The eccentric angle of a point on the ellipse | vedclass

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(B) The given equation of the ellipse is $x^2+3y^2=6$,which can be rewritten as $\frac{x^2}{6}+\frac{y^2}{2}=1$. Let the eccentric angle of the point

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) The given equation of the ellipse is $x^2+3y^2=6$,which can be rewritten as $\frac{x^2}{6}+\frac{y^2}{2}=1$. Let the eccentric angle of the point be $	heta$. The coordinates of any point on the ellipse are given by $(a \cos 	heta, b \sin 	heta)$,where $a^2=6$ and $b^2=2$. Thus,the coordinates are $(\sqrt{6} \cos 	heta, \sqrt{2} \sin 	heta)$. The distance of this point from the centre $(0,0)$ is $2$ units. So,$(\sqrt{6} \cos 	heta)^2 + (\sqrt{2} \sin 	heta)^2 = 2^2$. $6 \cos^2 	heta + 2 \sin^2 	heta = 4$. $6 \cos^2 	heta + 2(1 - \cos^2 	heta) = 4$. $4 \cos^2 	heta + 2 = 4$ $\Rightarrow 4 \cos^2 	heta = 2$ $\Rightarrow \cos^2 	heta = \frac{1}{2}$. $\cos 	heta = \pm \frac{1}{\sqrt{2}}$. Therefore,$	heta = \frac{\pi}{4}$ or $\frac{3\pi}{4}$.

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

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