Posted by Tabby on Saturday, April 21, 2012 at 2:41pm.
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.
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