2018 UNCO Math Contest II Problems/Problem 3

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

2018 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2018 UNCO Math Contest II Problems/Problem 3

Contents

Problem

Solution 1

Solution 2 (Grinding)

See also