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

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~[[User:Bxiao31415|Bxiao31415]]
 
~[[User:Bxiao31415|Bxiao31415]]
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== Solution 2 (Similar to Solution 1) ==
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We have that:
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<cmath>y^4+1=x^4+2y^2</cmath>
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<cmath>y^4-2y^2+1=x^4</cmath>
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<cmath>(y^2-1)^2=x^4</cmath>
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<cmath>y^2-1=x^2</cmath> <math>or</math> <cmath>-(y^2-1)=x^2</cmath>
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<cmath>y^2-x^2-1=</cmath> <math>or</math> <cmath>-y^2+1-x^2=0</cmath>
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<cmath>y^2-x^2=1</cmath> <math>or</math> <cmath>x^2+y^2=1</cmath>
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We realize that our final equations are <math>\boxed{\textbf{(D)}\ \text{a circle and a hyperbola}}</math>.
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 +
~iluvme
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2022|ab=B|num-b=7|num-a=9}}
 
{{AMC12 box|year=2022|ab=B|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:48, 9 December 2022

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

See also

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
All AMC 12 Problems and Solutions

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