Question

The ratio of the area enclosed by the locus of the midpoint of $P S$ 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 $P S$ is $(h, k)$
$\therefore h=\frac{a \cos \theta+a e}{2} $
$\Rightarrow \cos \theta=\frac{2 h-a e}{a}$
and $k=\frac{b \sin \theta}{2}$
Eliminating $\theta$, we get locus as
$\frac{(2 h-a e)^{2}}{a^{2}}+\frac{4 k^{2}}{b^{2}}=1 $
$\Rightarrow \frac{\left(x-\frac{a e}{2}\right)^{2}}{\frac{a^{2}}{4}}+\frac{y^{2}}{\frac{b^{2}}{4}}=1$
which is ellipse having enclosed area
$\therefore $ Area $=\pi \frac{a}{2} \cdot \frac{b}{2}=\frac{\pi a b}{4} $
Hence, required ratio $=\frac{1}{4}$