SOLUTION: Write an equation for the hyperbola with vertices at (2,5) and (2,1) and a conjugate axis of length 6 units.

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Question 458647: Write an equation for the hyperbola with vertices at (2,5) and (2,1) and a conjugate axis of length 6 units.
Answer by math-vortex(648) About Me  (Show Source):
You can put this solution on YOUR website!
A good way to start is to write down the standard form of the equation for a hyperbola. The transverse axis is vertical in this case since it passes through both vertices. (Plot the two vertices on a piece of graph paper if you don’t see why.)
The standard equation for a hyperbola with a vertical transverse axis is:
%28y-k%29%5E2%2Fa%5E2-%28x-k%29%5E2%2Fb%5E2=1
The point (h,k) is the center of the hyperbola; it is the midpoint of the line segment between the two vertices at (2,3). This means that h=2 and k=3.
The parameter a is the distance from the center to each vertex. So, a=2. (Plot the center on your graph if you don’t see why.)
The length of the conjugate axis is 2b. We are told that the conjugate axis is 6 units, so b=6/2=3.
Putting this all together, we have the equation,
%28y-3%29%5E2%2F2%5E2-%28x-2%29%5E2%2F3%5E2=1
Simplify:
%28y-3%29%5E2%2F4-%28x-2%29%5E2%2F9=1