9. (14 points) For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the left and right foci are $F_{1}, F_{2}$. The lines $A F_{1}, A F_{2}$ intersect the ellipse at $B, C$. Find the value of $\frac{A F_{1}}{F_{1} B}+\frac{A F_{2}}{F_{2} C}$.
9. Solution: Draw perpendiculars from $A$ and $B$ to the left directrix, with feet of the perpendiculars at $E$ and $F$ respectively. Let the left directrix intersect the $x$-axis at $D$. We have $p=\frac{b^{2}}{c}$, and 
$$
\begin{array}{l} 
\frac{A F_{1}}{F_{1} B}=\frac{A E}{B F}=\frac{D E}{D F}= \\
\frac{A E-D F_{1}}{D F_{1}-B F}= \frac{\frac{A F_{1}}{e}-p}{p-\frac{B F_{1}}{e}}=\frac{A F_{1}-e p}{e p-B F_{1}}, \\
\text { hence } \frac{A F_{1}}{F_{1} B}=\frac{A F_{1}-e p}{e p-B F_{1}}=\frac{2 A F_{1}-e p}{e p}=\frac{2 A F_{1}}{e p}-1 .
\end{array}
$$

Similarly, $\frac{A F_{2}}{F_{2} C}=\frac{2 A F_{2}}{e p}-1$.
From (1) and (2), we know that $\frac{A F_{1}}{F_{1} B}+\frac{A F_{2}}{F_{2} C}=\frac{4 a}{\frac{b^{2}}{a}}-1=\frac{4 a^{2}}{b^{2}}-1$.