$\text{ The ratio of the area enclosed by the locus of the midpoint of }PS\text{ and area of the ellipse }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ is}$

$( P\text{ be any point on the ellipse and S be iti focus)}$

$\text{Mid point of PS is (h, k)}$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}h=\frac{a\cos \theta +ae}{2}\Rightarrow \cos \theta =\frac{2h-ae}{a}\\ \text{ and }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}k=\frac{b\sin \theta }{2}\end{array}$

$\text{ Eliminating }\theta ,\text{ we get locus as }$

$\begin{array}{l}\frac{(2h-ae{)}^2}{a^2}+\frac{4k^2}{b^2}=1\\ \Rightarrow \frac{{\left(x-\frac{ae}{2}\right)}^2}{\frac{a^2}{4}}+\frac{y^2}{\frac{b^2}{4}}=1\end{array}$

$\text{which is ellipse having enclosed area }$

$\begin{array}{l}\therefore \text{ Area }=\pi \frac{a}{2}\cdot \frac{b}{2}=\frac{\pi ab}{4}\\ \text{ Hence, required ratio }=\frac{1}{4}\end{array}$