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March 27, 2015

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

(-1,1) with a focus at (-1,-7). Please and thank you.

- Hyperbolas -
**MathGuru**, Saturday, April 21, 2012 at 8:15pmSince 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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