11. (This sub-question is worth 20 points)

Given the parabola $y^{2}=2 p x(p>0)$, $A$ and $B$ are two moving points on the parabola different from the vertex $O$, and let $\angle A O B=\theta$ $\left(\theta \neq 90^{\circ}\right)$. If $S_{\triangle A O B}=m \tan \theta$, find the maximum value of $\tan \theta$ when $m$ is minimized.
Solution: Since $S_{\triangle A O B}=m \tan \theta$, we have $\frac{1}{2}|O A| \cdot|O B| \sin \theta=m \cdot \frac{\sin \theta}{\cos \theta}$.
Therefore, $m=\frac{1}{2} \overrightarrow{O A} \cdot \overrightarrow{O B}$
5 points
Let $A\left(2 p t_{1}^{2}, 2 p t_{1}\right)$ and $B\left(2 p t_{2}^{2}, 2 p t_{2}\right)\left(t_{1} t_{2} \neq-1,0\right)$, then
$$
m=\frac{1}{2}\left(4 p^{2} t_{1}^{2} t_{2}^{2}+4 p^{2} t_{1} t_{2}\right)=2 p^{2}\left(t_{1}^{2} t_{2}^{2}+t_{1} t_{2}\right) \text {. }
$$

Let $t=t_{1} t_{2}(t \neq-1,0)$, then $m=2 p^{2}\left[\left(t+\frac{1}{2}\right)^{2}-\frac{1}{4}\right]$ at this point, $t_{1} t_{2}=-\frac{1}{2}$. Without loss of generality, assume $t_{1}>0$, then
$$
\begin{array}{l}
\tan \theta=\frac{k_{O A}-k_{O B}}{1+k_{O A} \cdot k_{O B}}=\frac{\frac{1}{t_{1}}-\frac{1}{t_{2}}}{1+\frac{1}{t_{1} t_{2}}}=-\left(\frac{1}{t_{1}}-\frac{1}{t_{2}}\right)=-\left(\frac{1}{t_{1}}+2 t_{1}\right) \\
\leqslant-2 \sqrt{\frac{1}{t_{1}} \cdot 2 t_{1}}=-2 \sqrt{2} \quad \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \cdots \cdots \cdot 15 \text { points } \\
\end{array}
$$

When and only when $\frac{1}{t_{1}}=2 t_{1}$, i.e., $t_{1}=\frac{\sqrt{2}}{2}$, the equality holds.
Thus, when $m$ reaches its minimum value $-\frac{p^{2}}{2}$, $\tan \theta$ reaches its maximum value of $-2 \sqrt{2}$ 20 points