Find the vertex, focus, and directrix of the parabola.
x^2 - 2x + 8y + 9 = 0
x^2 - 2x +1 = -8y + 9+ 1
(x-1)^2 = -8(y-1.25)
vertex:(1,1.25)
focus:(1,-.75)
directrix: y=3.25
it looks good to me
My teacher said that I had sign errors. I do not see where I went wrong.
I don't either
To find the vertex, focus, and directrix of the given parabola, you can start by putting the equation in the standard form for the equation of a parabola: (x-h)^2 = 4p(y-k), where (h,k) is the vertex and p is the distance from the vertex to the focus and directrix.
First, let's rewrite the equation to isolate y:
x^2 - 2x + 8y + 9 = 0
x^2 - 2x = -8y - 9
(x^2 - 2x + 1) = -8(y + 9/8)
(x - 1)^2 = -8(y + 9/8)
Comparing this to the standard form, we have h = 1 and k = -9/8.
The vertex is given by the coordinates (h,k), so the vertex of the parabola is (1, -9/8) or (1, 1.25).
To find the focus and directrix, we need to determine the value of p. From the standard form, we see that 4p = -8, so p = -2.
For the focus, add p to the y-coordinate of the vertex, giving us (1, -9/8 - 2) = (1, -25/8) or (1, -3.125).
For the directrix, subtract p from the y-coordinate of the vertex, resulting in y = (-9/8) - (-2) = -9/8 + 2 = 7/8. Therefore, the equation of the directrix is y = 7/8.
In summary, the vertex of the parabola is at (1, 1.25), the focus is at (1, -3.125), and the directrix is y = 7/8.