(3) $O$ is the vertex of the parabola, $F$ is the focus, and $P Q$ is a chord passing through $F$. It is known that
$|O F|=a,|P Q|=b$, then the area of $\triangle O P Q$ is
(3) $a \sqrt{a b}$ Hint: Let the equation of the parabola be $y^{2}=$ $2 p x$.
$$
\frac{p}{2}=|O F|=a, p=2 a,
$$

Therefore, the equation of the parabola is $y^{2}=4 a x$.
Let the equation of the line passing through $F$ be $y=k(x-a)$,
From (1)(2) eliminating $x$ we get $k y^{2}-4 a y-4 a^{2} k=0$.
Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, then
$$
\begin{array}{c}
y_{1}+y_{2}=\frac{4 a}{k}, y_{1} y_{2}=-4 a^{2} . \\
\left(y_{1}-y_{2}\right)^{2}=\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}=\frac{16 a^{2}}{k^{2}}+16 a^{2},
\end{array}
$$

So
$$
\begin{array}{c}
b=|P Q|=\sqrt{1+\frac{1}{k^{2}}} \cdot\left|y_{1}-y_{2}\right| \\
=\sqrt{1+\frac{1}{k^{2}}} \cdot 4 a \sqrt{\frac{1}{k^{2}}+1}=4 a\left(1+\frac{1}{k^{2}}\right), \\
1+\frac{1}{k^{2}}=\frac{b}{4 a},
\end{array}
$$

Therefore
$$
\begin{array}{l}
\left|y_{1}-y_{2}\right|=4 a \sqrt{\frac{1}{k^{2}}+1}=4 a \sqrt{\frac{b}{4 a}}=2 \sqrt{a b} \\
S_{\triangle O P Q}=\frac{1}{2} \cdot|O F| \cdot\left|y_{1}-y_{2}\right|=\frac{1}{2} \cdot a \cdot 2 \sqrt{a b}=a \sqrt{a b}
\end{array}
$$
(When $k$ does not exist, it also holds)