Difference between revisions of "2022 AMC 12B Problems/Problem 8"
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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> | <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> | ||
| − | == Solution == | + | == Solution 1 == |
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>. | 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>. | ||
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~[[User:Bxiao31415|Bxiao31415]] | ~[[User:Bxiao31415|Bxiao31415]] | ||
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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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| + | ~mathman239458 | ||
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==Video Solution(1-16)== | ==Video Solution(1-16)== | ||
Revision as of 22:19, 13 July 2024
Problem
What is the graph of 
in the coordinate plane?
Solution 1
Since the equation has even powers of 
and 
, let 
and 
. Then 
. Rearranging gives 
, or 
. There are two cases: 
or 
.
If , taking the square root of both sides gives 
, and rearranging gives 
. Substituting back in 
and gives us 
, the equation for a circle.
Similarly, if , we take the square root of both sides to get 
, or 
, which is equivalent to 
, a hyperbola.
Hence, our answer is 
.
Solution 2 (Factoring)
We can subtract 
from both sides and factor the left side to get 
. If we take the square root of both sides, we are left with two equations: 
and 
. In the former case, we get 
, 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 , 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 .
~mathman239458
Video Solution(1-16)
~~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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
