9. (16 points) As shown in Figure 1, $A_{1}, A_{2}, A_{3}, A_{4}$ are four distinct points on the parabola $y^{2}=4 x$, and $A_{1}, F, A_{3}$ and $A_{2}, F, A_{4}$ are collinear respectively. Let the lines $A_{1} A_{2}$ and $A_{3} A_{4}$, $A_{1} A_{4}$ and $A_{2} A_{3}$ intersect at points $M$ and $N$, respectively, where $F(1,0)$. Find the minimum value of the area of $\triangle F M N$.
(1) Let $A_{1}\left(4 t_{1}^{2}, 4 t_{1}\right), A_{2}\left(4 t_{2}^{2}, 4 t_{2}\right)$, $A_{3}\left(4 t_{3}^{2}, 4 t_{3}\right), A_{4}\left(4 t_{4}^{2}, 4 t_{4}\right)$.
Then $l_{A_{1} A_{3}}:\left(t_{1}+t_{3}\right) y=x+4 t_{1} t_{3}$.
Since the line $A_{1} A_{3}$ passes through the point $F(1,0)$, we get $t_{3}=-\frac{1}{4 t_{1}}$.
Similarly, $t_{4}=-\frac{1}{4 t_{2}}$.
Thus, $l_{A_{1} A_{2}}:\left(t_{1}+t_{2}\right) y=x+4 t_{1} t_{2}$,
$l_{A_{3} A_{4}}:\left(t_{3}+t_{4}\right) y=x+4 t_{3} t_{4}$.
Transforming $l_{A_{3} A_{4}}$ yields
$l_{A_{3} A_{4}}:-\left(t_{1}+t_{2}\right) y=\left(4 t_{1} t_{2}\right) x+1$.
From the equations of $l_{A_{1} A_{2}}$ and $l_{A_{3} A_{4}}$, we get
$$
\left(4 t_{1} t_{2}+1\right)(x+1)=0 \text {. }
$$

If $4 t_{1} t_{2}+1=0$, then $t_{2}=t_{3}$, leading to points $A_{2}$ and $A_{3}$ coinciding, which is a contradiction.
Therefore, $4 t_{1} t_{2}+1 \neq 0, M\left(-1, \frac{4 t_{1} t_{2}-1}{t_{1}+t_{2}}\right)$.
Similarly, $N\left(-1, \frac{-4\left(t_{1}+t_{2}\right)}{4 t_{1} t_{2}-1}\right)$.
Thus, $\overrightarrow{F M} \cdot \overrightarrow{F N}$
$$
=\left(-2, \frac{4 t_{1} t_{2}-1}{t_{1}+t_{2}}\right) \cdot\left(-2, \frac{-4\left(t_{1}+t_{2}\right)}{4 t_{1} t_{2}-1}\right)
$$
$$
=0 \text {. }
$$
(2) Let the line $M N$ intersect the $x$-axis at point $K$. Then $|F K|=2$.

Let $M\left(-1, y_{1}\right), N\left(-1, y_{2}\right), y_{1} y_{2}=-4$. Then
$$
\begin{aligned}
S & =\frac{1}{2}|M N||F K|=\left|y_{1}-y_{2}\right| \\
& =\left|y_{1}\right|+\left|y_{2}\right| \geqslant 2 \sqrt{\left|y_{1}\right|\left|y_{2}\right|}=4 .
\end{aligned}
$$

When $y_{1}$ and $y_{2}$ take 2 and -2 respectively, $S_{\text {min }}=4$.