The eccentric angle of a point on the ellipse x 2 + 3 y 2 = 6 lying at a distance of 2 units from its centre…

The given equation of ellipse can be written in the form $\frac{x^2}{{\left(\sqrt{6}\right)}^2}+\frac{y^2}{{\left(\sqrt{2}\right)}^2}=1$.

Now let the eccentric angle of the point be $\theta$, then its co-ordinates are $\left(\sqrt{6}\cos \theta ,\sqrt{2}\sin \theta \right)$.

Since the distance of the point from the centre is $2$ units

$\therefore {\left(\sqrt{6}\cos \theta -0\right)}^2+{\left(\sqrt{2}\sin \theta -0\right)}^2=4$

$\Rightarrow 6{\cos }^2\theta +2\left(1-{\cos }^2\theta \right)=4\Rightarrow 4{\cos }^2\theta =2$

$\Rightarrow \cos \theta =\pm \frac{1}{\sqrt{2}} \therefore \theta =\frac{\pi }{4}$ or $\frac{3\pi }{4}$