Conic-Sections Question 554
Question: Let d be the perpendicular distance from the centre of the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ to the tangent drawn at a point P on the ellipse. If $ F_1 $ and $ F_2 $ be the foci of the ellipse, then $ {{(PF_1-PF_2)}^{2}}= $
Options:
A) $ 4a^{2}( 1-\frac{b^{2}}{d^{2}} ) $
B) $ a^{2}( 1-\frac{b^{2}}{d^{2}} ) $
C) $ 4a^{2}( 1-\frac{a^{2}}{d^{2}} ) $
D) $ b^{2}( 1-\frac{a^{2}}{d^{2}} ) $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let the point P be $ (acos\theta ,bsin\theta ) $ The equation of tangent at P is $ \frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1 $ ? (1) If d be the length of perpendicular from the centre $ C(0,0) $ of the ellipse to the tangent given by (1) then $ d=\frac{1}{\sqrt{\frac{{{\cos }^{2}}\theta }{a^{2}}+\frac{{{\sin }^{2}}\theta }{b^{2}}}} $
$ \Rightarrow \frac{1}{d^{2}}=\frac{{{\cos }^{2}}\theta }{a^{2}}+\frac{{{\sin }^{2}}\theta }{b^{2}} $
$ \Rightarrow \frac{b^{2}}{d^{2}}=\frac{b^{2}}{a^{2}}{{\cos }^{2}}\theta +1-{{\cos }^{2}}\theta $
$ \Rightarrow 1-\frac{b^{2}}{d^{2}}=( 1-\frac{b^{2}}{a^{2}} ){{\cos }^{2}}\theta =e^{2}{{\cos }^{2}}\theta $ ? (2) Now, $ {{(PF_1-PF_2)}^{2}}={{(2aecos\theta )}^{2}} $ $ =4a^{2}e^{2}{{\cos }^{2}}\theta =4a^{2}( 1-\frac{b^{2}}{d^{2}} ) $
BETA
