Conic-Sections Question 591
Question: The sum of the squares of the perpendiculars on any tangent to the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ from two points on the minor axis each at a distance $ \sqrt{a^{2}-b^{2}} $ from the centre is
Options:
A) $ 2a^{2} $
B) $ 2b^{2} $
C) $ a^{2}+b^{2} $
D) $ a^{2}-b^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \sqrt{a^{2}-b^{2}}=\pm ae(0,ae) $ So, the points are (ae, 0) and (0, -ae). Let $ \frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1 $ be a tangent then sum of squares of perpendicular from these points is $ =\frac{{{( 1-\frac{ae}{b}\sin \theta )}^{2}}+{{( 1+\frac{ae}{b}\sin \theta )}^{2}}}{\frac{{{\cos }^{2}}\theta }{a^{2}}+\frac{{{\sin }^{2}}\theta }{b^{2}}} $ $ =\frac{2( 1+\frac{a^{2}e^{2}}{b^{2}}{{\sin }^{2}}\theta )}{\frac{{{\cos }^{2}}\theta }{a^{2}}+\frac{{{\sin }^{2}}\theta }{b^{2}}} $ $ =2a^{2}( \frac{b^{2}+(a^{2}-b^{2})sin^{2}\theta }{b^{2}(1-sin\theta )+a^{2}{{\sin }^{2}}\theta } )=2a^{2} $
BETA
