1take a point (acosθ, bsinθ) ...
nahi hua kya waise ??
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) ]
-
UP 0 DOWN 0 1 3
3 Answers
11YEH TOH BAHUT SIMPLE LAG RAHA HAI
ITS RELATED TO AUXILLARY CIRCLE AND YOU GET THE ANSWER PROBABLY
1(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!!!!: )