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Difference between revisions of "2018 UNCO Math Contest II Problems/Problem 3"

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== Problem ==
 
== Problem ==
 +
Find all values of <math>B</math> that have the property that if <math>(x, y)</math> lies on the hyperbola <math>2y^2-x^2 = 1</math>,
 +
then so does the point <math>(3x + 4y, 2x + By)</math>.
  
 +
== Solution 1 ==
 +
We can write a system of equations -
 +
<cmath>2y^2-x^2 = 1</cmath>
 +
<cmath>2(2x + By)^2 - (3x+4y)^2 = 1</cmath>
  
 +
Expanding the second equation, we get <math>-x^2+8Bxy-24xy+2B^2y^2-16y^2=1</math>.
  
−
== Solution ==
+
Since we want this to look like <math>2y^2-x^2=1</math>, we plug in B's that would put it into that form. If we plug in <math>B=3</math>, things cancel, and we get <math>-x^2+24xy-24xy+18y^2-16y^2=1 \rightarrow 2y^2-x^2=1</math>. So <math>\boxed{B=3}</math>
 +
~Ultraman
  
 +
== Solution 2 (Grinding) ==
 +
As with Solution 1, we create a system of equations.
 +
<cmath>2y^2-x^2 = 1</cmath>
 +
<cmath>2(2x + By)^2 - (3x+4y)^2 = 1</cmath>
  
−
== See also ==
+
Through expanding the second equation, we get <math>-x^2+8Bxy-24xy+2B^2y^2-16y^2 = 1</math>. Since <math>2y^2-x^2 = 1</math>, we have <cmath>-x^2+8Bxy-24xy+2B^2y^2-16y^2 = 2y^2-x^2</cmath>
−
{{UNCO Math Contest box|year=2018|n=II|num-b=2 Question|num-a=4}}
+
The <math>-x^2</math> terms on each side cancel out, so the equation becomes
 +
<cmath>(8B-24)xy + (2B^2-16)y^2 = 2y^2</cmath>
 +
The coefficient of <math>xy</math> on the RHS is <math>0</math> and the coefficient of <math>y^2</math> is <math>2</math>. From these two observations, we now create two new equations.
 +
<cmath>8B-24 = 0</cmath>
 +
<cmath>2B^2-16 = 2</cmath>
 +
Solving either equation and then checking with the other will reveal that <math>\boxed{B=3}</math>.
 +
~kingme271
  
−
[[Category:]]
+
== See also ==
 +
{{UNCO Math Contest box|year=2018|n=II|num-b=2|num-a=4}}
 +
 
 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 02:18, 11 June 2022

Problem

Find all values of $B$ that have the property that if $(x, y)$ lies on the hyperbola $2y^2-x^2 = 1$, then so does the point $(3x + 4y, 2x + By)$.

Solution 1

We can write a system of equations - \[2y^2-x^2 = 1\] \[2(2x + By)^2 - (3x+4y)^2 = 1\]

Expanding the second equation, we get $-x^2+8Bxy-24xy+2B^2y^2-16y^2=1$.

Since we want this to look like $2y^2-x^2=1$, we plug in B's that would put it into that form. If we plug in $B=3$, things cancel, and we get $-x^2+24xy-24xy+18y^2-16y^2=1 \rightarrow 2y^2-x^2=1$. So $\boxed{B=3}$ ~Ultraman

Solution 2 (Grinding)

As with Solution 1, we create a system of equations. \[2y^2-x^2 = 1\] \[2(2x + By)^2 - (3x+4y)^2 = 1\]

Through expanding the second equation, we get $-x^2+8Bxy-24xy+2B^2y^2-16y^2 = 1$. Since $2y^2-x^2 = 1$, we have \[-x^2+8Bxy-24xy+2B^2y^2-16y^2 = 2y^2-x^2\] The $-x^2$ terms on each side cancel out, so the equation becomes \[(8B-24)xy + (2B^2-16)y^2 = 2y^2\] The coefficient of $xy$ on the RHS is $0$ and the coefficient of $y^2$ is $2$. From these two observations, we now create two new equations. \[8B-24 = 0\] \[2B^2-16 = 2\] Solving either equation and then checking with the other will reveal that $\boxed{B=3}$. ~kingme271

See also

2018 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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