Question

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

Answer 1

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 = (-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`