Standard form for the hyperbola with vertices ( – 12,0) and (12,0), and foci ( – 13,0) and (13,0) is 
Hyperbola is a plane curve generated by a point so moving that the difference of the distances from two fixed points is a constant.
Given,
The Vertices of the hyperbola = (-12,0) and (12,0)
Foci = (-13,0) and (13,0)
a=12
ac=13
c=
We know,
c=

The equation of the hyperbola is

Hence, the standard form for the hyperbola with vertices ( – 12,0) and (12,0), and foci ( – 13,0) and (13,0) is 
Learn more about hyperbola here
brainly.com/question/7098764
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