Posted by John on .
Find the center, foci, and vertices of this hyperbola 16x^2y^232x+8y+16=0

math 
Steve,
16x^2  32x + 16  y^2 + 8y  16 = 16 + 16 + 16
16(x1)^2  (y4)^2 = 16
(x1)^2  (y4)^2/16 = 1
h=1, k=4
a=1, b=4, c=√17
center: (1,4)
foci: (1√17,4) (1+√17,4)
vertices: (0,4) (2,4)