# 1985 AHSME Problems/Problem 19

## Problem

Consider the graphs of $y = Ax^2$ and $y^2+3 = x^2+4y$, where $A$ is a positive constant and $x$ and $y$ are real variables. In how many points do the two graphs intersect?

$\mathrm{(A) \ }\text{exactly }4 \qquad \mathrm{(B) \ }\text{exactly }2 \qquad$

$\mathrm{(C) \ }\text{at least }1,\text{ but the number varies for different positive values of }A \qquad$

$\mathrm{(D) \ }0\text{ for at least one positive value of }A \qquad \mathrm{(E) \ }\text{none of these}$

## Solution 1

Substituting $y = Ax^2$ into the equation $y^2+3 = x^2+4y$ gives \begin{align*}\left(Ax^2\right)+3 = x^2+4\left(Ax^2\right) &\iff A^2x^4+3 = x^2+4Ax^2 \\ &\iff A^2x^4-\left(4A+1\right)x^2+3 = 0 \\ &\iff x^2 = \frac{4A+1 \pm \sqrt{4A^2+8A+1}}{2A^2} \\ &\text{(using the quadratic formula)}.\end{align*} Now observe that since $A$ is positive, $4A^2+8A+1$ is also positive, so the square root will always give two distinct real values. Moreover, $$\left(4A+1\right)^2 = 16A^2+8A+1 > 4A^2+8A+1,$$ so $4A+1-\sqrt{4A^2+8A+1} > 0$, meaning that both solutions for $x^2$ are positive. Hence both solutions will give $2$ distinct values of $x$ (the positive and negative square roots), and each of these will correspond to a distinct point of intersection of the graphs, so there are $2 \cdot 2 = \boxed{\text{(A) exactly} \ 4}$ points of intersection.

## Solution 2

Firstly, note that $y = Ax^2$ is an upward-facing parabola (since $A > 0$) whose vertex is at the origin. We now manipulate the equation of the second graph as follows: \begin{align*}y^2+3 = x^2+4y &\iff y^2-4y+3-x^2 = 0 \\ &\iff y^2-4y+4-x^2 = 1 \\ &\iff \frac{(y-2)^2}{1}-\frac{(x-0)^2}{1} = 1,\end{align*} showing that it is a vertical (upward- and downward-opening) hyperbola with center $(0,2)$ and asymptotes $y=x+1$ and $y=-x+1$. It therefore remains to consider graphically where the parabola will intersect the hyperbola.

On the lower branch of the hyperbola, the maximum point is $(0,2-1) = (0,1)$, which is above the vertex of the parabola. Therefore, by continuity and the symmetry of both the parabola and the hyperbola in the $y$-axis, there are always exactly $2$ intersection points here.

For the top branch, as it approaches the asymptote $y = x+1$, its slope also approaches that of this asymptote, which is $1$. However, for any upward-opening parabola, the slope approaches infinity as $x$ does, so no matter how small $A$ is (i.e. how 'flat' the parabola is), the parabola will eventually overtake the hyperbola, giving a point of intersection with positive $x$-coordinate. As above, symmetry gives another point of intersection with negative $x$-coordinate, so that there are $2$ intersection points with this branch too.

Thus there are a total of $2+2 = \boxed{\text{(A) exactly} \ 4}$ intersection points.