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

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<math>\textbf{(D)}\ \text{a circle and a hyperbola} \qquad \textbf{(E)}\ \text{a circle and two parabolas}</math>
 
<math>\textbf{(D)}\ \text{a circle and a hyperbola} \qquad \textbf{(E)}\ \text{a circle and two parabolas}</math>
  
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== Solution 1 ==
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== Solution ==
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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.

Revision as of 17:12, 9 January 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

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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