# 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