9. (16 points) Given that $F$ is the focus of the parabola $y^{2}=4 x$, $Q$ is the intersection point of its directrix with the $x$-axis, and line $l$ passes through point $Q$. Let line $l$ intersect the parabola at points $A$ and $B$.
(1) Let the slopes of lines $A F$ and $B F$ be $k_{1}$ and $k_{2}$, respectively. Find the value of $k_{1}+k_{2}$;
(2) If there is a point $R$ on segment $A B$ such that
$$
\frac{|A R|}{|R B|}=\frac{|A Q|}{|Q B|} \text {, }
$$

find the locus of point $R$.
9. (1) Let the line $l: y=k(x+1)$,
$$
A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), F(1,0).
$$

Then $k_{1}+k_{2}=\frac{y_{1}}{x_{1}-1}+\frac{y_{2}}{x_{2}-1}$
$$
\begin{array}{l}
=\frac{k\left(x_{1}+1\right)}{x_{1}-1}+\frac{k\left(x_{2}+1\right)}{x_{2}-1} \\
=\frac{2 k\left(x_{1} x_{2}-1\right)}{\left(x_{1}-1\right)\left(x_{2}-1\right)}. \\
\text { From }\left\{\begin{array}{l}
y=k(x+1), \\
y^{2}=4 x
\end{array}\right. \\
\Rightarrow k^{2} x^{2}+\left(2 k^{2}-4\right) x+k^{2}=0 \\
\Rightarrow x_{1}+x_{2}=\frac{4-2 k^{2}}{k^{2}}, x_{1} x_{2}=1 .
\end{array}
$$

Substituting into equation (1) gives $k_{1}+k_{2}=0$.
(2) Let $R(x, y)$.
$$
\begin{array}{l}
\text { From } \frac{|A R|}{|R B|}=\frac{|A Q|}{|B Q|} \Rightarrow \frac{y_{1}-y}{y-y_{2}}=\frac{y_{1}}{y_{2}} \\
\Rightarrow y=\frac{2 y_{1} y_{2}}{y_{1}+y_{2}}=2 k .
\end{array}
$$

Also, $x=\frac{1}{k} y-1=1$, and from $\Delta>0$, we get
$-1<k<1$, and $k \neq 0$.
Thus, $y \in(-2,0) \cup(0,2)$.
Therefore, the trajectory equation of point $R$ is
$$
x=1(-2<y<2 \text {, and } y \neq 0).
$$