# How do you find the standard form of the hyperbola that satisfies the given conditions: Vertices at (0, +-2), foci at (0, +-4)?

Nov 12, 2016

${\left(y - 0\right)}^{2} / {2}^{2} - {\left(x - 0\right)}^{2} / {\left(\sqrt{12}\right)}^{2} = 1$

#### Explanation:

The y coordinate of the vertices is the one that changes, therefore, the hyperbola is the vertical transverse axis type. The general equation for this type is:

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

where:

The centerpoint is

$\left(h , k\right)$

The vertices are:

$\left(h , k - a\right)$ and $\left(h , k + a\right)$

And the foci are:

$\left(h , k - \sqrt{{a}^{2} + {b}^{2}}\right)$ and $\left(h , k + \sqrt{{a}^{2} + {b}^{2}}\right)$

Given that the vertices are $\left(0 , - 2\right)$ and #(0, 2), we can write the following equations:

$h = 0 \text{ [1]}$
$k - a = - 2 \text{ [2]}$
$k + a = 2 \text{ [3]}$

Use equations [2] and [3] to solve for a and k:

$2 k = 0 \text{ [2 + 3]}$

$k = 0$

$2 a = 4 \text{ [3 - 2]}$

$a = 2$

Given that one of foci is $\left(0 , - 4\right)$, allows us to write the following equation:

$- 4 = k - \sqrt{{a}^{2} + {b}^{2}}$

Substitute 0 for k and 2 for a:

$- 4 = 0 - \sqrt{{2}^{2} + {b}^{2}}$

Solve for b:

$16 = 4 + {b}^{2}$

${b}^{2} = 12$

$b = \sqrt{12}$

Substitute all of the computed values into the general form:

${\left(y - 0\right)}^{2} / {2}^{2} - {\left(x - 0\right)}^{2} / {\left(\sqrt{12}\right)}^{2} = 1$