9. In the ellipse $\Gamma$, $A$ is one endpoint of the major axis, $B$ is one endpoint of the minor axis, and $F_{1}, F_{2}$ are the two foci. If $\overrightarrow{A F_{1}} \cdot \overrightarrow{A F_{2}} + \overrightarrow{B F_{1}} \cdot \overrightarrow{B F_{2}} = 0$, find the value of $\tan \angle A B F_{1} \cdot \tan \angle A B F_{2}$.
Let $A(a, 0), B(0, b), F_{1}(-c, 0), F_{2}(c, 0)$, then
$$
\overrightarrow{F_{1} A}=(a+c, 0), \overrightarrow{F_{2} A}=(a-c, 0), \overrightarrow{F_{1} B}=(c, b), \overrightarrow{F_{1} B}=(-c, b),
$$

Thus, $\overrightarrow{A F_{1}} \cdot \overrightarrow{A F_{2}}+\overrightarrow{B F_{1}} \cdot \overrightarrow{B F_{2}}=(a+c)(a-c)-c^{2}+b^{2}=a^{2}+b^{2}-2 c^{2}=0$
$$
\Rightarrow\left\{\begin{array}{l}
a^{2}+b^{2}=2 c^{2}, \\
a^{2}=b^{2}+c^{2}
\end{array} \Rightarrow \left\{\begin{array}{l}
a^{2}=\frac{3}{2} c^{2}, \\
b^{2}=\frac{1}{2} c^{2}
\end{array}\right.\right.
$$

Since $\tan \angle A B O=\frac{a}{b}, \tan \angle F_{2} B O=\frac{c}{b}$, we have
$$
\begin{array}{l}
\tan \angle A B F_{1}=\tan \left(\angle A B O+\angle F_{2} B O\right)=\frac{\frac{a}{b}+\frac{c}{b}}{1-\frac{a c}{b^{2}}}=\frac{(a+c) b}{b^{2}-a c}, \\
\tan \angle A B F_{1}=\tan \left(\angle A B O-\angle F_{2} B O\right)=\frac{\frac{a}{b}-\frac{c}{b}}{1+\frac{a c}{b^{2}}}=\frac{(a-c) b}{b^{2}+a c},
\end{array}
$$

Therefore, $\tan \angle A B F_{1} \cdot \tan \angle A B F_{2}=\frac{\left(a^{2}-c^{2}\right) b^{2}}{b^{4}-a^{2} c^{2}}=\frac{\frac{1}{2} c^{2} \cdot \frac{1}{2} c^{2}}{\frac{1}{4} c^{4}-\frac{3}{2} c^{4}}=-\frac{1}{5}$.