26-37 Among the following four figures, it is known that one is the graph of the equations $m x+n y^{2}=0$ and $m x^{2}+n y^{2}=1(m \neq 0, n \neq 0)$ in the same coordinate system, it should be (China High School Mathematics League, 1985)
[Solution] When $m<0$, $m x^{2}+n y^{2}=1$ is a hyperbola with the $y$-axis as the real axis, and $m x+n y^{2}=0$ is a parabola opening to the right.

The asymptotes of the hyperbola $y= \pm \sqrt{\frac{-m}{n}} x$ intersect with the parabola $m x+y^{2}=0$. By the discriminant $\Delta=m^{2}>0$, it is determined that there are two intersection points, so figure (A) is correct.

When $m>0, n>0$, let $m=\frac{1}{a^{2}}, n=\frac{1}{b^{2}}$, the equation of the ellipse in figure (D) is
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text {, and } a>b>0, a>1 .
$$

But in this case, the parabola $\frac{x}{a^{2}}+\frac{y^{2}}{b^{2}}=0$ intersects with the line $y=-x$ at two points:
$$
(0,0),\left(-\frac{b^{2}}{a^{2}}, \frac{b^{2}}{a^{2}}\right), \text { and } -1<-\frac{b^{2}}{a^{2}}<0 .
$$

This fact is not shown in figure (D), so figure (D) is incorrect. Therefore, the correct choice is (A).