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

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What is the graph of <math>y^4+1=x^4+2y^2</math> in the coordinate plane?
 
What is the graph of <math>y^4+1=x^4+2y^2</math> in the coordinate plane?
  
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<math>\textbf{(A)}\ \text{two intersecting parabolas} \qquad \textbf{(B)}\ \text{two nonintersecting parabolas} \qquad \textbf{(C)}\ \text{two intersecting circles} \qquad</math>
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<math>\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}</math>
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<math>\textbf{(D)}\ \text{a circle and a hyperbola} \qquad \textbf{(E)}\ \text{a circle and two parabolas}</math>
 
  
 
== Solution 1 ==
 
== Solution 1 ==
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Since the equation has even powers of <math>x</math> and <math>y</math>, let <math>y'=y^2</math> and <math>x' = x^2</math>. Then <math>y'^2 + 1 = x'^2 + 2y'</math>. Rearranging gives <math>y'^2 - 2y' + 1 = x'^2</math>, or <math>(y'-1)^2=x'^2</math>. There are 2 cases: <math>y' \leq 1</math> or <math>y' > 1</math>.
+
Since the equation has even powers of <math>x</math> and <math>y</math>, let <math>y'=y^2</math> and <math>x' = x^2</math>. Then <math>y'^2 + 1 = x'^2 + 2y'</math>. Rearranging gives <math>y'^2 - 2y' + 1 = x'^2</math>, or <math>(y'-1)^2=x'^2</math>. There are two cases: <math>y' \leq 1</math> or <math>y' > 1</math>.
  
 
If <math>y' \leq 1</math>, taking the square root of both sides gives <math>1 - y' = x'</math>, and rearranging gives <math>x' + y' = 1</math>. Substituting back in <math>x'=x^2</math> and <math>y'=y^2</math> gives us <math>x^2+y^2=1</math>, the equation for a circle.
 
If <math>y' \leq 1</math>, taking the square root of both sides gives <math>1 - y' = x'</math>, and rearranging gives <math>x' + y' = 1</math>. Substituting back in <math>x'=x^2</math> and <math>y'=y^2</math> gives us <math>x^2+y^2=1</math>, the equation for a circle.
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~[[User:Bxiao31415|Bxiao31415]]
 
~[[User:Bxiao31415|Bxiao31415]]
  
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== Solution 2 (Similar to Solution 1) ==
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(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.)
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We have that:
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 +
== Solution 2 (Factoring) ==
  
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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>.
  
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<cmath>y^4+1=x^4+2y^2</cmath>
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~mathman239458
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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> or <cmath>-(y^2-1)=x^2</cmath>
 
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<cmath>y^2-x^2-1=</cmath> or <cmath>-y^2+1-x^2=0</cmath>
 
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<cmath>y^2-x^2=1</cmath> or <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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==Video Solution(1-16)==
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https://youtu.be/SCwQ9jUfr0g
  
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~iluvme
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~~Hayabusa1
  
 
== 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}}

Latest revision as of 22:54, 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

(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.)

Solution 2 (Factoring)

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

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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