Example 4 As shown in Figure 2, let the equation of the parabola be $x^{2}=2 p y$ $(p>0)$, and let $M$ be any point on the line $y=-2 p$. Draw tangents from $M$ to the parabola, with points of tangency at $A$ and $B$.
(1) Prove that the x-coordinates of points $A$, $M$, and $B$ form an arithmetic sequence.
(2) When point $M(2,-2 p)$, $|A B|=4 \sqrt{10}$, find the equation of the parabola at this time.
(3) Does there exist a point $M$ such that the point $C$'s reflection $D$ over the line $A B$ lies on the parabola $x^{2}=2 p y(p>0)$, where point $C$ satisfies $\overrightarrow{O C}=\overrightarrow{O A}+\overrightarrow{O B}(O$ is the origin)? If it exists, find all points $M$ that meet the condition; if not, explain the reason.
Proof: (1) Let $M\left(x_{0}, y_{0}\right)$. Then the equation of the line containing the chord $AB$ is
$$
x_{0} x=p\left(y+y_{0}\right) \text {. }
$$

Substituting into the parabola $x^{2}=2 p y$ and eliminating $y$ gives
$$
x^{2}-2 x_{0} x+2 p y_{0}=0 \text {. }
$$

Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. Then $x_{1}+x_{2}=2 x_{0}$.
Therefore, the x-coordinates of points $A$, $M$, and $B$ form an arithmetic sequence.
Solution: (2) When $x_{0}=2, y_{0}=-2 p$,
$$
x_{1}+x_{2}=4, x_{1} x_{2}=-4 p^{2}, k_{A B}=\frac{x_{0}}{p}=\frac{2}{p} \text {. }
$$

By the chord length formula,
$$
\begin{array}{l}
|A B|=\sqrt{1+k^{2}} \cdot \sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
=\sqrt{1+\frac{4}{p^{2}}} \cdot \sqrt{16+16 p^{2}} .
\end{array}
$$

Given $|A B|=4 \sqrt{10}$, we have $p=1$ or $p=2$.
Thus, the required parabola equations are $x^{2}=2 y$ or $x^{2}=4 y$.
(3) Let $D\left(x_{3}, y_{3}\right)$.

By the problem, $C\left(x_{1}+x_{2}, y_{1}+y_{2}\right)$, so the coordinates of the midpoint of $CD$ are
$$
Q\left(\frac{x_{1}+x_{2}+x_{3}}{2} ; \frac{y_{1}+y_{2}+y_{3}}{2}\right) .
$$

Since point $Q$ lies on the line $AB$, and noting that the point $\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$ also lies on the line $AB$, substituting into $x_{0} x=p\left(y+y_{0}\right)$ gives
$$
y_{3}=\frac{x_{0}}{p} x_{3} .
$$

If $D\left(x_{3}, y_{3}\right)$ lies on the parabola, then
$$
x_{3}^{2}=2 p y_{3}=2 x_{0} x_{3} \text {. }
$$

Therefore, $x_{3}=0$ or $x_{3}=2 x_{0}$, i.e.,
$D(0,0)$ or $D\left(2 x_{0}, \frac{2 x_{0}^{2}}{p}\right)$.
When $x_{0}=0$, $x_{1}+x_{2}=2 x_{0}=0$, in this case, point $M\left(0,-2 p\right)$ satisfies the condition:

When $x_{0} \neq 0$, since $x_{C}=x_{1}+x_{2}=2 x_{0}$, we have $x_{C}=x_{D}$. Thus, the line $CD$ is parallel to the $y$-axis, but $k_{A B}=\frac{x_{0}}{p} \neq 0$, which contradicts $A B \perp C D$. Therefore, there does not exist a point $M$ that satisfies the condition.

In summary, only the point $M(0,-2 p)$ satisfies the condition.