6.4 Q-8

Question Statement

A parabolic arch has a base of 100 meters and a height of 25 meters. Find the height of the arch at a point 30 meters from the center of the base.

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Background and Explanation

This problem involves finding the height of a parabolic arch at a specific horizontal distance from its center. The parabola is symmetric about its vertical axis, with the vertex (lowest point) at the origin.

Key points:

1. Equation of the Parabola: The standard form of a parabola opening upwards is:

$$x^2 = 4ay$$

where $a$ is the focal length. 2. The parabola’s dimensions are determined using the point corresponding to the vertex height and half the base width.

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Solution

Step 1: Define the Coordinate System

• Place the vertex of the parabola at the origin $(0, 0)$.
• The parabola opens upwards with the equation:

$$x^2 = 4ay$$

• The total base width is 100 meters, so the endpoints of the base are at $(-50, 0)$ and $(50, 0)$.
• The height of the arch at its center (vertex) is 25 meters.

Step 2: Use the Point $(50, 25)$ to Determine $a$

The point $(50, 25)$ lies on the parabola. Substituting into the equation $x^2 = 4ay$:

$$50^2 = 4a(25)$$

Simplify:

$$2500 = 100a$$

Solve for $a$:

$$a = 25$$

Step 3: Write the Equation of the Parabola

Substitute $a = 25$ into the standard equation:

$$x^2 = 4(25)y$$

Simplify:

$$x^2 = 100y$$

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Step 4: Find the Height at $x = 30$

Substitute $x = 30$ into the equation $x^2 = 100y$:

$$30^2 = 100y$$

Simplify:

$$900 = 100y$$

Solve for $y$:

$$y = 9 , \mathrm{m}$$

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Final Answer

The height of the arch at a point 30 meters from the center is:

$$\boxed{9 , \mathrm{m}}$$

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Key Formulas or Methods Used

1. Standard Equation of a Parabola:

$$x^2 = 4ay$$

2. Substitution: Use the known point $(50, 25)$ to calculate $a$, the focal distance.

3. Symmetry of the Parabola: The parabola is symmetric about the $y$-axis, so calculations for positive $x$ apply equally for negative $x$.

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Summary of Steps

1. Set up the parabola equation with the vertex at the origin.
2. Use the known dimensions of the arch (base width and height) to find the focal length $a$.
3. Write the equation of the parabola.
4. Substitute $x = 30$ into the equation to find the corresponding height $y$.
5. Solve for $y$ to determine the height of the arch.