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

Jan 18, 2017

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:

${\left(y - k\right)}^{2} / {a}^{2} - {\left(x - h\right)}^{2} / {b}^{2} = 1 \text{ }$

Its vertices are located at the points, $\left(h , k - a\right) , \mathmr{and} \left(h , k + a\right)$.

Its foci are located at the points, $\left(h , k - \sqrt{{a}^{2} + {b}^{2}}\right) , \mathmr{and} \left(h , k + \sqrt{{a}^{2} + {b}^{2}}\right)$.

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

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1 \text{ }$

Its vertices are located at the points, $\left(h - a , k\right) , \mathmr{and} \left(h + a , k\right)$.

Its foci are located at the points, $\left(h - \sqrt{{a}^{2} + {b}^{2}} , k\right) , \mathmr{and} \left(h + \sqrt{{a}^{2} + {b}^{2}} , k\right)$.

Given:

Foci: $\left(- 6 , 0\right) , \left(6 , 0\right)$
Vertices: $\left(- 4 , 0\right) , \left(4 , 0\right)$

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 . 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 .

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

$k = 0 \text{ }$
$h - a = - 4 \text{ }$
$h + a = 4 \text{ }$

To find the value of h, add equation  to equation :

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

To find the value of a, subtract equation  from equation :

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

Using, h = 0, a = 4, and the point $\left(6 , 0\right)$, we can write the equation:

$6 = 0 + \sqrt{{4}^{2} + {b}^{2}}$

and then solve for b:

$36 = 16 + {b}^{2}$
${b}^{2} = 20$
$b = \sqrt{20}$
$b = 2 \sqrt{5}$

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

${\left(x - 0\right)}^{2} / {4}^{2} - {\left(y - 0\right)}^{2} / {\left(2 \sqrt{5}\right)}^{2} = 1 \text{ }$

Equation  is the desired equation.