Question

The distance of a point $P$ on the ellipse $x^2+3y^2=6$ from the centre is $2,$ the eccentric angle of $P$ is

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

Answer 1

Answer: (C)

$\frac{\pi }{4}$

Given that point $P$ lies on ellipse

it will be in the form of $(\sqrt{6}\cos \theta ,\sqrt{2}\sin \theta )$

The center of ellipse is $(0,0)$ and its distance from $P$ is $2$

by using distance formula

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

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

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

therefore the value of $\theta =\frac{\pi }{4}$