9. Let the left and right vertices of the ellipse $C$ be $A, B(a, 0)$, and draw a non-horizontal line $l$ through the focus $F(1,0)$ intersecting the ellipse $C$ at points $P, Q$. Denote the slopes of the lines $A P, B Q$ as $k_{1}, k_{2}$. Prove that $\frac{k_{1}}{k_{2}}$ is a constant, and find this constant (expressed as a function of $a$).

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Proof: Let $l: x=t y+1$, substituting into the ellipse equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}-1}=1$ yields
$$
\left(\left(a^{2}-1\right) t^{2}+a^{2}\right) y^{2}+2\left(a^{2}-1\right) t y-\left(a^{2}-1\right)^{2}=0
$$

Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, then
$$
y_{1}+y_{2}=-\frac{2\left(a^{2}-1\right) t}{\left(a^{2}-1\right) t^{2}+a^{2}}, \quad y_{1} y_{2}=-\frac{\left(a^{2}-1\right)^{2}}{\left(a^{2}-1\right) t^{2}+a^{2}}
$$

Dividing the two equations gives
$$
\frac{y_{1}+y_{2}}{y_{1} y_{2}}=\frac{2}{a^{2}-1} t, \quad t y_{1} y_{2}=\frac{a^{2}-1}{2}\left(y_{1}+y_{2}\right)
$$

By the problem statement,
$$
k_{1}=\frac{y_{1}}{x_{1}+a}=\frac{y_{1}}{t y_{1}+a+1}, \quad k_{2}=\frac{y_{2}}{x_{2}-a}=\frac{y_{2}}{t y_{2}-a+1}
$$

Thus,
$$
\frac{k_{1}}{k_{2}}=\frac{y_{1}\left(t y_{2}-a+1\right)}{y_{2}\left(t y_{1}+a+1\right)}=\frac{\frac{\left(a^{2}-1\right)\left(y_{1}+y_{2}\right)}{2}-a y_{1}+y_{1}}{\frac{\left(a^{2}-1\right)\left(y_{1}+y_{2}\right)}{2}+a y_{2}+y_{2}}=\frac{\left(a^{2}-2 a+1\right) y_{1}+\left(a^{2}-1\right) y_{2}}{\left(a^{2}-1\right) y_{1}+\left(a^{2}+2 a+1\right) y_{2}}
$$

Since
$$
\frac{a^{2}-2 a+1}{a^{2}-1}=\frac{a-1}{a+1}=\frac{a^{2}-1}{a^{2}+2 a+1}
$$

Therefore,
$$
\frac{k_{1}}{k_{2}}=\frac{a-1}{a+1}
$$