1955 AHSME Problems/Problem 8

The graph of $x^2-4y^2=0$:

$\textbf{(A)}\ \text{is a hyperbola intersecting only the }x\text{-axis}\\ \textbf{(B)}\ \text{is a hyperbola intersecting only the }y\text{-axis}\\ \textbf{(C)}\ \text{is a hyperbola intersecting neither axis}\\ \textbf{(D)}\ \text{is a pair of straight lines}\\ \textbf{(E)}\ \text{does not exist}$

Solution

By difference of squares, we can rewrite the equation as $(x-2y)(x+2y) = 0$, which is just the union of the two lines $x - 2y = 0$ and $x + 2y = 0$. Therefore, our answer is $\boxed{\textbf{(D)}}$, and we are done.

See also

1955 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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