2022 AMC 12B Problems/Problem 8

Problem

What is the graph of $y^4+1=x^4+2y^2$ in the coordinate plane?

$\textbf{(A)}\ \text{two intersecting parabolas} \qquad \textbf{(B)}\ \text{two nonintersecting parabolas} \qquad \textbf{(C)}\ \text{two intersecting circles} \qquad$

$\textbf{(D)}\ \text{a circle and a hyperbola} \qquad \textbf{(E)}\ \text{a circle and two parabolas}$

Solution 1

Since the equation has even powers of $x$ and $y$ , let $y'=y^2$ and $x' = x^2$ . Then $y'^2 + 1 = x'^2 + 2y'$ . Rearranging gives $y'^2 - 2y' + 1 = x'^2$ , or $(y'-1)^2=x'^2$ . There are 2 cases: $y' \leq 1$ or $y' > 1$ .

If $y' \leq 1$ , taking the square root of both sides gives $1 - y' = x'$ , and rearranging gives $x' + y' = 1$ . Substituting back in $x'=x^2$ and $y'=y^2$ gives us $x^2+y^2=1$ , the equation for a circle.

Similarly, if $y' > 1$ , we take the square root of both sides to get $y' - 1 = x'$ , or $y' - x' = 1$ , which is equivalent to $y^2 - x^2 = 1$ , a hyperbola.

Hence, our answer is $\boxed{\textbf{(D)}\ \text{a circle and a hyperbola}}$ .

~Bxiao31415

Solution 2 (Similar to Solution 1)

We have that:

$y^4+1=x^4+2y^2$ $y^4-2y^2+1=x^4$ $(y^2-1)^2=x^4$ $y^2-1=x^2$ or $-(y^2-1)=x^2$ $y^2-x^2-1=$ or $-y^2+1-x^2=0$ $y^2-x^2=1$ or $x^2+y^2=1$

We realize that our final equations are $\boxed{\textbf{(D)}\ \text{a circle and a hyperbola}}$ .

~iluvme

2022 AMC 12B (Problems • Answer Key • Resources)
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
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2022 AMC 12B Problems/Problem 8

Contents

Problem

Solution 1

Solution 2 (Similar to Solution 1)

See also