10. (20 points) Given the parabola $\Gamma: y^{2}=8 x$ with focus $F$, a line $l$ through $F$ intersects the parabola $\Gamma$ at points $A$ and $B$. Tangents to the parabola $\Gamma$ at $A$ and $B$ intersect the $y$-axis at points $P$ and $Q$, respectively. Find the minimum value of the area of quadrilateral $A P Q B$.
10. Let \( A\left(\frac{y_{1}^{2}}{8}, y_{1}\right), B\left(\frac{y_{2}^{2}}{8}, y_{2}\right)\left(y_{1}>0, y_{2}<0\right) \), \( l_{A B}: x=m y+2 \).
Combining the above equation with the equation of the parabola \(\Gamma\) and eliminating \(x\) yields \( y^{2}-8 m y-16=0 \)
\[
\Rightarrow y_{1}+y_{2}=8 m, y_{1} y_{2}=-16 \text {. }
\]

The equation of the tangent line at point \(A\) is \( y y_{1}=4\left(x+x_{1}\right) \).
Let \(x=0\), we get
\[
y_{P}=\frac{4 x_{1}}{y_{1}}=4 \times \frac{\frac{y_{1}^{2}}{8}}{y_{1}}=\frac{y_{1}}{2} \text {. }
\]

Similarly, \( y_{Q}=\frac{y_{2}}{2} \).
Draw perpendiculars from points \(A\) and \(B\) to the \(y\)-axis, with the feet of the perpendiculars being \(E\) and \(D\) respectively.
\[
\begin{array}{l} 
\text { Then } S_{\text {quadrilateral } A P Q B}=S_{\text {quadrilateral } A E D B}-S_{\triangle A E P}-S_{\triangle B D Q} \\
= \frac{1}{2}\left(x_{1}+x_{2}\right)\left(y_{1}-y_{2}\right)- \\
\left(\frac{1}{2}\left|y_{1}-\frac{1}{2} y_{1}\right| x_{1}+\frac{1}{2}\left|y_{2}-\frac{1}{2} y_{2}\right| x_{2}\right) \\
= \frac{1}{2}\left(x_{1}+x_{2}\right)\left(y_{1}-y_{2}\right)-\frac{1}{4}\left(x_{1} y_{1}-x_{2} y_{2}\right) \\
= \frac{1}{32}\left(2\left(y_{1}^{2}+y_{2}^{2}\right)\left(y_{1}-y_{2}\right)-\left(y_{1}^{3}-y_{2}^{3}\right)\right) \\
=\left.\frac{1}{32}\left(\left(y_{1}+y_{2}\right)^{2}-3 y_{1} y_{2}\right)\right) \sqrt{\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}} \\
= \frac{1}{32}\left(64 m^{2}+48\right) \sqrt{64 m^{2}+64} .
\end{array}
\]

Notice that the function
\[
f(m)=\frac{1}{32}\left(64 m^{2}+48\right) \sqrt{64 m^{2}+64}
\]

is monotonically increasing.
Thus, when \(m=0\), the area is minimized, and the minimum value is 12.