+101 Hyperbola ebola has struck me again! The only cure is an explanation of ths Hyperbola question with steps on how it is done!
\(y^2-9x^2-4y+18x-14=0\)
find the:
center
foci
asymototes
and vertices
graph
Remember the Hyperbola ebola will never be cured without help that has no explained steps, otherwise the brain won't be able to learn the way to fight it off!
Thank you!!!
+118608 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/
