9. (16 points) Given the parabola $C_{1}: y=x^{2}, M$ is a moving point in the plane. Two tangent lines are drawn from $M$ to the parabola $C_{1}$, touching it at points $A$ and $B$. If the area of $\triangle M A B$ is a constant $2 c^{3}$, find the equation of the locus of point $M$.
9. Let $A\left(x_{1}, x_{1}^{2}\right), B\left(x_{2}, x_{2}^{2}\right)$.
Then $l_{A B}: y=\left(x_{1}+x_{2}\right) x-x_{1} x_{2}$.
The equations of the tangents $M A$ and $M B$ are
$$
\begin{array}{l}
y=2 x_{1}\left(x-x_{1}\right)+x_{1}^{2}=2 x_{1} x-x_{1}^{2}, \\
y=2 x_{2}\left(x-x_{2}\right)+x_{2}^{2}=2 x_{2} x-x_{2}^{2} .
\end{array}
$$

Solving the equations of $M A$ and $M B$ simultaneously, we get
the point $M\left(\frac{x_{1}+x_{2}}{2}, x_{1} x_{2}\right)$.
The distance from $M$ to the line $A B$ is
$$
\begin{aligned}
h & =\frac{\left|\frac{\left(x_{1}+x_{2}\right)^{2}}{2}-2 x_{1} x_{2}\right|}{\sqrt{1+\left(x_{1}+x_{2}\right)^{2}}} \\
& =\frac{\left|\frac{\left(x_{1}-x_{2}\right)^{2}}{2}\right|}{\sqrt{1+\left(x_{1}+x_{2}\right)^{2}}} .
\end{aligned}
$$

Thus, $S_{\triangle M A B}=\frac{1}{2} h|A B|$
$=\frac{1}{2} \cdot \frac{\left|\frac{\left(x_{1}-x_{2}\right)^{2}}{2}\right|}{\sqrt{1+\left(x_{1}+x_{2}\right)^{2}}} \sqrt{1+\left(x_{1}+x_{2}\right)^{2}}\left|x_{1}-x_{2}\right|$
$$
=\frac{1}{4}\left|x_{1}-x_{2}\right|^{3}=2 c^{3} \text {. }
$$

Therefore, $\left|x_{1}-x_{2}\right|=2 c$, which means
$$
\left(\frac{x_{1}+x_{2}}{2}\right)^{2}-x_{1} x_{2}=c^{2} \text {. }
$$

Hence, the equation of the required locus is $x^{2}-y-c^{2}=0$.