# algebra

posted by
**Marissa**
.

how would you solve this;

Write an equation for the hyperbola with vertices (-10,1) and (6,1) and foci (-12,1)and (8,1)

The centre is the midpoint between the vertices, which is ((-10+6)/2,(1+1)/2) or (-2,1)

so the equation looks like

(x+2)^2/a^2 - (y-1)^2/b^2 = 1

a is the distance from the centre to the vertex, so a = 8

c is the distance between the centre and the focal point, so c = 10

In a hyperbola a^2 + b^2 = c^2

so 64 + b^2 = 100

b^2 = 36

your equation would be

(x+2)^2/64 - (y-1)^2/36 = 1