Hyperbolas
posted by Tabby on .
Find the equation of the hyperbola whose vertices are at (1,5) and
(1,1) with a focus at (1,7). Please and thank you.

Since the vertices and focus are on the same vertical line (x = 1), this hyperbola has an equation in the form:
(y  k)^2/a^2  (x  h)^2/b^2 = 1
The center of the hyperbola is at the midpoint of the transverse axis:
h = [1 + (1)]/2 = 1
k = [5 + 1]/2 = 2
The distance between the center (1,2) and a vertex (1,5) is 3 units, so a = 3. The distance between the center (1,2) and a focus (1,7) is 5 units, so c = 5.
Using c^2 = a^2 + b^2, we have:
b^2 = c^2  a^2 = 25  9 = 16
Substituting 1 for h, 2 for k, 9 for a^2, and 16 for b^2:
[y  (2)]^2/9  [x  (1)]^2/16 = 1
Simplifying:
16(y + 2)^2  9(x + 1)^2 = 144
16(y^2 + 4y + 4)  9(x^2 + 2x + 1) = 144
16y^2 + 64y + 64  9x^2  18x  9 = 144
16y^2 + 64y + 64  9x^2  18x  9  144 = 0
16y^2 + 64y  9x^2  18x  89 = 0
If I haven't missed anything, this should give you an idea of how to do these kinds of problems.
I hope this helps.