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

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~[[User:Bxiao31415|Bxiao31415]]
 
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==Video Solution(1-16)==
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https://youtu.be/SCwQ9jUfr0g
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~~Hayabusa1
  
 
== See also ==
 
== See also ==
 
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{{AMC12 box|year=2022|ab=B|num-b=7|num-a=9}}
 
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Revision as of 13:32, 5 April 2023

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

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

Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

See also

2022 AMC 12B (ProblemsAnswer KeyResources)
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
All AMC 12 Problems and Solutions

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