Here's the question you clicked on:
GiggleSquid
Find the center of this hyperbola: -4x^2+y^2-8x-12y+36=0
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!
honestly though, can someone work through this with me?
http://www.dummies.com/how-to/content/how-to-graph-a-hyperbola.html
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
more please actually
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
-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
so its (x+1)^2+(y-6)^2=1 How do i get the center ? -1,6?
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