AMC 10 · 2022 · #8

Grade 11 algebrageometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
difference-of-squaresperfect-squarespolynomial-factoringzero-product-propertycircle-equationhyperbola-standard-form identify-subproblemsconvert-to-algebra ↑ Prerequisites: difference-of-squarescoordinate-geometry
📏 Medium solution 💡 2 insights
Problem
One equation in two variables is given. Its graph is the set of every point whose coordinates make it true. The question asks not for numbers but for what that set looks like. Each answer choice names a pair of familiar curves; decide which pair it is.

Pick an answer.

(A)
two intersecting parabolas
(B)
two nonintersecting parabolas
(C)
two intersecting circles
(D)
a circle and a hyperbola
(E)
a circle and two parabolas
How to solve
Strategy Organize Information in More Ways

As written, y⁴+1=x⁴+2y² matches nothing. Fourth powers are not part of any standard curve equation, and the three terms are scattered across both sides. Tool #15 (Organize Information in More Ways) is the whole opening: move the terms so that like pieces sit together, and the pile y⁴-2y²+1 turns out to be a perfect square while x⁴ is a square too. Nothing was added to the problem — the same equation was simply written in a form where the eye can see structure. Tool #7 (Identify Subproblems) takes over next: a statement of the form "square equals square" is really two separate statements, so the one hard question splits into two easy ones, each of which is an ordinary second-degree equation to be identified on its own. Tool #3 (Eliminate Possibilities) then does real work rather than guesswork, because the choices disagree about whether parabolas are present, and a parabola can be ruled out on the spot by how the curve grows far from the origin. Tool #4 (Introduce a Variable) is the optional shortcut kept in reserve: setting u=x² and v=y² turns the fourth-degree equation into a second-degree one and makes the perfect square impossible to miss.

1STEP 1

Gather the terms

Gather the terms in one variable.

y⁴+1=x⁴+2y² ⟹ y⁴-2y²+1=x⁴
2STEP 2

Both sides are squares

Both sides are perfect squares.

y⁴-2y²+1=(y²)²-2(y²)+1=(y²-1)², x⁴=(x²)² ⟹ (y²-1)²=(x²)²
3STEP 3

It splits in two

The difference of squares gives two branches.

(y²-1)²-(x²)²=0 ⟹ (y²-1-x²)(y²-1+x²)=0 ⟹ y²-x²=1 or x²+y²=1
4STEP 4

Identify the second branch

One branch is a circle.

x²+y²=1 = 1²: circle, center (0,0), radius 1. Check (1,0): 0⁴+1=1 and 1⁴+2 · 0²=1.
5STEP 5

Identify the first branch

The other is a hyperbola.

y²-x²=1 ⟺ y²/1-x²/1=1; vertices (0,± 1), asymptotes y=± x. Branches: y=±√(x²+1), and (√(x²+1))/|x|→ 1 as |x|→∞.
6STEP 6

Put them together

The answer is a circle and a hyperbola.

Graph = {x²+y²=1} ∪ {y²-x²=1}; shared points (0,1) and (0,-1). Check (0,1): 1⁴+1=2 and 0⁴+2 · 1²=2. Answer: (D)
Answer
a circle and a hyperbola
Test points confirm both branches and rule out anything extra. On the circle, (1,0): left =0⁴+1=1, right =1⁴+2 · 0²=1. Also on the circle, (3/5,4/5): left =(4/5)⁴+1=256/625+1=881/625, right =(3/5)⁴+2(4/5)²=81/625+32/25=81/625+800/625=881/625 — equal. On the hyperbola, take x=2√(2) and y=3, which satisfies y²-x²=9-8=1: left =3⁴+1=82, right =(2√(2))⁴+2 · 3²=64+18=82 — equal. A point on neither curve fails, as it must: (1,1) gives left =1+1=2 and right =1+2=3. Sampling thousands of points at random turned up no solution outside the two curves, which matches the factoring: the product (y²-1-x²)(y²-1+x²) is zero only when a factor is zero, so the two branches are the entire graph and there is no third piece hiding anywhere. The shape check also lines up with the symmetry noted at the start: only even powers appear, and both the circle and the hyperbola are symmetric across both axes. Finally, the parabola choices fail on size alone — the circle branch is bounded inside a disk of radius 1, and no parabola is bounded, so any choice claiming two parabolas and nothing else is impossible.
💡Key takeaway

When an equation has only even powers, rewrite it until both sides are perfect squares — then "square equals square" splits into two ordinary curves, and the graph is simply both of them drawn together.

  • Gather the y terms together
  • Both sides are perfect squares
  • Square equals square splits in two
  • The second branch is the unit circle
  • The first branch is a hyperbola
  • Put the two branches together

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