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(C) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The area $A_1 = \pi ab$.Let the focus be $S(ae, 0)$ and a point $P$ on the ellipse be

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$4 : 1$

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(C) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The area $A_1 = \pi ab$.Let the focus be $S(ae, 0)$ and a point $P$ on the ellipse be $(a \cos 	heta, b \sin 	heta)$.The midpoint $M(h, k)$ of $SP$ is given by $h = \frac{a \cos 	heta + ae}{2}$ and $k = \frac{b \sin 	heta + 0}{2}$.Thus,$\cos 	heta = \frac{2h - ae}{a}$ and $\sin 	heta = \frac{2k}{b}$.Using $\cos^2 	heta + \sin^2 	heta = 1$,we get $\frac{(2h - ae)^2}{a^2} + \frac{4k^2}{b^2} = 1$.This simplifies to $\frac{(h - ae/2)^2}{(a/2)^2} + \frac{k^2}{(b/2)^2} = 1$.This is an ellipse with semi-axes $a' = a/2$ and $b' = b/2$.The area $A_2 = \pi a' b' = \pi (a/2)(b/2) = \frac{\pi ab}{4}$.Therefore,$A_1 : A_2 = \pi ab : \frac{\pi ab}{4} = 4 : 1$.

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 the midpoint of the line segment joining the focus of the ellipse and a point $P$ on the given ellipse. Then $A_1 : A_2$ is equal to:

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