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September 20, 2014

September 20, 2014

Posted by **EVJ** on Tuesday, July 19, 2011 at 11:08pm.

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.

- analytic geometry -
**drwls**, Wednesday, July 20, 2011 at 12:22amFirst 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

- Trig -
**Reiny**, Wednesday, July 20, 2011 at 7:29amI 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

- analytic geometry -
**drwls**, Wednesday, July 20, 2011 at 8:51amYes, 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

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