AP EAMCET · Maths · Ellipse
Let \(A_1\) be the area of the given ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). Let \(A_2\) be the area of the region bounded by the curve which is the locus of mid point of the line segment joining the focus of the ellipse and a point \(P\) on the given ellipse, then \(A_1: A_2\) \(=\)
- A \(3: 2\)
- B \(a:b\)
- C \(4: 1\)
- D \(2a:3b\)
Answer & Solution
Correct Answer
(C) \(4: 1\)
Step-by-step Solution
Detailed explanation
\(A_1 = \pi ab\) Let \(F=(ae,0)\) and \(P=(a \cos \theta, b \sin \theta)\). \(x_m = \frac{a \cos \theta + ae}{2}, y_m = \frac{b \sin \theta}{2}\) \(\cos \theta = \frac{2x_m-ae}{a}, \sin \theta = \frac{2y_m}{b}\)…
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