Question

$P$ is the extremity of the latus rectum of ellipse $3x^2+4y^2=48$ in the first quadrant. The eccentric angle of $P$ is

• A. $\frac{\pi }{8}$
• B. $\frac{3\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{2\pi }{3}$

Answer 1

The correct option is C $\frac{\pi }{3}$

Equation of ellipse is $\frac{x^2}{16}+\frac{y^2}{12}=1$
$a=4,b=2\sqrt{3}{e}^2=1-\frac{12}{16}$
$\Rightarrow e=\frac{1}{2}$

$x$-coordinate of $P$ is $ae=2$
$y$-coordinate of $P$ is $\pm b\sqrt{1-e^2}=\pm \frac{b^2}{a}=3$

Let eccentric angle at $P$ be $t$
$2=a\cos t=4\cos t$
$3=b\sin t=2\sqrt{3}\sin t$
$\Rightarrow \tan t=\sqrt{3}$
$\Rightarrow t=\frac{\pi }{3}$