Q82 / 155 · WBJEE 2019 · Mathematics
$P$ is the extremity of the latusrectum of ellipse $3 x^{2}+4 y^{2}=48$ in the first quadrant. The eccentric angle of $P$ is
- A$\frac{\pi}{8}$
- B$\frac{3 \pi}{4}$
- Correct answerC$\frac{\pi}{3}$
- D$\frac{2 \pi}{3}$
Solution
Given equation of ellipse is
$$
\begin{array}{l}
\quad 3 x^{2}+4 y^{2}=48 \\
\Rightarrow \quad \frac{3 x^{2}}{48}+\frac{4 y^{2}}{48}=1 \quad \Rightarrow \quad \frac{x^{2}}{16}+\frac{y^{2}}{12}=1 \\
\text { Here, } \quad a=4 \text { and } b=2 \sqrt{3}
\end{array}
$$
$\therefore$
$e=\sqrt{1-\frac{b^{2}}{a^{2}}}$
$=\sqrt{1-\frac{12}{16}}=\sqrt{\frac{4}{16}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$
$\therefore$ Coordinates of $P$ are $=(2,3)$ $\left.(4 \cos \theta, 2 \sqrt{3} \sin \theta)=(2,3) \quad [\because p=\left(a e, \frac{b^{2}}{a}\right)\right]$
By comparing, we get
$$
4 \cos \theta=2 \text { and } 2 \sqrt{3} \sin \theta=3
$$
$\Rightarrow \quad \cos \theta=\frac{1}{2}$ and $\sin \theta=\frac{\sqrt{3}}{2}$
$\Rightarrow \quad \cos \theta=\cos \frac{\pi}{3}$ and $\sin \theta=\sin \frac{\pi}{3}$
$\theta=\frac{\pi}{3}$ and $\theta=\frac{\pi}{3}$
$\therefore$ eccentric angle of $P$ is $\frac{\pi}{3}$
Hence, the correct answer is option: 'C'.