Question

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 $\sqrt{5},$ units is/are

• A. $\frac{\pi }{6}$
• B. $\frac{3\pi }{6}$
• C. $\frac{5\pi }{6}$
• D. $\frac{11\pi }{6}$

Answer 1

Answer: (A), (C), (D)

$\frac{11\pi }{6}$

Any point on the ellipse $\frac{x^2}{6}+\frac{y^2}{2}=1$ is $P(\sqrt{6}\cos \theta ,\sqrt{2}\sin \theta )$

Now $CP=\sqrt{6{\cos }^2\theta +2{\sin }^2\theta }=\sqrt{5}$ where $C$ is centre
$\Rightarrow 6(1-{\sin }^2\theta )+2{\sin }^2\theta =5$
$\Rightarrow 4{\sin }^2\theta =1$
$\Rightarrow {\sin }^2\theta =\frac{1}{4}={\left(\frac{1}{2}\right)}^2$
$\sin \theta =n\pi \pm \left(\frac{\pi }{6}\right)$
$\therefore \theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$
$\because \theta \in [o,2\pi )$