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

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NotTim
 4 years ago
Best ResponseYou've already chosen the best response.1http://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
 4 years ago
Best ResponseYou've already chosen the best response.0ya, but this equasion seems to break halfway through

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0honestly though, can someone work through this with me?

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

NotTim
 4 years ago
Best ResponseYou've already chosen the best response.1http://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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dont you freaken post that....just backspace now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then no medal for you...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ZZ, i need help on like 4 hyperbola Q's, cause i can do the ellipses

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

NotTim
 4 years ago
Best ResponseYou've already chosen the best response.1http://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
 4 years ago
Best ResponseYou've already chosen the best response.0LOL! "Do you mean a parabola? the center of hyperbola is the letter r. " thats what chacha said

zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.24x^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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so its (x+1)^2+(y6)^2=1 How do i get the center ? 1,6?

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

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

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

zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.2o 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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no no its cool, its kinda hard without seeing it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i need help finding the Foci for that problem
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