11. (20 points) Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and a line $l$ that intersects the ellipse $C$ at points $P$ and $Q$, where point $S$ is the reflection of point $P$ across the $x$-axis. Find the maximum value of the area of $\triangle O S Q$.

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Here is the translation, maintaining the original format and line breaks.
11. When $P Q // x$-axis, points $S, O, Q$ are collinear, not forming a triangle. As shown in Figure 2, let the equation of line $l$ be
$$
x=m+k y .
$$

We will prove that $S Q$ passes through point $B\left(\frac{a^{2}}{m}, 0\right)$. To do this, we need to prove that $k_{B Q}=k_{S B}$.
Substituting equation (1) into the ellipse equation and rearranging, we get
$$
\left(b^{2} k^{2}+a^{2}\right) y^{2}+2 m k b^{2} y+b^{2} m^{2}-a^{2} b^{2}=0 .
$$

Thus, $y_{1}+y_{2}=-\frac{2 m k b^{2}}{b^{2} k^{2}+a^{2}}, y_{1} y_{2}=\frac{b^{2} m^{2}-a^{2} b^{2}}{b^{2} k^{2}+a^{2}}$.
Let the line $l$ intersect the ellipse $C$ at points $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, then $S\left(x_{1},-y_{1}\right)$, so
$$
k_{B Q}=k_{S P} \Leftrightarrow \frac{y_{2}}{x_{2}-\frac{a^{2}}{m}}=\frac{-y_{1}}{x_{1}-\frac{a^{2}}{m}},
$$

which means
$$
y_{2}\left(x_{1}-\frac{a^{2}}{m}\right)+y_{1}\left(x_{2}-\frac{a^{2}}{m}\right)=0 .
$$

Noting that $x_{1}=m+k y_{1}, x_{2}=m+k y_{2}$, equation (2) becomes
$$
\left(m-\frac{a^{2}}{m}\right)\left(y_{1}+y_{2}\right)+2 k y_{1} y_{2}=0
$$

which is
$$
-\left(m-\frac{a^{2}}{m}\right) \frac{2 m k b^{2}}{b^{2} k^{2}+a^{2}}+2 k \cdot \frac{b^{2} m^{2}-a^{2} b^{2}}{b^{2} k^{2}+a^{2}}=0. 
$$

This is true, hence the proof is complete.
Also, note that
$$
\begin{aligned}
S_{\triangle O S Q} & =\frac{1}{2}|O B|\left|y_{1}+y_{2}\right|=\frac{1}{2}\left|\frac{a^{2}}{m}\right| \cdot\left|\frac{2 m k b^{2}}{b^{2} k^{2}+a^{2}}\right| \\
& \leqslant \frac{1}{2}\left|\frac{a^{2}}{m}\right| \cdot\left|\frac{2 m k b^{2}}{2 a b k}\right|=\frac{1}{2} a b
\end{aligned}
$$

Therefore, the maximum value of $S_{\triangle O S Q}$ is $\frac{1}{2} a b$, which is achieved when $k= \pm \frac{a}{b}$.