5. Given the equation of the ellipse $\Gamma$ as $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$, a line passing through the left focus $F(-2,0)$ with a slope of $k_{1}\left(k_{1} \neq 0\right)$ intersects the ellipse at points $A$ and $B$. Let $R(1,0)$, and extend $A R$ and $B R$ to intersect the ellipse at points $C$ and $D$, respectively. The slope of line $C D$ is $k_{2}$. Then $\frac{k_{1}}{k_{2}}=$ $\qquad$ .
5. $\frac{4}{7}$.

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right)$, $D\left(x_{4}, y_{4}\right)$.
Then the line $l_{A R}: x=\frac{x_{1}-1}{y_{1}} y+1$.
Substituting into the ellipse equation and eliminating $x$ yields
$$
\frac{5-x_{1}}{y_{1}^{2}} y^{2}+\frac{x_{1}-1}{y_{1}} y-4=0 \text {. }
$$

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

Since $A, F, B$ are collinear, we have
$$
\begin{array}{l}
\frac{y_{1}}{x_{1}+2}=\frac{y_{2}}{x_{2}+2} \\
\Rightarrow y_{1} x_{2}-y_{2} x_{1}=2\left(y_{2}-y_{1}\right) .
\end{array}
$$

Therefore, $k_{2}=\frac{7}{4} \cdot \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{7}{4} k_{1} \Rightarrow \frac{k_{1}}{k_{2}}=\frac{4}{7}$.