How to write the equation of the hyperbola given Foci: (0,-5),(0, 5) and vertices (0,...

There are two different kinds of hyperbolas: one where the line connecting its vertices and foci is horizontal, and the other where the line 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:
${\displaystyle \frac{(y-k)²}{a²}-\frac{(x-h)²}{b²}=1}$
Observe that the x coordinate of the foci and the vertices is 0; this tells us that ${\displaystyle h=0}$
${\displaystyle \frac{(y-k)²}{a²}-\frac{(x-0)²}{b²}=1}$
The value of k is the sum of y coordinate of the vertices divided by two:${\displaystyle k=\frac{-3+3}{2}=0}$
${\displaystyle \frac{(y-0)²}{a²}-\frac{(x-0)²}{b²}=1}$
Please notice that when ${\displaystyle x=0,y=\pm 3}$; this allows us to find the value of ""a"":
${\displaystyle \frac{(3-0)²}{a²}-\frac{(0-0)²}{b²}=1}$
${\displaystyle \frac{(3-0)²}{a²}=1}$
${\displaystyle (3-0)²=(a²)}$
${\displaystyle a=3}$
${\displaystyle \frac{(y-0)²}{3²}-\frac{(x-0)²}{b²}=1}$
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:
${\displaystyle c²=a²+b²}$
${\displaystyle 5²=3²+b²}$
${\displaystyle b²=25-9}$
${\displaystyle b²=16}$
${\displaystyle b=4}$
${\displaystyle \frac{(y-0)²}{3²}-\frac{(x-0)²}{4²}=1}$
Simplifying a bit:
${\displaystyle \frac{y²}{3²}-\frac{x²}{4²}=1}$