Given the equation of the parabola:
[tex]\[
(x-1)^2 = 8(y+4)
\][/tex]
We recognize that this equation is in the standard form of a vertical parabola:
[tex]\[
(x-h)^2 = 4p(y-k)
\][/tex]
where [tex]\((h, k)\)[/tex] is the vertex and [tex]\(p\)[/tex] is the distance from the vertex to the focus.
To match the given equation with the standard form, we compare:
[tex]\[
(x-1)^2 = 8(y+4)
\][/tex]
with
[tex]\[
(x-h)^2 = 4p(y-k)
\][/tex]
From this comparison, we identify:
- [tex]\(h = 1\)[/tex]
- [tex]\(k = -4\)[/tex]
- [tex]\(4p = 8\)[/tex]
Now, let's solve for [tex]\(p\)[/tex]:
[tex]\[
4p = 8 \implies p = \frac{8}{4} = 2
\][/tex]
The vertex of the parabola is at [tex]\((h, k)\)[/tex], which gives us:
[tex]\[
\text{Vertex} = (1, -4)
\][/tex]
The focus of a parabola in this form [tex]\((x-h)^2 = 4p(y-k)\)[/tex] is located at [tex]\((h, k+p)\)[/tex].
So, the coordinates of the focus are:
[tex]\[
\text{Focus} = (1, -4 + 2) = (1, -2)
\][/tex]
The directrix of the parabola is the line [tex]\(y = k - p\)[/tex].
Substituting the values of [tex]\(k\)[/tex] and [tex]\(p\)[/tex]:
[tex]\[
\text{Directrix} = y = -4 - 2 = -6
\][/tex]
Hence, the focus and directrix of the given parabola are:
[tex]\[
\text{Focus}: (1, -2)
\][/tex]
[tex]\[
\text{Directrix}: y = -6
\][/tex]
