# How do you identify the vertices, foci, and direction of y^2/25-x^2/16=1?

Nov 27, 2017

See the explanation below

#### Explanation:

The equation is

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

$\implies$, ${\left(\frac{y}{5}\right)}^{2} - {\left(\frac{x}{4}\right)}^{2} = 1$

This is the equation of a hyperbola with a vertical transverse axis.

${y}^{2} / {a}^{2} - {x}^{2} / {b}^{2} = 1$

The center of the hyperbola is $C = \left(0 , 0\right)$

The vertices are $V = \left(0 , 5\right)$ and $V ' = \left(0 , - 5\right)$

The foci are $F = \left(0 , c\right)$ and $F ' = \left(0 , - c\right)$

where $c = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{25 + 16} = \sqrt{41}$

The foci are $F = \left(0 , \sqrt{41}\right)$ and $F ' = \left(0 , - \sqrt{41}\right)$

The asymptotes are $y = \pm \left(\frac{a}{b}\right) x$

The asymptotes are $y = \pm \left(\frac{5}{4}\right) x$

graph{y^2/25-x^2/16=1 [-32.47, 32.47, -16.25, 16.24]}