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

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 = (-3 + 3)/2 = 0#

#((y - 0)²)/(a²) - ((x - 0)²)/(b²) = 1#

Please notice that when #x = 0, y = +-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#