Difference between revisions of "1964 AHSME Problems/Problem 2"
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− | + | == Problem == | |
+ | |||
+ | The graph of <math>x^2-4y^2=0</math> is: | ||
+ | |||
+ | <math>\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}</math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME 40p box|year=1964|num-b=1|num-a=3}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 02:13, 23 July 2019
Problem
The graph of is:
Solution
In the equation , because the coefficients of and are of opposite sign, the graph is typically a hyperbola for most real values of . However, there is one exception. When , the equation can be factored as . This gives the graph of two lines passing though the origin: and . Thus, the answer is
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.