# How do you write the equation of the hyperbola given Foci: (0,-5),(0, 5) and vertices (0, -3), (0,3)?

Oct 8, 2016

Please see the explanation for the process. The equation is

(y²)/(3²) - (x²)/(4²) = 1

#### Explanation:

There are two types of hyperbolas, one where a line drawn through its vertices and foci is horizontal, and one where a line drawn through its vertices and foci is vertical. This hyperbola is the type where a line drawn through its vertices and foci is vertical. We know this by observing that it is the y coordinate that changes when we move from a focus point to a vertex.

The general equation for this type of hyperbola is:

((y - k)²)/(a²) - ((x - h)²)/(b²) = 1

Observe that the x coordinate of the foci and the vertices is 0; this tells us that $h = 0$

((y - k)²)/(a²) - ((x - 0)²)/(b²) = 1

The value of k is the sum of y coordinate of the vertices divided by two:

$k = \frac{- 3 + 3}{2} = 0$

((y - 0)²)/(a²) - ((x - 0)²)/(b²) = 1

Please notice that when $x = 0 , y = \pm 3$; this allows us to find the value of "a":

((3 - 0)²)/(a²) - ((0 - 0)²)/(b²) = 1

((3 - 0)²)/(a²) = 1

(3 - 0)² = (a²)

$a = 3$

((y - 0)²)/(3²) - ((x - 0)²)/(b²) = 1

Let c = the distance between the the center point and focus = 5. The equation for the square of this distance helps us to find the value of b:

c² = a² + b²

5² = 3² + b²

b² = 25 - 9

b² = 16

$b = 4$

((y - 0)²)/(3²) - ((x - 0)²)/(4²) = 1

Simplifying a bit:

(y²)/(3²) - (x²)/(4²) = 1