Question

A point on the ellipse is $\frac{x^2}{6}+\frac{y^2}{2}=1$ at a distance of $2$ from the centre of the ellipse has the eccentric angle

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

Answer 1

The correct option is A $\frac{\pi }{4}$

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 $(acos\theta ,bsin\theta )\Rightarrow P(\sqrt{6}cos\theta ,\sqrt{2}sin\theta )$
Distance of point P from center=2
$\Rightarrow \sqrt{(}(\sqrt{6}cos\theta -0{)}^2+(\sqrt{2}sin\theta -0{)}^2)=2$
$\Rightarrow (6co{s}^2\theta +2si{n}^2\theta )=4$
$\Rightarrow (3co{s}^2\theta +si{n}^2\theta )=2$
$\Rightarrow 2co{s}^2\theta +1=2$
$\Rightarrow cos\theta =\pm \frac{1}{\sqrt{2}}$
$\Rightarrow \theta =\frac{\pi }{4}or\frac{-\pi }{4}$
Option $A$ is correct