SOLUTION: Hi I need your help I'm struggling with solving this. Find the equation of the hyperbola, given vertices: (15,1), (-1,1); Endpoints of the conjugate axis: (7,7), (7,-5)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Hi I need your help I'm struggling with solving this. Find the equation of the hyperbola, given vertices: (15,1), (-1,1); Endpoints of the conjugate axis: (7,7), (7,-5)      Log On


   



Question 1186316: Hi I need your help I'm struggling with solving this. Find the equation of the hyperbola, given vertices: (15,1), (-1,1); Endpoints of the conjugate axis: (7,7), (7,-5)
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Previously solved....

The vertices are the endpoints of the transverse axis; the transverse axis is horizontal, so the branches of the hyperbola open left and right.

The general form of the equation of a hyperbola with center (h,k) and branches opening left and right is

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

To find the equation from the given information, you need to determine the center (h,k) and the values of a^2 and b^2.

(1) The center (h,k) is the midpoint of both the transverse axis and the conjugate axis
(2) The length of the transverse axis is 2a; the length of the conjugate axis is 2b

You now have all the pieces you need to write the equation.