Question

The ratio of the area enclosed by the locus of the midpoint of $PS$ and area of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $(P$ be any point on the ellipse and $S$ be its focus)

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{5}$
• D. $\frac{1}{4}$

Answer 1

Answer: (D)

Mid point of $PS$ is $(h,k)$
$\therefore h=\frac{a\cos \theta +ae}{2}$
$\Rightarrow \cos \theta =\frac{2h-ae}{a}$
and $k=\frac{b\sin \theta }{2}$
Eliminating $\theta$, we get locus as
$\frac{(2h-ae{)}^2}{a^2}+\frac{4k^2}{b^2}=1$
$\Rightarrow \frac{{\left(x-\frac{ae}{2}\right)}^2}{\frac{a^2}{4}}+\frac{y^2}{\frac{b^2}{4}}=1$
which is ellipse having enclosed area
$\therefore$ Area $=\pi \frac{a}{2}\cdot \frac{b}{2}=\frac{\pi ab}{4}$
Hence, required ratio $=\frac{1}{4}$