anonymous
  • anonymous
Need your help again rkfitz053!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
trying to write it in standard form. i got -4(x^2-8x) + (y^2-12y)=-36 so far
anonymous
  • anonymous
Start by gathering all like terms so the equation looks like this (you got this one.):\[-4x^2-8x+y^2-12y=-36\] Then start by completing the square (take 1/2 of the coefficient on the x term and square that, then add it to the end, make sure you also add it to the right side. And multiply that -4 to the 1 you added) for the x terms and the y terms: \[-4(x^2+2x+1)+(y^2-12y+36)=-4\] Now you have two perfect square trinomials and you can factor those and divide all terms by -4 to end up with: \[(x+1)^2 - (y-6)^2/4=1\] Can you figure out the centre from there?
anonymous
  • anonymous
ya its -1,6 right? thanks!

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anonymous
  • anonymous
Got it. No problem.
anonymous
  • anonymous
The trickiest part is always remembering to multiply what you add by the factor that you took out before (multiplying -4 to 1 when completing the sq)
anonymous
  • anonymous
how do i find the vertices? i'm better with ellipses :/
anonymous
  • anonymous
@rkfitz053
anonymous
  • anonymous
Your vertices are going to be a distance of 'a' away from the center. Remember the std eqn for hyperbolas: \[x^2/a^2-y^2/b^2=1\] so you just have to take the sqrt of what the first term is divided by. In your question a^2=1
anonymous
  • anonymous
the foci?
anonymous
  • anonymous
The foci are almost the same as ellipses too...it's just \[c=\sqrt{a^2+b^2}\] They're added now because the foci points are on the outside of the vertices, rather than inside an ellipse.
anonymous
  • anonymous
Vertices are the same as ellipses and foci you add instead of subtract.
anonymous
  • anonymous
Got it?
anonymous
  • anonymous
got it
anonymous
  • anonymous
good stuff

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