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Position of point wrt ellipse - class-XII
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The dist.of a point P on the ellipse $\cfrac{{{x^2}}}{{12}} + \cfrac{{{y^2}}}{4} = 1$ from centre is $\sqrt 6 $ then the eccentric angle of P is
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A
$\cfrac{\pi }{3}$
💡 Explanation:
Let the point be $P = (a\cos\theta , b\sin\theta )$
Distance of point $P$ from centre is $\sqrt 6$
$\therefore \sqrt { { a }^{ 2 }\cos^{ 2 }\theta + b^{ 2 }\sin^{ 2 }\theta} = \sqrt { 6 }$
$\Rightarrow { a }^{ 2 }\cos^{ 2 }\theta + b^{ 2 }\sin^{ 2 }\theta = 6$
$\Rightarrow 12\cos^{ 2 }\theta + 4\sin^{ 2 }\theta = 6 \cos^{ 2 }\theta + \sin^{ 2 }\theta = 1$
$\Rightarrow 8\cos^{ 2 }\theta = 2$
$\Rightarrow \cos\theta = \cfrac { 1 }{ 2 }$
Hence, $\theta = \cfrac { \pi }{ 3 }$