10. (20 points) 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 it at points $A$ and $B$ respectively, and the midpoint of segment $AB$ is $M$. Find
(1) the locus of point $M$;
(2) the minimum value of $|AB|$.
10. (1) From the problem, we have
$$
y=\frac{1}{2 p} x^{2} \Rightarrow y^{\prime}=\frac{1}{p} x \text {. }
$$

Let $A\left(x_{1}, \frac{1}{2 p} x_{1}^{2}\right), B\left(x_{2}, \frac{1}{2 p} x_{2}^{2}\right)$.
Thus, $k_{P A}=\frac{x_{1}}{p}, k_{P B}=\frac{x_{2}}{p}$.
Also, from $P(t, b)$, we have
$$
\frac{\frac{x_{1}^{2}}{2 p}-b}{x_{1}-t}=k_{P A}=\frac{x_{1}}{p} \text {, }
$$

which gives
$$
x_{1}^{2}-2 t x_{1}+2 p b=0 \text {. }
$$

Similarly, $x_{2}^{2}-2 t x_{2}+2 p b=0$.
Therefore, $x_{1} 、 x_{2}$ are the two roots of the equation $x^{2}-2 t x+2 p b=0$.
Hence, $x_{1}+x_{2}=2 t, x_{1} x_{2}=2 p b$.
Let $M(x, y)$. Then,
$$
\begin{array}{l}
x=\frac{x_{1}+x_{2}}{2}=t, \\
y=\frac{y_{1}+y_{2}}{2}=\frac{1}{4 p}\left(x_{1}^{2}+x_{2}^{2}\right)=\frac{t^{2}}{p}-b .
\end{array}
$$

Eliminating $t$ from the above two equations, we get the trajectory equation of $M$ as $x^{2}=p(y+b)$.
(2) Note that,
$$
\begin{array}{l}
|A B|=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\frac{\left(x_{1}^{2}-x_{2}^{2}\right)^{2}}{4 p^{2}}} \\
=\sqrt{\left(4 t^{2}-8 p b\right)\left(1+\frac{t^{2}}{p^{2}}\right)} .
\end{array}
$$

Thus, when $t=0$, $|A B|_{\min }=2 \sqrt{-2 p b}$.