Posted by **John** on Tuesday, July 10, 2012 at 12:22pm.

Find the center, foci, and vertices of this hyperbola 16x^2-y^2-32x+8y+16=0

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**Steve**, Tuesday, July 10, 2012 at 1:00pm
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)

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