Eccentric angle of a point on the ellipse {x^ | vedclass

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(d) Let the eccentric angle of the point be $	heta $, then its co-ordinates are $(\sqrt 6 \cos 	heta ,\sqrt 2 \sin 	heta ).$

Hence $6{\cos ^2}

Vedclass · Accepted Answer

$(a)$ and $(c)$ both

Vedclass · Answer

(d) Let the eccentric angle of the point be $	heta $, then its co-ordinates are $(\sqrt 6 \cos 	heta ,\sqrt 2 \sin 	heta ).$

Hence $6{\cos ^2}	heta + 2{\sin ^2}	heta = 4$ or ${\cos ^2}	heta = \frac{1}{2}$

Hence, $\cos 	heta = \pm \frac{1}{{\sqrt 2 }}$,

$	heta = \frac{\pi }{4}$or $\frac{{3\pi }}{4}$.

Eccentric angle of a point on the ellipse ${x^2} + 3{y^2} = 6$ at a distance $2$ units from the centre of the ellipse is

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