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Transcript of 27780660 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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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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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.
Let the equation of the ellipse bex2 a2 + y2 b2 =1
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.
Let the equation of the ellipse bex2 a2 + y2 b2 =1
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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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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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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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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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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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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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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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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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)
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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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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
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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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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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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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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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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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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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