10. 已知双曲线的方程是 $16 x^{2}-9 y^{2}=144, F_{1}, F_{2}$ 是其左、右焦点, 点 $P$ 在双曲线上, 且 $\left|P F_{1}\right| \cdot\left|P F_{2}\right|=32$, 求 $\angle F_{1} P F_{2}$ 的大小.
：解 双曲线的方程可化为 $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$,

$\therefore a^{2}=9, \quad b^{2}=16, \quad \therefore c=5$.

由双曲线的定义知 ||$P F_{1}|-| P F_{2}||=2 a=6$.

$\therefore \cos \angle F_{1} P F_{2}=\frac{\left|P F_{1}\right|^{2}+\left|P F_{2}\right|^{2}-\left|F_{1} F_{2}\right|^{2}}{2\left|P F_{1}\right| \cdot\left|P F_{2}\right|}$

$=\frac{\left(\left|P F_{1}\right|-\left|P F_{2}\right|\right)^{2}+2\left|P F_{1}\right| \cdot\left|P F_{2}\right|-4 c^{2}}{2\left|P F_{1}\right| \cdot\left|P F_{2}\right|}$.

又 $\left|P F_{1}\right| \cdot\left|P F_{2}\right|=32$,

$\therefore \cos \angle F_{1} P F_{2}=\frac{6^{2}+2 \times 32-4 \times 25}{2 \times 32}=0$.

$\therefore \angle F_{1} P F_{2}$ 的大小为 $90^{\circ}$.