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If $\theta $ and $\phi $ are eccentric angles of the ends of a pair of conjugate diameters of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, then $\theta - \phi $ is equal to
$ \pm \frac{\pi }{2}$
$ \pm \pi $
$0$
None of these
Solution
(a) Let $y = {m_1}x$ and $y = {m_2}x$ be a pair of conjugate diameters of an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
and let $P(a\cos \theta ,\,b\sin \theta )$ and $Q(a\cos \phi ,\,b\sin \phi )$ be ends of these two diameters.
Then ${m_1}{m_2} = – \frac{{{b^2}}}{{{a^2}}}$
$ \Rightarrow \frac{{b\sin \theta – 0}}{{a\cos \theta – 0}} \times \frac{{b\sin \phi – 0}}{{a\cos \phi – 0}} = – \frac{{{b^2}}}{{{a^2}}}$
==> $\sin \theta \sin \phi = – \cos \theta \cos \phi $
==> $\cos (\theta – \phi ) = 0$
$\Rightarrow \theta – \phi = \pm \frac{\pi }{2}$.
