Five. (20 points) Given the parabola $\Gamma_{1}: y=x^{2}$ and the circle $\Gamma_{2}:(x-3)^{2}+y^{2}=1$, the focus of $\Gamma_{1}$ is $F$, and point $A$ is on $\Gamma_{2}$. The line $F A$ intersects $\Gamma_{1}$ at points $B$ and $C$ (point $B$ is between point $F$ and point $A$). The tangents to $\Gamma_{1}$ at points $B$ and $C$ intersect at point $P$. If point $A$ moves around the circle $\Gamma_{2}$ once, find the equation of the trajectory of point $P$.
Clearly, the focus of $\Gamma_{1}$ is $F\left(0, \frac{1}{4}\right)$. If the slope of line $F A$ is $k$, then the equation of line $F A$ is
$$
y=\frac{1}{4}+k x \text {. }
$$

Next, we find the range of $k$.
Since point $A(x, y)$ satisfies both equation (1) and $\Gamma_{2}$, substituting equation (1) into $\Gamma_{2}$ yields
$$
(x-3)^{2}+\left(\frac{1}{4}+k x\right)^{2}=1 \text {, }
$$

which simplifies to $\left(1+k^{2}\right) x^{2}+\left(\frac{k}{2}-6\right) x+\frac{129}{16}=0$.
Let $\Delta=\left(\frac{k}{2}-6\right)^{2}-4\left(1+k^{2}\right) \frac{129}{16} \geqslant 0$, solving this gives
$$
\frac{-3-\sqrt{129}}{32} \leqslant k \leqslant \frac{-3+\sqrt{129}}{32} \text {. }
$$

Let $B\left(x_{B}, x_{B}^{2}\right)$ and $C\left(x_{C}, x_{C}^{2}\right)$. Substituting equation (1) into $\Gamma_{1}$ yields
$$
x^{2}=k x+\frac{1}{4} \text {, }
$$

which simplifies to $4 x^{2}-4 k x-1=0$.
Then $x_{B} \backslash x_{C}\left(x_{C}<x_{B}\right)$ are the two real roots of equation (3), and
$$
\left\{\begin{array}{l}
x_{B}+x_{C}=k, \\
x_{B} x_{C}=-\frac{1}{4} .
\end{array}\right.
$$

It is easy to see that the equation of the tangent line $C P$ is
$$
y-x_{C}^{2}=2 x_{C}\left(x-x_{C}\right),
$$

and the equation of the tangent line $B P$ is
$$
y-x_{B}^{2}=2 x_{B}\left(x-x_{B}\right) \text {. }
$$
Subtracting (6) from (5) gives
$$
x_{B}^{2}-x_{C}^{2}=2 x\left(x_{C}-x_{B}\right)+2\left(x_{B}^{2}-x_{C}^{2}\right),
$$

which simplifies to $x=\frac{x_{B}+x_{C}}{2}=\frac{k}{2}$,
$$
y=x_{C}^{2}+2 x_{C} \cdot \frac{x_{B}-x_{C}}{2}=x_{C} x_{B}=-\frac{1}{4} \text {. }
$$

Therefore, the intersection point of $C P$ and $B P$ is $P\left(\frac{k}{2},-\frac{1}{4}\right)$, meaning the trajectory of point $P$ is a horizontal line segment, with the trajectory equation being
$$
y=-\frac{1}{4}\left(\frac{-3-\sqrt{129}}{64} \leqslant x \leqslant \frac{-3+\sqrt{129}}{64}\right) .
$$