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 ellipse is $2$, is:

• A. $-\frac{\pi }{4}$
• B. $\frac{\pi }{4}$
• C. $\frac{3\pi }{2}$
• D. $\frac{5\pi }{3}$

Answer 1

Answer: (A), (B)

The correct options are
A $-\frac{\pi }{4}$
B $\frac{\pi }{4}$

Given, equation of ellipse as $\frac{x^2}{6}+\frac{y^2}{2}=1$(where length of major axis=$\sqrt{6}$,length of minor axis=$\sqrt{2}$)
center of ellipse is $(0,0)$ which is parallel to horizontal axis and with eccentricity $e$
Any point on the ellipse will be as $(a\cos \theta ,b\sin \theta )\Rightarrow P(\sqrt{6}\cos \theta ,\sqrt{2}\sin \theta )$
Distance of point $P$ from center$=2$
$\Rightarrow \sqrt{\left(\right(\sqrt{6}\cos \theta -0{)}^2+(\sqrt{2}\sin \theta -0{)}^2)}=2$
$\Rightarrow (6{\cos }^2\theta +2{\sin }^2\theta )=4$
$\Rightarrow (3{\cos }^2\theta +{\sin }^2\theta )=2$
$\Rightarrow 2{\cos }^2\theta +1=2$
$\Rightarrow \cos \theta =\pm \frac{1}{\sqrt{2}}$
$\Rightarrow \theta =\frac{\pi }{4}or\frac{-\pi }{4}$