14. Let $O$ be the vertex of the parabola, $F$ the focus, and $P Q$ a chord passing through $F$. Given $|O F|=a,|P Q|=$ $b$, find the area of $\triangle O P Q$.
Given the parabola equation $y^{2}=4 a x$, and the line $P Q: x=n y+a$, combining $\left\{\begin{array}{l}y^{2}=4 a x, \\ x=n y+a\end{array} \Rightarrow y^{2}-4 a n y-4 a^{2}=0\right.$ $\Rightarrow|P Q|=\sqrt{1+n^{2}} \cdot \sqrt{16 a^{2} n^{2}+16 a^{2}}=4 a\left(1+n^{2}\right)$. Thus, by $|P Q|=b \Rightarrow 1+n^{2}=\frac{b}{4 a}$.

The distance from the origin $O$ to the line $P Q$ is $d=\frac{a}{\sqrt{1+n^{2}}}=\frac{2 a \sqrt{a b}}{b}$, so the area of $\triangle O P Q$ is $S_{\triangle O P Q}=\frac{1}{2}|P Q| \cdot d=a \sqrt{a b}$.