9. Given the ellipse $\Gamma: \frac{x^{2}}{9}+\frac{y^{2}}{5}=1$, a line passing through the left focus $F(-2,0)$ of the ellipse $\Gamma$ with a slope of $k_{1}\left(k_{1} \notin\{0\right.$, $\infty\})$ intersects the ellipse $\Gamma$ at points $A$ and $B$. Let point $R(1,0)$, and extend $A R$ and $B R$ to intersect the ellipse $\Gamma$ at points $C$ and $D$ respectively. The slope of line $C D$ is $k_{2}$. Write $\frac{k_{1}^{2}}{k_{2}^{2}}$ as a reduced fraction $\frac{a}{b}$ (where $a$ and $b$ are coprime positive integers). Then $a^{2}+b=$
9. 305 .

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, $C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$,
$l_{A R}: x=\frac{x_{1}-1}{y_{1}} y+1$.
Substitute into the equation of the ellipse $\Gamma$, eliminate $x$ to get
$$
\frac{5-x_{1}}{y_{1}^{2}} y^{2}+\frac{x_{1}-1}{y_{1}} y-4=0 \text {. }
$$

By Vieta's formulas, we have
$$
y_{1} y_{3}=-\frac{4 y_{1}^{2}}{5-x_{1}}\left(y_{1} \neq 0\right) \Rightarrow y_{3}=\frac{4 y_{1}}{x_{1}-5} \text {. }
$$

Substitute into the equation of the line $A R$ to get $x_{3}=\frac{5 x_{1}-9}{x_{1}-5}$.
Similarly, $x_{4}=\frac{5 x_{2}-9}{x_{2}-5}, y_{4}=\frac{4 y_{2}}{x_{2}-5}$.
Thus, $k_{2}=\frac{y_{3}-y_{4}}{x_{3}-x_{4}}$
$$
=\frac{4\left(y_{1} x_{2}-5 y_{1}-y_{2} x_{1}+5 y_{2}\right)}{16\left(x_{2}-x_{1}\right)} \text {. }
$$

Since points $A, F, B$ are collinear, we have
$$
\begin{array}{l}
\frac{y_{1}}{x_{1}+2}=\frac{y_{2}}{x_{2}+2} \\
\Rightarrow y_{1} x_{2}-y_{2} x_{1}=2\left(y_{2}-y_{1}\right) \\
\Rightarrow k_{2}=\frac{7}{4} \cdot \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{7}{4} k_{1} \\
\Rightarrow \frac{a}{b}=\frac{k_{1}^{2}}{k_{2}^{2}}=\frac{16}{49} \Rightarrow a=16, b=49 \\
\Rightarrow a^{2}+b=305 .
\end{array}
$$