Example 1 Let $O$ be the vertex of a parabola, $F$ the focus, and $P Q$ a chord passing through point $F$. Given $|O F|=a,|P Q|=$ $b$. Find the area of $\triangle O P Q$.
(1991, National High School Mathematics Competition)
Explanation: As shown in Figure 1, establish a polar coordinate system with $F$ as the pole and $F x$ as the polar axis. Then the equation of the parabola is
$$
\rho=\frac{2 a}{1-\cos \theta} .
$$

Let point $P\left(\rho_{1}, \theta\right)(0<\theta<\pi)$. Then
$$
\begin{array}{l}
Q\left(\rho_{2}, \theta+\pi\right) . \\
\text { Hence }|P Q|=\rho_{1}+\rho_{2} \\
=\frac{2 a}{1-\cos \theta}+ \\
\frac{2 a}{1-\cos (\theta+\pi)} \\
=\frac{4 a}{\sin ^{2} \theta},
\end{array}
$$

which means $\frac{4 a}{\sin ^{2} \theta}=b$.
$$
\begin{array}{l}
\text { Therefore, } \sin \theta=2 \sqrt{\frac{a}{b}} . \\
\text { Also, } S_{\triangle O P F}=\frac{1}{2} a|F P| \sin \theta, \\
S_{\triangle O Q F}=\frac{1}{2} a|F Q| \sin \theta,
\end{array}
$$

Thus,
$$
\begin{array}{l}
S_{\triangle O P Q}=S_{\triangle O P F}+S_{\triangle O Q F} \\
=\frac{1}{2} a(|F P|+|F Q|) \sin \theta=\frac{1}{2} a b \sin \theta \\
=\frac{1}{2} a b \cdot 2 \sqrt{\frac{a}{b}}=a \sqrt{a b} .
\end{array}
$$