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

To find the center and latus rectum of the equation 3y^2 - 2y + x + 1 = 0, we need to first rewrite it in the standard form of a parabolic equation.

The standard form of a parabolic equation is given by (y - k)^2 = 4a(x - h), where (h, k) represents the coordinates of the vertex or center of the parabola, and 4a represents the length of the latus rectum.

To rewrite the equation in the standard form, let's complete the square for the y terms:

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

Now, we want to create a perfect square trinomial for the y terms. To do this, we take half of the coefficient of y, square it, and add and subtract it within the parentheses:

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

Expanding this expression gives us:

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

Comparing this equation with the standard form, we can see that (h, k) = (1, 1/3) represents the coordinates of the center of the parabola.

To find the length of the latus rectum, we need to determine the value of 4a. In the standard form of a parabolic equation, a represents the distance from the vertex to the focus and the directrix. Hence, the length of the latus rectum is 4a.

From the equation, it is evident that 4a = 3. Therefore, the length of the latus rectum is 3.

In summary, the center of the parabola is at (1, 1/3), and the length of the latus rectum is 3.