Question 1171510: Show the mathematics of designing and constrcting a parabolic arch that is 10 meters across the base and 10 meters tall (measured from the ground to the vertex at the bottom of the keystone). Remember: -4a(y-k)=(x-h)^2
a) Place an axis system on this parabola
b) Find the function for this parabola
c) Make a chart of values for functional heights spaced 1 meter apart
d)How far down from the vertex is the focal point?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down the design and construction of the parabolic arch step-by-step.
**a) Placing an Axis System**
* We'll place the vertex of the parabola at the origin (0, 0).
* The x-axis will run along the base of the arch.
* The y-axis will run vertically through the center of the arch.
* The base of the arch is 10 meters across, so the endpoints of the base are at (-5, -10) and (5, -10).
* The arch is 10 meters tall, so the vertex is at (0, 0).
**b) Finding the Function for the Parabola**
* We'll use the standard form of a vertical parabola: -4a(y - k) = (x - h)²
* Vertex (h, k) = (0, 0)
* The equation becomes: -4ay = x²
* We know a point on the parabola is (5, -10). Let's plug this in:
* -4a(-10) = 5²
* 40a = 25
* a = 25/40 = 5/8
* Substitute 'a' back into the equation:
* -4(5/8)y = x²
* -5/2 y = x²
* y = (-2/5)x²
**c) Chart of Values for Functional Heights (1 Meter Spacing)**
| x (meters) | y (meters) |
| :--------- | :--------- |
| -5 | -10 |
| -4 | -6.4 |
| -3 | -3.6 |
| -2 | -1.6 |
| -1 | -0.4 |
| 0 | 0 |
| 1 | -0.4 |
| 2 | -1.6 |
| 3 | -3.6 |
| 4 | -6.4 |
| 5 | -10 |
**d) Focal Point**
* The distance from the vertex to the focal point is 'a'.
* We found a = 5/8 meters.
* Therefore, the focal point is 5/8 meters (0.625 meters) below the vertex.
**Summary**
* **Axis System:** Vertex at (0, 0), base endpoints at (-5, -10) and (5, -10).
* **Parabola Function:** y = (-2/5)x²
* **Chart of Values:** (See table above)
* **Focal Point:** 5/8 meters (0.625 meters) below the vertex.
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