Question

The eccentric angle of the point $(2,\sqrt{3})$ lying on $\frac {x^2}{16}+\frac{y^2}{4}-1$ is _________

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

Answer 1

Answer: (A)

$\frac {\pi}{3}$

Answer 2

Given equation of an ellipse is
$\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ and point $P(2, \sqrt{3})$
Let $\theta$ be the eccentric angle.
The parametric coordinate of an ellipse is
$\begin{cases} x=4 \cos \theta \\\ y=2 \sin \theta \end{cases}$...(i)
Given that, eccentric angle at $P$ is,
$2=4 \cos \theta \Rightarrow \cos \theta=\frac{1}{2}$
$\sqrt{2}=2 \sin \theta \Rightarrow \sin \theta=\frac{\sqrt{3}}{2}$
Hence, $\theta=\frac{\pi}{3}$