• # question_answer If $\theta$ and $\varphi$ 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 $\theta -\varphi$ is equal to A)            $\pm \frac{\pi }{2}$                B)            $\pm \pi$ C)            0      D)            None of thesew

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 $Q(a\cos \varphi ,\,b\sin \varphi )$ be ends of these two diameters. Then ${{m}_{1}}{{m}_{2}}=-\frac{{{b}^{2}}}{{{a}^{2}}}$                   $\Rightarrow \frac{b\sin \theta -0}{a\cos \theta -0}\times \frac{b\sin \varphi -0}{a\cos \varphi -0}=-\frac{{{b}^{2}}}{{{a}^{2}}}$                    Þ$\sin \theta \sin \varphi =-\cos \theta \cos \varphi$Þ$\cos (\theta -\varphi )=0\Rightarrow \theta -\varphi =\pm \frac{\pi }{2}$.