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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
${\tan ^{ - 1}}\left( { \pm \frac{{ae}}{b}} \right)$
${\tan ^{ - 1}}\left( { \pm \frac{{be}}{a}} \right)$
${\tan ^{ - 1}}\left( { \pm \frac{b}{{ae}}} \right)$
${\tan ^{ - 1}}\left( { \pm \frac{a}{{be}}} \right)$
Solution
(c) 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 $\left( {ae,\, \pm \frac{{{b^2}}}{a}} \right).$
$\therefore a\cos \theta = ae$ and $b\sin \theta = \pm \frac{{{b^2}}}{a}$
$⇒ \tan \theta = \pm \frac{b}{{ae}} $
$\Rightarrow \theta = {\tan ^{ – 1}}\left( { \pm \frac{b}{{ae}}} \right)$.
