# How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola y^2/16-x^2/25=1?

Jun 7, 2018

Vertices are at $\left(0 , 4\right) \mathmr{and} \left(0 , - 4\right)$ , Focii are at
$\left(0 , 6.4\right) \mathmr{and} \left(0 , - 6.4\right)$ Asymptotes are $y = \pm \frac{4}{5} x$

#### Explanation:

${y}^{2} / 16 - {x}^{2} / 25 = 1$

The standard equation of vertical hyperbola is

${\left(y - k\right)}^{2} / {a}^{2} - {\left(x - h\right)}^{2} / {b}^{2} = 1 \therefore h = 0 , k = 0 , a = 4 , b = 5$

Center isat $0 , 0$ Vertices are at $\left(0 , a\right) \mathmr{and} \left(0 , - a\right)$ or

Vertices are at $\left(0 , 4\right) \mathmr{and} \left(0 , - 4\right)$

${c}^{2} = {a}^{2} + {b}^{2} = {4}^{2} + {5}^{2} = 41 \therefore c = \pm \sqrt{41} \approx \pm 6.4$

Foci are $c$ units from the center. Therefore,

Focii are at $\left(0 , 6.4\right) \mathmr{and} \left(0 , - 6.4\right)$ equation of asymptotes

of vertical hyperbola are $y - k = \pm \frac{a}{b} \left(x - h\right)$

Asymptotes are : $y - 0 = \pm \frac{4}{5} \left(x - 0\right) \mathmr{and} y = \pm \frac{4}{5} x$

graph{y^2/16-x^2/25=1 [-12.67, 12.64, -6.33, 6.33]}