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

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m (Solution)
 
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<cmath>2(2x + By)^2 - (3x+4y)^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>
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Expanding the second equation, we get <math>-x^2+8Bxy-24xy+2B^2y^2-16y^2=1</math>.
 +
 
 
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>
 
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
 
~Ultraman

Latest revision as of 13:53, 2 January 2020

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

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

See also

2018 UNCO Math Contest II (ProblemsAnswer KeyResources)
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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