I need help finding the center and latus rectum of the equation: 3y^2 - 2y + x + 1 = 0
Ah, searching for the center and latus rectum, huh? That sounds like quite a mathematical quest! Well, fear not, for I, the Clown Bot, am here to assist you. Let's tackle this together!
First things first, let's rearrange the equation to a more standard form. We have 3y^2 - 2y + x + 1 = 0. To make it easier, let's group the terms with x and the constant separately:
x = -3y^2 + 2y - 1
Now, we can identify the coefficients of y^2 and y, which are -3 and 2 respectively. To find the x-coordinate of the center, we can use the formula:
x = -h
where h is the coefficient of y divided by 2. In this case, h = 2/(-6) = -1/3. So, the x-coordinate of the center is x = 1/3.
As for the latus rectum, we know that for a parabola in the form x = ay^2, the length of the latus rectum is given by 4a. In this case, a = -3, so the length of the latus rectum is 4(-3) = -12. However, since length cannot be negative, we take the absolute value and get 12.
So, in summary:
- The center of the parabola is (1/3, something), where "something" represents the y-coordinate.
- The length of the latus rectum is 12.
Remember, math can be a circus at times, but with a little clowning around, we'll solve any problem!
To find the center and latus rectum of the given equation, we need to rewrite it in the standard form of a parabola.
The standard form of a parabola is (y - k)^2 = 4a(x - h), where (h, k) represents the coordinates of the vertex, and 4a represents the length of the latus rectum.
Let's begin by completing the square for the given equation:
3y^2 - 2y + x + 1 = 0
Rearrange the terms:
3y^2 - 2y = -x - 1
Divide the equation by 3 to make the coefficient of y^2 equal to 1:
y^2 - (2/3)y = -(1/3)x - 1/3
Now, we complete the square for the y terms. To do this, we take half of the coefficient of y and square it. Then, we add and subtract the result within the parentheses:
y^2 - (2/3)y + (1/3)^2 - (1/3)^2 = -(1/3)x - 1/3 + (1/3)^2
Simplify the expression within the parentheses:
y^2 - (2/3)y + (1/9) = -(1/3)x - 1/3 + 1/9
y^2 - (2/3)y + (1/9) = -(1/3)x - 3/9 + 1/9
y^2 - (2/3)y + (1/9) = -(1/3)x - 2/9
Now, we can rewrite the equation in the standard form:
(y - 1/3)^2 = -(1/3)(x + 2/9)
Comparing with the standard form (y - k)^2 = 4a(x - h), we can find the values of the vertex and latus rectum:
The vertex coordinates (h, k) are (-2/9, 1/3). So, the center of the parabola is (-2/9, 1/3).
The value of 4a is 1/3. The latus rectum is the distance between the vertex and the directrix, which is equal to 4a.
Therefore, the length of the latus rectum is 4 * (1/3) = 4/3.
In summary:
- The center of the parabola is (-2/9, 1/3).
- The length of the latus rectum is 4/3.
To find the center and latus rectum of a given equation, we need to determine its conic section. Let's start by rearranging the given equation:
3y^2 - 2y + x + 1 = 0
We can rewrite the equation in the standard form of a conic section by grouping the x and y terms:
x = -3y^2 + 2y - 1
Now, we can see that the equation represents a parabola because it is a quadratic equation in y. However, we need to confirm whether it is a vertical or horizontal parabola.
To determine the orientation of the parabola, let's look at the coefficient of the y^2 term. In this case, the coefficient is positive (3), indicating a vertical parabola.
Now, let's find the vertex of the parabola, which represents the center of the conic section. The vertex of a parabola can be found using the formula:
h = -b/2a
k = f(h), where f(h) is the corresponding x-value when x = h in the equation.
In our case, the equation is already in vertex form, so we can directly identify the values of h and k. Comparing it to the standard vertex form, we have:
(x - h)^2 = 4a(y - k)
Therefore, h = -1 and k = 1/4.
So, the center of the parabola is at (-1, 1/4).
Next, to find the latus rectum, we can use the formula:
latus rectum = 4a
In this case, a = 1/4, so
latus rectum = 4 * (1/4) = 1
Therefore, the latus rectum of the given parabola is 1 unit.
To summarize:
- The center of the parabola is (-1, 1/4).
- The latus rectum of the parabola is 1 unit.
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)