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

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Five, (20 points)
Let the parabola $y^{2}=2 p x$ inscribe an equilateral triangle with the coordinates of its 3 vertices being $p_{i}\left(2 p t_{i}^{2}, 2 p t_{i}\right)(i=1,2, 3)$. The coordinates of the center of the triangle $p$ are $(x, y)$.
Then $\left\{\begin{array}{l}x=\frac{2}{3} p\left(t_{1}^{2}+t_{2}^{2}+t_{3}^{2}\right) \\ y=\frac{2}{3} p\left(t_{1}+t_{2}+t_{3}\right)\end{array}\right.$
Without loss of generality, we can assume that $\Delta p_{1} p_{2} p_{3}$ is arranged counterclockwise.
Then $\overrightarrow{p_{1} p_{3}}=\overrightarrow{p_{1} p_{2} e^{\frac{\pi}{3}}}$.
Thus $2 p\left(t_{3}^{2}-t_{1}^{2}\right)+2 p\left(t_{3}-t_{1}\right) i=\left[2 p\left(t_{2}^{2}-t_{1}^{2}\right)+2 p\left(t_{2}-t_{1}\right) i\right] e^{i \frac{\pi}{3}}$.
By equating the real parts, we get: $2 t_{3}^{2}-t_{1}^{2}-t_{2}^{2}=-\sqrt{3}\left(t_{2}-t_{1}\right)$,
So $\left(2 t_{3}^{2}-t_{1}^{2}-t_{2}^{2}\right)\left(t_{1}+t_{2}\right)=-\sqrt{3}\left(t_{2}^{2}-t_{1}^{2}\right)$
Similarly, by equating the imaginary parts, we get: $\sqrt{3}\left(t_{2}^{2}-t_{1}^{2}\right)=2 t_{3}-t_{1}-t_{2}$
From (3) and (4), we get: $\left(2 t_{3}^{2}-t_{1}^{2}-t_{2}^{2}\right)\left(t_{1}+t_{2}\right)=-\left(2 t_{3}-t_{1}-t_{2}\right)$

That is $\left(3 t_{3}^{2}-t_{1}^{2}-t_{2}^{2}-t_{3}^{2}\right)\left(t_{1}+t_{2}\right)=-\left(3 t_{3}-t_{1}-t_{2}-t_{3}\right)$
Substituting (1) and (2) into (5) gives $\left(t_{3}^{2}-\frac{x}{2 p}\right)\left(\frac{3 y}{2 p}-t_{3}\right)=\frac{y}{2 p}-t_{3}$.

Similarly: From $\overrightarrow{p_{3} p_{2}}=\overrightarrow{p_{3}} \overrightarrow{p_{1}} e^{-\frac{\pi}{3}}$ we get $\left(t_{2}^{2}-\frac{x}{2 p}\right)\left(\frac{3 y}{2 p}-t_{2}\right)=\frac{y}{2 p}-t_{2}$.
From $\overrightarrow{p_{2} p_{1}}=\overrightarrow{p_{2} p_{3}} e^{i \frac{\pi}{3}}$ we get $\left(t_{1}^{2}-\frac{x}{2 p}\right)\left(\frac{3 y}{2 p}-t_{1}\right)=\frac{y}{2 p}-t_{1}$.
Therefore, $t_{1}, t_{2}, t_{3}$ are the roots of the equation $t^{3}-\left(\frac{3 y}{2 p}\right) t^{2}-\left(1+\frac{x}{2 p}\right) t+\left(\frac{y}{2 p}+\frac{3 x y}{4 p^{2}}\right)=0$.
By Vieta's formulas, we get: $t_{1} t_{2}+t_{2} t_{3}+t_{3} t_{1}=\left(1+\frac{x}{2 p}\right)$, and since $t_{1}^{2}+t_{2}^{2}+t_{3}^{2}+2$ $\left(t_{1} t_{2}+t_{2} t_{3}+t_{3} t_{1}\right)=\left(t_{1}+t_{2}+t_{3}\right)^{2}$, we thus get $9 y^{2}=2 p x-8 p^{2}$.