Question

The 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

• A. 4.00
• B. 4.0
• C. 4

Answer 1

Answer: (A), (B), (C)

Let $P$ and $Q$ be the end points of the chord.
Let one end of the chord is $P\equiv (2\sqrt{2}\cos \theta ,2\sin \theta )$
As eccentric angles of its extremities differ by $\frac{\pi }{2}$, so
$Q\equiv (2\sqrt{2}\cos (\frac{\pi }{2}+\theta ),2\sin (\frac{\pi }{2}+\theta ))\Rightarrow Q=(-2\sqrt{2}\sin \theta ,2\cos \theta )$

Now,
$P{Q}^2=8(\cos \theta +\sin \theta {)}^2+4(\sin \theta -\cos \theta {)}^2\Rightarrow PQ=\sqrt{12+4\sin 2\theta }\therefore (PQ{)}_{max}=4$