## GiggleSquid 3 years ago Find the center of this hyperbola: -4x^2+y^2-8x-12y+36=0

1. GiggleSquid

ya, but this equasion seems to break half-way through

2. NotTim

arghhh don't complex me with the simplicities of maths!

3. GiggleSquid

I try my best.

4. GiggleSquid

honestly though, can someone work through this with me?

5. NotTim

bahhahahaha

6. NotTim
7. GiggleSquid

go away. lol

8. NotTim
9. GiggleSquid

stoppit

10. NotTim
11. GiggleSquid

dont you freaken post that....just backspace now

12. NotTim

no more?

13. GiggleSquid

14. NotTim

I'm all out.

15. GiggleSquid

then no medal for you...

16. GiggleSquid

ZZ, i need help on like 4 hyperbola Q's, cause i can do the ellipses

17. NotTim
18. NotTim
19. GiggleSquid

LOL! "Do you mean a parabola? the center of hyperbola is the letter r. " thats what chacha said

20. NotTim

EEK

21. zzr0ck3r

sec

22. zzr0ck3r

-4x^2+y^2-8x+36 = 0 -4x^2-8x-4+y^2-12y+36+4=0 -4(x^2+2x+1) + (y-6)^2 = -4 (x+1)^2^2+(y-6)^2 = 1

23. zzr0ck3r

got the rest?

24. NotTim

25. GiggleSquid

so its (x+1)^2+(y-6)^2=1 How do i get the center ? -1,6?

26. zzr0ck3r

yep

27. GiggleSquid

ok, so how do i get vertices? now in standard form: (x+1)^2/1 - (y-6)^2/4=1

28. zzr0ck3r

where did th e/4 come from?

29. GiggleSquid

thats what it gives me though. (x+1)^2/1 - (y-6)^2/4 how do i find the vertices?

30. zzr0ck3r

o yeah sorry I forgot something -4x^2+y^2-8x+36 = 0 -4x^2-8x-4+y^2-12y+36+4=0 -4(x^2+2x+1) + (y-6)^2 = -4 (x+1)^2+(y-6)^2/-4 = (x+1)^2-(y-6)^2/4=1

31. zzr0ck3r

I forget vertices, one sec

32. GiggleSquid

no no its cool, its kinda hard without seeing it

33. GiggleSquid

i need help finding the Foci for that problem