we have the equation
[tex]16y^2-x^2+x-4y-9=0[/tex]
step 1
Group similar terms and move the constant term to the right side
[tex](16y^2-4y)+(-x^2+x)=9[/tex]
step 2
Factor 16 in the first term and factor -1 in the second term
[tex]16(y^2-\frac{y}{4})-(x^2-x)=9[/tex]
step 2
Complete the square twice
[tex]16(y^2-\frac{y}{4}+\frac{1}{64}-\frac{1}{64})-(x^2-x+\frac{1}{4}-\frac{1}{4})=9[/tex][tex]16(y^2-y\/4+1\/64)-(x^2-x+1\/4)=9+\frac{1}{4}-\frac{1}{4}[/tex]
step 3
Rewrite as perfect squares
[tex]16(y-\frac{1}{8})^2-(x-\frac{1}{2})^2=9[/tex]
step 4
Divide both sides by 9
[tex]\frac{16(y-1\/8)^2}{9}-\frac{(x-1\/2)^2}{9}=1[/tex]
therefore
The answer is Hyperbola
