The eccentric angle of a point on the ellipse | vedclass

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(D) The given equation of the ellipse is $x^2 + 3y^2 = 6$,which can be written as $\frac{x^2}{6} + \frac{y^2}{2} = 1$.Here,$a^2 = 6$ and $b^2 = 2$,so

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Both $A$ and $C$

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(D) The given equation of the ellipse is $x^2 + 3y^2 = 6$,which can be written as $\frac{x^2}{6} + \frac{y^2}{2} = 1$.Here,$a^2 = 6$ and $b^2 = 2$,so $a = \sqrt{6}$ and $b = \sqrt{2}$.Any point on the ellipse can be represented as $(a \cos 	heta, b \sin 	heta) = (\sqrt{6} \cos 	heta, \sqrt{2} \sin 	heta)$,where $	heta$ is the eccentric angle.The distance of this point from the centre $(0, 0)$ is given as $2$ units.So,$\sqrt{(\sqrt{6} \cos 	heta)^2 + (\sqrt{2} \sin 	heta)^2} = 2$.Squaring both sides,we get $6 \cos^2 	heta + 2 \sin^2 	heta = 4$.Using $\sin^2 	heta = 1 - \cos^2 	heta$,we have $6 \cos^2 	heta + 2(1 - \cos^2 	heta) = 4$.$4 \cos^2 	heta + 2 = 4 \implies 4 \cos^2 	heta = 2 \implies \cos^2 	heta = \frac{1}{2}$.Thus,$\cos 	heta = \pm \frac{1}{\sqrt{2}}$.This gives $	heta = \frac{\pi}{4}$ or $	heta = \frac{3\pi}{4}$.

The eccentric angle of a point on the ellipse $x^2 + 3y^2 = 6$ at a distance of $2$ units from the centre of the ellipse is:

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