Difference between revisions of "1964 AHSME Problems/Problem 2"

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Since there is an x^2 and a y^2, it is a hyperbola. Therefore, it is E.
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== Problem ==
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The graph of <math>x^2-4y^2=0</math> is:
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<math>\textbf{(A)}\ \text{a parabola} \qquad
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\textbf{(B)}\ \text{an ellipse} \qquad
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\textbf{(C)}\ \text{a pair of straight lines}\qquad \\
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\textbf{(D)}\ \text{a point}\qquad
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\textbf{(E)}\ \text{None of these}</math>   
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==Solution==
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In the equation <math>x^2 - 4y^2 = k</math>, because the coefficients of <math>x^2</math> and <math>y^2</math> are of opposite sign, the graph is typically a hyperbola for most real values of <math>k</math>. However, there is one exception.  When <math>k=0</math>, the equation can be factored as <math>(x - 2y)(x+2y) = 0</math>.  This gives the graph of two lines passing though the origin:  <math>x=2y</math> and <math>x=-2y</math>. Thus, the answer is <math>\boxed{\textbf{(C)}}</math>
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==See Also==
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{{AHSME 40p box|year=1964|num-b=1|num-a=3}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 02:13, 23 July 2019

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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All AHSME Problems and Solutions

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