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  • The eccentric angle of a point on the ellipse <img alt="\mathrm{\frac{x^2}{6}+\frac{y^2}{2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7B6%7D&plus;%

The eccentric angle of a point on the ellipse \mathrm{\frac{x^2}{6}+\frac{y^2}{2}=1} whose distance from the centre of the ellipse is 2, is

Option: 1

\mathrm{\pi / 4}


Option: 2

\mathrm{3 \pi / 2}


Option: 3

\mathrm{5 \pi / 3}


Option: 4

7 \pi / 6


Answers (1)

best_answer

Let \theta be the eccentric angle of the point P. Then the coordinates of P are \mathrm{(\sqrt{6} \cos \theta, \sqrt{2} \sin \theta)} The centre of the ellipse is at the origin, It is given that \mathrm{O P=2}
\mathrm{\Rightarrow \sqrt{6 \cos ^2 \theta+2 \sin ^2 \theta}=2 \Rightarrow 6 \cos ^2 \theta+2 \sin ^2 \theta=4 \Rightarrow 3 \cos ^2 \theta+\sin ^2 \theta=2 \Rightarrow 2 \sin ^2 \theta=1}
\mathrm{\Rightarrow \quad \sin ^2 \theta=\frac{1}{2} \Rightarrow \sin \theta= \pm \frac{1}{\sqrt{2}} \Rightarrow \theta= \pm \pi / 4}

Posted by

Divya Prakash Singh

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