To find the equation for the parabolic shape of the gate, we start by understanding the geometry of the gate. The gate is a parabola that is 80 feet wide and 25 feet tall.
### Step-by-Step Solution:
1. Vertex of the Parabola:
We can assume that the highest point of the arch, which is the vertex of the parabola, is located at the origin with a vertical translation. This gives us the vertex at [tex]\((0, 25)\)[/tex], where 25 is the maximum height.
2. Standard Form of the Parabola:
The general form for a parabola that opens downwards is [tex]\(x^2 = -4p(y - k)\)[/tex], where [tex]\((0, k)\)[/tex] is the vertex.
3. Substitute the Vertex:
Plugging in the vertex [tex]\((0, 25)\)[/tex], the equation becomes:
[tex]\[
x^2 = -4p(y - 25)
\][/tex]
4. Using the Width Information:
The total width of the gate is 80 feet, so at the base of the gate when [tex]\(y = 0\)[/tex], the [tex]\(x\)[/tex]-coordinates will be [tex]\(\pm 40\)[/tex] feet (half-width on each side of the origin).
5. Solve for 'p':
Substitute [tex]\(y = 0\)[/tex] and [tex]\(x = \pm 40\)[/tex] into the equation [tex]\(x^2 = -4p(y - 25)\)[/tex]:
[tex]\[
40^2 = -4p(0 - 25)
\][/tex]
Simplify and solve for [tex]\(p\)[/tex]:
[tex]\[
1600 = 100p
\][/tex]
[tex]\[
p = 16
\][/tex]
6. Final Equation:
With [tex]\(p = 16\)[/tex], the equation of the parabola becomes:
[tex]\[
x^2 = -64(y - 25)
\][/tex]
Therefore, the equation that describes the parabolic shape of the gate is:
[tex]\[
\boxed{x^2 = -64(y - 25)}
\][/tex]
So, the correct option is:
C. [tex]\(x^2 = -64(y - 25)\)[/tex]
