# How do you write a standard form equation for the hyperbola with vertices at (-2, 0) and (2, 0) and foci at (-6, 0) and (6, 0)?

May 26, 2017

Because the vertices are horizontal, we know that the standard form is,

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

, the vertices are $\left(h \pm a , k\right)$

and the foci are $\left(h \pm \sqrt{{a}^{2} + {b}^{2}} , k\right)$

#### Explanation:

Using the form of the vertices and the given vertices we can write the following equations:

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

Solving the first two equations we have:

$h = 0$
$a = 2$
$k = 0$

Using the form of the foci and one of the given foci we can write:

$6 = h + \sqrt{{a}^{2} + {b}^{2}}$

Substitute in the known values for h and a:

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

Solve for b:

$36 = 4 + {b}^{2}$

$b = \sqrt{32}$

Substituting the know values into equation [1]:

${\left(x - 0\right)}^{2} / {2}^{2} - {\left(y - 0\right)}^{2} / {\left(\sqrt{32}\right)}^{2} = 1 \text{ [2]}$

Equation [2] is the answer.