## anonymous one year ago HELP!!!! Find the foci of 16y^2-36x^2-576=0. Then graph the hyperbole.

1. Loser66

move -576 to the right side, what do you get?

2. anonymous

Positive 576

3. Loser66

yes, now, divide both sides by 576, what do you get?

4. anonymous

1/36y^2-1/16x^2

5. Loser66

=??

6. anonymous

=1?

7. Loser66

$\dfrac{y^2}{6^2}-\dfrac{x^2}{4^2}=1$ right?

8. anonymous

yeah

9. Loser66

Hence, its b =6, a =4 ok?

10. anonymous

yes

11. Loser66

foci c, $$c=\sqrt{a^2+b^2}=??$$

12. anonymous

thanks can you stay with me for just a second i need to write this down

13. anonymous

ok is it square root of 52

14. Loser66

You need know where c is. Your equation has y first, hence its graph is vertical, right? Hence foci will be (0, -c) and (0,c) . Yes for c =sqrt 52

15. Loser66

|dw:1437522466412:dw|

16. anonymous

do you put the foci into decimal form for the graph or do you just leave it as c=sqrt52

17. Loser66

|dw:1437522532448:dw|

18. Loser66

leave it as it is.

19. anonymous

okay so I would just put the sqrt52 into the graph vertically and be done with the graphing

20. anonymous

okay im gonna assume its done