$26 \cdot 27$ Consider the graphs of $y=A x^{2}$ and $y^{2}+3=x^{2}+4 y$, where $A$ is a positive constant, and $x$ and $y$ are real variables. The number of points of intersection of these two curves is
(A) exactly 4.
(B) exactly 2.
(C) at least 1, and the number of intersections varies with different values of $A$.
(D) for at least one positive value of $A$, the number of intersections is 0.
(E) none of the above conclusions is correct.
(36th American High School Mathematics Examination, 1985)
[Solution] Let $y=A x^{2}(A>0)$, then $x^{2}=\frac{y}{A}$. Substituting into the other equation gives $y^{2}+3=\frac{y}{A}+4 y$, which simplifies to $y^{2}-\left(4+\frac{1}{A}\right) y+3=0$.
$\cdot$ Since $\Delta=\left(4+\frac{1}{A}\right)^{2}-12>0$ (noting that $A>0$), and the coefficient of the linear term is less than 0, and the constant term is greater than 0, by Vieta's formulas, it is known that there are two positive roots, denoted as $y_{1}, y_{2}$. Therefore, $x= \pm \sqrt{\frac{y_{1}}{A}}$, or $\pm \sqrt{\frac{y_{2}}{A}}$.
$\therefore$ The two curves intersect at four points. Hence, the answer is $(A)$.