14. As shown in Figure 3, given the parabola $C: x^{2}=2 p y$ $(p>0)$, $O$ is the origin, and $M(0, b)(b>0)$ is a moving point on the $y$-axis. A line through $M$ intersects the parabola $C$ at points $A$ and $B$. Let $S_{\triangle A O B}=m \tan \angle A O B$. Find the minimum value of $m$.
14. Let $l_{A B}: y=k x+b(k \neq 0) . l_{A B}$ intersects with the parabola $C$ at

$$
\left\{\begin{array}{l}
y=k x+b, \\
x^{2}=2 p y .
\end{array}\right.
$$

Thus, $x^{2}-2 p k x-2 p b=0$.
Then $x_{1}+x_{2}=2 p k, x_{1} x_{2}=-2 p b$.
Hence $y_{1} y_{2}=\left(k x_{1}+b\right)\left(k x_{2}+b\right)$
$$
\begin{array}{l}
=k^{2} x_{1} x_{2}+k b\left(x_{1}+x_{2}\right)+b^{2} \\
=k^{2}(-2 p b)+k b(2 p k)+b^{2}=b^{2} .
\end{array}
$$

When $A B \perp y$ axis, the above equation still holds.
$$
\begin{array}{l}
\text { By } S_{\triangle A O B}=\frac{1}{2}|O A| \cdot|O B| \sin \angle A O B \\
=\frac{1}{2}|O A| \cdot|O B| \cos \angle A O B \cdot \tan \angle A O B \\
=m \tan \angle A O B,
\end{array}
$$

we get $m=\frac{1}{2}|O A| \cdot|O B| \cos \angle A O B=\frac{1}{2} O A \cdot O B$
$$
\begin{array}{l}
=\frac{1}{2}\left(x_{1} x_{2}+y_{1} y_{2}\right)=\frac{1}{2}\left(-2 p b+b^{2}\right) \\
=\frac{1}{2}(b-p)^{2}-\frac{1}{2} p^{2} \geqslant-\frac{1}{2} p^{2} .
\end{array}
$$

When $b=p$, $m_{\text {min }}=-\frac{1}{2} p^{2}$.