9. Given the parabola $\Gamma: y^{2}=8 x$ with focus $F$, a line $l$ through $F$ intersects the parabola $\Gamma$ at points $A, B$. Tangents to the parabola $\Gamma$ at $A, B$ intersect the $y$-axis at points $P, Q$ respectively. Find the minimum value of the area of quadrilateral $A P Q B$.

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9 Analysis: It is known that the focus of the parabola is $F(2,0)$. By symmetry, let's assume $A\left(2 x_{1}^{2}, 4 x_{1}\right)\left(x_{1}>0\right), B\left(2 x_{2}^{2}, 4 x_{2}\right)\left(x_{2}<0\right)$, then $\frac{4 x_{1}-0}{2 x_{1}^{2}-2}=\frac{4 x_{2}-0}{2 x_{2}^{2}-2}$, simplifying gives $x_{1} x_{2}=-1 \quad(AB$ has no slope when $x_{1}=1, x_{2}=-1$, this equation also holds). Therefore, $x_{1}-x_{2} \geq 2 \sqrt{x_{1} \cdot\left(-x_{2}\right)} \geq 2$. The tangent line equation at $A$ is $y=\frac{1}{x_{1}} x+2 x_{1}$, thus $P\left(0,2 x_{1}\right)$, similarly, we get $Q\left(0,2 x_{2}\right)$. Draw perpendiculars from $A, B$ to the $y$-axis, with feet at $C, D$ respectively. Let $t=x_{1}-x_{2} \geq 2$, then
$$
\begin{aligned}
S_{\text {quadrilateral }} A B Q P= & S_{\text {trapezoid }} A B D C-S_{\triangle} A P C-S_{\Delta} B Q D \\
& =\frac{\left(2 x_{1}^{2}+2 x_{2}^{2}\right) \cdot\left(4 x_{1}-4 x_{2}\right)}{2}-\left(\frac{x_{1}^{2} \cdot 4 x_{1}}{2}+\frac{x_{2}^{2} \cdot\left(-4 x_{2}\right)}{2}\right) \\
& =4 t\left(t^{2}-2\right)-2 t\left(t^{2}-3\right)=2 t\left(t^{2}-1\right) \geq 12
\end{aligned}
$$

Equality holds when $t=2$, i.e., $x_{1}=1, x_{2}=-1$, which means $AB$ is perpendicular to the $x$-axis, at which point the maximum area is 12.