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Q.
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
Conic Sections
Solution:
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} $
$(\because 0 \leq \theta < 2 \pi)$