12. Given that $A$ and $B$ are two moving points on the parabola $C: y^{2}=4 x$, point $A$ is in the first quadrant, and point $B$ is in the fourth quadrant. Lines $l_{1}$ and $l_{2}$ pass through points $A$ and $B$ respectively and are tangent to the parabola $C$. $P$ is the intersection point of $l_{1}$ and $l_{2}$.
(1) If line $A B$ passes through the focus $F$ of the parabola $C$, prove that the moving point $P$ lies on a fixed line, and find the equation of this line;
(2) Let $C$ and $D$ be the intersection points of lines $l_{1}$ and $l_{2}$ with the line $x=4$. Find the minimum value of the area of $\triangle P C D$.
12. (1) Let $A\left(\frac{y_{1}^{2}}{4}, y_{1}\right), B\left(\frac{y_{2}^{2}}{4}, y_{2}\right)\left(y_{1}>0>y_{2}\right)$. It is easy to know that the slope of line $l_{1}$ exists, let's assume it is $k_{1}$. Then $l_{1}: y-y_{1}=k_{1}\left(x-\frac{y_{1}^{2}}{4}\right)$.
Substituting the above equation into the parabola equation and simplifying, we get
$$
k_{1} y^{2}-4 y+4 y_{1}-k_{1} y_{1}^{2}=0 \text {. }
$$

Since line $l_{1}$ is tangent to the parabola $C$,
$$
\Delta=16-4 k_{1}\left(4 y_{1}-k_{1} y_{1}^{2}\right)=0 \text {. }
$$

Thus, $k_{1}=\frac{2}{y_{1}}$.
Therefore, $l_{1}: y=\frac{2}{y_{1}} x+\frac{1}{2} y_{1}$.
Similarly, $l_{2}: y=\frac{2}{y_{2}} x+\frac{1}{2} y_{2}$.
Solving the equations of $l_{1}$ and $l_{2}$ simultaneously, we get point $P\left(\frac{y_{1} y_{2}}{4}, \frac{y_{1}+y_{2}}{2}\right)$.
Since $k_{A B}=\frac{y_{1}-y_{2}}{\frac{y_{1}^{2}}{4}-\frac{y_{2}^{2}}{4}}=\frac{4}{y_{1}+y_{2}}$, we have
$$
l_{A B}: y-y_{1}=\frac{4}{y_{1}+y_{2}}\left(x-\frac{y_{1}^{2}}{4}\right) \text {. }
$$

Since line $A B$ passes through the focus $F(1,0)$ of the parabola $C$, then $-y_{1}=\frac{4}{y_{1}+y_{2}}\left(1-\frac{y_{1}^{2}}{4}\right)$, $y_{1} y_{2}=-4$.
Thus, $x_{P}=\frac{y_{1} y_{2}}{4}=-1$, which means point $P$ lies on the fixed line $x=-1$.