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

Answer: 4

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 $Q(-2\sqrt{2}\sin \theta ,2\cos \theta )$
$\begin{array}{l}\therefore (PQ=8(\cos \theta +\sin \theta +4(\sin \theta -\cos \theta \\ =12+4\sin 2\theta \le 16\\ \therefore (PQ=4\end{array}$