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

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> 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      Log On

Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

   

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(12144) About Me  (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:

%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1

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

%28x-2%29%5E2%2F3%5E2-%28y-%28-5%29%29%5E2%2F3%5E2=1

%28x-2%29%5E2%2F9-%28y%2B5%29%5E2%2F9=1

%28x-2%29%5E2-%28y%2B5%29%5E2=9

x%5E2-4x%2B4-%28y%5E2%2B10y%2B25%29=9

x%5E2-4x%2B4-%28y%5E2%2B10y%2B25%29=9

x%5E2-y%5E2-4x-10y%2B4-25-9=0

x%5E2-y%5E2-4x-10y-30=0

The correct form is actually:

Ax%5E2+%2B+Bxy+%2B+Cy%5E2+%2B+Dx+%2B+Ey+%2B+F+=+0, 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.