Question

The distance of a point on the ellipse $\frac{x^2}{6}+\frac{y^2}{2}=1$ from the centre is $2$, then the eccentric angle is-

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B $\frac{\pi }{4}$

Given ellipse $\frac{x^2}{6}+\frac{y^2}{2}=1$

Let $\theta$ be the eccentric angle of the point $p$

coordinate of $p$ $(\sqrt{6}cos\theta ,\sqrt{2}sin\theta )$

Given distance $=2$

$\therefore$ $OP=2$

$\sqrt{6co{s}^2\theta +2si{n}^2\theta }=2\Rightarrow 6co{s}^2\theta +2si{n}^2\theta =4$

$3co{s}^2\theta +si{n}^2\theta =2$

$2si{n}^2\theta =1$

$si{n}^2\theta =\frac{1}{2}\Rightarrow sin\theta =\pm \frac{1}{\sqrt{2}}$

$\therefore$ eccentric angle $\theta =\pm \frac{\pi }{4}$