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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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.
ya, but this equasion seems to break half-way through
arghhh don't complex me with the simplicities of maths!

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Other answers:

I try my best.
honestly though, can someone work through this with me?
bahhahahaha
http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx
go away. lol
http://www.dummies.com/how-to/content/how-to-graph-a-hyperbola.html
stoppit
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
dont you freaken post that....just backspace now
no more?
more please actually
I'm all out.
then no medal for you...
ZZ, i need help on like 4 hyperbola Q's, cause i can do the ellipses
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
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
LOL! "Do you mean a parabola? the center of hyperbola is the letter r. " thats what chacha said
EEK
sec
-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
got the rest?
Yayy. Real answer.
so its (x+1)^2+(y-6)^2=1 How do i get the center ? -1,6?
yep
ok, so how do i get vertices? now in standard form: (x+1)^2/1 - (y-6)^2/4=1
where did th e/4 come from?
thats what it gives me though. (x+1)^2/1 - (y-6)^2/4 how do i find the vertices?
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
I forget vertices, one sec
no no its cool, its kinda hard without seeing it
i need help finding the Foci for that problem

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