11. (20 points) Through a point $P$ on the directrix $l$ of the parabola $C: y^{2}=4 x$, draw two tangent lines to the parabola $C$, touching at points $A$ and $B$ respectively, and let $M$ be the intersection of the directrix $l$ with the $x$-axis.
(1) Let the focus of the parabola be $F$, prove:
$$
|P F|^{2}=|A F||B F| \text {; }
$$
(2) If $\tan \angle A M B=2 \sqrt{2}$, find the coordinates of point $P$ and the equation of the line $A B$.
11. (1) It is known that the equation of the directrix $l$ of the parabola $C$ is $x=-1$.
Let $P\left(-1, y_{0}\right), A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Then $l_{A B}: y_{0} y=2(x-1)$.
When $y_{0} \neq 0$, the slope of line $A B$ is $k=\frac{2}{y_{0}}$.
It is known that $F(1,0)$.
Then the slope of line $P F$ is $k^{\prime}=-\frac{y_{0}}{2}$.
Thus, $k k^{\prime}=-1 \Rightarrow P F \perp A B$.
Assume without loss of generality that $y_{1}>0>y_{2}$.
From $\left\{\begin{array}{l}y^{2}=4 x, \\ y=k(x-1)\end{array}\right.$
$\Rightarrow k^{2} x^{2}-2\left(k^{2}+2\right) x+k^{2}=0$.
Using Vieta's formulas, we get
$x_{1}+x_{2}=2\left(1+\frac{2}{k^{2}}\right), x_{1} x_{2}=1$.
Then $y_{1} y_{2}=k^{2}\left(x_{1}-1\right)\left(x_{2}-1\right)$
$$
\begin{array}{l}
=k^{2}\left[x_{1} x_{2}-\left(x_{1}+x_{2}\right)+1\right]=-4, \\
\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
=\frac{4 \sqrt{1+k^{2}}}{k^{2}} .
\end{array}
$$

Thus, $k_{P A} k_{P B}=\frac{2}{y_{1}} \cdot \frac{2}{y_{2}}=-1$.
Therefore, $P A \perp P B$.
When $y_{0}=0$, point $P$ coincides with $M$, at this time, $\triangle P A B$ is an isosceles right triangle, and we also have
$P F \perp A B, P A \perp A B$.
In summary, $P Q$ is the altitude on the hypotenuse $A B$ of the right triangle $\triangle P A B$.
By the projection theorem, we get
$|P F|^{2}=|A F||B F|$.
(2) When $y_{0}=0$,
$\angle A M B=90^{\circ} \Rightarrow \tan \angle A M B \neq 2 \sqrt{2}$.
Therefore, $y_{0} \neq 0$.
When $x_{1}>x_{2}$, $k>0$;
When $x_{1}<x_{2}$, $k<0$.
Thus, $k\left(x_{1}-x_{2}\right)=\frac{4 \sqrt{1+k^{2}}}{|k|}$.
Hence, $\tan \angle A M B$
$$
\begin{array}{l}
=\frac{\tan \angle A M F+\tan \angle B M F}{1-\tan \angle A M F \cdot \tan \angle B M F} \\
=\frac{\frac{y_{1}}{x_{1}+1}+\frac{-y_{2}}{x_{2}+1}}{1-\frac{y_{1}}{x_{1}+1} \cdot \frac{-y_{2}}{x_{2}+1}} \\
=\frac{x_{2} y_{1}-x_{1} y_{2}+y_{1}-y_{2}}{\left(x_{1}+1\right)\left(x_{2}+1\right)+y_{1} y_{2}} \\
=\frac{2 k\left(x_{1}-x_{2}\right)}{x_{1}+x_{2}+x_{1} x_{2}+1+y_{1} y_{2}} \\
=2|k| \sqrt{1+k^{2}} \\
\Rightarrow 2|k| \sqrt{1+k^{2}}=2 \sqrt{2} \\
\Rightarrow k= \pm 1, y_{0}= \pm 2 .
\end{array}
$$

Therefore, $P(-1, \pm 2)$, and the equation of line $A B$ is $y=x-1$ or $y=-x+1$.