Find the equation of the hyperbola if y=2x−5 and y=−2x+3 are the asymptotes, and it is known that one of the vertices is on the y-axis.

The first thing you want to do is find the center of the hyperbola. Since you have the two equations of the asymptotes, their intersection must be what you want. setting 2x - 5 equal to -2x + 3, you get that x = 2, meaning that 2 is the x-coord of your center. Plugging this value of x into the equations, you get that y = -1. Now that you have your center, you have the information that one of the vertices is on the y-axis. Since the distance from the center to the vertice must be a straight line, you know for sure that this hyperbola is a horizontal one (a vertical one would not have a vertice on the y-axis given this center). Knowing this, you have the equation form (x-2)^2/a^2 - (y+1)^2/b^2 = 1. I got a little confused here, but remember that you now know the center and the equations of the asymptotes. y = 2x - 5 is equal to 2(x - 2) - 1. The form of the asymptotes of a horizontal hyperbola is b/a(x-h) + k, which means that b/a must be equal to 2. This also means that 2a = b, and 4a^2 = b^2. Now you can substitute b^2 for 4a^2, leaving you with the equation (x-2)^2/a^2 - (y+1)^2/ 4a^2 = 1. Remember that vertice on the y-axis? Since it must be at the same y-coordinate (since you know that it must be a horizontal line from the center to the vertice), you know the vertice has to be (0, -1). You can plug these values into the previous equation, which gives you (0 - 2)^2 / a^2 - 0 = 1. This leaves you with 4/a^2 = 1, which means that a must be 2. Since you already know that b is 2a, b must be equal to 4. Plugging those values into the equation, you get your final answer, (x-2)^2/4 - (y+1)^2/16 = 1.