# How do you find the equation for a hyperbola centered at the origin with a horizontal transverse axis of lengths 8 units and a conjugate axis of lengths 6 units?

Dec 8, 2016

The equation is: ${\left(x - 0\right)}^{2} / {4}^{2} - {\left(y - 0\right)}^{2} / {3}^{2} = 1$

#### Explanation:

Using this reference Conics: Hyperbola, the general equation for a hyperbola with a transverse horizontal axis is:

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

where $\left(h , k\right)$ is the center, $2 a$ is the length of transverse axis, and $2 b$ is the length of the conjugate axis.

We are given that the center is, $\left(0 , 0\right)$; substitute this into equation [1]:

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

We are given that the length of the horizontal transverse axis is 8; this allows us to write this equation:

$2 a = 8$

$a = 4$

Substitute 4 for "a" in equation [2]:

${\left(x - 0\right)}^{2} / {4}^{2} - {\left(y - 0\right)}^{2} / {b}^{2} = 1 \text{ [3]}$

We are given that the length of the horizontal transverse axis is 6; this allows us to write this equation:

$2 b = 6$

$b = 3$

Substitute 3 for "b" in equation [3]:

${\left(x - 0\right)}^{2} / {4}^{2} - {\left(y - 0\right)}^{2} / {3}^{2} = 1 \text{ [4]}$