$26 \cdot 36 P Q$ is any chord passing through the focus of the parabola $y^{2}=2 p x$, $M N$ is the projection of $P Q$ on the directrix $l$, the area of the surface of revolution obtained by rotating $P Q$ around $l$ is $S_{1}$, and the area of the sphere with $M N$ as its diameter is $S_{2}$. Among the following conclusions, the correct one is
(A) $S_{1}>S_{2}$.
(B) $S_{1}<S_{2}$.
(C) $S_{1}=S_{2}$.
(D) Sometimes $S_{1}>S_{2}$, sometimes $S_{1}=S_{2}$, sometimes $S_{1}<S_{2}$.
(China High School Mathematics League, 1985)
[Solution] As shown in the right figure, let $C$ be the focus of the parabola, then
$$
P M=P C, Q N=Q C .
$$

Thus, $\quad S_{1}=\pi \cdot P Q(P M+Q N)=\pi \cdot P Q^{2}$, similarly $S_{2}=\pi \cdot M N^{2}$.
Since $P Q \geqslant M N$, it follows that $S_{1} \geqslant S_{2}$. Therefore, the answer is (C).