Question

Eccentric angle of a point on the ellipse $x^2+3y^2=6$ at a distance $2$ units from the centre of the ellipse is

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

Answer 1

Answer: (A), (C)

The equation of ellipse can be written in the form
$\frac{x^2}{(\sqrt{6}{)}^2}+\frac{y^2}{(\sqrt{2}{)}^2}=1$
Let the eccentric angle of the point be $\theta$, then its coordinates are $(\sqrt{6}\cos \theta ,\sqrt{2}\sin \theta )$.
Since the distance of the point from the centre is $2$ units
$\therefore (\sqrt{6}\cos \theta -0{)}^2+(\sqrt{2}\sin \theta -0{)}^2=4$
$\Rightarrow 6{\cos }^2\theta +2\left(1-{\cos }^2\theta \right)=4$
$\Rightarrow 4{\cos }^2\theta =2$
$\Rightarrow \cos \theta =\pm \frac{1}{\sqrt{2}}$.
$\therefore \theta =\frac{\pi }{4}$ or $\frac{3\pi }{4}$