anonymous
  • anonymous
Find the center of this hyperbola: -4x^2+y^2-8x-12y+36=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
NotTim
  • NotTim
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.
anonymous
  • anonymous
ya, but this equasion seems to break half-way through
NotTim
  • NotTim
arghhh don't complex me with the simplicities of maths!

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More answers

anonymous
  • anonymous
I try my best.
anonymous
  • anonymous
honestly though, can someone work through this with me?
NotTim
  • NotTim
bahhahahaha
NotTim
  • NotTim
http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx
anonymous
  • anonymous
go away. lol
NotTim
  • NotTim
http://www.dummies.com/how-to/content/how-to-graph-a-hyperbola.html
anonymous
  • anonymous
stoppit
NotTim
  • NotTim
http://www.khanacademy.org/math/algebra/conic-sections?k?k http://www.khanacademy.org/math/algebra/conic-sections?k http://www.khanacademy.org/math/algebra/conic-sections/v/foci-of-a-hyperbola http://www.khanacademy.org/math/algebra/conic-sections/v/proof--hyperbola-foci
anonymous
  • anonymous
dont you freaken post that....just backspace now
NotTim
  • NotTim
no more?
anonymous
  • anonymous
more please actually
NotTim
  • NotTim
I'm all out.
anonymous
  • anonymous
then no medal for you...
anonymous
  • anonymous
ZZ, i need help on like 4 hyperbola Q's, cause i can do the ellipses
NotTim
  • NotTim
http://www.chacha.com/question/how-do-you-find-the-center-of-a-hyperbola http://www.chacha.com/question/how-to-you-find-the-foci-in-a-math-problem
NotTim
  • NotTim
http://answers.yahoo.com/question/index;_ylt=AlJSRWoxoFIzUov5Y_cVRD0jzKIX;_ylv=3?qid=20070301195802AAf0iGI http://answers.yahoo.com/question/index;_ylt=AtB52KBTjlHnp7_4iNEWfNMjzKIX;_ylv=3?qid=20120502105534AA7PoIm http://answers.yahoo.com/question/index;_ylt=AlqXH42YECjvzFYn1jmbZ.0jzKIX;_ylv=3?qid=20090118162832AAvNLzR http://answers.yahoo.com/question/index;_ylt=AkvVrtHbT4PqKgZZLMgraTkjzKIX;_ylv=3?qid=20090118154214AAAqyjS http://answers.yahoo.com/question/index;_ylt=Alc7Vf0UvFo6bvnrZBJ6fsAjzKIX;_ylv=3?qid=20101122130528AAL40iV
anonymous
  • anonymous
LOL! "Do you mean a parabola? the center of hyperbola is the letter r. " thats what chacha said
NotTim
  • NotTim
EEK
zzr0ck3r
  • zzr0ck3r
sec
zzr0ck3r
  • 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
zzr0ck3r
  • zzr0ck3r
got the rest?
NotTim
  • NotTim
Yayy. Real answer.
anonymous
  • anonymous
so its (x+1)^2+(y-6)^2=1 How do i get the center ? -1,6?
zzr0ck3r
  • zzr0ck3r
yep
anonymous
  • anonymous
ok, so how do i get vertices? now in standard form: (x+1)^2/1 - (y-6)^2/4=1
zzr0ck3r
  • zzr0ck3r
where did th e/4 come from?
anonymous
  • anonymous
thats what it gives me though. (x+1)^2/1 - (y-6)^2/4 how do i find the vertices?
zzr0ck3r
  • 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
zzr0ck3r
  • zzr0ck3r
I forget vertices, one sec
anonymous
  • anonymous
no no its cool, its kinda hard without seeing it
anonymous
  • anonymous
i need help finding the Foci for that problem

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