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?

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Answer 1 · 2016-10-23T21:54:22

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

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 $a = b$

$4 = 2a^2$

$2 = a^2$

$a = sqrt(2)$

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