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Q.
Eccentric angle of a point on the ellipse $x^{2}+3 y^{2}=6$ at a distance $2$ units from the centre of the ellipse is
Conic Sections
Solution:
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 $\frac{3 \pi}{4}$