# How do we find the eccentric angle of ellipse x^2/16+y^2/4=1?

##### 1 Answer
Nov 27, 2016

Eccentric angle is $\frac{\pi}{3}$

#### Explanation:

The equation of ellipse is ${x}^{2} / 16 + {y}^{2} / 4 = 1$ or ${x}^{2} + 4 {y}^{2} = 16$ and point $\left(2 , \sqrt{3}\right)$ lies in first quadrant.

Equation of auxiliary circle will be ${x}^{2} + {y}^{2} = 16$.

Now the point $P \left(2 , \sqrt{3}\right)$ and corresponding point on auxiliary circle is

$Q \left(2 , \sqrt{16 - {2}^{2}}\right)$ or $Q \left(2 , 2 \sqrt{3}\right)$ and

eccentric angle $\theta = {\tan}^{- 1} \left(\frac{2 \sqrt{3}}{2}\right) = {\tan}^{- 1} \sqrt{3} = \frac{\pi}{3}$