# How do you write a standard form equation for the hyperbola with foci at (3,11) and (3,-5) and vertices at (3,7) and (3,-1)?

Dec 26, 2017

${\left(y - 3\right)}^{2} / 16 - {\left(x - 3\right)}^{2} / 48 = 1$

#### Explanation:

The midpoint of the segment connecting the vertices (or the foci) is the center, $\left(h , k\right) \setminus \rightarrow \left(3 , 3\right)$.

The distance from the center to a focus is $c \setminus \rightarrow c = 8$.
The distance from the center to a vertex is $a \setminus \rightarrow a = 4$.

In a hyperbola we have the relationship${c}^{2} = {a}^{2} + {b}^{2}$ and we know both $a$ and $c$ so we can solve for ${b}^{2} \setminus \rightarrow {b}^{2} = 64 - 16 = 48$.

Since the vertices are on the same vertical line, $y$ comes first in the equation, so the equation has the form:

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

We can fill in to get:

${\left(y - 3\right)}^{2} / 16 - {\left(x - 3\right)}^{2} / 48 = 1$