4. Among the following four figures, it is known that one is the graph of the equations $m x+n y^{2}=0$ and $m s^{2}+n y^{2}=1(m \neq 0, n \neq 0)$ in the same coordinate system, it should be
(A)
(B)
(C)
(D)
4. (A)
$\because m x^{2}+n y^{2}=1$, and $m \cdot n \neq 0$,
$\therefore m, n$ cannot both be negative.
If $m>0, n>0$,
$\because$ the major axis of the ellipse is on the $x$-axis,
$\therefore \quad 0<m<n$.
At this time, the parabola $m x+n y^{2}=0$,
i.e., $y^{2}=-\frac{m}{n} x$ opens to the left, so the answer is clearly not (C).
Since $0<m<n, \therefore \quad 0<\frac{m}{n}<1$,

Thus, the parabola $y^{2}=-\frac{m}{n} x$ and the line $y=-x$ should intersect when $-1<x<0$. Since the graph in (D) does not match this, the answer is also not (D). (The graph in (D) is marked with unit 1).
If $m, n$ have different signs, then the parabola
$$
y^{2}=-\frac{m}{n} x
$$

opens to the right, so the answer definitely is not (B).
Therefore, it is known that (A) is correct,
for example, taking $n=2, m=-2$ can yield graph (A).