SOLUTION: **Sorry if this appears twice, it doesn't seem to have worked the first time. A hyperbola with the transverse axis on the line y=-5, length of transverse axis = 6, conjugate axi

Question 126660: **Sorry if this appears twice, it doesn't seem to have worked the first time.
A hyperbola with the transverse axis on the line y=-5, length of transverse axis = 6, conjugate axis on the line x = 2, and length of conjugate axis = 6.
I need to use this information and put this hyperbola into an equation in the Ax2+By^2+cxy+Dx+Ey+F=0
So far, I know that the center is at (2,-5) and both a and b = 3. (at least I think)
I am really having problems putting it into the form that they ask. Help would be greatly appreciated!!
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
So far, so good.
The center is at (2,-5) because that is where the transverse and conjugate axes intersect, and both a and b = 3 because a is the length of the transverse axis divided by 2 and b is the length of the conjugate axis divided by 2.

Since the transverse axis is horizontal, this is an 'east-west' opening hyperbola. The formula for such a hyperbola centered at (h,k) is:

Substituting the coordinates of the center and the a and b values we get:

The correct form is actually:

, but it doesn't matter in this case because the coefficient on the xy term is 0. You only get a non-zero coefficient on the xy term if the transverse and conjugate axes are other than parallel to the coordinate axes.

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