Question

The eccentric angle $\theta$ of a point $(\sqrt{6}\cos \theta ,\sqrt{2}\sin \theta )$ on the ellipse $\frac{x^2}{6}+\frac{y^2}{2}=1$ whose distance from the centre of the ellipse is $2$ is

• A. $\pi /2$
• B. $\pi /6$
• C. $\pi /4$
• D. $\pi /3$

Answer 1

The correct option is C

$\pi /4$

Let $\theta$ be the eccentric angle of the point $P$. Then the coordinates of $P$ are $(\sqrt{6}\cos \theta ,\sqrt{2}\sin \theta )$. Centre of ellipse is origin and $OP=2$

$\therefore \sqrt{6{\cos }^2+2{\sin }^2\theta }=2\Rightarrow {\sin }^2\theta =\frac{1}{2}\therefore \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$