# 1964 AHSME Problems/Problem 2

## Problem

The graph of $x^2-4y^2=0$ is: $\textbf{(A)}\ \text{a parabola} \qquad \textbf{(B)}\ \text{an ellipse} \qquad \textbf{(C)}\ \text{a pair of straight lines}\qquad \\ \textbf{(D)}\ \text{a point}\qquad \textbf{(E)}\ \text{None of these}$

## Solution

In the equation $x^2 - 4y^2 = k$, because the coefficients of $x^2$ and $y^2$ are of opposite sign, the graph is typically a hyperbola for most real values of $k$. However, there is one exception. When $k=0$, the equation can be factored as $(x - 2y)(x+2y) = 0$. This gives the graph of two lines passing though the origin: $x=2y$ and $x=-2y$. Thus, the answer is $\boxed{\textbf{(C)}}$

## See Also

 1964 AHSC (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 All AHSME Problems and Solutions

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