WBJEE · Maths · Ellipse
\(P\) 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 & Solution
Correct Answer
(C) \(\frac{\pi}{3}\)
Step-by-step Solution
Detailed explanation
Given equation of ellipse is \[ \begin{array}{l} \quad 3 x^{2}+4 y^{2}=48 \\ \Rightarrow \quad \frac{3 x^{2}}{48}+\frac{4 y^{2}}{48}=1 \quad \Rightarrow \quad \frac{x^{2}}{16}+\frac{y^{2}}{12}=1 \\ \text { Here, } \quad a=4 \text { and } b=2 \sqrt{3} \end{array} \]…
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