SOLUTION: 16x squared minus y squared minus 128x plus 240 = 0 It's a Hyperbola identify the center, vertices, and foci.

Question 811230: 16x squared minus y squared minus 128x plus 240 = 0
It's a Hyperbola identify the center, vertices, and foci.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
16x squared minus y squared minus 128x plus 240 = 0
It's a Hyperbola identify the center, vertices, and foci.
***
Given hyperbola has a horizontal transverse axis
Its standard form of equation: (x-h)^2-(y-k)^2=1, (h,k)=(x,y) coordinates of cente.
..
16x^2-y^2-128x+240=0
16x^2-128x-y^2=-240
16(x^2-8x+16)-y^2=-240+256
16(x-4)^2-y^2=16
(x-4)^2/1-y^2/16=1
Center:(4,0)
a^2=1
a=1
b^2=16
b=4
c^2=a^2+b^2=1+16=17
c=√17≈4.12
..
Vertices:(4�a,0)=(4�1,0)=(3,0) and (5,0)
Foci:(4�c,0)=(4�4.12,0)=(-.12,0) and (8.12,0)
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