Example 2 Let $F$ be the focus of the parabola $y^{2}=4 x$, and $A, B$ be two points on the parabola different from the origin $O$, satisfying $\overrightarrow{F A} \cdot \overrightarrow{F B}=0$. Extend $A F, B F$ to intersect the parabola at points $C, D$ respectively. Find the minimum value of the area of quadrilateral $A B C D$.
Solve: Taking the focus $F$ as the pole and the axis of symmetry $Fx$ as the polar axis to establish a polar coordinate system. Since the distance from the focus to the corresponding directrix is 2, the polar equation of the conic section is
$$
\rho=\frac{2}{1-\cos \theta} \text {. }
$$

Let $\angle A F x=\theta$.
By symmetry, assume $0<\theta<\frac{\pi}{2}$.
Then $|F A|=\frac{2}{1-\cos \theta},|F C|=\frac{2}{1+\cos \theta}$.
Thus, $|A C|=|F A|+|F C|$
$$
=\frac{2}{1-\cos \theta}+\frac{2}{1+\cos \theta}=\frac{4}{\sin ^{2} \theta} \text {. }
$$

Since $\overrightarrow{F A} \cdot \overrightarrow{F B}=0$, it follows that $F A \perp F B$.
Substituting $\theta$ with $\theta+\frac{\pi}{2}$, we get
$$
|B D|=\frac{4}{\sin ^{2}\left(\theta+\frac{\pi}{2}\right)}=\frac{4}{\cos ^{2} \theta} \text {. }
$$

Therefore, $S_{\text {quadrilateral } A B C D}=\frac{1}{2}|A C||B D|$
$$
=\frac{8}{\sin ^{2} \theta \cdot \cos ^{2} \theta}=\frac{32}{\sin ^{2} 2 \theta} \geqslant 32 \text {. }
$$

The equality holds if and only if $\theta=\frac{\pi}{4}$.
Thus, the minimum area of quadrilateral $A B C D$ is 32.