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

The correct option is B $\frac{\pi }{4}$

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.
$\therefore 10{\cos }^2\theta +8{\sin }^2\theta =9=9({\sin }^2\theta +{\cos }^2\theta )$
$\Rightarrow {\tan }^2\theta =1$

$\Rightarrow \theta =\frac{\pi }{4}$