Let's determine the focus and directrix of the given parabola equation:
[tex]\[
(x-1)^2 = 8(y+4)
\][/tex]
The standard form of a parabola that opens vertically is:
[tex]\[
(x - h)^2 = 4p(y - k)
\][/tex]
where [tex]\((h, k)\)[/tex] is the vertex of the parabola and [tex]\(p\)[/tex] is the distance from the vertex to the focus (or from the vertex to the directrix).
By comparing the given equation [tex]\((x - 1)^2 = 8(y + 4)\)[/tex] with the standard form [tex]\((x - h)^2 = 4p(y - k)\)[/tex], we can identify the following parameters:
- [tex]\( h = 1 \)[/tex]
- [tex]\( k = -4 \)[/tex]
- The coefficient [tex]\( 4p = 8 \)[/tex], thus [tex]\( p = 2 \)[/tex].
### Focus of the Parabola
The coordinates of the focus for a parabola that opens upwards or downwards are given by:
[tex]\[
(h, k + p)
\][/tex]
Given that [tex]\( h = 1 \)[/tex], [tex]\( k = -4 \)[/tex], and [tex]\( p = 2 \)[/tex], we substitute these values in:
[tex]\[
\text{Focus} = (1, -4 + 2) = (1, -2)
\][/tex]
### Directrix of the Parabola
The equation of the directrix for a parabola that opens upwards or downwards is:
[tex]\[
y = k - p
\][/tex]
Substituting the values [tex]\( k = -4 \)[/tex] and [tex]\( p = 2 \)[/tex]:
[tex]\[
\text{Directrix} = y = -4 - 2 = -6
\][/tex]
### Summary
Thus, the focus of the parabola is at [tex]\((1, -2)\)[/tex] and the directrix is the line [tex]\(y = -6\)[/tex].
[tex]\[
\text{Focus: } (1, -2)
\][/tex]
[tex]\[
\text{Directrix: } y = -6
\][/tex]
