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
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.
3 answers
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
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
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
2 answers