Question

Convert ellipse to its standard equation.
x^2-6x+3y^2-6y=0

(the standard equation is (x-h)^2/a^2 - (y-k)^2/b^2 = 1

Answer 1

x^2-6x+9-9+3(y^2-2y+1-1)=0
(x-3)^2-9+3((y-1)^2-1)=0
(x-3)^2-9+3(y-1)^2-3=0
(x-3)^2+3(y-1)^2=12
(x-3)^2/12+3(y-1)^2/12=1
(x-3)^2/12+(y-1)^2/4=1

Answer 2

what I got was:

(x^2-6x) + (3y^2-6y) = 0
(x-3)^2 - 3(y^2+2y) = 0
(x-3)^2 - 3(y+1)^2

how did you get your answer?

Answer 3

Brush up on Completing the Square. You can't just throw numbers in there. You must fix it after you break it.

(x^2-6x) + (3y^2-6y) = 0

(x^2-6x + -----) + (3y^2-6y) = 0 + _____

(x^2-6x + 9) + (3y^2-6y) = 0 + 9

(x-3)^2 + (3y^2-6y) = 9

(x-3)^2 + 3(y^2-2y + _____) = 9 + 3(_____)

(x-3)^2 + 3(y^2-2y + 1) = 9 + 3(1)

(x-3)^2 + 3(y-1)^2 = 9 + 3

(x-3)^2 + 3(y-1)^2 = 12

\(\dfrac{(x-3)^{2}}{12} + \dfrac{(y-1)^{2}}{4} = 1\)