Example 10 In $\triangle P M N$ with area 1, $\tan M=\frac{1}{2}, \tan N=-2$. Establish an appropriate coordinate system and find the equation of the ellipse with foci at $M, N$ and passing through point $P$.

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Solve: Establish a rectangular coordinate system with the line $MN$ as the $x$-axis and the perpendicular bisector of segment $MN$ as the $y$-axis.
On one hand, $\tan P = -\tan (M+N) = \frac{\tan M + \tan N}{\tan M \cdot \tan N - 1} = \frac{3}{4}$.
On the other hand, $\tan P = \frac{2 \tan \frac{P}{2}}{1 - \tan^2 \frac{P}{2}}$, thus
$\frac{2 \tan \frac{P}{2}}{1 - \tan^2 \frac{P}{2}} = \frac{3}{4}$, which simplifies to $3 \tan^2 \frac{P}{2} + 8 \tan \frac{P}{2} - 3 = 0$.
Solving this, we get $\tan \frac{P}{2} = \frac{1}{3}$ or $\tan \frac{P}{2} = -3$ (discard the latter).
By property 5 (II), we know $b^2 = S_{\triangle PMV} \cdot \cot \frac{P}{2} = 1 \cdot 3 = 3$.
Construct $PQ \perp MN$, with the foot of the perpendicular at $Q$. Let $|PQ| = h$ and $|NQ| = m$. From $\tan M = \frac{h}{2c + m} = \frac{1}{2}$ and $\tan \angle PNQ = \frac{h}{m} = 2$, it is easy to find $h = \frac{4}{3}c$. Also, $S_{\triangle PMN} = \frac{1}{2} \cdot 2c \cdot \frac{4c}{3} = 1$, which gives $c^2 = \frac{3}{4}$, and thus $a^2 = b^2 + c^2 = \frac{15}{4}$. Therefore, the equation of the ellipse is $\frac{4x^2}{15} + \frac{y^2}{3} = 1$.