## anonymous one year ago Find an equation in standard form for the hyperbola with vertices at (0, ±6) and foci at (0, ±9).

1. anonymous

@phi

2. phi

the center is the average of the vertices

3. anonymous

So (0,0)

4. phi

the vertices are on the y-axis, so the hyperbola looks like a frown/smile combo that means the "y' goes first in the standard equation $\frac{(y-k)^2}{a^2}- \frac{(x-h)^2}{b^2}=1$ a is the distance from the center to the focus "c" is the distance from the center to the vertex we use a and c to find b a^2 + b^2 = c^2 we already know (h,k) is (0,0)

5. phi

oops, got that swapped: a is the distance from the center to the vertex "c" is the distance from the center to the focus

6. anonymous

alright.

7. anonymous

so a is 6 and c is 9

8. phi

yes, but in the equation you use a^2 = 36 and c^2=81 what is b^2 ?

9. anonymous

$3\sqrt{13}$

10. anonymous

b^2 is 117

11. phi

a^2 + b^2 = c^2 36 + b^2 = 81

12. anonymous

13. anonymous

b^2 = 45 b=3sqrt5

14. phi

now fill in the numbers to get the equation

15. anonymous

thanks. I still have a few more