# How do you find the standard form given vertices (-6, -2) and (6,-2) and foci of (-8, -2) and (8, -2)?

Apr 8, 2017

There are two standard forms for hyperbolas:
${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1 \text{ [1]}$
${\left(y - k\right)}^{2} / {a}^{2} - {\left(x - h\right)}^{2} / {b}^{2} = 1 \text{ [2]}$
Where $\left(h , k\right)$ is the center point.

#### Explanation:

Because the vertices have the same y coordinate but different x coordinates, we know that we must use equation [1].

For equation [1] here are the general formulas for the vertices are:

${x}_{1} = h - a$
${x}_{2} = h + a$
${y}_{1} = {y}_{2} = k$

Substitute -6 for ${x}_{1}$, 6 for ${x}_{2}$ and -2 for ${y}_{1}$:

$- 6 = h - a$
$6 = h + a$
$- 2 = k$

It is obvious that $h = 0$, $k = - 2$, and $a = 6$.

Substitute these values into equation [1]:

${\left(x - 0\right)}^{2} / {6}^{2} - {\left(y - - 2\right)}^{2} / {b}^{2} = 1 \text{ [3]}$

For equation [1] here are the general formulas for the foci are:

${x}_{1} = h - \sqrt{{a}^{2} + {b}^{2}}$
${x}_{2} = h + \sqrt{{a}^{2} + {b}^{2}}$
${y}_{1} = {y}_{2} = k$

Substitute 8 for ${x}_{2}$, 0 for h and 6 for a:

$8 = 0 + \sqrt{{6}^{2} + {b}^{2}}$

Solve for b:

$64 = 36 + {b}^{2}$

${b}^{2} = 28$

$b = \sqrt{28}$

Substitute this into equation [3]:

${\left(x - 0\right)}^{2} / {6}^{2} - {\left(y - - 2\right)}^{2} / {\left(\sqrt{28}\right)}^{2} = 1 \text{ [4]}$

Equation [4] is the answer.