C ar e t aken on t he ellipse

x 2 y2 + = 1 wit h accent r ic angles   +  and  a2 b2

+ 2, t hen (a) t he ar ea of ABC is independent of  (c) t he maxmum value of ar ea is

3 ab 4

(b) t he ar ea of ABC is independent of  (d) t he maximum value value of ar ea is

EL L IPSE

3 3 ab 4

94

ZION TUTORIALS

18.

I f t he nor mal at t he end of lat usr ect um of t he ellipse

x 2 y2 + = 1 passes t hr ough (0, – b), t hen a2 b2

e4 + e2 (wher e e is eccent r icit y) equals (a) 1 (c)

19.

5 1 2

I f t he line x + 2y + 4 = 0 cut t ing t he ellipse

(b)

2

(d)

5 +1 2

x 2 y2 + = 1 at point s whose eccent r ic angles ar e 30° a2 b2

A I

and 60° subt end r ight angle at t he or igin t hen it s equat ion is

20.

x2 y2 + =1 (b) 16 4

x2 y2 + =1 (c) 4 16

(d) none of t hese

(a) (–, )

U T

(b) (0, )

(c) ( )

22.

O T

2 , t hen t he r ange of values of  for which t he point  on t he ellipse x2 + 4y2= 4 falls 3

inside t he cir cle x2 + y2 + 4x + 3 = 0 is

21.

R

x2 y2 + =1 (a) 8 4

I f cos  =

S L

(d) ( – ,  + )

N

I f  and  ar e eccent r ic angles of t he ends of a focal chod of t he ellipse

O I

t an

 is equal t o 2

(a)

1 e 1+ e

(c)

e+ 1 e 1

Z

(b)

e 1 e+ 1

(d)

e 1 e+ 3

The dist ances fr om t he foci of P(a, b) on t he ellipse

x 2 y2 = 1 ar e + 9 25

(a) 4 ±

5 b 4

(b) 5 ±

(c) 5 ±

4 b 5

(d) none of t hese EL L IPSE

4 a 5

 x 2 y2 + 2 = 1, t hen t an 2 2 a b

95

ZION TUTORIALS

23.

24.

25.

I f C be t he cent r e of t he ellipse 9x2 + 16y2 = 144 and S is one focus, t he r at io of CS t o major axis is (a)

7 : 16

(b)

(c)

5:

(d) none of t hese

7

7: 4

S L

The lengt h of t he lat usr ect um of an ellipse is one t hir d of t he major axis, it s eccent r icit y would be (a)

2 3

(b)

(c)

1 2

(d)

1 3

FG 2 IJ H 3K

A I

R

An ellipse is descr ibed by using an endless st r ing which is passed over t wo pins. I f t he axes ar e 6 cm and 4 cm, t he necessar y lengt h of t he st r ing and t he dist ance bet ween t he pins r espect ively in cms ar e (a) 6, 2 5

O T

(b) 6,

(c) 4, 2 5

U T

5

(d) none of t hese

• MULTIPLE CORRECT CHOICE TYPE

Each of t h ese q u est i on s h as 4 ch oi ces (a ), (b), (c) an d (d ) f or i t s an sw er , ou t of w h i ch ON E OR M ORE i s/ar e cor r ect . 1.

conic is

2.

N

I f (5, 12) and (24, 7) ar e t he foci of a conic passing t hr ough t he or igin t hen t he eccent r icit y of

(a)

386 12

(c)

386 25

Z

O I

I f P is any point lying on t he ellipse

(b)

386 38

(d)

2

x 2 y2 + = 1, whose foci ar e S and S´. L et PSS´ =  and a2 b2

PS´S = , t hen (a) PS + PS´ = 2a, if a > b (c) t an

  1 e t an = 2 2 1+ e

(b) PS + PS´ = 2b, if a < b (d) t an

EL L IPSE

  tan = 2 2

LM N

OP Q

a 2  b2 a  a2  b2 when a > b b2

96

3.

4.

ZION TUTORIALS

The equat ion

x 2 + ( y 1) 2  x2 + ( y+ 1) 2 = K will r epr esent a hyper bola for

(a) K  (0, 2)

(b) K  (0, 1)

(c) K  (1, )

(d) K  (0, )

I f a quadr ilat er al for med by four t angent s t o t he ellipse 3x2 + 4y2 = 12 is a squar e t hen (a) The ver t ices of t he squar e lie on is y = ±x (b) The ver t ices of t he squar e lie on x2 + y2 = 7 (c) The ar ea of all such squar es is const ant

A I

(d) Only t wo such squar es ar e possible

S L

5.

The locus of ext r emit ies of t he lat us r ect um of t he family of ellipses b2x2 + y2 = a2b2 wher e b is

6.

a par amet er (b2 < 1), is (a) x2 + a2y2 = a2 (b) x2 + ay + a2 2 2 (c) x – ay = a (d) x2 – a2y = a2 I f t he st r aight line 3x + 4y = 24 int er sect s t he axes at A and B and t he st r aight line 4x + 3y = 24 at C and D, t hen point s A, B, C, D lies on

7.

(b) par abola

(c) ellipse

(d) hyper bola

U T

I f lat us r ect um of t he ellipse x2 t an 2 + y2 sec2  = 1 is 1/2, t hen  (0 <  < ) is equal t o (a) /2

(b) /6

(c) 5/12 8.

O T

(a) cir cle

R

(d) none of t hese

I n t he ellipse

N

25x2 + 9y2 – 150x – 90y + 225 = 0

O I

(a) foci ar e at (3, 1), (3, 9) (c) cent r e is (5, 3) 9.

Z

(b) e = 4/5 (d) major axis is 6

The eccent r ic angle of a point on t he ellipse x2 + 3y2 = 6 at point on t he ellipse x2 + 3y2 = 6 at a dist ance 2 unit fr om or igin is

10.

(a)

 4

(c)

5 4

(b)

3 4

(d)

7 4

A t angent t o t he ellipse 4x2 + 9y2 = 36 is cut by t he t angent at t he ext r emit ies of t he major axis at T and T´. The cir cle on TT´ as diamet er passes t hr ough t he point (a) (– 5 , 0)

(b) ( 5 , 0)

(c) (0, 0)

(d) (3, 2) EL L IPSE

97

ZION TUTORIALS

11.

The pr oduct of eccent r icit ies of t wo conics is unit y, one of t hem can be a/an (a) par abola

(b) ellipse

(c) hyper bola

(d) cir cle

• COMPREHENSION TYPE

S L

T h i s sect i on con t ai n s gr ou p s of q u est i on s. Each gr ou p i s f ol l ow ed b y som e m u l t i p l e ch oi ce q u est i on b ased on a p ar agr ap h . Each q u est i on h as 4 ch oi ces (a ), (b), (c) an d (d ) f or i t s an sw er , ou t of w h i ch ON E i s cor r ect .

A I

PASSAGE-1

Consider a hyper bola whose cent r e is at or igin. A line x + y = 2 t ouches t his hyper bola at P(1, 1) and int er sdect s t he asympt ot es at A and B such t hat AB = 6 2 unit s. (you can use t he concept t hat in case

R

of hyper bola por t ion of t angent int er cept ed bet ween asympt ot es is bisect ed at t he point of cont act ). 1.

2.

Equat ion of asympt ot es ar e

(b) 3x2 + 4y2 + 6xy = 0

(c) 2x2 + 2y2 – 5xy = 0

(d) none of t hese

Angle subt ended by AB at t he cent r e of t he hyper bola is (a) sin –1 (c) sin –1

3.

O T

(a) 2x2 + 2y2 + 5xy = 0

U T

2 (b) sin –1 5

4 5 3 5

N

Equat ion of t he t angent t o t he hyper bola at (a) 5x + 2y = 2 (c) 3x + 4y = 1

Z

O I

(d) t an –1

4 5

FG 1, 7 IJ H 2K

is

(b) 3x + 2y = 4 (d) x + 2y = 6

PASSAGE-2

A sequence of ellipses E 1, E 2, ..., E n is const r uct ed as follows : Ellipse E n is dr awn so as t o t ouch ellipse E n – 1 as t he ext r emit es of t he major axis of E n – 1 and t o have it s foci at t he ext r emit ies of t he minor axis of E n – 1. On t h e ba si s of a bove i n f or m a t i on , a n sw er t h e f ol l ow i n g qu est i on s : 1.

I f E n is independent of n, t hen t he eccent r icit y of ellipse E n (a)

(c)

F3 5I GH 2 JK 2 3 2

(b)

(d) EL L IPSE

F GH

5 1 2 3 1 2

I JK

– 2

is

98

2.

3.

4.

5.

ZION TUTORIALS

I f eccent r icit y of ellipse E n is en t hen t he locus of (en 2, e2n–1) is (a) a par abola

(b) an ellipse

(c) a hyper bola

(d) a r ect angular hyper bola

I f eccent r icit y of ellipse E 1 is

3 /2 t hen t he eccent r icit y of ellipse E 4 is

(a)

1 5

(b)

(c)

3 23

(d) none of t hese

2 3

A I

x 2 y2 = 1, t hen t he equat ion of ellipse E 3 is + 9 16

I f equat ion of ellipse E 1 is (a)

x2 y2 + =1 9 16

(b)

(c)

x2 y2 + =1 25 41

(d)

x 2 y2 + =1 25 49

O T x2 y2 + =1 16 25

S L

R

I f eccent r icit y E n is independent of n, t hen t he locus of mid point of chor ds of slope – 1 of E n is (I f axis of E n along y-axis). (a) ( 5 – 1) x = 2y (c) (3 – 5 )x = 2y

• ASSERTION & REASON

O I

U T

(b) ( 5 + 1) x = 2y

N

(d) (3 + 5 )x = 2y

The following quest ions consist of two st at ement s, one labelled as ‘Asser tion (A)’and the ot her labelled as ‘Reason (R)’. You examine t hese two st atement s car efully and decide if the Asser tion (A) and the Reason (R) are individually tr ue and if so, whet her Reason (R) is the cor rect explanat ion for the gi ven Assert ion (A). Select your answer t o these items using t he codes given below t hen select t he corr ect opt ion. Cod es :

(A) (B) (C) (D )

Z

Bot h A a n d R a r e i n d i vi d u a l l y t r u e a n d R i s t h e cor r ect exp l a n a t i on of A Bot h A a n d R a r e i n d i vi d u a l l y t r u e bu t R i s n ot t h e cor r ect exp l a n a t i on of A A i s t r u e bu t R i s f a l se A i s f a l se bu t R i s t r u e

1.

Asser t i on (A) : The dist ance bet ween t he foci of an ellipse is always less t he sum of focal dist ances of any point on it . Reason (R) : The eccent r icit y of an ellipse is less t han 1.

2.

Asser t i on (A) : The equat ion Reason (R)

x2 y2 + = 1 r epr esent s a r eal ellipse if 8 < p < 12. 12  p p 8

: Eccent r icit y of an ellipse is less t han 1.

EL L IPSE

99

ZION TUTORIALS

3.

Asser t i on (A) : Number of int egr al point s on t he ellipse

Reason (R) 4.

: The eccent r icit y of t he ellipse is

x 2 y2 + = 1 is 4 5 4

5 3

Asser t i on (A) : The lat us r ect um of t he hyper bola is 6 a. Reason (R) 5.

S L

The t r ansver se axis of a hyper bola is given 2a and it s ver t ex bisect s t he dist ance bet ween t he cent r e and focus.

: The eccent r icit y of t he hyper bola is

A I

2

Asser t i on (A) : The equation x2 + 2xy + y2 + 2x + 2y + 4 = 0 repr esents an ellipse if  (– 1, 1).

R

Reason (R) : The gener al equat ion of second degr ee Repr esent s an ellipse if   0, h 2 < ab. 6.

O T

Asser t i on (A) : The equat ion 13x2 – 18xy + 37y2 + 2x + 14y – 2 = 0 r epr esent s an ellipse. Reason (R) : The squar e of t he coefficient of xy is less t han t he pr oduct of coefficient of x2 and y2 is only condit ion.

• MATCHING TYPE

U T

Gi v en b el ow ar e M at ch i n g T y p e Qu est i on s, w i t h t w o col u m n s (each h av i n g som e i t em s) each . Each i t em of Col u m n I h as t o b e m at ch ed w i t h t h e i t em s of Col u m n I I , b y en ci r cl i n g t h e cor r ect m at ch (es). 1.

N

L et t h e ci r cl e (x – 1) 2 + (y – 2) 2 =25 cu t s a r ect an gu l ar h y p er b ol a w i t h t r an sv er se ax i s

O I

al on g y = x at f ou r p oi n t s A, B , C an d D h av i n g coor d i n at es (x 1, y1), i = 1, 2, 3, 4 r esp ect i v el y . O b ei n g t h e cen t r e of t h e h y p er b ol a. N ow m at ch t h e en t r i es f r om t h e

Z

f ol l ow i n g t w o col u m n s :

Col u m n -I

Col u m n -I I

(A)

x1 + x2 + x3 + x4 is equal t o

p. 2

(B)

x21 + x22 + x23 + x24 is equal t o

q. 4

(C)

y21 + y22 + y23 + y24 is equal t o

r . 44

(D)

OA 2 + OB 2 + OC2 + OD 2 is equal t o

s. 56 t . 100

EL L IPSE

100

2.

ZION TUTORIALS

Ob ser v e t h e f ol l ow i n g col u m n s : Col u m n -I (A)

Col u m n -I I

I f E : 2x2 + y2 = 2 and dir ect or cir cle of E is C1, dir ect or cir cle of C1 is C2, dir ect or cir cle of C2 is

(P) r 21, r 22, r 23 ... ar e in GP wit h common r at io 2.

C3 and so on. I f r 1, r 2, r 3 .... ar e t he r adii of C1, C2, C3 ... r espect ively. Then (B)

I f E : 3x2 + 2y2 = 6 and dir ect or cir cle of E is C1, dir ect or cir cle of C1 is C2, dir ect or cir cle of C2 is C3 and so on. I f r 1, r 2, r 3, ... ar e t he r adii of C1,

(C)

C2, C3, ... r espect ively. Then

I f E : 3x2 + 4y2 = 12 and dir ect or cir cle of E is C1, dir ect or cir cle of C1 is C2, dir ect or cir cle of

C2 is C3 and so on. I f r 1, r 2, r 3, ... ar e t he r adii of C1, C2, C3, ... r espect ively. Then

Z

O I

N

S L

(Q) GM of r 21, r 22, r 23 is 6.

(R) r 21 + r 22 + r 23 + ... + r 2n = r 21 (2n – 1)

A I

(S) GM of r 21, r 22, r 23 is 10

O T

R

(T) GM of r 21, r 22, r 23 is 14

U T

EL L IPSE

101

ZION TUTORIALS

EXERCISE–II • SINGLE CHOICE Each of t h ese q u est i on s h as 4 ch oi ces (a ), (b), (c) an d (d ) f or i t s an sw er , ou t of w h i ch ON L Y ON E i s cor r ect . 1.

2.

(a)

3 2

(b)

(c)

1 2

(d)

3 4

A I

1 2

R

The sum of t he squar es of t he per pendicular s on any t angent t o t he ellipse point s on t he minor axis each at a dist ance

3.

S L

I n an ellipse, if t he lines joining a focus t o t he ext r emit ies of t he minor axis make an equilat er al t r iangle wit h t he minor axis, t hen t he eccent r icit y of t he ellipse is

O T

a 2  b2 fr om t he cent r e is

(a) 2a2

(b) 2b2

(c) a2 + b2

(d) a2 – b2

Fr om any point on t he hyper bola

x 2 y2 + = 1 fr om t wo a2 b2

U T

x 2 y2 x 2 y2   = 1, t angent s ar e dr awn t o t he hyper bola = 2. a2 b2 a2 b2

The ar ea cut off by t he chor d of cont act on t he asympt oes is equal t o (a)

ab 2

(c) 2ab 4.

Tangent s ar e dr awn t o t he ellipse for med is (a) 27

(c)

5.

O I

N

27 4

Z

(b) ab (d) 4ab

x 2 y2 + = 1 at ends of lat er ar ect a. The ar ea of quadr ilat er al so 9 5

(b)

27 2

(d)

27 55

I f CF is per pendicular fr om t he cent r e of t he ellipse

x 2 y2 + = 1 t o t he t angent at P and G is t he a2 b2

point wher e t he nor mal at P meet s t he major axis, t hen CF × PG = (a) a2

(b) b2

(c) a2b2

(d) a2 + b2 EL L IPSE

102

6.

ZION TUTORIALS

 x2 y + = 1. The least A t angent is dr awn at t he point (3 3 cos , sin ); 0 <  < of an ellipse 2 27 1

value of t he sum of t he int er cept s on t he coor dinat e axes by t his t angent is at t ained at  =

7.

8.

(a)

 6

(b)

 3

(c)

 8

(d)

 4

I f  is t he angle bet ween t he diamet er t hr ough any point on a st andar d ellipse and t he nor mal at t hat point , t hen t he gr eat est value of t an  is (a)

2 ab a 2 + b2

(b)

a2 + b2 ab

(c)

a2  b2 2 ab

(d)

b2 a2

O T

A I

R

PQ and RS ar e t wo per pendicular chor ds of t he r ect angular hyper bola xy = c2. I f C is t he cent r e of t he r ect angular hyper bola. Then t he pr oduct of t he slopes of CP, CQ, CR and CS is equal t o. (a) – 1

U T

(b) 1

(c) 0 9.

S L

(d) None of t hese

I f t he nor mals at P, Q, R on t he r ect angular hyper bola xy = c2 int er sect at a point S on t he hyper bola, t hen cent r oid of t he t r iangle PQR is at t he ________ of t he hyper bola. (a) cent r e (c) ver t ex

10.

N

(b) focus (d) dir ect or cir cle

PM and PN ar e t he per pendicular s fr om any point P on t he r ect angular hyper bola xy = c2 t o t he

O I

asympt ot es. The locus of t he mid-point of M N is a hyper bola wit h eccent r icit y

11.

(a)

2

(c)

1 2

Z

I f e is t he eccent r icit y of t he hyper bola

Then cos

(b) 2 (d) 2 2 x 2 y2  = 1 and  is t he angle bet ween t he asympt ot es. a2 b2

 is equal t o 2

(a)

e

(b)

e 1+ e

(c)

1 e

(d)

1 e

EL L IPSE

103

ZION TUTORIALS

12.

I f t he ellipse

x2 y2 + y2 = 1 meet s t he ellipse x2 + 2 = 1 in four dist inct point s and a = b2 – 5b + 4 a

7, t hen b does not lie in

13.

(a) [4, 5]

(b) (– , 2)  (3, )

(c) (–, 0)

(d) [2, 3]

S L

L et P be any point on any dir ect r ix of an ellipse. Then chor ds of cont act of point P wit h r espect t o t he ellipse and it s auxiliar y cir cle int er sect at (a) some point on t he major axis depending upon t he posit ionof point P

(b) mid point of t he line segment joining t he cent r e t o t he cor r esponding focus

A I

(c) cor r esponding focus (d) none of t hese 14.

The maximum dist ance of t he cent r e of t he ellipse per pendicular t angent s of t he ellipse is

15.

(a)

144 5

(c)

9 5

(b)

O T 16 5

(d) 5

x 2 y2  A nor mal t o t he hyper bola = 1, has equal int er cept s on t he posit ive x and y axes. I f t his 4 1

O I

nor mal t ouches t he ellipse (a) 5 (c) 16 16.

U T

R

x 2 y2 + = 1 fr om chor d of cont act of mut ually 16 9

Z

N

x 2 y2 + = 1, t hen a2 + b2 is equal t o a2 b2

(b) 25 (d)

25 3

Fr om a point on t he axis of x common t angent s ar e dr awn t o t he par abola y2 = 4x and t he ellipse x 2 y2 + = 1 (a > b > 0). I f t hese t angent s for m an equilat er al t r iangle wit h t heir chor d of cont act a2 b2

w.r .t . par abola, t hen set of exhaust ive values of a i s (a) (0, 3)

(b)

FG 0, 3 IJ H 2K

FG 3 , 3IJ H2 K

(d)

FG1, 3 IJ H 2K

(c)

EL L IPSE

104

17.

18.

19.

20.

ZION TUTORIALS

The ar ea of a t r iangle inscr ibed in an ellipse bear s a const ant r at io t o t he ar ea of t he t r iangle for med by joining point s on t he auxiliar y cir cle cor r esponding t o t he ver t ices of t he fir st t r iangle. This r at io is (a)

b a

(b)

2a b

(c)

a2 b2

(d)

b2 a2

S L

The set of values of a for which (13x – 1)2 + (13y – 2)2 = a(5x + 12y – 1)2 r epr esent s an ellipse, if (a) 1 < a < 2

(b) 0 < a < 1

(c) 2 < a < 3

(d) None of t hese

A I

R

Tangent s ar e dr awn fr om t he point s on t he line x – y – 5 = 0 t o x2 + 4y2 = 4, t hen all t he chor ds of cont act pass t hr ough a fixed point , whose coor dinat e ar e (a)

FG 1 , H5



2 5

(c)

FG 2 , H5



1 5

IJ K

(b)

IJ K

U T

O T FG 4 , H5



1 5

IJ K

(d) none of t hese

The set of posit ive value of m for which a line wit h slope m is a common t angent t o ellipse x 2 y2 + = 1 and par abola y2 = 4ax is given by a2 b2

(a) (2, 0) (c) (0, 1) 21.

(b) (3, 5) (d) none of t hese

I f CP and CD ar e semi-conjugat e diamet er s of t he ellipse (a) a + b (c) a2 – b2

22.

O I

N

Z

x 2 y2 + = 1, t hen CP2 + CD 2 is equal a2 b2

(b) a2 + b2 (d)

I f t he nor mal at t he point P() t o t he ellipse

a 2 + b2

x 2 y2 + = 1 int er sect s it again at t he point Q(2, ), 14 5

t hen cos  is equal t o (a)

2 3

(b) 

2 3

(c)

3 2

(d) 

3 2

EL L IPSE

105

ZION TUTORIALS

23.

24.

An ellipse slides bet ween t wo per pendicular st r aight lines. Then t he locus of it s cent r e is a/an (a) par abola

(b) ellipse

(c) hyper bola

(d) cir cle

The lengt h of t he chor d of t he ellipse

(a)

(c)

25.

26.

1 10

8061 10

A I

(d) none of t hese

For t he ellipse

FG 1 , 2 IJ H 2 3K

8161 10

(b)

is

S L

R

x 2 y2 + = 1, t he equat ion of t he diamet er conjugat e t o ax – by = 0 is a2 b2

O T

(a) bx + ay = 0

(b) bx – ay = 0

(c) a3y + b3x = 0

(d) a3y – b3x = 0

L et E be t he ellipse

U T

x 2 y2 + = 1 and C be t he cir cle x2 + y2 = 9. L et P and Q be t he point s (1, 2) 9 4

and (2, 1) r espect ively. Then (a) Q lies inside C but out side E

N

(c) P lies inside bot h C and E 27.

x 2 y2 + = 1, wher e mid point is 25 16

O I

(b) Q lies out side bot h C and E (d) P lies inside C but out side E

The t angent at a point P(a cos , b sin ) of an ellipse

x 2 y2 + = 1 meet s it s auxiliar y cir cle in a2 b2

t wo point s, t he chor d joining which subt ends a r ight angle at t he cent r e, t hen t he eccent r icit y of t he ellipse is

Z

(a) (1 + sin 2 )–1

(b) (1 + sin 2 )–1/2

(c) (1 + sin 2 )–3/2

(d) (1 + sin 2 )–2

• MULTIPLE CORRECT CHOICE TYPE Each of t h ese q u est i on s h as 4 ch oi ces (a ), (b), (c) an d (d ) f or i t s an sw er , ou t of w h i ch ON E OR M ORE i s/ar e cor r ect . 1.

I f t he nor mals at (xi , yi ) i = 1, 2, 3, 4 t o t he r ect angular hyper bola xy = 2 meet at t he point (3, 4), t hen

(a) x1 + x2 + x3 + x4 = 3

(b) y1 + y2 + y3 + y4 = 4

(c) x1 x2 x3 x4 = – 4

(d) y1 y2 y3 y4 = 4 EL L IPSE

106

2.

ZION TUTORIALS

I f t he nor mal at P t o t he r ect angular hyper bola x2 – y2 = 4 meet s t he axis in G and g and C is t he cent r e of t he hyper bola, t hen

3.

(a) PG = PC

(b) Pg = PC

(c)

(d) Gg = 2PC

2 PG = Pg

I f t he nor mals t o concur r ent , t hen 1 a2

(a) l 1l 2 = – (c) l 1l 2 = 4.

S L

x 2 y2 + = 1 at t he ends of t he chor d l 1x + m 1y = 1 and l 2x + m 2y = 1 ar e a2 b2

(b) m 1m 2 = –

1 a2

(d) m 1m 2 =

A I

1 b2

1 b2

O T

R

I f t wo concent r ic ellipses be such t hat t he foci of one be on t he ot her and t heir major axes ar e equal. L et e1 and e2 be t heir eccent r icit ies, t hen

(a) t he quadr ilat er al for med by joining t he foci of t he t wo ellipses is a par allelogr am

U T

1 1 1 (b) t he angle  bet ween t heir axes is given by, cos  = e2 + e2  e2 e2 1 2 1 2

(c) if

e12

+

e22

= 1 t hen t he angle bet ween t he axes of t he t wo ellipses is 90°

N

2 2 2 2 (d) ar ea of t he quadr ilat er al = a2 e1 + e2 + e1 e2  1

5.

O I

I f fr om (1, ) t wo t angent s ar e dr awn on exact ly one br anch of a hyper bola

Z

x 2 y2  = 1, t hen 4 1

value of  may sat isfy

LM 1 , 1OP N2 Q F 1 1I (c) GH  , JK 2 2 (a)

6.

(b) (d)

A lat us r ect um of an ellipse is a line (a) passing t hr ough a focus

(b) t hr ough t he cent r e

(c) per pendicular t o t he major axis 7.

LM 1 , 1 OP N 4 4Q LM 3 , 3 OP N 4 4Q

2

(d) par allel t o t he minor axis 2

2

I f lat us r ect um of t he ellipse x t an  + y sec2 = 1 is 1/2, t hen (0 <  < ) is equal t o (a) /12

(b) /6

(c) 5/12

(d) none of t hese EL L IPSE

107

ZION TUTORIALS

8.

9.

10.

Equat ion of t angent t o t he ellipse x2/9 + y2/4 = 1 which cut off equal int er cept s on t he axes is (a) y = x + (13)

(b) y = – x + (13)

(c) y = x – (13)

(d) y = – x – (13)

The equation of tangent to t he ellipse x2 + 3y2 = 3 which is perpendicular t o the line 4y = x – 5 is (a) 4x + y + 7 = 0

(b) 4x + y – 7 = 0

(c) 4x + y + 3 = 0

(d) 4x + y – 3 = 0

0 is/ar e

11.

(a) (5, 2)

(b) (2, 5)

(c) (1, 3)

(d) (– 5, – 2)

O T

(b) PT bisect s F 1PF 2

(c) PT bisect s angle (180° – F 1PF 2)

(d) none of t he above

I f (5, 12) and (24, 7) ar e t he focii of a conic passing t hr ough t he or igin, t hen t he eccent r icit y of

(a)

386 38

(c)

386 13

N

U T (b)

386 12

(d)

386 25

The t angent s fr om which of t he following point s t o t he ellipse 5x2 + 4y2 = 20 ar e per pendicular (a) (1, 2 2 ) (c) (2, 5 )

14.

R

(a) PN bisect s F 1PF 2

conic is

13.

A I

L et F 1, F 2 be t wo focii of t he ellipse and PT and PN be t he t angent and t he nor mal r espect ively t o t he ellipse at point P. Then

12.

S L

The point s, wher e t he nor mals t o t he ellipse x2 + 3y2 = 37 be par allel t o t he line 6x – 5y + 7 =

Z

O I

(b) (2 2 , 1) (d) ( 5 , 2)

I f t he t angent at t he point t o t he ellipse 16x2 + 11y2 = 256 is also a t angent t o t he cir cle x2 + y2 – 2x = 15, t hen  equals (a)

 3

(c) – 15.

 3

(b)

2 3

(d)

5 3

I f t he t angent t o t he ellipse x2 + 4y2 = 16 at t he point P() is a nor mal t o t he cir cle x2 + y2 – 8x – 4y = 0, t hen  equals (a) /2

(b) /4

(c) 0

(d) –/4 EL L IPSE

108

ZION TUTORIALS

• COMPREHENSION TYPE T h i s sect i on con t ai n s gr ou p s of q u est i on s. Each gr ou p i s f ol l ow ed b y som e m u l t i p l e ch oi ce q u est i on b ased on a p ar agr ap h . Each q u est i on h as 4 ch oi ces (a ), (b), (c) an d (d ) f or i t s an sw er , ou t of w h i ch ON E i s cor r ect .

PASSAGE-1 Read t h e f ol l ow i n g w r i t eu p car ef u l l y :

S L

I f ax2 + by2 + 2hxy+ 2gx + 2fy + c = 0 r epr esent s an ellipse, t hen h 2 < ab and abc + 2fgh – af 2 – bg2 – ch 2  0. I f for ever y point (x1, y1) sat isfying above equat ion (2h – x1, 2k – y1) also sat isfy it , t hen (h, k) is cent r e of it . The lengt h of semi major axis is t he maximum and minimum value of t he dist ances of point s lying on t he cur ve fr om it s cent r e.

A I

N ow an sw er t h e f ol l ow i n g q u est i on (1–5) : 1.

2

2

(a)

 12

(b)

3 (c) 8

2.

O T  8

5 (d) 8

U T

The equat ion of t angent t o 2x2 – 2xy + 4y2 – (3 + 2 ) = 0 such t hat sum of per pendicular s dr opped fr om foci is 2 unit s, is

3.

R

For t he ellipse 2x – 2xy + 4y – (3 + 2 ) = 0, t he inclinat ion of major axis of it wt h x-axis is

(a) y cos

3 3 – x sin = 1 4 4

(c) x cos

  – y sin = 1 8 8

O I

N

(b) y sin

3 3 – x cos = 1 8 8

(d) y cos

5 5 + x sin = 1 8 8

The pr oduct of per pendicular s fr om t he foci t o any t angent t o above given ellipse is (a) 4 (c) 1

Z

(b) 2 (d)

1 2

PASSAGE-2 Consider t he st andar d equat ion of an ellipse whose focus and cor r esponding foot of dir ect r ix ar e ( 7 , 0) and

FG 16 , 0IJ H 7 K

and a cir cle wit h equat ion x2 + y2 = r 2. I f in t he fir st quadr ant , t he common t angent

t o a cir cle of t his family and t he above ellipse meet s t he coor dinat e axes at A and B.

EL L IPSE

109

ZION TUTORIALS

On t h e ba si s of a bove i n f or m a t i on , a n sw er t h e f ol l ow i n g qu est i on s : 1.

2.

The equat ion of t he ellipse is (a) 16x2 + 9y2 = 144

(b) 9x2 + 16y2 = 144

(c) 16x2 + y2 = 144

(d) x2 + 9y2 = 144

L et P be a var iable point on t he ellipse wit h foci at S and S´. I f  be t he ar ea of t r iangle PSS´, t hen t he maximum value of  is (a)

3.

4.

7 sq unit

(b) 2 7 sq unit

(c) 3 7 sq unit (d) 4 7 sq unit I f mid point of A and B is (x1, y1) and slope of common t angent be m, t hen (a) 2mx1 + y1 = 0

(b) 2my1 + x1 = 0

(c) my1 + x1 = 0

(d) mx1 + y1 = 0

The locus of mid point of A and B is

(a) y = x

F r 9 I GH 16  r JK

(c) y = x

F 16  r I GH r + 9 JK

2

2

5.

O T

(b) y = x

2

U T

(d) y = x

2

A I

S L

R

F 16 + r I GH 9  r JK 2

2

F r 9 I GH r  16 JK 2

2

The domain of value of r (r > 0) for which t he above quest ion is t r ue is (a) r  (1, 2) (c) r  (4, 5)

O I

• MATCHING TYPE

N

(b) r  (3, 4) (d) r  (5, 6)

Gi v en b el ow ar e M at ch i n g T y p e Qu est i on s, w i t h t w o col u m n s (each h av i n g som e i t em s) each . Each i t em of Col u m n I h as t o b e m at ch ed w i t h t h e i t em s of Col u m n I I , b y en ci r cl i n g t h e cor r ect m at ch (es). 1.

Z

Ob ser v e t h e f ol l ow i n g col u m n s : Col u m n -I (A)

If

Col u m n -I I

3 bx + ay = 2ab t ouches t he ellipse

x 2 y2 + = 1 at a a2 b2

p. 0

point whose eccent r ic angle is  t hen cosec  equals (B)

L et e(k) be t he eccent r icit y of (x – 3) (y + 2) = k 2, t hen

q. 1

e(2) – e(3) equals (C)

I f x2 + y2 = a2 is dr awn wit hout int er sect ing t he cur ve xy = 9 t hen int egr al value of a equals EL L IPSE

r. 2

110

ZION TUTORIALS

(D)

I f xy = 1 + sin 2 ( being par amet er ) be a family of

s. 3

r ect angular hyper bolas and  is t he ar ea of t r iangle for med by any t angent wit h coodinat e axes t hen  can be equal t o 2.

Ob ser v e t h e f ol l ow i n g col u m n s : Col u m n -I (A)

The minimum and maximum dist ance of a point (2, 6) 2

(P) L + G = 10

2

fr om t he ellipse 9x + 8y – 36x – 16y – 28 = 0 ar e L and G, t hen (B)

A I

The minimum and maximum dist ance of a point (1, 2) 2

(Q) L + G = 6

2

fr om t he ellipse 4x + 9y + 8x – 36y + 4 = 0 ar e L and G, t hen (C)

S L

Col u m n -I I

R

FG 9 , 12 IJ H5 5 K

O T

The minimum and maximum dist ance of a point

(R) G – L = 6

fr om t he ellipse 4(3x + 4y)2 + 9(4x – 3y)2 = 900 ar e L and G, t hen

Z

O I

N

U T

EL L IPSE

(S) G – L = 4 (T) L G + GL = 6