SOLUTION: Dear sir/Ma'am, Please help me with this math problem: center(1,-2), transverse axis parallel to the x-axis, transverse axis= 6, conjugate axis = 10. Find the equation of th

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Dear sir/Ma'am, Please help me with this math problem: center(1,-2), transverse axis parallel to the x-axis, transverse axis= 6, conjugate axis = 10. Find the equation of th      Log On


   

Question 812914: Dear sir/Ma'am,
Please help me with this math problem:
center(1,-2), transverse axis parallel to the x-axis, transverse axis= 6, conjugate axis = 10. Find the equation of the hyperbola.
Thank you so much!
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
The standard forms for equations of ellipses are:
%28x-h%29%5E2%2Fa%5E2+-+%28y-k%29%5E2%2Fb%5E2+=+1 for horizontally-oriented hyperbolas and
%28y-k%29%5E2%2Fa%5E2+-+%28x-h%29%5E2%2Fb%5E2+=+1 for vertically-oriented hyperbolas
Since the transverse axis is parallel to the x-axis in this problem, this hyperbola is horizontally-oriented. So we will be using the first form.

In both forms, the coordinates of the center are represented by the 'h' and the 'k'. So with a center of (1, -2) out 'h' is 1 and our 'k' is -2.

In both forms the 'a' represents the distance from the center to a vertex on the transverse axis. Since the center is halfway between the two vertices on the transverse axis, 'a' is 1/2 of the length of the transverse axis. This makes our 'a' 1/2 of 6 or, more simply, 3.

With similar logic the 'b' is 1/2 the length of the conjugate axis. So our 'b' is 1/2 of 10 or just 5.

With our 'h', 'k', 'a' and 'b' we are now ready to write the equation. Inserting the values we have found into %28x-h%29%5E2%2Fa%5E2+-+%28y-k%29%5E2%2Fb%5E2+=+1 we get:
%28x-%281%29%29%5E2%2F%283%29%5E2+-+%28y-%28-2%29%29%5E2%2F%285%29%5E2+=+1
which simplifies to:
%28x-1%29%5E2%2F9+-+%28y%2B2%29%5E2%2F25+=+1