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 simplify the equation and identify its properties.
First, let's rearrange the equation:
3y^2 - 2y + x + 1 = 0
To identify the properties, we need to compare this equation to the standard form of the equation of a parabola:
(y - k)^2 = 4a(x - h)
Here, (h, k) represents the coordinates of the vertex, and 4a represents the distance between the vertex and the focus (which is called the latus rectum).
Comparing the two forms, we can see that "x + 1" in our equation matches with "(x - h)" in the standard form, and therefore, "h = -1". Similarly, we can see that "3y^2 - 2y" matches with "(y - k)^2" in the standard form, indicating that the equation represents a parabola with its vertex at (-1, k).
To find the value of k (the y-coordinate of the vertex), we need to complete the square for the y terms. Let's do that:
3y^2 - 2y + x + 1 = 0
3y^2 - 2y = -x - 1
3(y^2 - (2/3)y) = -x - 1
Now, we want to complete the square for the y terms. To do that, we complete the square for the expression inside the parentheses, which is y^2 - (2/3)y.
The coefficient of y is (-2/3), and to complete the square, we need to divide it by 2 and square the result:
(-2/3) ÷ 2 = -1/3
(-1/3)^2 = 1/9
Adding and subtracting 1/9 inside the parentheses, we have:
3(y^2 - (2/3)y + 1/9 - 1/9) = -x - 1
3[(y - 1/3)^2 - 1/9] = -x - 1
Expanding and simplifying:
3(y - 1/3)^2 - 1 = -x - 1
3(y - 1/3)^2 = -x
Now, we can see that our equation matches the standard form:
4a(x - h)
Where "4a" represents the distance between the focus and the vertex (or the latus rectum). In our case, a = 1/3.
Therefore, the latus rectum is 4(1/3) = 4/3.
To find the center of the parabola, we take the values of h and k, which are -1 and 1/3, respectively.
Therefore, the center of the parabola is (-1, 1/3), and the length of the latus rectum is 4/3.