What is the standard form of the hyperbola given the foci (6, 1+2 sqrt 58) , (6, 1-2 sqrt 58) and points of the hyperbola are 28 units closer to one focus than the other?

Answer 1

c = 2√58
For any P=(x,y) on the hyperbola,
PF1 - PF2 = 28
If the center were at (0,0), this would mean
√(x^2 + (y+2√58)^2) - √(x^2 + (y-2√58)^2) = 28
I get that to mean
y^2/14^2 - x^2/6^2 = 1
Moving that to the center at (6,1), that gives
(y-1)^2/14^2 - (x-6)^2/6^2 = 1
Check my math.