Find the equation of the hyperbola whose vertices are at (-1,-5) and

centre must be the midpoint of the vertices

centre is (-1, -2)
after making a sketch we can see the major axis is vertical and
a = 3, c = 5
a^2 + b^2 = c^2
9+b^2 = 25
b^2 = 16

(x+1)^2 /6 - (y+2)^2 /9 = -1

Thank you so much!

To find the equation of a hyperbola given its vertices and one focus, you can use the standard equation of a hyperbola:

(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1

Where (h, k) are the coordinates of the center, and a and b are the lengths of the semi-major and semi-minor axes, respectively.

Given that the center of the hyperbola is at (-1, -3), we can determine the values of a and b by using the distance formula:

For the vertices:
Distance from (-1, -3) to (-1, -5) = a
Distance from (-1, -3) to (-1, 1) = 2b

Using the distance formula, we have:
a = √((-1 - (-1))^2 + (-3 - (-5))^2) = √(0^2 + 2^2) = 2
2b = √((-1 - (-1))^2 + (-3 - 1)^2) = √(0^2 + 4^2) = 4

Simplifying, we find:
b = 4 / 2 = 2

Now we have the values of a and b, so we can substitute these into the standard equation to get the final equation:

(x + 1)^2 / 2^2 - (y + 3)^2 / 4^2 = 1

Simplifying further, we have:
(x + 1)^2 / 4 - (y + 3)^2 / 16 = 1

And that's the equation of the hyperbola.