(7) Through a fixed point $F(2,0)$, draw a line $l$ intersecting the $y$-axis at point $Q$. Through point $Q$, draw $QT \perp FQ$ intersecting the $x$-axis at point $T$. Extend $TQ$ to point $P$ such that $|TQ|=|QP|$. Then the equation of the locus of point $P$ is $\qquad$.
(7) $y^{2}=8 x$ Hint: Let the equation of line $l$ be $y=k(x-2)$, then the coordinates of point $Q$ are $(0,-2 k)$. When $k \neq 0$, the equation of line $Q T$ is $y=-\frac{1}{k} x-2 k$, so

the coordinates of point $T$ are $\left(-2 k^{2}, 0\right)$. Thus, the coordinates of point $P$ are $\left(2 k^{2},-4 k\right)$. If we set the coordinates of $P$ as $(x, y)$, then $\left\{\begin{array}{l}x=2 k^{2}, \\ y=-4 k,\end{array}\right.$ eliminating $k$ yields the trajectory equation of point $P$ as $y^{2}=8 x$. It can be verified that the condition is also satisfied when $k=0$.