Question 614450
find the center? 
9y^2-16x^2-36y-108=0
complete the square
9(y^2-4y+4)-16x^2=108+36
9(y-2)^2-16x^2=144
(y-2)^2/16-x^2/9=1
This is an equation of a hyperbola with vertical transverse axis.
Its standard form:(y-k)^2/a^2-(x-h)^2/b^2, (h,k)=(x,y) coordinates of center.
For given equation:(y-2)^2/16-x^2/9=1
center: (0,2)