11. Given the parabola $y^{2}=2 p x(p>0), A, B$ are two moving points on the parabola different from the vertex $O$. 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.
Since $S_{\triangle A O B}=\frac{1}{2}|\overrightarrow{O A}| \cdot|\overrightarrow{O B}| \sin \theta=\frac{1}{2} \overrightarrow{O A} \cdot \overrightarrow{O B} \cdot \tan \theta \Rightarrow m=\frac{1}{2} \overrightarrow{O A} \cdot \overrightarrow{O B}$. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then $x_{1}=\frac{y_{1}^{2}}{2 p}, x_{2}=\frac{y_{2}^{2}}{2 p}$.
Thus, $m=\frac{1}{2}\left(x_{1} x_{2}+y_{1} y_{2}\right)=\frac{1}{2}\left(\frac{y_{1}^{2} y_{2}^{2}}{4 p^{2}}+y_{1} y_{2}\right)=\frac{1}{8 p^{2}}\left[\left(y_{1} y_{2}\right)^{2}+4 p^{2}\left(y_{1} y_{2}\right)\right]$ $=\frac{1}{8 p^{2}}\left[\left(y_{1} y_{2}+2 p^{2}\right)^{2}-4 p^{4}\right] \geqslant-\frac{p^{2}}{2}$, equality holds when $y_{1} y_{2}=-2 p^{2}$.
Therefore, the minimum value of $m$ is $-\frac{p^{2}}{2}$. At this point, assume point $A$ is in the first quadrant, by the angle formula we get $\tan \theta=\frac{k_{O A}-k_{O B}}{1+k_{O A} k_{O B}}=\frac{\frac{2 p}{y_{1}}-\frac{2 p}{y_{2}}}{1+\frac{4 p^{2}}{y_{1} y_{2}}}=\frac{2 p\left(y_{2}-y_{1}\right)}{y_{1} y_{2}+4 p^{2}}=-\frac{y_{1}-y_{2}}{p}$ $\leqslant-\frac{2 \sqrt{\left|y_{1} y_{2}\right|}}{p}=-2 \sqrt{2}$, equality holds when $y_{1}=-y_{2}=\sqrt{2} p$. In summary, when $m$ reaches its minimum value $-\frac{p^{2}}{2}$, the maximum value of $\tan \theta$ is $-2 \sqrt{2}$.