Question

Maximum length of chord of the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1,$ such that eccentric angles of its extremities differ by $\frac{\pi }{2}$ is________ $.$

Answer 1

Given ellipse is $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$
Two points on the ellipse whose eccentric angles differ by $\frac{\pi}{2}$ are $P(2 \sqrt{2} \cos \theta, 2 \sin \theta)$ and
$ Q\left(2 \sqrt{2} \cos \left(\theta+\frac{\pi}{2}\right), 2 \sin \left(\theta+\frac{\pi}{2}\right)\right) $
or $Q(-2 \sqrt{2} \sin \theta, 2 \cos \theta)$
$ \begin{array}{l} \therefore(P Q=8(\cos \theta+\sin \theta+4(\sin \theta-\cos \theta \\ =12+4 \sin 2 \theta \leq 16 \\ \therefore(P Q=4 \end{array} $