Calculus: Conic Sections?

Question

For the following problem we are asked to find an equation for the conic that satisfies the given conditions:

Hyperbola, foci (2,2) and (6,2), asymptotes y = x – 2 and y = 6 – x

I know that the center is (4,2), the halfway point between the two foci, and thus that c = 2. However, I am not sure how to use the given asymptotes to find a and b to get the equation. Please explain how to do this part of the problem step-by-step.

Answer 1

You are right so far.

c = 2
center (h, k) = (4, 2)

Notice that the line of symmetry thru the foci is horizontal. So the hyperbolas open sideways. So we have an equation of the form:

(x - h)²/a² - (y - k)²/b² = 1
(x - 4)²/a² - (y - 2)²/b² = 1

We can use the asymptotes to solve for a and b. The slopes of the asymptotes are:

m = ±∆y/∆x = ±b/a = ±1

a = ±b
a² = b²

c² = a² + b² = 2a²
2² = 2a²
4 = 2a²
2 = a²

b² = a² = 2

So the equation of the hyperbolas is:

(x - 4)²/2 - (y - 2)²/2 = 1