Complete the standard form of the equation of a hyperbola that has vertices at (-10, - 15) and (70, - 15) and one of its foci at (-11, - 15).

The vertices and the foci lie on the line y = - 15.

Then the center has x-coordinate in the middle between - 10 and 70, this is (-10+70) / 2 = 30. And y = - 15. Then, center is (30,-15).

semi axis a = distance from the center to the vertex = |-10 - 30| = |-40| = 40

c = distance from the focus to the center = |-11 - 30| = |-41| = 41

semi axis b is calculated from c^2 = a^2 + b^2 ⇒ b^2 = c^2 - a^2 = 41^2 - 40^2 = 81

Now the standarf form is

[x - h]/a^2 - [y - k}/b^2 = 1

where (h, k) is the center = (30,-15)

Then,

[x - 30]^2 / 1600 - [y + 15]^2 / 6561 = 1 is the equation searched.