I need help finding the center and latus rectum of the equation: 3y^2 - 2y + x + 1 = 0

Answer 1

3 y^2 - 2 y = - x -1

y^2 -(2/3)y = -x/3 -1/3

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

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

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

Answer 2

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