Question

The eccentric angle of a point on the ellipse $\frac{x^2}{6}+\frac{y^2}{2}=1$ whose distances from the centre of the ellipse is 2, is

• A. $\frac{\pi}{4}$
• B. $\frac{3\pi}{2}$
• C. $\frac{5\pi}{3}$
• D. $\frac{7\pi}{6}$

Answer 1

Answer: (A)

$\frac{\pi}{4}$

Answer 2

Let $\theta$ be the eccentric angle. $\therefore pt$. is $\left(\sqrt{6} \, cos\,\theta, \sqrt{2} \,sin\,\theta\right)$ $\therefore \left(\sqrt{6}\, cos\,\theta -0\right)^{2} +\left(\sqrt{2}\, sin\,\theta -0\right)^{2} =\left(2\right)^{2}$ $\Rightarrow 6\, cos^{2}\, \theta+2\, sin^{2}\, \theta=4$ $\Rightarrow 6\left(1-sin^{2} \theta\right)+2\, sin^{2} \, \theta = 4$ $\Rightarrow 4\, sin^{2}\, \theta= 2$ $\Rightarrow sin^{2}\, \theta= \frac{1}{2}$ $\Rightarrow \theta = \frac{\pi}{4}$