Question

a hyperbola has vertices (+-5,0) and one focus (6,0) what is the standard form equation of the hyperbola

Answer 1

Hyperbola again.
Looking at http://www.clausentech.com/lchs/dclausen/algebra2/hyperbolas.htm
it says that the vertices are at
(h+a, k) and (h-a, k)
the average of these is (h+a+ h-a, 2k)/2= (2h,2k)/2 = (h,k)
(algebra again)

they give us the vertices as (5,0) and (-5,0). add together: (5-5,0+0)= (0,0)
divide by 2 to get the average: (0,0)/2 = (0,0) (divide by 2 didn't change anything when the center is at (0,0)

Answer 2

so we know, so far,
the center is at (0,0), the transverse axis is horizontal (it goes through the 2 vertices)
the vertices are at (-5,0) and (5,0)
one focus is at (6,0)
the vertex is at (h+a,k). h=0 and k=0, so this is (a,0). this means a=5 (to match (5,0)