Difference between revisions of "2022 AMC 12B Problems/Problem 8"

Revision as of 22:53, 1 November 2024

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 two 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 (Factoring)

(Solutions 1 and 2 are in essence the same; Solution 1 lets $(x',y')=\left(x^2,y^2\right)$ for convenience, but the two solutions are otherwise identical.)

We can subtract $2y^2$ from both sides and factor the left side to get $(y^2-1)^2 = x^4$ . If we take the square root of both sides, we are left with two equations: $y^2-1 = x^2$ and $y^2-1 = -x^2$ . In the former case, we get $y^2-x^2=1$ , which, is the formula for a hyperbola. This means that a hyperbola will appear in the graph of the equation. In the latter case, we get $x^2+y^2=1$ , which is the equation of a circle which will also appear in the graph. Looking at our options, this is the only valid answer. Hence, the answer is $\boxed{\textbf{(D)}\ \text{a circle and a hyperbola}}$ .

~mathman239458

Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

@@ Line 16: / Line 16: @@
 == Solution 2 (Factoring) ==
+(Solutions 1 and 2 are in essence the same; Solution 1 lets <math>(x',y')=\left(x^2,y^2\right)</math> for convenience, but the two solutions are otherwise identical.)
 We can subtract <math>2y^2</math> from both sides and factor the left side to get <math>(y^2-1)^2 = x^4</math>. If we take the square root of both sides, we are left with two equations: <math>y^2-1 = x^2</math> and <math>y^2-1 = -x^2</math>. In the former case, we get <math>y^2-x^2=1</math>, which, is the formula for a hyperbola. This means that a hyperbola will appear in the graph of the equation. In the latter case, we get <math>x^2+y^2=1</math>, which is the equation of a circle which will also appear in the graph. Looking at our options, this is the only valid answer. Hence, the answer is <math>\boxed{\textbf{(D)}\ \text{a circle and a hyperbola}}</math>.
 ~mathman239458
 ==Video Solution(1-16)==

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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