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October 30, 2014

October 30, 2014

Posted by **Miley** on Monday, June 9, 2008 at 3:40pm.

A hyperbola with foci: (0,5) (0,-5)and vertices: (0,3) (0,-3). So i got y^2/9-x^2/16=1. but how do you get the 16?

- Algebra 2 Mannn -
**Damon**, Monday, June 9, 2008 at 3:49pmcenter is at (0,0) so the form is:

y^2/a^2 -x^2/b^2 = 1

distance center to focus = 5 = sqrt(a^2+b^2)

distance center to vertex = 3 = a

so

5^2 = (3^2 + b^2)

25 = 9 + b^2

b^2 = 16

- Algebra 2 Mannn -
**Miley**, Monday, June 9, 2008 at 4:08pmthanks man

- Algebra 2 Mannn -

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