# How do you write the equation for a hyperbola given vertices (9,-3) and (-5,-3), foci (2+-sqrt53,-3)?

Feb 21, 2018

The equation of hyperbola is ${\left(x - 2\right)}^{2} / 49 - {\left(y + 3\right)}^{2} / 4 = 1$

#### Explanation:

Vertices are $\left(9 , - 3\right) \mathmr{and} \left(- 5 , - 3\right)$

Foci are $\left(2 + \sqrt{53} , - 3\right) \mathmr{and} \left(2 - \sqrt{53} , - 3\right)$

By the Midpoint Formula, the center of the hyperbola occurs at the

point (2,-3); h=2, k=-3 :. a= 9-2=7; a^2=49 ;

$c = 2 + \sqrt{53} - 2 = \sqrt{53} \therefore {c}^{2} = 53$

${b}^{2} = {c}^{2} - {a}^{2} = 53 - 49 = 4 \therefore b = 2$ . So, the hyperbola has a

horizontal transverse axis and the standard form of the

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

${\left(x - 2\right)}^{2} / {7}^{2} - {\left(y + 3\right)}^{2} / {2}^{2} = 1$ or

${\left(x - 2\right)}^{2} / 49 - {\left(y + 3\right)}^{2} / 4 = 1$

The equation of hyperbola is ${\left(x - 2\right)}^{2} / 49 - {\left(y + 3\right)}^{2} / 4 = 1$ [Ans]