Question

\(P\) 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) \(\frac{3 \pi}{4}\)
(C) \(\frac{\pi}{3}\)
(D) \(\frac{2 \pi}{3}\)

Answer 1

Ans: (C)
Hint \(: 3 x^{2}+4 y^{2}=48 \Rightarrow \frac{x^{2}}{16}+\frac{y^{2}}{12}=1 .\) Here \(a=4, b=2 \sqrt{3}\) and \(e=\frac{1}{2} \therefore(\) co-ordinates of \(P\) are \((2,3)\)
\(\therefore(4 \cos \theta, 2 \sqrt{3} \sin \theta) \equiv(2,3) \Rightarrow \theta=\frac{\pi}{3}\)

Answer 2

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Answer 3

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