2022 AMC 12B Problem 8

Attempt Problem 8 of the 2022 AMC 12B 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 2022 AMC 12B solutions, or check the answer key.

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

8.

What is the graph of y4+1=x4+2y2y^4 + 1 = x^4 + 2y^2 in the coordinate plane?
two intersecting parabolas
two nonintersecting parabolas
two intersecting circles
a circle and a hyperbola
a circle and two parabolas
Answer: D
Concepts:difference of squarescirclehyperbola
Small hint:
Move everything to one side to get (y2−1)2=(x2)2(y^2 - 1)^2 = (x^2)^2
Big hint:
Factor as a difference of squares, giving two separate equations
Solution:
Rearranging, y4−2y2+1=x4,y^4 - 2y^2 + 1 = x^4, so (y2−1)2=(x2)2.(y^2 - 1)^2 = (x^2)^2. This factors as (y2−1−x2)(y2−1+x2)=0. (y^2 - 1 - x^2)(y^2 - 1 + x^2) = 0.
Thus either y2−x2=1,y^2 - x^2 = 1, which is a hyperbola, or x2+y2=1,x^2 + y^2 = 1, which is a circle.
Thus, the correct answer is D.
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