write equation of hyperbola with cebter c(4,2) vertex (4,5) equation of one asymptote 4y-3x=-4

Answer 1

looks like the hyperbola is vertical, so
(x-4)^2 /a^2 - (y-2)^2/b^2 = -1
we know b = 3 , (the distance between (4,2) and (4,5) )

4y-3x=-4
3x - 4y = 4
slope of asymptote = 3/4
that is b/a = 3/4, but b = 3, so a = 4

(x-4)^2 / 16 - (y-2)^2 /9 = -1

confirmation:
http://www.wolframalpha.com/input/?i=vertex+of+(x-4)%5E2+%2F+16+-+(y-2)%5E2+%2F9+%3D+-1