11. (20 points) Find the equation of the locus of the center of an equilateral triangle inscribed in the parabola $y^{2}=2 p x$.


11. Let the parabola $y^{2}=2 p x$ inscribe an equilateral triangle with the coordinates of its three vertices being $P_{i}\left(2 p t_{i}^{2}, 2 p t_{i}\right)(i=1,2,3)$, and the center of the triangle be $M(x, y)$. Then
$$
\left\{\begin{array}{l}
x=\frac{2}{3} p\left(t_{1}^{2}+t_{2}^{2}+t_{3}^{2}\right), \\
y=\frac{2}{3} p\left(t_{1}+t_{2}+t_{3}\right) .
\end{array}\right.
$$

Assume $\triangle P_{1} P_{2} P_{3}$ is in a counterclockwise direction.
Then $\overrightarrow{P_{1} P_{3}}=\overrightarrow{P_{1} P_{2}}\left(\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}\right)$.
Thus $2 p\left(t_{3}^{2}-t_{1}^{2}\right)+2 p\left(t_{3}-t_{1}\right) \mathrm{i}$
$$
=\left[2 p\left(t_{2}^{2}-t_{1}^{2}\right)+2 p\left(t_{2}-t_{1}\right) \mathrm{i}\right] \mathrm{e}^{\frac{\pi}{3} \mathrm{i}} \text {. }
$$

By equating the real parts, we get
$$
\begin{array}{l}
2 t_{3}^{2}-t_{1}^{2}-t_{2}^{2}=-\sqrt{3}\left(t_{2}-t_{1}\right) \\
\Rightarrow\left(2 t_{3}^{2}-t_{1}^{2}-t_{2}^{2}\right)\left(t_{2}+t_{1}\right)=-\sqrt{3}\left(t_{2}^{2}-t_{1}^{2}\right) .
\end{array}
$$

Similarly, by equating the imaginary parts, we get
$$
\sqrt{3}\left(t_{2}^{2}-t_{1}^{2}\right)=2 t_{3}-t_{1}-t_{2} \text {. }
$$

From equations (3) and (4), we have
$$
\begin{array}{l}
\left(2 t_{3}^{2}-t_{1}^{2}-t_{2}^{2}\right)\left(t_{2}+t_{1}\right)=-\left(2 t_{3}-t_{1}-t_{2}\right) \\
\Rightarrow\left(3 t_{3}^{2}-t_{1}^{2}-t_{2}^{2}-t_{3}^{2}\right)\left(t_{2}+t_{1}\right) \\
\quad=-\left(3 t_{3}-t_{1}-t_{2}-t_{3}\right) .
\end{array}
$$

Substituting equations (1) and (2) into the above equation, we get
$$
\left(t_{3}^{2}-\frac{x}{2 p}\right)\left(\frac{3 y}{2 p}-t_{3}\right)=\frac{y}{2 p}-t_{3} \text {. }
$$

Similarly,
$$
\begin{array}{l}
\left(t_{1}^{2}-\frac{x}{2 p}\right)\left(\frac{3 y}{2 p}-t_{1}\right)=\frac{y}{2 p}-t_{1}, \\
\left(t_{2}^{2}-\frac{x}{2 p}\right)\left(\frac{3 y}{2 p}-t_{2}\right)=\frac{y}{2 p}-t_{2} .
\end{array}
$$

Thus, $t_{1}, t_{2}, t_{3}$ are the roots of the equation
$$
t^{3}-\frac{3 y}{2 p} t^{2}-\left(1+\frac{x}{2 p}\right) t+\left(\frac{y}{2 p}+\frac{3 x y}{4 p^{2}}\right)=0
$$

By Vieta's formulas, we have
$$
\begin{array}{l}
t_{1} t_{2}+t_{2} t_{3}+t_{3} t_{1}=-\left(1+\frac{x}{2 p}\right) . \\
\text { Also, } t_{1}^{2}+t_{2}^{2}+t_{3}^{2}+2\left(t_{1} t_{2}+t_{2} t_{3}+t_{3} t_{1}\right) \\
=\left(t_{1}+t_{2}+t_{3}\right)^{2},
\end{array}
$$

Thus $9 y^{2}=2 p x-8 p^{2}$.