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 $\boxed{\textbf{(C)}}$

1964 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1964 AHSME Problems/Problem 2

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