## anonymous one year ago Find an equation in standard form for the hyperbola with vertices at (0, ±8) and asymptotes at y equals inverse of plus minus 1 divided by 2x.

1. anonymous

@jim_thompson5910

2. jim_thompson5910

what is the center of this hyperbola?

3. anonymous

0,0

4. jim_thompson5910

yes

5. jim_thompson5910

so (h,k) = (0,0) h = 0 k = 0

6. anonymous

hey that makes a cool face 0,0

7. anonymous

sorry off topic

8. jim_thompson5910

lol yes it does how far do go from (0,0) to either vertex?

9. anonymous

8 units

10. jim_thompson5910

so a = 8 a = half of the length of the transverse axis the transverse axis connects the two vertices and goes through the center

11. anonymous

alright

12. anonymous

SO what is the formula that we need to plug these into?

13. jim_thompson5910

how is the hyperbola oriented? is it opening left/right? or up/down?

14. anonymous

up/down

15. jim_thompson5910

according to this http://www.purplemath.com/modules/hyperbola.htm it says that if the hyperbola opens up/down, then the basic form is $\Large \frac{(y-k)^2}{a^2} + \frac{(x-h)^2}{b^2} = 1$

16. anonymous

yup

17. jim_thompson5910

on that same page it says that the asymptotes have a slope of +-a/b focus on say the positive slope slope = 1/2 a/b = 1/2 8/b = 1/2 b = ??

18. anonymous

16

19. jim_thompson5910

$\Large \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$ $\Large \frac{(y-0)^2}{8^2} - \frac{(x-0)^2}{16^2} = 1$ $\Large \frac{y^2}{64} - \frac{x^2}{256} = 1$

20. jim_thompson5910

I had a + in there for some reason. It should have been -

21. anonymous

Ok thanks! I have 4 more questions of similar nature if you dont't mind :)

22. anonymous

I'll make new questions

23. jim_thompson5910

try the next one out and I'll take a look at the work you have so far on it

24. anonymous

Not the same type of question but the same chapter in my book