Question

If 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

Answer: (B)

Equation of the ellipse is $\frac{x^2}{6}+\frac{y^2}{2}=1$.
Here $a^2=6$ and $b^2=2\Rightarrow a=\sqrt{6}$ and $b=\sqrt{2}$.
Let ' $\theta$ ' be the eccentric angle of the point so that the coordinates of the point are $(\sqrt{6}\cos \theta ,\sqrt{2}\sin \theta )$.
Since distance of this point from the centre $C(0,0)$ is $2$.
$\therefore \sqrt{6{\cos }^2\theta +2{\sin }^2\theta }=2$
$\Rightarrow 6{\cos }^2\theta +2{\sin }^2\theta =4$
$\Rightarrow 6{\cos }^2\theta +2\left(1-{\cos }^2\theta \right)=4$ or $4{\cos }^2\theta =2$
$\Rightarrow \cos \theta =\pm \frac{1}{\sqrt{2}}$
$\therefore \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$
$(\because 0\le \theta <2\pi )$