Hi Dom, lets see if i can help you cure your hyperbola ebola.
\(y^2-9x^2-4y+18x-14=0\\ y^2-4y-9x^2+18x=14\\ (y^2-4y)-9(x^2-2x)=14\\ (y^2-4y+4)-9(x^2-2x+1)=14+4-9\\ (y^2-2)^2-9(x-1)^2=9\\ \frac{(y^2-2)^2}{3^2}-\frac{(x-1)^2}{1^2}=1\\ \)
Now it is in standard equation form for a hyperbola with a vertical transverse axis.
For the rest see here.
http://www.sparknotes.com/math/precalc/conicsections/section4/