Question
Find the center, foci, and vertices of this hyperbola 16x^2-y^2-32x+8y+16=0
Answers
Steve
16x^2 - 32x + 16 - y^2 + 8y - 16 = -16 + 16 + 16
16(x-1)^2 - (y-4)^2 = 16
(x-1)^2 - (y-4)^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)
16(x-1)^2 - (y-4)^2 = 16
(x-1)^2 - (y-4)^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)