write an equation of a hyperbola with vertices (3, -2) and (-9,-2) and foci (7,-2) -13,-2)

I will be glad to critique your work.

To write the equation of a hyperbola, we need to determine the center, the distances between the center and the vertices, and the distances between the center and the foci.

Step 1: Finding the Center:
The center of the hyperbola is the midpoint between the vertices. We can find the midpoint by averaging the x-coordinates and the y-coordinates separately.

x-coordinate of the center = (3 + (-9))/2 = -6/2 = -3
y-coordinate of the center = (-2 + -2)/2 = -4/2 = -2

Therefore, the center of the hyperbola is (-3, -2).

Step 2: Finding the Distance between the Center and the Vertices:
The distance between the center and the vertices is called the "transverse axis length." In this case, the transverse axis is horizontal.

transverse axis length = distance between the x-coordinates of the vertices = |3 - (-9)| = 12

Step 3: Finding the Distance between the Center and the Foci:
The distance between the center and the foci is called the "focal distance." In this case, the focal distance is also horizontal.

focal distance = distance between the x-coordinates of the foci = |7 - (-13)| = 20

Step 4: Determine the Equation:
Now, we can write the equation of the hyperbola in the standard form:

For a horizontally-oriented hyperbola:
(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1

Where:
(h, k) represents the center of the hyperbola.
a represents the semi-major axis.
b represents the semi-minor axis.

Given that the center is (-3, -2), the transverse axis length is 12 (2a), and the focal distance is 20 (2c):

h = -3
k = -2
a = 12 / 2 = 6
c = 20 / 2 = 10

To find the value of b, we can use the relationship between a, b, and c in a hyperbola:

c^2 = a^2 + b^2

Substituting the known values:
10^2 = 6^2 + b^2
100 = 36 + b^2
b^2 = 100 - 36
b^2 = 64
b = √64 = 8

Finally, we can substitute the values into the equation:

(x - (-3))^2 / 6^2 - (y - (-2))^2 / 8^2 = 1

Simplifying, the equation of the hyperbola is:

(x + 3)^2 / 36 - (y + 2)^2 / 64 = 1