12. 已知双曲线过点 $(3,-2)$ 且与椭圆 $4 x^{2}+9 y^{2}=36$ 有相同的焦点.

(1)求双曲线的标准方程;

(2)若点 $M$ 在双曲线上, $F_{1} 、 F_{2}$ 为左、右焦点, 且 $\left|M F_{1}\right|+\left|M F_{2}\right|$ $=6 \sqrt{3}$, 试判断 $\triangle M F_{1} F_{2}$ 的形状.
：解 (1)陏圆方程可化为 $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, 焦点在 $x$ 轴上, 且 $c=\sqrt{9-4}=$ $\sqrt{5}$, 依据题意可设双曲线方程为 $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, 则有 $\left\{\begin{array}{l}\frac{9}{a^{2}}-\frac{4}{b^{2}}=1, \\ a^{2}+b^{2}=5 .\end{array}\right.$解得 $a^{2}=3, b^{2}=2$. 故双曲线方程为 $\frac{x^{2}}{3}-\frac{y^{2}}{2}=1$.

(2)不妨设点 $M$ 在双曲线的右支上,

则有 $\left|M F_{1}\right|-\left|M F_{2}\right|=2 \sqrt{3}$.

又 $\left|M F_{1}\right|+\left|M F_{2}\right|=6 \sqrt{3}$,

$\therefore\left|M F_{1}\right|=4 \sqrt{3},\left|M F_{2}\right|=2 \sqrt{3}$.

又 $\left|F_{1} F_{2}\right|=2 \sqrt{5}$,

$\therefore$ 在 $\triangle M F_{1} F_{2}$ 中, $\left|M F_{1}\right|$ 边最长.

由余弦定理得
$\cos \angle M F_{2} F_{1}=\frac{\left|M F_{2}\right|^{2}+\left|F_{1} F_{2}\right|^{2}-\left|M F_{1}\right|^{2}}{2\left|M F_{2}\right| \cdot\left|F_{1} F_{2}\right|}$

$=\frac{12+20-48}{2 \times 2 \sqrt{3} \times 2 \sqrt{5}}<0$.

$\therefore \angle M F_{2} F_{1}$ 为钝角. 故 $\triangle M F_{1} F_{2}$ 为钝角三角形.