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GiggleSquid

Find the center of this hyperbola: -4x^2+y^2-8x-12y+36=0

  • one year ago
  • one year ago

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  1. NotTim
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    http://answers.yahoo.com/question/index?qid=20070301195802AAf0iGI http://www.purplemath.com/modules/hyperbola2.htm http://www.purplemath.com/modules/hyperbola3.htm Related questions. Try and derive I guess.

    • one year ago
  2. GiggleSquid
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    ya, but this equasion seems to break half-way through

    • one year ago
  3. NotTim
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    arghhh don't complex me with the simplicities of maths!

    • one year ago
  4. GiggleSquid
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    I try my best.

    • one year ago
  5. GiggleSquid
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    honestly though, can someone work through this with me?

    • one year ago
  6. NotTim
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    bahhahahaha

    • one year ago
  7. NotTim
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    http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx

    • one year ago
  8. GiggleSquid
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    go away. lol

    • one year ago
  9. NotTim
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    http://www.dummies.com/how-to/content/how-to-graph-a-hyperbola.html

    • one year ago
  10. GiggleSquid
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    stoppit

    • one year ago
  11. GiggleSquid
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    dont you freaken post that....just backspace now

    • one year ago
  12. NotTim
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    no more?

    • one year ago
  13. GiggleSquid
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    more please actually

    • one year ago
  14. NotTim
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    I'm all out.

    • one year ago
  15. GiggleSquid
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    then no medal for you...

    • one year ago
  16. GiggleSquid
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    ZZ, i need help on like 4 hyperbola Q's, cause i can do the ellipses

    • one year ago
  17. GiggleSquid
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    LOL! "Do you mean a parabola? the center of hyperbola is the letter r. " thats what chacha said

    • one year ago
  18. NotTim
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    EEK

    • one year ago
  19. zzr0ck3r
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    sec

    • one year ago
  20. zzr0ck3r
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    -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

    • one year ago
  21. zzr0ck3r
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    got the rest?

    • one year ago
  22. NotTim
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    Yayy. Real answer.

    • one year ago
  23. GiggleSquid
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    so its (x+1)^2+(y-6)^2=1 How do i get the center ? -1,6?

    • one year ago
  24. zzr0ck3r
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    yep

    • one year ago
  25. GiggleSquid
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    ok, so how do i get vertices? now in standard form: (x+1)^2/1 - (y-6)^2/4=1

    • one year ago
  26. zzr0ck3r
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    where did th e/4 come from?

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

    • one year ago
  28. zzr0ck3r
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    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

    • one year ago
  29. zzr0ck3r
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    I forget vertices, one sec

    • one year ago
  30. GiggleSquid
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    no no its cool, its kinda hard without seeing it

    • one year ago
  31. GiggleSquid
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    i need help finding the Foci for that problem

    • one year ago
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