Example 7 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 $\triangle O P Q$.
(1991, National High School Mathematics Competition)
Analysis 1: To find the area of $\triangle ABC$, common formulas are $S=\frac{1}{2} a h_{a}$ or $S=\frac{1}{2} a b \sin C$. If the former is used to find the area of $\triangle OPQ$, the key is to find the distance from point $O$ to $PQ$, which requires establishing a Cartesian coordinate system and finding the equation of the line on which chord $PQ$ lies, which is relatively complex; if the latter is used, it can be solved by establishing a polar coordinate system, which is very simple.

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

Let $P\left(\rho_{P}, \theta\right)$, then $Q\left(\rho_{Q}, \pi+\theta\right)$. Thus, $|PQ|=\rho_{P}+\rho_{Q}$
$$
=\frac{2 a}{1-\cos \theta}+\frac{2 a}{1-\cos (\pi+\theta)}=\frac{4 a}{\sin ^{2} \theta},
$$

i.e., $\frac{4 a}{\sin ^{2} \theta}=b$. Solving for $\sin \theta$ yields $\sin \theta=2 \sqrt{\frac{a}{b}}$.
Therefore, $S_{\triangle OPQ}=\frac{1}{2} a b \sin \theta=a \sqrt{a b}$.
Analysis 2: Establish a Cartesian coordinate system so that the parabola has the standard equation $y^{2}=4 a x$. Substituting the parametric equation of line $PQ$ $\left\{\begin{array}{l}x=a+t \cos \theta, \\ y=t \sin \theta\end{array}\right.$ into the parabola equation, we get
$$
t^{2} \sin ^{2} \theta-4 a t \cos \theta-4 a^{2}=0.
$$

We have $b=|PQ|=\left|t_{1}-t_{2}\right|=\frac{4 a}{\sin ^{2} \theta}$.
Thus, $\sin \theta=2 \sqrt{\frac{a}{b}}$.
Therefore, $S_{\triangle OPQ}=\frac{1}{2} a b \sin \theta=a \sqrt{a b}$.