# How do you find the vertex, focus, and directrix of a hyperbola  (y-2)^2 - (x^2)/4 = 1?

Jul 12, 2015

The vertices are at ($0 , 3$) and ($0 , 1$).
The foci are at ($0 , 2 - \sqrt{5}$) and ($0 , 2 + \sqrt{5}$).
The directrixes are at $y = - \frac{\sqrt{5}}{5}$ and $y = \frac{\sqrt{5}}{5}$

#### Explanation:

The standard form of the equation for a hyperbola with a vertical transverse axis is

(y-k)^2/a^2 - (x-h)^2/b^2 –= 1

(y-2)^2 –x^2/4 = 1 or

(y-2)^2/1^2 –(x–0)^2/2^2 = 1

So $a = 1$, $b = 2$, $k = 2$, and $h = 0$

Direction

The negative sign in front of the $x$-variables shows that the transverse axis is vertical.

Your graph will look like the one below.

(from www.shmoop.com)

Vertices

The coordinates of the vertices are ($h , k + a$) and ($h , k - a$).

So your hyperbola has vertices at ($0 , 3$) and ($0 , 1$)

Foci

The coordinates of the foci are ($h , k - c$) and ($h , k + c$), where

$c = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{{1}^{2} + {2}^{2}} = \sqrt{1 + 4} = \sqrt{5}$

So the foci are at ($0 , 2 - \sqrt{5}$) and ($0 , 2 + \sqrt{5}$).

Directrixes

The equation for the directrixes is y =k ± 1/c = 1± sqrt5

$y = 2 - \frac{\sqrt{5}}{5}$

$y = 2 + \frac{\sqrt{5}}{5}$