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February 1, 2015

February 1, 2015

Posted by **becky** on Thursday, January 4, 2007 at 9:41pm.

write the equation of the hyperbola that meets each set of conditions.

Is the asterisk supposed to be an 8? It has to be some number. I will set this up for you later of you respond. The fomulas you need are in most calculus and analytic geometry texts.

yeah it is an 8

your vertices are (2,1) and (2,7), so your centre is (2,4), the midpoint.

Since your hyperbola is vertical it must have equation:

(x-2)^2 over a^2 - (y-4)^2 over b^2 = -1

the distance from (2,4) to (2,7) is 3, so b=3

the distance from (2,4) to your focus (2,8) is 4, so c=4

recall that in a hyperbola a^2 + b^2 = c^2, substitute to get a^2 = 7

Therefore (x-2)^2 over 7 - (y-4)^2 over 9 = -1

Nice job, reiny! Thanks for helping us out here.

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