Question

The eccentric angle in the first quadrant of a point on the ellipse $\frac{x^{2}}{10}+\frac{y^{2}}{8}=1$ at a distance $3$ units from the centre of the ellipse is

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

Answer 1

Answer: (B)

$\frac{\pi}{4}$

Answer 2

Let $P(\sqrt{10} \cos \theta, \sqrt{8} \sin \theta)$ be the required point
on $\frac{x^{2}}{10}+\frac{y^{2}}{8}=1$
Whose distance from centre $(0,0)$ is 3 units.
$\therefore 10 \cos ^{2} \theta+8 \sin ^{2} \theta=9=9\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$
$\Rightarrow \tan ^{2} \theta=1 \Rightarrow \theta=\frac{\pi}{4}$