Alright, let's identify the focus and directrix of the graph of the equation [tex]\( y = -2x^2 \)[/tex], given that the vertex is located at the origin.
First, recall that the standard form of a vertical parabola 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 and also from the vertex to the directrix.
Given the equation [tex]\( y = -2x^2 \)[/tex], we can rewrite it as
[tex]\[
x^2 = -\frac{1}{2}y
\][/tex]
Now, compare this with the standard form [tex]\( (x - h)^2 = 4p(y - k) \)[/tex]:
- Here, [tex]\(h = 0\)[/tex] and [tex]\(k = 0\)[/tex] since the vertex is at the origin.
- We have [tex]\((x - 0)^2 = 4p(y - 0)\)[/tex]
So, we have
[tex]\[
x^2 = 4p \cdot y
\][/tex]
Comparing this with [tex]\( x^2 = -\frac{1}{2}y \)[/tex]:
[tex]\[
4p = -\frac{1}{2}
\][/tex]
Solving for [tex]\(p\)[/tex]:
[tex]\[
p = -\frac{1}{8}
\][/tex]
The focus [tex]\((h, k + p)\)[/tex] will be:
[tex]\[
(0, 0 + p) = (0, -\frac{1}{8})
\][/tex]
The directrix, given by [tex]\( y = k - p \)[/tex], will be:
[tex]\[
y = 0 - (-\frac{1}{8}) = \frac{1}{8}
\][/tex]
Thus, the focus is [tex]\((0, -\frac{1}{8})\)[/tex] and the directrix is [tex]\( y = \frac{1}{8} \)[/tex].
So, the correct answer is:
[tex]\[\boxed{F(0, -\frac{1}{8}), y = \frac{1}{8}}\][/tex]
