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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The eccentric angle of a point on the ellipse $\large\frac{x^2}{6} +\frac{y^2}{2}$$=1$ Whose distance from the centre of ellipse is 2 is

$\begin{array}{1 1}(A)\;\frac{\pi}{4} \\(B)\;\frac{3 \pi}{2} \\(C)\;\frac{5 \pi}{3} \\(D)\;\frac{7 \pi}{6} \end{array}$

Can you answer this question?
 
 

1 Answer

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Let $\theta$ be the eccentric angle of the point P.
Then the co-ordinate of P are $(\sqrt 6 \cos \theta, \sqrt 2 \sin \theta)$
The center of the ellipse is at the point of origin.
It is given that
$\therefore OP=2$
=> $ \sqrt {6 \cos^2 \theta + 2\sin ^2 \theta}=2$
=> $ 6 \cos ^2 \theta +2 \sin ^2 \theta=4$
=> $ 3 \cos ^2 \theta + \sin ^2 \theta=2$
=> $ 2 \sin ^2 \theta=1$
=> $\sin ^2 \theta=\large\frac{1}{2}$
=> $\sin \theta= \pm \large\frac{1}{\sqrt 2}$
$\theta= \pm \large\frac{\pi}{4}$
Hence A is the correct answer.
answered Apr 10, 2014 by meena.p
 

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