If theta and phi are eccentric angles of th | vedclass

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Question

(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 le

Vedclass · Accepted Answer

$ \pm \frac{\pi }{2}$

Vedclass · Answer

(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 $P(a\cos 	heta ,\,b\sin 	heta )$ 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 	heta - 0}}{{a\cos 	heta - 0}} 	imes \frac{{b\sin \phi - 0}}{{a\cos \phi - 0}} = - \frac{{{b^2}}}{{{a^2}}}$

==> $\sin 	heta \sin \phi = - \cos 	heta \cos \phi $

==> $\cos (	heta - \phi ) = 0$

$\Rightarrow 	heta - \phi = \pm \frac{\pi }{2}$.

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

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