9. As shown in the figure, given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with the right focus at $F(c, 0)(c>0)$, let the line $l$ passing through point $F$ intersect the ellipse at points $A, B$, and intersect the $y$-axis at point $P, \overrightarrow{P A}=\lambda_{1} \overrightarrow{A F}, \overrightarrow{P B}=\lambda_{2} \overrightarrow{B F}$, find the value of $\lambda_{1}+\lambda_{2}$.
9. Solution: The slope of $l$ exists, let the equation of $l$ be $y=k(x-c), A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$,
$$
\begin{array}{l}
\text { By }\left\{\begin{array}{l}
y=k(x-c) \\
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
\end{array} \text { we get }\left(b^{2}+a^{2} k^{2}\right) x^{2}-2 a^{2} k^{2} c x+a^{2} k^{2} c^{2}-a^{2} b^{2}=0\right. \text {, } \\
\therefore x_{1}+x_{2}=\frac{2 a^{2} k^{2} c}{b^{2}+a^{2} k^{2}}, x_{1} x_{2}=\frac{a^{2} k^{2} c^{2}-a^{2} b^{2}}{b^{2}+a^{2} k^{2}} \text {, } \\
\therefore \lambda_{1}+\lambda_{2}=\frac{x_{1}-0}{c-x_{1}}+\frac{x_{2}-0}{c-x_{2}}=\frac{c\left(x_{1}+x_{2}\right)-2 x_{1} x_{2}}{c^{2}-c\left(x_{1}+x_{2}\right)+x_{1} x_{2}} \\
=\frac{\frac{2 a^{2} k^{2} c^{2}}{b^{2}+a^{2} k^{2}}-\frac{2 a^{2} k^{2} c^{2}-2 a^{2} b^{2}}{b^{2}+a^{2} k^{2}}}{c^{2}-\frac{2 a^{2} k^{2} c^{2}}{b^{2}+a^{2} k^{2}}+\frac{a^{2} k^{2} c^{2}-a^{2} b^{2}}{b^{2}+a^{2} k^{2}}}=\frac{2 a^{2} b^{2}}{c^{2} b^{2}-a^{2} b^{2}}=-\frac{2 a^{2}}{b^{2}} . \\
\end{array}
$$