SOLUTION: I know this equation is an ellipse, I'm just trying to figure out how to get it to the ellipse equation: 2x^2+12x+18+y^2=3(2+y^2)+4y

Question 554858: I know this equation is an ellipse, I'm just trying to figure out how to get it to the ellipse equation:
2x^2+12x+18+y^2=3(2+y^2)+4y
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
I know this equation is an ellipse, I'm just trying to figure out how to get it to the ellipse equation:
2x^2+12x+18+y^2=3(2+y^2)+4y
distribute:
2x^2+12x+18+y^2=6+3y^2+4y
combine like terms:
2x^2+12x-2y^2-4y=-12
complete the square:
2(x^2+6x+9)-2(y^2+2y+1)=-12-2+18
2(x+3)^2-2(y+1)^2=4
Divide by 4
(x+3)^2/2-(y+1)^2/2=1
This is an equation of a hyperbola with horizontal transverse axis of the standard form:
(x-h)^2/a^2-(y-k)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
..
note: Before deciding what kind of conic the equation represents, it is a good idea to complete the square first.
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