Pre-calculus/trig

This hyperbola will have its center at (0,0)

Since  (√13, 0)  and (-√13,0) are the foci....the hyperbola will have the form   

x^2         y^2

___  -   ____  =   1

a^2         b^2

Since  the difference  in  distances from any point on the hyperbola to the focal points  = 6 , we have this property that

2a  =  6       divide both sides by  2

a =  3

a^2  = 9

And     a^2 + b^2  = c^2          where c = √13

So

3^2 + b^2  = (√13)^2

9 + b^2  = 13

b^2 = 13 - 9

b^2  = 4

So....the correct  form is

x^2        y^2

___  -    ___     =  1

 9            4