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GiggleSquid
Group Title
Find the center of this hyperbola: 4x^2+y^28x12y+36=0
 one year ago
 one year ago
GiggleSquid Group Title
Find the center of this hyperbola: 4x^2+y^28x12y+36=0
 one year ago
 one year ago

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NotTim Group TitleBest ResponseYou've already chosen the best response.1
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

GiggleSquid Group TitleBest ResponseYou've already chosen the best response.0
ya, but this equasion seems to break halfway through
 one year ago

NotTim Group TitleBest ResponseYou've already chosen the best response.1
arghhh don't complex me with the simplicities of maths!
 one year ago

GiggleSquid Group TitleBest ResponseYou've already chosen the best response.0
I try my best.
 one year ago

GiggleSquid Group TitleBest ResponseYou've already chosen the best response.0
honestly though, can someone work through this with me?
 one year ago

NotTim Group TitleBest ResponseYou've already chosen the best response.1
bahhahahaha
 one year ago

NotTim Group TitleBest ResponseYou've already chosen the best response.1
http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx
 one year ago

GiggleSquid Group TitleBest ResponseYou've already chosen the best response.0
go away. lol
 one year ago

NotTim Group TitleBest ResponseYou've already chosen the best response.1
http://www.dummies.com/howto/content/howtographahyperbola.html
 one year ago

GiggleSquid Group TitleBest ResponseYou've already chosen the best response.0
stoppit
 one year ago

NotTim Group TitleBest ResponseYou've already chosen the best response.1
http://www.khanacademy.org/math/algebra/conicsections?k?k http://www.khanacademy.org/math/algebra/conicsections?k http://www.khanacademy.org/math/algebra/conicsections/v/fociofahyperbola http://www.khanacademy.org/math/algebra/conicsections/v/proofhyperbolafoci
 one year ago

GiggleSquid Group TitleBest ResponseYou've already chosen the best response.0
dont you freaken post that....just backspace now
 one year ago

GiggleSquid Group TitleBest ResponseYou've already chosen the best response.0
more please actually
 one year ago

NotTim Group TitleBest ResponseYou've already chosen the best response.1
I'm all out.
 one year ago

GiggleSquid Group TitleBest ResponseYou've already chosen the best response.0
then no medal for you...
 one year ago

GiggleSquid Group TitleBest ResponseYou've already chosen the best response.0
ZZ, i need help on like 4 hyperbola Q's, cause i can do the ellipses
 one year ago

NotTim Group TitleBest ResponseYou've already chosen the best response.1
http://www.chacha.com/question/howdoyoufindthecenterofahyperbola http://www.chacha.com/question/howtoyoufindthefociinamathproblem
 one year ago

NotTim Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

GiggleSquid Group TitleBest ResponseYou've already chosen the best response.0
LOL! "Do you mean a parabola? the center of hyperbola is the letter r. " thats what chacha said
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.2
4x^2+y^28x+36 = 0 4x^28x4+y^212y+36+4=0 4(x^2+2x+1) + (y6)^2 = 4 (x+1)^2^2+(y6)^2 = 1
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.2
got the rest?
 one year ago

NotTim Group TitleBest ResponseYou've already chosen the best response.1
Yayy. Real answer.
 one year ago

GiggleSquid Group TitleBest ResponseYou've already chosen the best response.0
so its (x+1)^2+(y6)^2=1 How do i get the center ? 1,6?
 one year ago

GiggleSquid Group TitleBest ResponseYou've already chosen the best response.0
ok, so how do i get vertices? now in standard form: (x+1)^2/1  (y6)^2/4=1
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.2
where did th e/4 come from?
 one year ago

GiggleSquid Group TitleBest ResponseYou've already chosen the best response.0
thats what it gives me though. (x+1)^2/1  (y6)^2/4 how do i find the vertices?
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.2
o yeah sorry I forgot something 4x^2+y^28x+36 = 0 4x^28x4+y^212y+36+4=0 4(x^2+2x+1) + (y6)^2 = 4 (x+1)^2+(y6)^2/4 = (x+1)^2(y6)^2/4=1
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.2
I forget vertices, one sec
 one year ago

GiggleSquid Group TitleBest ResponseYou've already chosen the best response.0
no no its cool, its kinda hard without seeing it
 one year ago

GiggleSquid Group TitleBest ResponseYou've already chosen the best response.0
i need help finding the Foci for that problem
 one year ago
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