Question

How do you write the equation of the hyperbola given Foci: (-6,0),(6,0) and vertices (-4,0), (4,0)?

Answer 1

There are two standard Cartesian forms for the equation of a hyperbola. I will explain how one knows which one to use and how to use it in the explanation.

Explanation:

The standard Cartesian form for the equation of a hyperbola with a vertical transverse axis is:

`(y - k)^2/a^2 - (x - h)^2/b^2 = 1" [1]"`

Its vertices are located at the points, `(h, k - a), and (h, k + a)`.

Its foci are located at the points, `(h, k - sqrt(a^2 + b^2)), and (h, k + sqrt(a^2 + b^2))`.

The standard Cartesian form for the equation of a hyperbola with a horizontal transverse axis is:

`(x - h)^2/a^2 - (y - k)^2/b^2 = 1" [2]"`

Its vertices are located at the points, `(h- a, k), and (h + a, k)`.

Its foci are located at the points, `(h - sqrt(a^2 + b^2), k), and (h + sqrt(a^2 + b^2), k)`.

Given:

Foci: `(-6, 0), (6,0)`
Vertices: `(-4, 0), (4,0)`

Please observe that it is the x coordinates of the vertices and foci that changes, while the y coordinate remains fixed; this matches the the horizontal transverse type, equation [2]. This is how one knows which equation to use. If it were the y coordinates that were changing and the x coordinates were fixed, then it would be equation [1].

We can use `(-4, 0), (4,0)` and `(h- a, k), (h + a, k)` to write the following equations:

`k = 0" [3]"`
`h - a = -4" [4]"`
`h + a = 4" [5]"`

To find the value of h, add equation [4] to equation [5]:

`h - a + h + a = -4 + 4`
`2h = 0`
`h = 0`

To find the value of a, subtract equation [4] from equation [5]:

`-h + a + h + a = 4 + 4`
`2a = 8`
`a = 4`

Using, h = 0, a = 4, and the point `(6,0)`, we can write the equation:

`6 = 0 + sqrt(4^2 + b^2)`

and then solve for b:

`36 = 16 + b^2`
`b^2 = 20`
`b = sqrt(20)`
`b = 2sqrt(5)`

Substitute the values for h, k, a, and b into equation [2]:

`(x - 0)^2/4^2 - (y - 0)^2/(2sqrt(5))^2 = 1" [6]"`

Equation [6] is the desired equation.