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.
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
To find the equation of a hyperbola given the foci and asymptotes, you can follow these steps:
1. Identify the center of the hyperbola. The center is the midpoint between the two foci. In this case, the foci are at (-4, 0) and (4, 0), so the center will be at the midpoint of these coordinates, which is (0, 0).
2. Determine the distance between the center and one of the foci. The distance between the center and each focus is called the "c" value. In this case, the distance between the center (0, 0) and either of the foci is 4 units.
3. Find the equation for the asymptotes. The equation for the asymptotes of a hyperbola can be written in the form y = mx, where "m" is the slope. In this case, the asymptote equation is given as y = -x, so the slope is -1.
4. Determine the distance between the center and one of the vertices. The distance between the center and each vertex is called the "a" value. Since the asymptotes of this hyperbola are given by y = -x, which passes through the center, the distance will be the same as "c", which is 4 units.
5. Calculate the value of "b". The value of "b" can be found using the equation: c^2 = a^2 + b^2. Substitute the known values: (4^2) = (4^2) + b^2. Solving for "b", we get b = 0.
6. Write the equation of the hyperbola. The equation of a hyperbola in standard form is: (x - h)^2 / a^2 - (y - k)^2 / b^2 = 1, where (h, k) is the center of the hyperbola, "a" is the distance from the center to each vertex, and "b" is the distance from the center to each co-vertex. In this case, the center is (0, 0), "a" is 4, and "b" is 0. Therefore, the equation of the hyperbola is: x^2 / 16 - y^2 / 0 = 1.
However, it is important to note that dividing by 0 is undefined. In this case, since "b" is equal to 0, the equation of the hyperbola will become y^2 = 0, which simplifies to y = 0. So, the equation of the hyperbola with the given foci and asymptote is x^2 / 16 - y^2 / 0 = 1, which simplifies to x^2 / 16 = 1.
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