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An engineer is designing an arch-shaped gate for the entrance to an amusement park. The gate must be 80 feet wide and 25 feet tall. What will be the equation of the parabolic shape of the gate?

A. [tex]x^2=-16(y-25)[/tex]
B. [tex](x-16)^2=-4(y-25)[/tex]
C. [tex]x^2=-64(y-25)[/tex]
D. [tex](x-25)^2=-16(y-16)[/tex]
E. [tex]x^2=-40(y-25)[/tex]

Sagot :

To determine the equation of the parabolic shape of the gate, we need to find the appropriate parabolic equation given the dimensions: 80 feet wide and 25 feet tall.

We start by noting that the width of 80 feet means the gate extends 40 feet on either side of the origin, i.e., from [tex]\( x = -40 \)[/tex] to [tex]\( x = 40 \)[/tex].
The highest point of the arch (the vertex) will be at [tex]\( (0, 25) \)[/tex], since the gate is 25 feet tall.

The vertex form of a parabola is given by:

[tex]\[ y = a(x - h)^2 + k \][/tex]

Here, [tex]\( (h, k) \)[/tex] is the vertex. Substituting the coordinates of the vertex, we get:

[tex]\[ y = a(x - 0)^2 + 25 \][/tex]
[tex]\[ y = ax^2 + 25 \][/tex]

We also know that at the points where the gate meets the ground, [tex]\( y = 0 \)[/tex] (the base of the gate). This happens at [tex]\( x = \pm 40 \)[/tex]. So we use these points to find [tex]\( a \)[/tex]. Substitute [tex]\( x = 40 \)[/tex] and [tex]\( y = 0 \)[/tex] into the equation:

[tex]\[ 0 = a(40)^2 + 25 \][/tex]
[tex]\[ 0 = 1600a + 25 \][/tex]

To solve for [tex]\(a\)[/tex]:

[tex]\[ 1600a = -25 \][/tex]
[tex]\[ a = -\frac{25}{1600} \][/tex]
[tex]\[ a = -\frac{1}{64} \][/tex]

Thus, the equation of the parabola in vertex form is:

[tex]\[ y = -\frac{1}{64}x^2 + 25 \][/tex]

To convert this into the standard parabolic form [tex]\( x^2 = 4p(y - k) \)[/tex]:

[tex]\[ y - 25 = -\frac{1}{64}x^2 \][/tex]

Multiplying both sides by [tex]\(-64\)[/tex] to eliminate the fraction:

[tex]\[ -64(y - 25) = x^2 \][/tex]

Or, rewriting it:

[tex]\[ x^2 = -64(y - 25) \][/tex]

Thus, the equation of the parabolic shape of the gate is:

[tex]\[ x^2 = -64(y - 25) \][/tex]

Therefore, the correct answer is:

C. [tex]\( x^2 = -64(y - 25) \)[/tex]