13.7.5 ** Given the parabola $y^{2}=2 p x(p>0)$, two perpendicular chords $F A$ and $F B$ are drawn through the focus $F$ of the parabola. Find the minimum value of the area of $\triangle F A B$.
Given the focus $F$ of a parabola as the pole and the $x$-axis as the polar axis, establish a polar coordinate system. The polar equation of the parabola is $\rho-\frac{p}{1-\cos \theta}$. Let $\Lambda\left(\rho_{1}, \theta\right), B\left(\rho_{2}, \theta+\frac{\pi}{2}\right), \theta \in(0, \pi]$, then $\rho_{1}=\frac{p}{1-\cos \theta}, \rho_{2}=\frac{p}{1+\sin \theta}$. Therefore,
$$
\begin{aligned}
S_{\triangle F A B} & =\frac{1}{2} \rho_{1} \rho_{2}-\frac{p^{2}}{2(1-\cos \theta)(1+\sin \theta)} \\
& =\frac{p^{2}}{(\sin \theta-\cos \theta+1)^{2}} \\
& =\frac{p^{2}}{\left[\sqrt{2} \sin \left(\theta-\frac{\pi}{4}\right)+1\right]^{2}} .
\end{aligned}
$$

Thus, when $\theta=\frac{3 \pi}{4}$, the minimum value of the area is $(3-2 \sqrt{2}) p^{2}$.