1964 AMC 12 Problem 2

Attempt Problem 2 of the 1964 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1964 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

2.

The graph of x2−4y2=0x^2-4y^2=0 is:
a parabola
an ellipse
a pair of straight lines
a point
none of these
Answer: C
Concepts:factoringdifference of squarescoordinate geometry
Small hint:
Factor the left side as a difference of squares
Big hint:
Set each linear factor equal to zero
Solution:
Factoring gives x2−4y2=(x−2y)(x+2y).x^2-4y^2=(x-2y)(x+2y). Thus every point lies on x=2yx=2y or x=−2y,x=-2y, a pair of straight lines.
Therefore, the correct answer is C.
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