SOLUTION: Write an equation for the hyperbola with vertices (2,5),(2,-3) and conjugate axis of length 10

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Question 1112780: Write an equation for the hyperbola with vertices (2,5),(2,-3) and conjugate axis of length 10
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Your hyperbola has one vertex above the other, both on the vertical line x=2 ,
so we say it has a vertical transverse axis.
That hyperbola, with its asymptotes and transverse axis would look like this:
.
Hyperbolas, like ellipses, have constants a , b , and c ,
that determine their shapes, and their equations.
Adding those constant, the conjugate axis, and the foci, we have
, with equation %28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1 .
In your ellipse the distance between the vertices is 2a=5-%28-3%29=8 ,
so a=8%2F2=4 .
The center is the midpoint of the transverse axes,
with h=2 like the vertices) and k=%285%2B%28-3%29%29%2F2=2%2F2=1 .
The conjugate axis length is 2b=10 , so b=10%2F2=5 .
Plugging all those numbers into the equation "formula" above,
highlight%28%28y-2%29%5E2%2F4%5E2-%28x-2%29%5E2%2F5%5E2=1%29 or highlight%28%28y-2%29%5E2%2F16-%28x-2%29%5E2%2F25=1%29 .

For this problem, you do not need c=sqrt%28a%5E2%2Bb%5E2%29 ,
which is the focal distance (the distance from the center to a focus).