MHT CET · Maths · Ellipse
The eccentric angle of the point \(\mathrm{P}(-6,2)\) of the ellipse \(\frac{x^2}{48}+\frac{y^2}{16}=1\) is
- A \(30^{\circ}\)
- B \(135^{\circ}\)
- C \(150^{\circ}\)
- D \(120^{\circ}\)
Answer & Solution
Correct Answer
(C) \(150^{\circ}\)
Step-by-step Solution
Detailed explanation
\(a^2=48, b^2=16 \implies a=4\sqrt{3}, b=4\) \(a \cos \theta = -6 \implies 4\sqrt{3} \cos \theta = -6 \implies \cos \theta = -\frac{\sqrt{3}}{2}\)
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