JEE-95 Question

Let d be the perpendicular distance from the center of the ellipse
x2/a2 + y2/b2 = 1
to the tangent drawn at a point P on the ellipse. If F' and F'' are the two foci of the ellipse , then show that

(PF' - PF'')2= 4a2[ 1- (b2/a2) ]

3 Answers

1
skygirl ·

take a point (acosθ, bsinθ) ...

nahi hua kya waise ??

1
Amit Choraria ·

(x/a)cos (theta)+(y/b)sin (theta)=1

Foci F1=(ae,0) ; Foci F2=(-ae,0)

d=1/(√(cos^2(theta)/a^2)+√(sin^2(theta)/b^2))

d=ab/(√(a^2sin^2(theta)+b^2cos ^2(theta)))

Simplify it a bit and you will get:

d=4cos^2(theta)(a^2-b^2)

4a^2e^2cos^2(theta)=(2aecos(theta))^2

=[(a-aecos(theta))-(a+aecos(theta))]^2

=(PF1-PF2)^2

Hence,Proved.

I hope it helps!!!!: )

11
pawanseerwani seerwani ·

YEH TOH BAHUT SIMPLE LAG RAHA HAI
ITS RELATED TO AUXILLARY CIRCLE AND YOU GET THE ANSWER PROBABLY

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