9. (14 points) As shown in Figure 1, given the parabola $x^{2}=2 p y$ $(p>0)$ and the line $y=b$ $(b<0)$, point $P(t, b)$
moves on the line. Two tangent lines are drawn from $P$ to the parabola, touching the parabola at points $A$ and $B$, respectively, and the midpoint of segment $A B$ is $M$.
(1) Find the trajectory of point $M$;
(2) Find the minimum value of $|A B|$.
(1) From the problem, we have $y=\frac{1}{2 p} x^{2}$, then $y^{\prime}=\frac{1}{p} x$. Let $A\left(x_{1}, \frac{1}{2 p} x_{1}^{2}\right)$ and $B\left(x_{2}, \frac{1}{2 p} x_{2}^{2}\right)$. Thus, $k_{P A}=\frac{x_{1}}{p}, k_{P B}=\frac{x_{2}}{p}$.
Therefore, $\frac{x_{1}}{p}=\frac{\frac{x_{1}^{2}}{2 p}-b}{x_{1}-t}$, which simplifies to $x_{1}^{2}-2 t x_{1}+2 p b=0$.
Similarly, $x_{2}^{2}-2 t x_{2}+2 p b=0$.
Hence, $x_{1}$ and $x_{2}$ are the two roots of the equation $x^{2}-2 t x+2 p b=0$. Then $x_{1}+x_{2}=2 t, x_{1} x_{2}=2 p b$.
Let $M(x, y)$. Thus,
$$
\begin{array}{l}
x=\frac{x_{1}+x_{2}}{2}=t, \\
y=\frac{y_{1}+y_{2}}{2}=\frac{1}{4 p}\left(x_{1}^{2}+x_{2}^{2}\right)=\frac{t^{2}}{p}-b .
\end{array}
$$

By eliminating $t$ from the above two equations, we get the trajectory equation of point $M$ as $x^{2}=p(y+b)$.
(2) $|A B|=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\frac{x_{1}^{2}-x_{2}^{2}}{4 p^{2}}}$
$$
=\sqrt{\left(4 t^{2}-8 p b\right)\left(1+\frac{t^{2}}{p^{2}}\right)} \text {. }
$$

From $b<0$, we know that when $t=0$,
$$
|A B|_{\text {min }}=2 \sqrt{-2 p b} \text {. }
$$