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Q.
The distance of a point on ellipse $\frac{x^2}{6}+\frac{y^2}{2}= 1$ from centre is 2. The eccentric angle of the point will be
Conic Sections
Solution:
Let the eccentric angle of the point be $\theta$ , then its coordinates are $(\sqrt{6} \,cos \,\theta, \sqrt{2}\,sin \,\theta)$ Centre of a given ellipse is $(0, 0) $
According to the equation,
$\sqrt{\left(\sqrt{6} cos\,\theta -0\right)^{2} + \left(\sqrt{2} sin\, \theta -0\right)^{2}} = 2$
$ \Rightarrow 6\, cos^{2}\, \theta +2\,sin^{2}\,\theta = 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}} $
$ \Rightarrow \theta = \frac{\pi}{4}$ or $\frac{3\pi}{4}$