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

Question

nottim · Answer

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 · Answer

ya, but this equasion seems to break half-way through

nottim · Answer

arghhh don't complex me with the simplicities of maths!

anonymous · Answer

I try my best.

anonymous · Answer

honestly though, can someone work through this with me?

nottim · Answer

bahhahahaha

nottim · Answer

http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx

anonymous · Answer

go away. lol

nottim · Answer

http://www.dummies.com/how-to/content/how-to-graph-a-hyperbola.html

anonymous · Answer

stoppit

nottim · Answer

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 · Answer

dont you freaken post that....just backspace now

nottim · Answer

no more?

anonymous · Answer

more please actually

nottim · Answer

I'm all out.

anonymous · Answer

then no medal for you...

anonymous · Answer

ZZ, i need help on like 4 hyperbola Q's, cause i can do the ellipses

nottim · Answer

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 · Answer

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 · Answer

LOL! "Do you mean a parabola? the center of hyperbola is the letter r. "  thats what chacha said

nottim · Answer

EEK

zzr0ck3r · Answer

sec

zzr0ck3r · Answer

-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 · Answer

got the rest?

nottim · Answer

Yayy. Real answer.

anonymous · Answer

so its (x+1)^2+(y-6)^2=1   How do i get the center ? -1,6?

zzr0ck3r · Answer

yep

anonymous · Answer

ok, so how do i get vertices? now in standard form: (x+1)^2/1 - (y-6)^2/4=1

zzr0ck3r · Answer

where did th e/4 come from?

anonymous · Answer

thats what it gives me though.   (x+1)^2/1 - (y-6)^2/4 how do i find the vertices?

zzr0ck3r · Answer

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 · Answer

I forget vertices, one sec

anonymous · Answer

no no its cool, its kinda hard without seeing it

anonymous · Answer

i need help finding the Foci for that problem