# How do you find the standard form of the equation of the hyperbola given the properties foci (+-5,0), length of the conjugate axis 6?

Jul 1, 2017

#### Explanation:

Because the foci are given to be $\left(- 5 , 0\right)$ and $\left(5 , 0\right)$, we know that the hyperbola is the horizontal transverse axis type and has the general equation:

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

We know that $2 b$ is the length of the conjugate axis; this allows us to write the equation:

$2 b = 6$

$b = 3$

Substitute the value for "b" into equation :

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {3}^{2} = 1 \text{ }$

We know that the general forms for the foci are:

$\left(h - \sqrt{{a}^{2} + {b}^{2}} , k\right)$ and $\left(h + \sqrt{{a}^{2} + {b}^{2}} , k\right)$

This allows us to write the following equations:

$k = 0 \text{ }$
$- 5 = h - \sqrt{{a}^{2} + {3}^{2}} \text{ }$
$5 = h + \sqrt{{a}^{2} + {3}^{2}} \text{ }$

Substitute the value for k into equation :

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - 0\right)}^{2} / {3}^{2} = 1 \text{ }$

Find the value of h by adding equations  and :

$2 h = 0$

$h = 0$

Substitute into equation :

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

Use equation  to find the value of "a":

$5 = 0 + \sqrt{{a}^{2} + {3}^{2}}$

#25 = a^2+9

${a}^{2} = 16$

$a = 4$

Substitute the value of "a" into equation 

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

Equation  is the equation of the hyperbola.