inding vertex and focus of parabola y^2 +4y +8x - 12 =0
y^2 + 4 y = -8 x + 12
y^2 + 4 y + 4 = -8 x + 16
(y+2)^2 = -8 (x-2)
vertex at (2,-2) opens left etc
To find the vertex and focus of the given parabola, we can start by rearranging the equation into the standard form of a parabola, which is given by:
(y - k)^2 = 4a(x - h)
where (h, k) represents the coordinates of the vertex, and 'a' represents the distance between the vertex and the focus.
Let's rearrange the given equation accordingly:
y^2 + 4y + 8x - 12 = 0
First, we complete the square for the y terms by adding and subtracting the square of half the coefficient of 'y'. In this case, half of 4 is 2, so we have:
(y^2 + 4y + 4) + 8x - 12 = 4
Now, we can simplify the equation:
(y + 2)^2 + 8x - 8 = 4
Rearranging the terms, we get:
(y + 2)^2 = -8x + 12
Comparing this with the standard form equation, we find:
- h = 0 (since there is no x^2 term in the equation)
- k = -2
- 4a = -8
- a = -2
Therefore, the vertex is located at (h, k) = (0, -2).
To find the focus, we use the formula:
Focus = (h + a, k)
In this case, substituting the values calculated earlier, we get:
Focus = (0 - 2, -2) = (-2, -2)
So, the vertex of the parabola is located at (0, -2), and the focus is located at (-2, -2).