4. $P$ is a point on the central conic section $\frac{x^{2}}{m}+\frac{y^{2}}{n}=1(m>$ $|n|>0)$, and the lines connecting $P$ to the two foci $F_{1}$ and $F_{2}$ are perpendicular to each other. Then the area of $\mathrm{Rt} \triangle P F_{1} F_{2}$ is ( ).
(A) $|n|$
(B) $m$
(C) $n^{2}$
(D) cannot be determined, depends on the shape

直接输出翻译结果。
4.A.

When $n>0$, the curve is an ellipse, with the semi-major axis $a=\sqrt{m}$, the semi-minor axis $b=\sqrt{n}$, and the semi-focal distance $c=\sqrt{a^{2}-b^{2}}=\sqrt{m-n}$. According to the problem, we have
$$
\begin{array}{l}
P F_{1}+P F_{2}=2 \sqrt{m}, \\
P F_{1}^{2}+P F_{2}^{2}=4(m-n) .
\end{array}
$$
(1) ${ }^{2}$ - (2), we get $2 P F_{1} \cdot P F_{2}=4 n$. Therefore,
$$
S_{\triangle P F_{1} F_{2}}=\frac{1}{2} P F_{1} \cdot P F_{2}=n=|n| \text {. }
$$

When $n<0$, the curve is a hyperbola, with the real semi-axis $a=\sqrt{m}$, the imaginary semi-axis $b=\sqrt{-n}$, and the semi-focal distance $c=\sqrt{a^{2}+b^{2}}=\sqrt{m-n}$. According to the problem, we have
$$
\begin{array}{l}
\left|P F_{1}-P F_{2}\right|=2 \sqrt{m}, \\
P F_{1}^{2}+P F_{2}^{2}=4(m-n) .
\end{array}
$$
(5) - (4) ${ }^{2}$, we get $2 P F_{1} \cdot P F_{2}=-4 n$. Therefore,
$$
S_{\triangle P F_{1} F_{2}}=-n=|n| \text {. }
$$

In summary, the area of the right triangle $\triangle P F_{1} F_{2}$ is $|n|$.
Analysis: This problem tells us that for both ellipses and hyperbolas, the answer is the same. When written in standard form, it is always $b^{2}$.