9. (16 points) Given a line $l$ passing through point $P(3,0)$ with a slope of $k$ intersects the right branch of the hyperbola $C: x^{2}-\frac{y^{2}}{3}=1$ at points $A$ and $B$, and $F$ is the right focus of the hyperbola $C$, with $|A F|+|B F|=16$, find the value of $k$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Then the line $l: y=k(x-3)$.
By the problem, we know the eccentricity $e=2$.
Combining the hyperbola equation with the line equation, we get
$$
\begin{array}{l}
\left(k^{2}-3\right) x^{2}-6 k^{2} x+9 k^{2}+3=0 \\
\Rightarrow x_{1}+x_{2}=\frac{6 k^{2}}{k^{2}-3} \\
\Rightarrow \quad 16=|A F|+|B F| \\
\quad=\left(2 x_{1}-1\right)+\left(2 x_{2}-1\right) \\
\quad=2 \times \frac{6 k^{2}}{k^{2}-3}-2 \\
\Rightarrow k= \pm 3 .
\end{array}
$$