$26 \cdot 25$ Curve $C_{1}: x^{2}-y^{2}+4 y-3=0$, Curve $C_{2}: y=a x^{2}$ ( $a>0, a$ is a constant), the number of intersection points of $C_{1}$ and $C_{2}$ is
(A) 2.
(B) 4.
(C) 6.
(D) Uncertain.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] Notice that $x^{2}-y^{2}+4 y-3=0$ is equivalent to $(y-2)^{2}-x^{2}=1$, which represents a hyperbola with the center at $(0,2)$ and the real axis along the $y$-axis.

Also, the graph of $y=a x^{2}(a>0)$ is a parabola with the vertex at the origin and opening upwards.
Thus, they intersect at 4 points.
Therefore, the answer is $(B)$.