Four, (20 points)
Let a parabola be intersected by a line with a fixed slope at $M_{1}, M_{2}$, and a point $P$ on the plane satisfies $2 \left|P M_{1}\right|^{2}+2\left|P M_{2}\right|^{2}-\left|M_{1} M_{2}\right|^{2}$ being a constant $a(a \geqslant 0)$. Find the locus of point $P$.
Let the midpoint of $M_{1} M_{2}$ be $O$.
Let $\angle M_{1} O P=\theta$.
Then $2\left|P M_{1}\right|^{2}+2\left|P M_{2}\right|^{2}-\left|M_{1} M_{2}\right|^{2}$
$$
\begin{aligned}
= & 2\left(|P O|^{2}+\left|O M_{1}\right|^{2}-2|P O| \cdot\left|O M_{1}\right| \cos \theta+|P O|^{2}+\left|O M_{2}\right|^{2}+2\right. \\
& \left.|P O| \cdot\left|O M_{2}\right| \cos \theta\right)-4\left|O M_{1}\right|^{2}
\end{aligned}
$$
$=4|P O|^{2}=a \quad$ Thus, $|P O|=\frac{\sqrt{a}}{2}$ is a constant.
Considering the trajectory of $O$, let the line be $y=k x+l$ (where $k$ is a constant), and let $M_{1}\left(x_{1}, y_{1}\right)$ and $M_{2}\left(x_{2}, y_{2}\right)$ be points on the parabola $y^{2}=2 p x$.
Then, combining $y=k x+1$ and $y^{2}=2 p x$,
we have $k^{2} x^{2}+(2 k l-2 p) x+l^{2}=0$,
$\frac{x_{1}+x_{2}}{2}=\frac{2 p-2 k l}{2 k^{2}}=\frac{p}{k^{2}}-\frac{l}{k}$
$\frac{y_{1}+y_{2}}{2}=k \cdot\left(\frac{p}{k^{2}}-\frac{l}{k}\right)+l=\frac{p}{k}$ is a constant.
Therefore, the trajectory of point $O$ is a ray (emanating from a point on the parabola).
Thus, the trajectory of $P$ is the shaded region as shown in the figure.