Sketch those lines slopes of +/- 1
center axis halfway between 1 and 5 so at y = 3
the lines cross at (2,3), so that is center of hyperbola
so
(x-2)^2/a^2 - (y-3)^2/b^2 = 1
slopes of asymptotes = +/- b/a so b = a
(x-2)^2/a^2 - (y-3)^2/a^2 = 1
put that point in now
(4-2)^2/a^2 - (3-3)^2/a^2 = 1
a^2 = 4
a = 2
(x-2)^2/4 - (y-3)^2/4 = 1
HELP! I am not sure how I would even start this problem or solve it.
A hyperbola with a horizontal transverse axis contains the point at (4, 3). The equations of the asymptotes are y-x=1 and y+x=5 Write an equation for the hyperbola.
1 answer
1 answer