Write an equation for the hyperbola that satisfies the given set of conditions.

Vertices (9, -3) and (-5, -3) foci (2square root of 53,-3)

what the formula to do this, 10points to who ever give me a easy way to do this cuz im stuck

Because the vertices have the same y coordinate, we know that that the hyperbola has a transverse axis that is aligned with the x axis so the general equation is:

{(x - h)/a}² - {(y - k)/b}² = 1

We know that k = the y coordinate of the vertices = -3 which will change the sign in the equation as follows:

{(x - h)/a}² - {(y + 3)/b}² = 1

We know that h = the center line which is the midpoint between the x coordinates of the vertices = (9 - 5)/2 = 2 so the equation becomes:

{(x - 2)/a}² - {(y + 3)/b}² = 1

We know that a = is the distance from the center line (h) to either vertex which is 7 so the equation becomes:

{(x - 2)/7}² - {(y + 3)/b}² = 1

The reference pages introduce another parameter (c) which is the distance from the foci to the center line which equals the square root of the sum of the squares of "a" and "b":

c = √{a² + b²}

In this case c = 2√{53} - 2 so we can substitute 7 for "a", set the right sides equal, and then solve for b:

√{7² + b²} = 2√{53} - 2

49 + b² = 212 - 4√{53} + 4

49 + b² = 212 - 4√{53} + 4

b² = 167 - 4√{53}

b ≈ 11.74

The equation is done:

{(x - 2)/7}² - {(y + 3)/11.74}² = 1

The vertices are horizontally aligned (how do you know this?).
The center is halfway between foci, at (2, -3).

For a horizontal hyperbola:
     (x−h)²/a² − (y−k)²/b² = 1
     center (h, k)
     vertices (h±a, k)
     foci (h±c, k), where c² = a²+b²

Apply your data.
Center (h, k) = (2, -3)
h = 2
k = -3

vertices (2±a, -3) = (2±7, -3)
a = 7

focus (2+c, -3) = (2√53, -3)
c = 2√53 - 2
c² = a² + b²
4·53 - 8√53 + 4 = 7² + b²
b² = 216 - 8√53 ≅ 108.8

The equation of the hyperbola becomes
     (x−2)²/49 − (y+3)²/108.8 = 1