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

(Created page with "Since there is an x^2 and a y^2, it is a hyperbola. Therefore, it is E.")
 
 
Line 1: Line 1:
Since there is an x^2 and a y^2, it is a hyperbola. Therefore, it is E.
+
== 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 03:13, 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
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. AMC logo.png