1955 AHSME Problems/Problem 8

Revision as of 10:58, 21 July 2020 by Duck master (talk | contribs) (created page w/ solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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
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 41 42 43 44 45 46 47 48 49 50
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