Expert: Sherman D. - 5/5/2007

Question
i need to find the center, vertices, foci and asymptotes of the hyperbola: 16y^2 - x^2 - 6x - 128y + 231 = 0

Answer
16y^2 - x^2 - 6x - 128y + 231 = 0
16y^2 - 128y - x^2 - 6x + 231 = 0
(16y^2 - 128y) - (x^2 + 6x) + 231 = 0
16(y^2 - 8y) - (x^2 + 6x + 9 - 9) + 231 = 0
16(y^2 - 8y + 16 - 16) - ((x + 3)^2 - 9) + 231 = 0
16((y - 4)^2 - 16) - (x + 3)^2 + 9 + 231 = 0
16(y - 4)^2 - 256 - (x + 3)^2 + 240 = 0
16(y - 4)^2 - (x + 3)^2 - 16 = 0
16(y - 4)^2 - (x + 3)^2 = 16
(y - 4)^2 - (((x + 3)^2)/16) = 1

a^2 + b^2 = c^2
16 + 1 = c^2
c^2 = 17
c = sqrt(17)

Foci
(-3,k + c) and (-3,k - c)
(-3,4 + sqrt(17)) and (-3,4 - sqrt(17))

Vertices
(y - 4)^2 - (((x + 3)^2)/16) = 1
(y - 4)^2 - ((-3 + 3)^2/16) = 1
(y - 4)^2 = 1
y - 4 = 1 or -1
y = 5 or 3

Asymptotes
y - k = ±(b/a)(x - h)

y - 4 = ±(1/4)(x - (-3))
y - 4 = (±1/4)(x + 3)
y = (±1/4)(x + 3) + 4
y = (±1/4)x ± (3/4) + 4
y = (±1/4)x ± (19/4)
y = (1/4)x + (19/4) and (-1/4)x + (13/4)

Center : (-3,4)
Foci : (-3,4 + sqrt(17)) and (-3,4 - sqrt(17))
Vertices : (-3,5) and (-3,3)
Asymptotes : y = (1/4)x + (19/4) and y = (-1/4)x + (13/4)