1964 AHSME Problems/Problem 2

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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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