# How do you write an equation for the hyperbola with foci is at (2,2) and (6,2). and the asymptotes are y=x-2 and y=6-x?

Oct 23, 2016

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

#### Explanation:

Because the foci have the same y coordinate, we know that the equation is of the form:

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

The center, $\left(h , k\right)$ is the midpoint between the foci $\left(4 , 2\right)$

${\left(x - 4\right)}^{2} / {a}^{2} - {\left(y - 2\right)}^{2} / {b}^{2} = 1$

The equations for the foci are:

$\left(h - c , k\right) \mathmr{and} \left(h + c , k\right)$

$\left(6 , 2\right) = \left(4 + c , 2\right)$

$\therefore c = 2$

We know that

${c}^{2} = {a}^{2} + {b}^{2}$

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

Because the slope of the asymptotes are 1 and -1, we know that $a = b$

$4 = 2 {a}^{2}$

$2 = {a}^{2}$

$a = \sqrt{2}$

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