P is the extremity of the latus rectum of the | vedclass

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(C) The given equation of the ellipse is $3x^{2} + 4y^{2} = 48$.Dividing by $48$, we get $\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$.Here, $a^{2} = 16$

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$\frac{\pi}{3}$

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(C) The given equation of the ellipse is $3x^{2} + 4y^{2} = 48$.Dividing by $48$, we get $\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$.Here, $a^{2} = 16$ and $b^{2} = 12$, so $a = 4$ and $b = 2\sqrt{3}$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{12}{16}} = \sqrt{\frac{4}{16}} = \frac{1}{2}$.The coordinates of the extremity of the latus rectum $P$ in the first quadrant are $(ae, \frac{b^{2}}{a}) = (4 	imes \frac{1}{2}, \frac{12}{4}) = (2, 3)$.Let the eccentric angle of $P$ be $	heta$. Then $P = (a \cos 	heta, b \sin 	heta) = (4 \cos 	heta, 2\sqrt{3} \sin 	heta)$.Equating the coordinates: $4 \cos 	heta = 2 \Rightarrow \cos 	heta = \frac{1}{2}$ and $2\sqrt{3} \sin 	heta = 3 \Rightarrow \sin 	heta = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.Since $\cos 	heta = \frac{1}{2}$ and $\sin 	heta = \frac{\sqrt{3}}{2}$, we have $	heta = \frac{\pi}{3}$.

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

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