To find the equation of a parabola with a focus and a directrix, we can follow these steps:
1. Determine the vertex of the parabola: The vertex lies halfway between the focus and the directrix. Given the focus [tex]\((-2, -3)\)[/tex] and the directrix [tex]\(y = -5\)[/tex]:
- The midpoint or average of the y-coordinates of the focus and the directrix gives the y-coordinate of the vertex.
[tex]\[
k = \frac{-3 + (-5)}{2} = \frac{-8}{2} = -4
\][/tex]
- The x-coordinate of the vertex is the same as the x-coordinate of the focus.
[tex]\[
h = -2
\][/tex]
Thus, the vertex [tex]\((h, k)\)[/tex] is [tex]\((-2, -4)\)[/tex].
2. Calculate the distance [tex]\( p \)[/tex]: This is the distance from the vertex to either the focus or the directrix.
- The distance between [tex]\(-3\)[/tex] and [tex]\(-4\)[/tex] is:
[tex]\[
p = |-3 - (-4)| = |-3 + 4| = 1
\][/tex]
3. Form the standard equation: The standard form of the equation of a parabola that opens upwards or downwards is:
[tex]\[
y = \frac{1}{4p} (x - h)^2 + k
\][/tex]
Given that [tex]\( p = 1 \)[/tex], [tex]\( h = -2 \)[/tex], and [tex]\( k = -4 \)[/tex]:
- Substitute these values into the standard form equation:
[tex]\[
y = \frac{1}{4 \cdot 1} (x - (-2))^2 + (-4)
\][/tex]
Simplify:
[tex]\[
y = \frac{1}{4} (x + 2)^2 - 4
\][/tex]
Comparing this with the given options:
A) [tex]\( y = \frac{1}{4}(x-2)^2 - 4 \)[/tex]
B) [tex]\( y = \frac{1}{4}(x+2)^2 - 8 \)[/tex]
C) [tex]\( y = \frac{-1}{16}(x+2)^2 - 4 \)[/tex]
D) [tex]\( y = \frac{1}{4}(x+2)^2 - 4 \)[/tex]
The correct equation that matches our derived equation [tex]\( y = \frac{1}{4}(x + 2)^2 - 4 \)[/tex] is option D.
Therefore, the correct answer is:
[tex]\[
\boxed{4}
\][/tex]
