5. If the ellipse $\frac{x^{2}}{m}+\frac{y^{2}}{p}=1$ and the hyperbola $\frac{x^{2}}{n}-\frac{y^{2}}{p}=1$ $(m, n, p>0)$ have common foci $F_{1}$ and $F_{2}$, and their intersection point is $Q$. Then, the area of $\triangle Q F_{1} F_{2}$ is ( ).
(A) $m+n$
(B) $\frac{m+n}{2}$
(C) $p$
(D) $\frac{p}{2}$
5.C.

According to the problem, we get $m-n=2p$, and $m+n=2(m-p)$.
From the system of equations $\left\{\begin{array}{l}\frac{x^{2}}{m}+\frac{y^{2}}{p}=1, \\ \frac{x^{2}}{n}-\frac{y^{2}}{p}=1\end{array}\right.$
we solve to get $x^{2}=\frac{2 m n}{m+n}$, and then we find $y= \pm \frac{\sqrt{2} p}{\sqrt{m+n}}$.
Also, $\left|F_{1} F_{2}\right|=2 \sqrt{m-p}=2 \sqrt{n+p}$, so,
$S_{\triangle Q F_{1} F_{2}}=\frac{1}{2} \times 2 \sqrt{m-p} \times \frac{\sqrt{2} p}{\sqrt{m+n}}=p$.