How do you write an equation for the hyperbola with foci is at (2,2) and (6,2). and the asymptotes are y=x-2 and y=6-x?

1 Answer
Oct 23, 2016

Answer:

#(x - 4)^2/(sqrt(2))^2 - (y - 2)^2/(sqrt(2))^2 = 1#

Explanation:

Because the foci have the same y coordinate, we know that the equation is of the form:

#(x - h)^2/a^2 - (y - k)^2/b^2 = 1#

The center, #(h,k)# is the midpoint between the foci #(4,2)#

#(x - 4)^2/a^2 - (y - 2)^2/b^2 = 1#

The equations for the foci are:

#(h - c, k) and (h + c, k)#

#(6, 2) = (4 + c, 2)#

#:. c = 2#

We know that

#c^2 = a^2 + b^2#

#4 = a^2 + b^2#

Because the slope of the asymptotes are 1 and -1, we know that #a = b#

#4 = 2a^2#

#2 = a^2#

#a = sqrt(2)#

#(x - 4)^2/(sqrt(2))^2 - (y - 2)^2/(sqrt(2))^2 = 1#