Question

The dist.of a point P on the ellipse $\frac{x^2}{12}+\frac{y^2}{4}=1$ from centre is $\sqrt{6}$ then the eccentric angle of P is

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

Answer 1

The correct option is D $\frac{\pi }{3}$

Let the point be $P=(a\cos \theta ,b\sin \theta )$

Distance of point $P$ from centre is $\sqrt{6}$

$\therefore \sqrt{a^2{\cos }^2\theta +b^2{\sin }^2\theta }=\sqrt{6}$

$\Rightarrow a^2{\cos }^2\theta +b^2{\sin }^2\theta =6$

$\Rightarrow 12{\cos }^2\theta +4{\sin }^2\theta =6{\cos }^2\theta +{\sin }^2\theta =1$

$\Rightarrow 8{\cos }^2\theta =2$

$\Rightarrow \cos \theta =\frac{1}{2}$

Hence, $\theta =\frac{\pi }{3}$