Question

Eccentric angle of a point on the ellipse $x^2+3y^2=6$ at a distance $2$ units from the center of the ellipse is

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

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{\pi }{4}$
C $\frac{3\pi }{4}$

The equation of ellipse can be written in the form $\frac{x^2}{{\left(\sqrt{6}\right)}^2}+\frac{y^2}{{\left(\sqrt{2}\right)}^2}=1.$

Let the eccentric angle of the point be $\theta ,$ then its co-ordinates are $(\sqrt{6}\cos \theta ,\sqrt{2}\sin \theta ).$

Since the distance of the point from the centre is $2$ units

$\therefore {(\sqrt{6}\cos \theta -0)}^2+{\left(\sqrt{2}\sin \theta -0\right)}^2=4$

$\Rightarrow 6{\cos }^2\theta +2(1-{\cos }^2\theta )=4\Rightarrow 4{\cos }^2\theta =2$

$\Rightarrow \cos \theta =\pm \frac{1}{\sqrt{2}}.$ $\therefore \theta =\frac{\pi }{4}$ or $\frac{3\pi }{4}.$