Question

The distance of a point on thc ellipse $x^2+3y^2=6$ from its centre is $\sqrt{2}$, then the eccentric angle of the point is:

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is A $\frac{\pi }{2}$

Gven, $\frac{x^2}{6}+\frac{y^2}{\left(\sqrt{2}\right)2}=1$
$h^2+k^2=2$
$h^2=(2-k^2)$
$\frac{2-k^3}{6}+\frac{k^2}{2}=1$
$2-k^2+3k^2=6$
$2+2k^2=6$
$k^2=2$
$k=\sqrt{2}$
$\sqrt{2}=\sqrt{2}\sin \theta$
$\sin \theta =1$
$\therefore \theta =\frac{\pi }{2}$