Q. The eccentric angle of a point on the ellipse $\frac{x^{2}}{6} + \frac{y^{2}}{2} = 1$, whose distance from the centre of the ellipse is 2, is
A) $\frac{\pi}{4}$
B) $\frac{3\pi}{2}$
C) $\frac{5\pi}{3}$
D) $\frac{7\pi}{6}$
Explanation
(a)
Let point is $(\sqrt{6}\cos{\theta,\ \sqrt{2}\sin\theta})$ and let it’s distance $d$ from origin
$\therefore d = \sqrt{\sqrt{6\cos^{2}{\theta + 2\sin^{2}\theta}})}$
$\Rightarrow 2 = \sqrt{2 + 4\cos^{2}\theta}$
$\Rightarrow 2 + 4\cos^{2}\theta = 4$
$\Rightarrow \cos^{2}\theta = \frac{1}{2}$
$\Rightarrow \cos{\theta = \pm \frac{1}{\sqrt{2}} \Rightarrow \theta = \frac{\pi}{4}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.