SOLUTION: Find the equation of the hyperbola with the following properties. Foci at (5,-4) and (13,-4) Vertices at (6,-4) and (12,-4)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the equation of the hyperbola with the following properties. Foci at (5,-4) and (13,-4) Vertices at (6,-4) and (12,-4)       Log On


   

Question 1144011: Find the equation of the hyperbola with the following properties.
Foci at (5,-4) and (13,-4)
Vertices at (6,-4) and (12,-4)
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


(1) Foci at (5,-4) and (13,-4)

So...

Transverse axis is horizontal, so branches open right and left.
The general equation is %28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1.
The center (h,k) is halfway between the foci, at (9,-4).
c=4.

(2) Vertices at (6,-4) and (12,-4)

So the length of the transverse axis is 6, which means a=3.

Use c^2=a^c+b^2 to determine a^2; then write the equation.