27780660 mahesh tutorials science

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Mahesh Tutorials Science

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ELLIPSE.Q-2) GROUP-(A) CLASS WORK EXAMPLES Q-1)The equation of ellipse is

Find the lengths of the axis, eccentricity, co-ordinates of the foci, equation of directrices, length of latus-rectum, co-ordinates of the ends of latus rectumfor each of ellipse(i) x2 y2 + =1 25 9 x2 y2 + =1 25 9

9x 2 + 4y 2 = 36. Find the (i) lengths ofaxes (ii) eccentricity (iii) co-ordinate of foci (iv) equation of directrices v)length of L.R.

Ans.

9x 2 + 4y 2 = 36.

x y + =1 4 9+ y2 b2 =1

2

2

Comparing with2 2

x2 a2

Ans.

(i)

a =4&b =9 a =2&b=3 a b l (major axis)= 2a =10, l (minor axis) = 2b = 6Eccentricity = e= a 2 b2 a

2

=

25 9 16 4 = = 25 25 5

(ii) Eccentricity:

a 2 = b2 1 e 2 4 = 9 1 e 2 4 = 1 e2 9

(

)

(


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)

4 ae = 5 = 4, foci ( ae,0 ) ( 4,0 ) 5

Equations of directrices are x = i.e. x = 25 4 2b2 2 ( 9 ) 18 = = a 5 5

a e

e2 =

94 5 5 e2 = e = 9 9 3

iii) Co ordinates of Foci : 5 foci ( 0, be ) Foci 0, 3 3 Foci 0, 5

latus rectum =

Extremities of latus rectum are b2 9 ae, = 4, and a 5 b2 9ae,

=4,

a5b2

9ae,

=4,

and

a5

b2 9 ae, = 4, a 5

(

)

iv) Equation of directrices: Equation of directrices are y = b/e 3 9 y= y= 5 /3 5

v) Length of L.R. = 2a 2 2 4 8 = = units 3 3 b

Ellipse


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87


Q 3)

Find equation of ellipse in standard form of its vertices are ( 4,0 ) and lengt

h of minor axis is 6.

Q 5)

Find the equation of the ellipse (referred to its principal axes) whose minor axis = 8 and eccentricity=3 5

Ans.

Let equation be

x2 a2

+

y2 b2

= 1 ( a > b )

Vertices ( a,0 ) ( 4,0 ) a = 4 length of minor axis = 2b = 6. b = 3 Required equation is x y + =1 16 92 2

Ans.

Let the equation of the ellipse bex2 a2 + y2 b2 =1

x2 42

+

y2 32

=1

Since minor axis = 8 2b = 8 b = 4 eccentricity = e =

3 5

b2 = a 2 1 e 2 9 16a 2 16 = a 2 1 a 2 = 25. = 25 25 The equationof the ellipsex2 a2 y2 b2

(

)

Q 4)


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Find the equation of the following ellipse whose foci are ( 5,0 ) and5 eccentricity . 8

x2 y2 + =1 25 16

Ans.

Let equation of ellipse be foci ( ae,0 ) ( 5,0 )

+

=1

Q 6)

Find the equation of the ellipse (referred to its principal axes) such that distance between foci = 2 and

5 5 a = 5 a = 8 ae = 5 and e = 8 825 b2 = a 2 1 e2 b2 = 82 1 64

vertices are ( 2,0 ) Ans.Let equation of the ellipsex2 a2 + y2 b2 =1

(

)

..(i)

64 25 2 b2 = 64 b = 39 64 y2 Equation of ellipse 2 + =1 39 8 x2

Given 2ae = 2

ae = 1

b 2 = a 2 1 e 2 = a 2 a 2e 2 = 4 1 = 3 Also a = 2 a 2 = 4,b 2 = 3put these values in (i), the equation of the ellipse isx2 y2 + =1 4 3

(

)

.

x y + =1 64 39

2

2

Ellipse


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88

Q 7)

Find the equation of the ellipse (referred to its principal axes) Distance betwe

en directrices = 321 and eccentricity = 2

Q 9)

Find the equation of the ellipse passing through

(

15, 1

)

and

distance between whose foci is 8. Ans.Distance between foci = 8

Ans.


2 ae = 8ae = 4

(i) But e = 1 2

Ellipse passes through

(

15, 1 ,

)

Given 2 a=8

a a = 32 = 16 e e

x2 a2

+

y2 b2

=1

15


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Here13 b2 = a 2 1 e2 = 64 1 = 64 = 48 44

a

2

++

1

b2a2

=11 =1

(

)

15

a

2

a 2 = 64 and b2 = 48

(1 e )

2

from (i) the equation of the ellipse is x2 y2 + =1 64 48

15

a

2

+

1

16 a 2 1 2 a 1

=1

15

a

2

+

Q 8)


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Find the equation of the ellipse (referred to its principal axes) Distance between foci = minor axis, latus rectum = 10.

(a

2

16

)

=1

15 a 2 16 + a 2 = a 2 a 2 16 15a 2 240 + a 2 = a 4 16a 2

(

)

(

)

Ans.


a 4 32a 2 + 240 = 0 a 4 20a 2 12a 2 + 240 = 0

(i) (ii) (iii)

(a

2

20 a 2 12 = 0

)(

)

2ae = 2b and 2b = 10 a2

ae = b

a 2 = 20 or a 2 = 12

b2 = a 2 1 e 2

(

)

From (ii) and (iii), we get a 2 = 100 and b2 = 50 From (i), the equation of the

ellipse isy2 x2 + =1 100 50


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b 2 = a 2 a 2e 2When a 2 = 20 ,

b 2 = 20 16=4

equation of ellipse is,

x 2 y2 + =1 20 4 If a 2 = 12 ,

Ellipse


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89


b 2 = 12 16 b 2 = 4 But b 2 4 Equation of ellipse is, x 2 y2 + =1 20 4 b2= 27. Put these values in (i) the equation of the ellipse is x2 y2 + =1 36 27

Q

13) Q

11) Find the equation of the ellipse (referred to its principal axes) such that latus rectum = 39/4 and eccentricity = 5/8. Ans.Let equation of the ellipse isx2 a2 + y2 b2 =1

Find eccentricity of the ellipse, if its latus rectum = (1/2) major axis.

Ans.

Let the equation of the ellipse bex2 a2 + y2 b2 = 1 major axis = 2a,

eccentricity = e; Latus rectum =

2b2 a

(i) 39 4 ..(ii) ...(iii) and

2b2 1 a2 = ( 2a ) b2 = a 2 2

Since latus rectum = 5 e= 8 2b 39 = and a 42

b2 = a 2 1 e 2

(

)

a2 = a 2 1 e2 2

(

)

a2 1 = a 2 1 e2 = 1 e 2 2 2

(

)

(

)

e2 = e2 =

25 2 39 b2 = a 2 1 =a 64 64


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1 1 e= 2 2 1 1 e= 3 3

We get a 2 = 64 and b2 = 39 . Equation of the ellipse is x2 y2 + =1 64 39

Q 14)

Find eccentricity of the ellipse (referred to its principal axes) such its Dista

nce between directrices=3(distance between its foci)

Q 12)

Find the equation of the ellipse (referred to its principal axes) such Focus at(3,0) and whose directrix is x = 12. Ans.

Ans.

Let equation of the ellipse isx2 a2 + y2 b2 =1

2a 1 = 3 ( 2ae ) 3e 1 = 3e 2 e 3 1 1 e2 = e = 3 3 Here

(i) a = 12 e (iii)

Given ae = 3 (ii) and

from (ii) and (iii) we get a 2 = 36 andEllipse


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90

Q 15)

Find focal distance of the point

A 5,4 3

Q 16)

The length of latus rectum of the parabola y 2 = 4x is equal to length of m in or axis of ellipse. If ( 3,0 ) is one vertex of this ellipse, find equation of ellipse. Also find its eccentricity.

(

)

on ellipse

16x 2 + 25y 2 = 1600

Ans.

The equation of ellipse is x2 y2 16x + 25y = 1600 + =1 100 642 2

a 2 = 100 & b2 = 64 a = 10 &b = 8 a > b

Ans.

The equation of parabola is y 2 = 4x 4A = 4 A = 1

b2 = a 2 1 e 2 64 = 100 1 e2 ;

(

)

length of latus rectum = 4A = 4 x 1 = 4 Length of minor axis = 4 2b = 4 b2 2

(

64 ) 100 = (1 e )

=2 Also (3,0) is one vertex of ellipse a = 3 Equation of ellipse is x2 y2 + 1 94 x2 32 + y2 22 =1

64 = 1 100 e2 = 36 2 9 3 ;e = ;e = 100 25 5

(

a > b)

a Equation of directrices are x = and e a x= e x=


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10 10 50 and x = x= 3/5 3/5 3

b2 = a 2 1 e 2 ; 4 = 9 1 e 2

(

)

(

)

4 4 5 5 = 1 e2 ;e2 = 1 ;e2 = ; e = 9 9 9 3 5 3

eccentricity =

and x =

50 3

Q 17)

P is any point on the ellipsex2 y2 + = 1, S and S

are its foci. 25 9

50 50 x = 0 and x + =0 3 3

SP = ePM =

3 5

5

50 3 = 7 units 2 1

Find the perimeter of SPS

. Ans.d

M

A

O(h,0) x= a/e x= a/e

Y d P (x,y) M A S X

S

P = ePM

3 = . 5

5+

50 3 = 3 15 + 50 = 13 units 2 5 3 1

Given equation of ellipse is

x2 y2 + =1 25 9

S and S are foci. Here a 2 = 25,b2 = 9 ;

Ellipse


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91


e=

a2 b 2 a

2

=

25 9 4 = 25 5

Q r2 cos 90 + , r2 sin 90 + ( r2 sin + r2 cos )

(

(

)(

))

e = 5.

4 =4 5

S (4,0) nd S ( 4,0)

Now P lies on the ellipse

2 sin2 2 cos =1 r1 +

2 b2

x 2 y2 + = 1, 2 b2

If P is ny point on ellipse SP = ePM SP = ePM & SP

= ePM

SP + S

P = e ( PM + PM

) = e ( ZZ

) =e 2 = 2 = 2 5 = 10 e

cos 2 sin2 1 + = 2 2 2 b r1(i)

4 SS

= 2 e = 2 5 = 8 5 Perimeter of SPS

= SP + S

P + SS

= 10 + 8 = 18 units.

1 cos 2 sin2 = + OP 2 2 b2

Simil rly Q lies on the ellipse2 x 2 y2 cos 2 2 sin =1 + 2 = 1, r2 + 2 2 b b2

Q 18)

P nd Q re two points on the

ellipse


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x2 2

+

y2 b2

sin2 cos 2 1 + = 2 2 2

b r2

= 1, such th t seg PQ

subtends right ngle t the centre O of the ellipse. Show th t1 2 + 1 Q2Y

1 sin2 cos 2 = +

Q 2 2 b2 Adding (i) nd (ii), we get

(ii)

=1 2

+

1 b2

.

Ans.

Y Q

(90+ )

P X

1 1 cos 2 + sin2 sin2 + cos 2 + = + 2 Q 2 2 b2 1 1 1 1 + = 2+ 2 22 Q b

Q-19)

P () is point on ellipse

x 2 y2 + = 1, 2 b2

whose foci re S & S ' prove th tSeg PQ subtends right ngle t the centre

, the line

P

Q. If line

P m kes n ngle with the x- xis,

(i) SP .S ' P = 2 sin2 + b 2 cos 2 (ii) SP + S ' P = 2 Ans.Y d' M' A' Z'(- /e,0)

d P M

S' S A Z X( /e,0)

then

Q m kes n ngle 90 + with the x- xis.


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(

P ) = r1 & (

Q) = r2 . ThenP ( r1 cos ,r1 sin ) nd

The e u tion of the ellipse is

Ellipse


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M hesh Tutori ls Science

92

x 2 y2 + =1 2 b2 Its foci re S ( e,0 ) nd S ( e ,0 ) . The equ tion of its directrices re

Q

20)

S is focus of the ellipse

x 2 y2 + =1 4 3

corresponding to directrix x = 4 . Find the equ tion of the circle which p ssesthrough S, centre of the ellipse nd the point (3,3) . Ans.Comp ring we get 2 = 4 , b 2 = 3e2 = 2 b2 4 3 1 1 = = e = 2 4 4 2

x = / e

Let PM

nd PM be the length of the from P on the directrices. cos Then, PM =

1

e nd

cos + PM ' = 1

e

1

e = 2 = 1 2

Focus S ( e,0 ) (1,0 ) Let re uired e u tion of the circle bex 2 + y 2 + 2gx + 2 fy + c = 0 (i) Since it p sses through centre

(0,0); focus S(1,0) nd (3,3) c = 0 (ii) 1 + 0 + 2g + 0 + 0 = 0 2g = 1

By the focus directrix property of the ellipse.SP = ePM & S ' P = ePM ' where e < 1

SP = e cos / e = e cos 1

= (1 e cos )S P = e cos + / e = e cos + 1 = (1 + e cos ) (i) SP .S ' P = (1 e cos ) = (1 + e cos ) = 2 1 e 2 cos 2 = 2 2e 2 cos 2 = 2 2 b 2 cos 2 ( e = b )2 2 2 2

g=

1 (iii) 2

9 + 9 + 6g + 6 f + 0 = 0 6g + 6 f = 0 f =

(

)

(


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)

5 2

Equ tion of required circle 15 is x 2 + y 2 + 2x + 2y + 0 = 0 2 2

= 2 2 cos 2 + b 2 cos 2 = 2 1 cos 2 + b 2 cos 2

(

)

x 2 + y 2 x 5y = 0

= 2 sin2 + b 2 cos 2 .ii)

SP + S ' P = (1 e cos ) + (1 + e cos )= e cos + + e cos = 2

Ellipse


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93


Q-21)

Find the C rtesi n co-ordin tes of the point P 60 ellipse

( )

lying on the

( x1 3)2 + y12 = 64 16( x1 + 3)2 + y12 + ( x1 + 3 )2 + y12x12 6x1 + 9 + y12 = 64 16

x 2 y2 + = 1 nd the eccentric 25 9

5 3 , ngle of the point Q 2 2

Ans.

(i) Comp ring = 5, b = 3 ; Given th t

( x1 + 3 )2 + y12

+ 6x12 + 6x1 + 9 + y12

= 60x = cos = 5 cos 60 =y = b sin = 3 sin 60 =

12x1 64 = 16 x12 + 6x1 + 9 + y12

5 & 2

Dividing by 4, 3x1 + 16 = 4 x12 + 6x1 + 9 + y12 Squ ring both sides,9x12 + 96x12 + 256 = 16 x12 + 6x12 + 9 + y12

3 3 2

5 3 3 C rtesi n co ordin tes re , 2 2 5 (ii) x = cos = 5 cos = 2 cos = 1 2 3 1 sin = 2 2

(

)

9x11 + 96x12 + 256 = 16x12 + 96x1 + 144 + 16y12 7x12 + 16y12 = 112 x12 y12 + =116 7

y = b sin = 3 sin = Hence = 45

e u tion of locus of P is n ellipse,x 2 y2 + =1 16 7

Q-22)

If A nd B re two fixed points such th t l (AB)=6. Then show the locus of the point P which moves so th t l(PA) + l(PB) = 8 is n ellipse


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G

UP-(B): CLASSW

K P

BLEMS Q-1) Find e u tion of the t ngent to the ellipse x2 + 4y 2 = 100 t ( 8,3 ) Ans.E u tion of the ellipse x2 y2 + =1 100 25

Ans.

Let A ( 3,0 )

nd B ( 3,0 ) Let P ( x1, y1 ) l(PA) + l(PB) = 8

Equ tion of the t ngent to the given ellipse t (8,3) is

( x1 3)2 + (y1 0 )2 + ( x1 + 3 )2

xx1 yy1 + 2 =1 2 b

+ ( y1 0 ) = 8 =8

2

i.e.

8x 3y + =1 100 25

( x1,3)2 + y12

( x1 + 3)2 + y12

2x 3y + =1 25 25

Squ ring both sides,

2x + 3y = 25

Ellipse


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94

Q 2)

Find equ tion of the t ngent to the ellipse 9x 2 + 16y 2 = 144 t

x 7 + 16 y

9 4 =1

(4 cos , 3 sin ).Ans.E u tion of the ellipse is x 2 y2 + =1 16 9

7x y + =1 16 4 7x + 4y = 16

E u tion of the t ngent to the given ellipse t (4 cos , 3 sin ).

xx1 yy1 + 2 =1 2 b

Q-4)

A t ngent to b 2 x 2 + 2y 2 = 2b 2 cuts the co-ordin te xis A nd B nd touches the ellipse in the first u dr nt t the mid-point of AB. Show th t its e u

tion is

4 cos .x 3 sin + =1 16 9 3cos x + 4sin y = 12

Q-3)

Find e u tion of t ngent to ellipse

9x 2 + 16y 2 = 144

t point L. Where L

bx + y = b 2 .

Ans.

The e u tion of ellipse is Let e u tion of t ngent t

x2 2

+

y2 b2

=1

is end of L tus

ectum in 1st Qu dr nt Ans.x 2 y2 9x 2 + 16y 2 = 144 + =1 16 9 = 16 & b = 9 = 4 & b = 32 2

P ()

x cos y sin + = 1 (i) b

T ngent meet x xis t A ,0 cos

b2 = 2 1 e 2 9 = 16 16e 2 16e 2 = 7 7 e = 162


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(

)

b T ngent meet y xis t B 0, sin

P (

cos ,b sin ) is mid point of AB b +0 0+ cos sin & b sin = cos = 2 2 cos = b & b sin = 2 cos 2 sin

e=

7 4

b2 L e , 9 7, 4

cos2 = cos =

1 1 & sin2 = 2 21 1 & sin =

2 2

E u tion of tgf isxx1

2

but P lies in first u dr nt is cutecos = 1 2 & sin = 1 2

+

yy1 b2

=1

Hence e u tion of t ngent becomes

Ellipse


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95


x 1 y 1 . + . =1 2 b 2 x 2 + y b 2 = 1 bx + y b 2 = 1

2x 1

b + 2y = 1 1

2

2

bx + y = b 2 This is e u tion of re uired t ngent

2 b2 2 b2 + = 4 2 + 2 = 4 2 2 4x1 4y1 x1 y1

E u tion of locus of M (x1 ,y1 ) is 2 b2 + =4 x2 y2

Q-5)

Show th t the e u tion of the locus of the mid point of the portion of t ngentto the ellipsex2 2 + y2 b2 =1

Q-6)

A t ngent to ellipse

x2

2

+

y2 b2

=1

intercepted by the co-ordin te xis 2 b2 is 2 + 2 = 4 . x y

meets the co-ordin te xes t L nd M respectively. If t ngent to

Ans.

The e u tion of ellipse is

x y + 2 =1 2 b

2

2

ellipse

x2

2

+


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y2 b2

= 1 meets the

Let P () be ny point on ellipse. E u tion of t ngent t P is x cos y sin + =1 b

co-ordin te xes t L nd M respectively. If CL = p , CM = where C is centre of ellipse. Show th t

2 p2 + b2 2 =1x2 2 y2 b2

This t ngent meet x- xis suppose t A nd y- xis t B. b A cos , 0 nd B 0, sin

Ans.

The e u tion of ellipse is

+

=1

E u tion of t ngent t P ( x1 , y1 ) is xx1

2

Let M (x1 ,y1 ) be point on locus M is mid point of AB By mid point formul

b +0 0+ sin cos nd y1 = x1 = 2 2x1 = b nd y1 = 2 cos 2 sin

+

yy1 b2

=1

It meet x xis in point L. put y = 0 xx1 2 = 1 x = 2 2 L ,0 x x1 1

T ngent lso meet Y- xis put x = 0 in the e u tion of t ngent yy1 b2 =1 y = b2 b2 M 0, y y1 1

b cos = nd sin = 2 x1 2 y1

cos 2 + sin 2 = 1

Centre ( 0,0 )

Ellipse


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96

p = CL = 2 p2 b2 2

2 b2 ;

= CM = x1 y1= 2 2 x 12

+

+

b2 b2 y1 2

=

x12 2

+

y12 b2

2m 2 + b 2 e + P12 + P22 = m2 +1

=

2 b2 + =1 p2 q 2

2m 2 + b 2 + e m2 + 1 +

2

2m 2 + b 2 + 2 e 2m 2 + b 2 + 2e 2 m2 + 1m2 + 1

(x1,y1 )2 2

x2 y2 lies on ellipse 2 + 2 = 1 b=

2 2m 2 + 2b 2 + 2 2e 2

x1 y + 12 = 1 Hence proved. 2 b

2 2m 2 + 2 2 1 e 2 + 2 2e 2 =m2 + 12 2m 2 + 2 2 2 2e 2 + 2 2e 2

(

)

Q 7)


25/40

Prove th t the sum of the squ res of the perpendicul rs dr wn on ny t ngent tothe ellipse=

m2 + 1

2

2 m 2 + 1 =m +12

b + x + y = b , from thepoint (o, e ) nd (o, e ) is const nt. Ans.For ellipse b 2 + x 2 + 2y 2 = 2b 2 , Equ tion of t ngent is,

2

2

2 22 2

(

)

= 2 2

P12 + P22 = const nt

Q 8)

P nd Q re two points on the

ellipse

y = mx + 2m 2 + b 2Let P1 be the r dist nce between (0, e) nd t ngent, P1 = m ( 0 ) ( e ) + 2m 2 + b 2 m2 +1 2m 2 + b 2 e m2 +1

x2 y2 + = 1 such th t their 25 16

eccentric ngles differ by

. Show 2

that the locus of the oint of intersection of the tangents drawn from P and Q is also an elli se given by Ans.x2 y2 + = 2. 25 16 x2 y2 + = 1 (i) 25 16

=

Let P2 be the r distance between (0, ae) and tangent,

Equation of elli se is

a 2 = 25, b 2 = 16 a = 5, b = 4 Let P () nd Q + be points on 2


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P2 =

m ( 0 ) + ( e ) + 2m 2 + b 2 m2 + 1 m + b + e m +12 2 2 2

ellipse such th t their eccentric ngle

=

differ by

. 2

Elli se


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97


Equation of tangent at P and Q are x cos y sin + . =1 b x cos + y sin + 2 + 2 . =1 nd b

Q-11)

Show th t the line x + 3 2y = 9 is t ngent to the ellipsex2 y2 + =1 9 4

Ans.

Here 2 = 9, b2 = 4 E u tion of the line x + 3 2y = 9

x cos y sin + . = 1 (ii) 5 4 x sin y cos + . = 1 (iii) 5 4

y=c= 9

x 3 2 =

+ 3

9 3 2

m=

1 3 2

and

and

cos 2 + = sin ;sin 2 + = cos To find the locus of point ofintersection of t ngents (ii) nd (iii) we h ve to elimin te , S u ring nd

dding e u tions (ii) nd (iii) x cos y sin 2 + . 4 5 x sin y cos + + =2 4 52

3 2

2

Now y = mx + c is t ngent to the ellipse if c2 = 2m2 + b2 3 9 c = nd = 2 22 2

1 4 9 1 m + b = 9 +4 = + = 2 1 2 3 22 2 2

2

Hence c2 = a 2m2 + b2

x 2 cos2 2xy.cos .sin y 2 sin2 + + 25 20 16


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x 2 sin 2 2xy. sin . cos y 2 cos 2 + + =2 25 20 16 x2 y2 cos 2 + sin 2 + sin 2 + cos 2 = 2 25 16 x2 y2 + = 2 sin2 + cos 2 = 1 25 16 +

The line is t ngent to the ellipse.

Q-12)

Find k , if the line x + y + k = 0 touches the ellipse x 2 + 4y 2 = 20.

[

]

[

]

Ans.

The line x + y + k = 0(

)

put x = y k in x 2 + 4y 2 = 20. ( y k ) + 4y 2 = 20. y 2 + 2ky + k2 + 4y 2 20 = 02

This is equation of required locus.

Q 9 and Q 10 is given in notes

5y 2 + 2ky + k 2 20 = 0Since the line touch the elli se This equation has two equal roots

(

)

=0

( 2k )2 4 5 ( k 2 20 ) = 04k 2 20k 2 + 400 = 0 16k 2 = 400 k 2 = 25 k = 5Ellipse


29/40


98

Q 13)

Find equation of tangent to ellipse

1 x2 + y 2 = 1 having slope is 4 2

Q 15)

Find the equation of the tangents to the ellipsex2 y2 + = 1 making 64 36

Ans.

Equation of ellipse is

x2 y2 + =1 4 1 a 2 = 4,b2 = 1 a = 2,b = 1 1 2

equal intercepts on the co ordinate axes. Ans.x2 y2 + = 1 a 2 = 64,b2 = 36, 64 36 Tangents making equal intercepts

slope of tangent m =

Equation of ellipse is in the formx a2 2

( k,0 ) & ( 0,k ) Slope =

on the co

ordinate axes.

+

y

2

b2

=1

k 0 = 1 0k

then equation of tangents is

The equation of the tangent line is

y = mx a 2m2 + b2y = 1 1 x 4 +1 2 2 2

y = mx a 2m2 + b2y = x 64 ( 1) + 36 y = x 1002

y = y =


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1 x 1+1 2 1 x 2 2

y = x 10 Equations of tangents be

x + y + 10 = 0 and x + y 10 = 0

2y + x = 2 2

Q 16) Q 14) Find equations of tangent to thex2 y2 ellipse + = 1, parallel to 144 25

If the line y = mx + a 2m2 + b2 touches the ellipseb2 x 2 + a 2 y 2 = a 2b2 at P ( a cos ,b sin )

x + y 3 = 0Ans.Slo e of the line x + y 3 = 0 (i) is 1 Slo e of the tangent line arallel to(i) is 1

show that tan = Ans.

b amx2 a2+

The equation of elli se is

y2 b2

=1

and P (a cos , b sin ) is on ellipse. E

u

tion of t

ngent to ellipse

t P is xcos y sin + =1 b (i)

E u tion of the t ngent line isy = mx

2m2 + b2

m = 1,a 2 = 144,b2 = 25 y = x 144 + 25 y = x 169 y = x 13

cos b slo e of tangent = a = cot . sin b But e u tion of t ngent tP is

x + y + 13 = 0 nd x + y 13 = 0Elli se


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99


y = mx + a 2m2 + b2 (ii)

slo e of tangent = m

Equation (i) and (ii) re

resent same tangent

Area of OAB= l (OA ) l (OB ) =square units. 1 2 1 42 = 4 2

slopes are equal m=b am 1 cot = b t n b am

Q 18)

The line x y 5 = 0 touches the elli

se whose foci are S (3,0 ) &tan =

S ( 3,0 ) . Find the equation of theelli se

Q 17)

A tangent having slo e

1 to the 2

Ans.

Let the equation of the elli se bex2 a2+

y2 b2

elli se 3x 2 + 4y 2 = 12 interacts the x and y axes in the oint A andB res ectively. If O is the origin,

=1

(i)

The foci of the elli se are ( ae,0 ) ae = 3

(ii)

find the area of the OAB . Ans.The equation of the ellipse is x2 y2 + = 1 a 2 = 4, b2 = 3 ..(i) 4 3

b 2 = a 2 1 e 2 = a 2 a 2e 2 = a 2 9 Now slope of the given tangent is m =1

(


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)

1 Given slope of the tangent is m = 2The equation of the tangents with slope m are y = mx a 2m 2 + b 21 1 Here y = x 4 + 3 2 22

a2m2 + b2 = a2 (1) + a2 9 = 2a2 92

(

)

But line x y 5 = 0 condition for tangency

c 2 = a 2m 2 + b 2 2a 2 = 34 a 2 = 17from b 2 = 17 9 = 8 Equation of the ellipse is x 2 y2 + =1 17 8

y=y=

1 1 x 1+ 3 y = x 4 2 2

1 x 2 2y = x 4 2

x + 2y = 4Let the tangent x + 2y = 4 meet the x axis in A and y axis in B.

x = 4 A = ( 4,0 ) 2y = 4 B = ( 0, 2)

Ellipse


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100

Q 19)

Find the equation of tangent to the ellipse

x 2 y2 + = 1 from (3,

2). 7 4 x 2 y2 + =1 7 4

Q 20)

Show that the locus of the foot of the perpendicular drawn from the ellipse b 2x 2 + a 2y 2 = a 2b 2 to any tangent is x 2 + y 2

Ans.

The equation of ellipse is

(

)

2

= a 2 x 2 + b 2y 2 .

a 2 = 7, b 2 = 4 a = 7 ,b = 2 Equation of ellipse is in the form x 2 y2 + =1a 2 b2 Let equation of tangents be y = mx a 2m2 + b2 This passes through (3,

2) 2 = 3m 7m2 + 4

Ans.

The equation of ellipse is

x 2 y2 + =1 a 2 b2

Equation of a tangent with slope m is, y = mx a 2m 2 + b 2 Let P (x1,y1 ) be the foot of the perpendicular from centre (0,0 ) on tangent.

P (x1,y1 ) lies on tangenty1 = mx1 a 2m 2 + b 2

2 3m = 7m 2 + 4

(i)

4 + 12m + 9m 2 = 7m 2 + 4 2m 2 + 12m = 0 m 2 + 6m = 0 m (m + 6) = 0

slope of tangent = m slope of OP = 1 m1 x m 1 x1 m (ii)

Equation of OP is y =

m = 0 or m = 6If m = 0 and line passing through (3, 2) then equation of tangent is

P (x1,y1 ) lies on OP y1 =m= x1 y1


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y (2) = 6(x 3)

Put in (i)

y +2=0If m = 6 and line passing through (3, 2) then equation of tangent is

x x y1 = 1 x1 + a 2 1 + b2 y1 y1 y1 =

2

y (2) = 6(x 3) y + 2 = 6x + 18 6x + y = 16

a 2 x12 + b2 y12 x12 + y1 y12

y12 = x12 + a 2 x12 + b2 y12 x12 + y12 = a 2 x12 + b2 y12 x12 + y12

Required equations are y + 2 = 0and 6x + y = 16

(

)

2

= a 2 x12 + b2 y12

Equation of locus of P ( x1, y1 ) is

(xEllipse

2

+ y2

)

2

= a 2 x 2 + b2 y 2 .


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101


Q 21)

At the point P on the circle

x 2 + y 2 = a 2 and the point Q on the

Q 24)

Find the equation of the tangents to the ellipse 2x 2 + 3y 2 = 5 which are perpendicular to the line

ellipse

x2 a2

+

y2 b2

= 1 tangents drawn

3x + 2y + 7 = 0Ans.Equation of Ellipses,2x 2 + 3y 2 = 5

to respective curves. If xcoordinates of P and Q are the same, Prove that two tangents will intersect at the point on the x axis Ans.Let P (a cos ,sin ) nd

Q (

cos ,b sin )

2x 2 3y 2 + =1 5 5 x 2 y2 + =1 5 5 2 3

e u tion of t ngent t P to circle,x cos + y sin = x cos + y sin = 1

2 = 5 , b2 = 5 2 3Slope of 3x + 2y + 7 = 0 is 3 2

..(i)

equation of tangent at Q to elli se.x cos y sin + =1 b

Slope of t ngents is, m = e u tion of t ngent is, y = mx 2m 2 + b 2

y=

2 3

.. (ii)

Subtr ct E u tion (i) nd (ii),y sin y sin =0 a b1 1 y sin = 0 a b

2 5 4 5 x + 3 2 9 3 2 10 5 x + 3 9 3 2 10 + 15 x 3 9


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y=

y sin = 0 y =0 t ngents dr wn to circle t point P nd intersects the t ngent dr wn to ellipse t point Q t point on x- xis

y=

y=

2 5 x3 3

3y = 2x 5 2x 3y 5 = 0 .. equation of tangent

Q 22 and Q 23 is given in notes

Elli se


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102

Q 25)

If P and Q are two oints on the elli se

x2 a2 + y2 b2 = 1 such that PQ

Q 26)

Find the locus of oint of intersection of the two tangents drawn from to the elli se such that i) sum of slo es = 2 ii) cot 1 + cot 2 = 3 where 1 ndx2 2 + y2 b2 =1

p sses through centre of the ellipse. If

is ny other point on the ellipse, prove th t

( Slope of P

) ( Slope of Q

) = Const nt

Ans. Ans.2 re inclin tions of t ngents.E u tion of t ngent

y = mx 2m 2 + b 2

P ( x1, y1 )

y1 mx1 = a 2m 2 + b 2Squaring Let P (a cos , sin ) nd Q ( a cos , a sin ) Let

( cos , sin ) Slope of P

= b ( sin sin ) ( cos cos ) b ( sin + sin ) ( cos + cos ) m1.m2 =

(y1 mx1 )2 = a 2m 2 + b 2

(x

2 1

a 2 m 2 2x1y1.m + y12 b 2 = 0

)

(

)

Let above quadratic equations hasroots m1 and m2 , which are slo es of tangents, m1 + m2 = 2x1y1 x12 a 2 y12 b 2 x12 a 2

Slo e of QR =

Slo e of PR Slo e of QR= b ( sin sin ) ( cos cos ) ( cos + cos ) 2 sin2 sin2 =

2 2

b ( sin + sin )


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i) Sum of Slopes = 2 m1 + m2 = 2 2x1y1 x12 a 2 =2

( (cos

cos

2

) ))

=

2 2 b 2 1 cos 1 + cos

(

x1y1 = x12 a 2

a2 b2 a2b

cos2 cos2 2

e u tion of locus of point is,xy = x 2 a 2 x 2 xy a 2 = 0

=

(cos

2

cos

2

)

cos2 cos2

=

a2

ii) If cot 1 + cot 2 = 31 1 + =3 t n 1 t n 2

= const nts

Ellipse


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103


1 1 + =3 m1 m2

[m1 = t n 1 nd m2 = t n 2 ]

m1 + m2 =3 m1m2

2x1y1 x12 a 2 y12 b 2 x12 a 2 2x1y1 y12 b 2

=3

=3

2x1y1 = 3y12 3b 2

equation of locus of Point P is,2xy = 3y 2 3b 2

*****

Elli se


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