Conic Sections Question 130
Question: The eccentric angles of the extremities of latus recta of the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ are given by
Options:
A) $ {{\tan }^{-1}}( \pm \frac{ae}{b} ) $
B) $ {{\tan }^{-1}}( \pm \frac{be}{a} ) $
C) $ {{\tan }^{-1}}( \pm \frac{b}{ae} ) $
D) $ {{\tan }^{-1}}( \pm \frac{a}{be} ) $
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Answer:
Correct Answer: C
Solution:
Coordinates of any point on the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ whose eccentric angle is $ \theta $ are $ (a\cos \theta ,b\sin \theta ). $
The coordinates of the end points of latus recta are $ ( ae,\pm \frac{b^{2}}{a} ). $
$ \therefore a\cos \theta =ae $ and $ b\sin \theta =\pm \frac{b^{2}}{a} $
therefore $ \tan \theta =\pm \frac{b}{ae}\Rightarrow \theta ={{\tan }^{-1}}( \pm \frac{b}{ae} ) $ .
BETA
