10. 设双曲线 $\frac{x^{2}}{4}-\frac{y^{2}}{9}=1, F_{1}, F_{2}$ 是两个焦点, 点 $M$ 在双曲线上,若 $\angle F_{1} M F_{2}=90^{\circ}$, 求 $\triangle F_{1} M F_{2}$ 的面积.
：解 由题意知 $a^{2}=4, b^{2}=9, \therefore c^{2}=13$.

设 $\left|M F_{1}\right|=r_{1},\left|M F_{2}\right|=r_{2} ，$

则由双曲线定义知 $\left|r_{1}-r_{2}\right|=2 a=4$,

$\therefore\left(r_{1}-r_{2}\right)^{2}=r_{1}^{2}+r_{2}^{2}-2 r_{1} r_{2}=16$. (1)

又 $\because \angle F_{1} M F_{2}=90^{\circ}$,

$\therefore r_{1}^{2}+r_{2}^{2}=\left|F_{1} F_{2}\right|^{2}=4 c^{2}=52$.(2)

$\therefore$ 由(1)(2)得 $r_{1} r_{2}=18$.

$\therefore S \triangle F_{1} M F_{2}=\frac{1}{2} r_{1} r_{2}=9$.