Difference between revisions of "2022 AMC 12B Problems/Problem 8"
Bxiao31415 (talk | contribs) (Created page with "== Problem == What is the graph of <math>y^4+1=x^4+2y^2</math> in the coordinate plane? <math>\textbf{(A)}\ \textbf{Two intersecting parabolas} \qquad \textbf{(B)}\ \textbf{T...") |
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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? | ||
− | <math>\textbf{(A)}\ | + | <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 == | |
− | + | 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>. | |
− | == Solution | ||
− | 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 | ||
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. | ||
Similarly, if <math>y' > 1</math>, we take the square root of both sides to get <math>y' - 1 = x'</math>, or <math>y' - x' = 1</math>, which is equivalent to <math>y^2 - x^2 = 1</math>, a hyperbola. | Similarly, if <math>y' > 1</math>, we take the square root of both sides to get <math>y' - 1 = x'</math>, or <math>y' - x' = 1</math>, which is equivalent to <math>y^2 - x^2 = 1</math>, a hyperbola. | ||
− | Hence, our answer is <math>\ | + | |
+ | Hence, our answer is <math>\boxed{\textbf{(D)}\ \text{a circle and a hyperbola}}</math>. | ||
~[[User:Bxiao31415|Bxiao31415]] | ~[[User:Bxiao31415|Bxiao31415]] | ||
+ | |||
+ | ==Video Solution(1-16)== | ||
+ | https://youtu.be/SCwQ9jUfr0g | ||
+ | |||
+ | ~~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}} |
Revision as of 03:36, 26 November 2023
Problem
What is the graph of in the coordinate plane?
Solution
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 .
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.