SOLUTION: Vertices: (13, 0), (-1, 0); Asymptotes: 𝑦 = 𝑥 − 6 & 𝑦 = −𝑥 + 6

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Vertices: (13, 0), (-1, 0); Asymptotes: 𝑦 = 𝑥 − 6 & 𝑦 = −𝑥 + 6      Log On


   

Question 1186314: Vertices: (13, 0), (-1, 0); Asymptotes: 𝑦 = 𝑥 − 6 & 𝑦 = −𝑥 + 6
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


We are given the equations of the asymptotes, so the conic section is a hyperbola.

The segment joining the two vertices is the transverse axis; it is horizontal, so the two branches of the hyperbola open left and right.

The center (h,k) of the hyperbola is the center of the transverse axis. You can also find the center as the intersection of the two asymptotes. To learn as much as you can from this, determine the center both ways and verify that you get the same result.

The general form of the 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

You have determined h and k; to complete the equation you need to find a and b. Two facts will let you do that:

(1) The length of the transverse axis is 2a
(2) The slopes of the asymptotes (1 and -1) are b/a and -b/a

You should be able to write the equation now.