centre is the midpoint of the vertices:
C(-1,1)
our a value is 6, and we don't know b
but we know b/a = 1/3
b/6 = 1/3
3b = 6
b = 2
(x+1)^2 / 36 - (y-1)^2/4 = 1
or
(x+1)^2 - 9(y-1)^2 = 36
verification:
http://www.wolframalpha.com/input/?i=plot+(x%2B1)%5E2+-+9(y-1)%5E2+%3D+36
Find the equation of the hyperbola satisfying the given conditions.
Vertices at (5, 1) and (-7, 1); asymptotes (y - 1) = 1/3(x + 1) and (y - 1) = -1/3(x + 1)
2 answers
what is the standard equation given asymptotes are y=1/3(x+5) and y=-1/3(x-7) with foci (1,12)
1 answer