Question 974268
what is the center of the hyperbola 
4y^2-12x^2-72x-16y-44=0 ?
4y^2-16y-12x^2-72x=44
complete the square:
4(y^2-4y+4)-12(x^2+6x+9)=44+16-108=-48
mult. both sides of equation by (-1)
-4(y^2-4y+4)+12(x^2+6x+9)=48
12(x^2+6x+9)-4(y^2-4y+4)=48
12(x+3)^2-4(y-2)^2=48
Equation of given hyperbola:
{{{(x+3)^2/4-(y-2)^2/12=1}}}
center: (-3,2)