2022 AMC 12B Problem 8

Below is the professionally curated solution for Problem 8 of the 2022 AMC 12B, 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 of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:difference of squarescirclehyperbola

Difficulty rating: 1440

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

Solution:

Rearranging, y42y2+1=x4,y^4 - 2y^2 + 1 = x^4, so (y21)2=(x2)2.(y^2 - 1)^2 = (x^2)^2. This factors as (y21x2)(y21+x2)=0. (y^2 - 1 - x^2)(y^2 - 1 + x^2) = 0.

Thus either y2x2=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.

Problem 8 in Other Years