Trig
posted by EVJ on .
Find an equation for the hyperbola described.
Foci at (4,0) and (4,0); asymptote the line y= x.
Please explain the steps of this problem. My book does not get an example of this type of problem and I'm not having any luck working this problem.

First of all, this is not a trigonometry question. No matter what your course title may be.
Hyperbolas have two asymptotes. The other one, in this case, would be y = x. Did they forget to tell you that or did you just forget to include it?
The general equation for a hyperbola aligned with the x axis is
x^2/a^2 y^2/b^2 = 1
In your case, a = 4 and b/a is the asymptote slope, which is 1. Therefore
x^2/16  y^2/16 = 1 
I will have to disagree.
I think drwls read it as the vertex is (4,0)
but the focus is (4,0)
so c = 4
a^2 + b^2 = c^2, but for an asymptote of y = x , a = b
2a^2 = 16
a^2 = 8
equation:
x^2/8 + y^2/8 = 1 
Yes, I did confuse the vertex with the focus. Thanks to Reiny for catching that.
There appears to be a sign error in Reiny's final equation, however. As written, it is the equation of a circle of radius sqrt8. Try instead
x^2/8  y^2/8 = 1