Example 8 (1991 National High School League Question) Let $O$ be the vertex of a parabola, $F$ the focus, and $P Q$ a chord passing through $F$. Given $|O F|=a,|P Q|=b$, find the area of $\triangle O P Q$.

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Solution 1: Taking $F$ as the pole and the reverse extension of ray $F O$ as the polar axis to establish a polar coordinate system, the equation of the parabola is
$$
\rho=\frac{2 a}{1-\cos \theta}
$$

Let the polar angle of point $P$ be $\theta(\theta \in(0, \pi))$, then the polar angle of point $Q$ is $\pi+\theta$, so
$$
b=|P Q|=\rho_{P}+\rho_{Q}=\frac{2 a}{1-\cos \theta}+\frac{2 a}{1-\cos (\pi+\theta)}=\frac{4 a}{1-\cos ^{2} \theta}=\frac{4 a}{\sin ^{2} \theta},
$$

thus $\sin \theta=2 \sqrt{\frac{a}{b}}$. Therefore,
$$
\begin{aligned}
S_{\triangle O P Q} & =S_{\triangle O F P}+S_{\triangle O F Q}=\frac{1}{2} \cdot|O F| \cdot|P F| \cdot \sin \theta+\frac{1}{2}|O F| \cdot|F Q| \sin (\pi-\theta) \\
& =\frac{1}{2}|O F| \cdot(|P F|+|P Q|) \sin \theta=\frac{1}{2} \cdot|O F| \cdot|P Q| \cdot \sin \theta \\
& =\frac{1}{2} \cdot a \cdot b \cdot 2 \sqrt{\frac{a}{b}}=a \sqrt{a b} .
\end{aligned}
$$

Solution 2: Taking $O$ as the origin and $O F$ as the $x$-axis to establish a Cartesian coordinate system. Since $|O F|=a$ and $F$ is the focus, the equation of the parabola is $y^{2}=4 a x$. Let the parametric equation of line $P Q$ be
$$
\left\{\begin{array}{l}
x=a+t \cos \theta, \\
y=t \sin \theta,
\end{array}\right.
$$
where $\theta$ is the inclination angle of $P Q$, $\theta \in(0, \pi)$. Substituting into the equation of the parabola and rearranging, we get
$$
t^{2} \sin ^{2} \theta-4 a t \cos \theta-4 a^{2}=0 \text {. }
$$

Let its two roots be $t_{1}, t_{2}$, then
$$
\begin{aligned}
b & =|P Q|=|P F|+|F Q|=\left|t_{1}-t_{2}\right|=\sqrt{\left(t_{1}+t_{2}\right)^{2}-4 t_{1} t_{2}} \\
& =\sqrt{\left(\frac{4 a \cos \theta}{\sin ^{2} \theta}\right)^{2}-4\left(-\frac{4 a^{2}}{\sin ^{2} \theta}\right)}=\frac{4 a}{\sin ^{2} \theta} .
\end{aligned}
$$

The rest is the same as Solution 1.