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(C) Any point on the ellipse $\frac{x^2}{6} + \frac{y^2}{2} = 1$ is given by $(\sqrt{6} \cos \varphi, \sqrt{2} \sin \varphi)$,where $\varphi$ is the e

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$\frac{\pi}{4}, \frac{3\pi}{4}$

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(C) Any point on the ellipse $\frac{x^2}{6} + \frac{y^2}{2} = 1$ is given by $(\sqrt{6} \cos \varphi, \sqrt{2} \sin \varphi)$,where $\varphi$ is the eccentric angle.The distance of this point from the center $(0, 0)$ is $2$.Therefore,$(\sqrt{6} \cos \varphi)^2 + (\sqrt{2} \sin \varphi)^2 = 2^2$.$6 \cos^2 \varphi + 2 \sin^2 \varphi = 4$.Dividing by $2$,we get $3 \cos^2 \varphi + \sin^2 \varphi = 2$.Using $\sin^2 \varphi = 1 - \cos^2 \varphi$,we get $3 \cos^2 \varphi + 1 - \cos^2 \varphi = 2$.$2 \cos^2 \varphi = 1 \implies \cos^2 \varphi = \frac{1}{2}$.$\cos \varphi = \pm \frac{1}{\sqrt{2}}$.Thus,$\varphi = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.Given the options,the correct set is $\frac{\pi}{4}, \frac{3\pi}{4}$.

If the distance of a point on the ellipse $\frac{x^2}{6} + \frac{y^2}{2} = 1$ from the center is $2$,find its eccentric angle $\varphi$.

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