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

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

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

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

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

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

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

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

GiggleSquid Group TitleBest ResponseYou've already chosen the best response.0
stoppit
 2 years 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
 2 years ago

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

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

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

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

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

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

NotTim Group TitleBest ResponseYou've already chosen the best response.1
Yayy. Real answer.
 2 years 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?
 2 years 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
 2 years ago

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

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

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

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