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GiggleSquidBest ResponseYou've already chosen the best response.0
trying to write it in standard form. i got 4(x^28x) + (y^212y)=36 so far
 one year ago

rkfitz053Best ResponseYou've already chosen the best response.1
Start by gathering all like terms so the equation looks like this (you got this one.):\[4x^28x+y^212y=36\] 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: \[4(x^2+2x+1)+(y^212y+36)=4\] Now you have two perfect square trinomials and you can factor those and divide all terms by 4 to end up with: \[(x+1)^2  (y6)^2/4=1\] Can you figure out the centre from there?
 one year ago

GiggleSquidBest ResponseYou've already chosen the best response.0
ya its 1,6 right? thanks!
 one year ago

rkfitz053Best ResponseYou've already chosen the best response.1
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)
 one year ago

GiggleSquidBest ResponseYou've already chosen the best response.0
how do i find the vertices? i'm better with ellipses :/
 one year ago

rkfitz053Best ResponseYou've already chosen the best response.1
Your vertices are going to be a distance of 'a' away from the center. Remember the std eqn for hyperbolas: \[x^2/a^2y^2/b^2=1\] so you just have to take the sqrt of what the first term is divided by. In your question a^2=1
 one year ago

rkfitz053Best ResponseYou've already chosen the best response.1
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.
 one year ago

rkfitz053Best ResponseYou've already chosen the best response.1
Vertices are the same as ellipses and foci you add instead of subtract.
 one year ago
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