Example 7 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 $C$, and $P$ 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 $\triangle P C D$.
(2013, National High School Mathematics League Fujian Province Preliminary)
(1) Proof: Let
$$
A\left(\frac{y_{1}^{2}}{4}, y_{1}\right), B\left(\frac{y_{2}^{2}}{4}, y_{2}\right)\left(y_{1}>0>y_{2}\right) \text {. }
$$

Then $l_{1}: y_{1} y=2\left(x+\frac{y_{1}^{2}}{4}\right)$,
$$
l_{2}: y_{2} y=2\left(x+\frac{y_{2}^{2}}{4}\right) \text {. }
$$

Let $P\left(x_{0}, y_{0}\right)$. Then
$$
y_{1} y_{0}=2\left(x_{0}+\frac{y_{1}^{2}}{4}\right), y_{2} y_{0}=2\left(x_{0}+\frac{y_{1}^{2}}{4}\right) \text {. }
$$

Therefore, the coordinates of points $A$ and $B$ satisfy the equation
$$
y y_{0}=2\left(x_{0}+x\right) \text {. }
$$

Since the line $AB$ passes through the point $F(1,0)$, we have
$$
0=2\left(x_{0}+1\right) \Rightarrow x_{0}=-1
$$
$\Rightarrow$ Point $P\left(-1, y_{0}\right)$ lies on the fixed line $x=-1$.
(2) Solution: From (1), we know
$$
\begin{array}{l}
C\left(4, \frac{8}{y_{1}}+\frac{y_{1}}{2}\right), D\left(4, \frac{8}{y_{2}}+\frac{y_{2}}{2}\right) \\
\Rightarrow C D=\left|\left(\frac{8}{y_{1}}+\frac{y_{1}}{2}\right)-\left(\frac{8}{y_{2}}+\frac{y_{2}}{2}\right)\right| \\
\quad=\left|\frac{\left(y_{1} y_{2}-16\right)\left(y_{1}-y_{2}\right)}{2 y_{1} y_{2}}\right| \\
\Rightarrow S_{\triangle P C D}=\frac{1}{2}\left|4-\frac{y_{1} y_{2}}{4}\right|\left|\frac{\left(y_{1} y_{2}-16\right)\left(y_{1}-y_{2}\right)}{2 y_{1} y_{2}}\right| .
\end{array}
$$

Let $y_{1} y_{2}=-t^{2}(t>0),\left|y_{1}-y_{2}\right|=m$.
$$
\begin{array}{l}
\text { By }\left(y_{1}+y_{2}\right)^{2}=\left(y_{1}-y_{2}\right)^{2}+4 y_{1} y_{2} \\
=m^{2}-4 t^{2} \geqslant 0,
\end{array}
$$

we know $m \geqslant 2 t$, with equality holding in (1) if and only if $y_{1}+y_{2}=0$.
$$
\begin{array}{l}
\text { Hence } S_{\triangle P C D}=\frac{1}{2}\left|4+\frac{t^{2}}{4}\right|\left|\frac{\left(-t^{2}-16\right) m}{-2 t^{2}}\right| \\
=\frac{m\left(t^{2}+16\right)^{2}}{16 t^{2}} \\
\geqslant \frac{2 t\left(t^{2}+16\right)^{2}}{16 t^{2}}=\frac{\left(t^{2}+16\right)^{2}}{8 t} .
\end{array}
$$

Let $f(t)=\frac{\left(t^{2}+16\right)^{2}}{8 t}$. Then
$$
\begin{array}{l}
f^{\prime}(t)=\frac{2\left(t^{2}+16\right) \cdot 2 t^{2}-\left(t^{2}+16\right)^{2}}{8 t^{2}} \\
=\frac{\left(t^{2}+16\right)\left(3 t^{2}-16\right)}{8 t^{2}} .
\end{array}
$$

Thus, when $0\frac{4 \sqrt{3}}{3}$, $f^{\prime}(t)>0$.
Therefore, $f(t)$ is decreasing on the interval $\left(0, \frac{4 \sqrt{3}}{3}\right)$ and increasing on the interval $\left[\frac{4 \sqrt{3}}{3},+\infty\right)$.
Hence, when $t=\frac{4 \sqrt{3}}{3}$, $f(t)$ attains its minimum value $\frac{128 \sqrt{3}}{9}$.
In summary, when $y_{1}+y_{2}=0, y_{1} y_{2}=-\frac{16}{3}$, i.e., $y_{1}=\frac{4 \sqrt{3}}{3}$, $y_{2}=-\frac{4 \sqrt{3}}{3}$, the area of $\triangle P C D$ attains its minimum value $\frac{128 \sqrt{3}}{9}$.