# How do you write the equation for a hyperbola given vertices (-5,0) and (5,0), conjugate axis of lengths 12 units?

Nov 11, 2016

${\left(x - 0\right)}^{2} / {5}^{2} - {\left(y - 0\right)}^{2} / {6}^{2} = 1$

#### Explanation:

The two vertices have the same y coordinate, therefore, the hyperbola is the horizontal transverse axis type. The standard equation for the horizontal transverse axis type is:

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

We know that the general form for the vertices of this type are:

$\left(h + a , k\right)$ and $\left(h - a , k\right)$

Using the given vertices $\left(- 5 , 0\right)$ and $\left(5 , 0\right)$, we can write the following equations:

$k = 0$
$h + a = 5$
$h - a = - 5$

Solve for h and a:

$a = 5$
$h = 0$

From the reference, the length of the conjugate axis is equal to 2b:

$2 b = 12$

$b = 6$

We know the values of h, k, a, and b, substitute them into the general form:

${\left(x - 0\right)}^{2} / {5}^{2} - {\left(y - 0\right)}^{2} / {6}^{2} = 1$