Vertices are of the form (±a, 0), hence it is a horizontal hyperbola.
So, the equation is of the form: x2/a2 – y2/b2 = 1 …..(1)
Given: vertices of hyperbola (±6, 0)
Here a = 6
Again,
Foci is of the form (±c, 0)
Given: foci of hyperbola at (±8, 0)
⇒ c = 8
Find b:
We know, c2 = a2 + b2
64 = 36 + b2
or b2 = 64 – 36 = 28
Equation (1)⇒
x2/36 – y2/28 = 1
Which is required equation.