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

rkfitz053
 2 years ago
Best ResponseYou've already chosen the best response.1Start 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?

GiggleSquid
 2 years ago
Best ResponseYou've already chosen the best response.0ya its 1,6 right? thanks!

rkfitz053
 2 years ago
Best ResponseYou've already chosen the best response.1The 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)

GiggleSquid
 2 years ago
Best ResponseYou've already chosen the best response.0how do i find the vertices? i'm better with ellipses :/

rkfitz053
 2 years ago
Best ResponseYou've already chosen the best response.1Your 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

rkfitz053
 2 years ago
Best ResponseYou've already chosen the best response.1The 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.

rkfitz053
 2 years ago
Best ResponseYou've already chosen the best response.1Vertices are the same as ellipses and foci you add instead of subtract.
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