Question

An elliptical arch of a bridge has a horizontal span of 50 ft and a maximum height of 20 ft at the center. The bridge road is constructed 4 ft above the highest point of the arch.
If a point on the arch is 15 ft away from the center horizontally, find the vertical distance of this point from the bridge road.

• A. 5 ft
• B. 8 ft
• C. 10 ft
• D. 24 ft
• E. None of these

Hint:

In problems involving geometric shapes like ellipses or parabolas, the first step is always to set up a coordinate system and write down the standard equation. Placing the center at the origin (0,0) often simplifies calculations. Pay close attention to what each dimension in the problem represents (e.g., span vs. semi-axis).

Answer 1

The Correct Option is B

Step 1: Understanding the Question: We need to model the elliptical arch with a mathematical equation.
The horizontal span gives us the major axis, and the maximum height gives us the semi-minor axis.
We then need to find the height of the arch at a specific horizontal distance from the center.
Finally, we calculate the difference between the height of the road and the height of the arch at that point.

Step 2: Key Formula or Approach:
The standard equation of an ellipse centered at the origin (0,0) is:
[ frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ]
Here, (2a) is the length of the major axis (horizontal span) and (b) is the length of the semi-minor axis (maximum height).

Step 3: Detailed Explanation:
1. Determine the parameters of the ellipse:
The horizontal span is 50 ft, which corresponds to the major axis (2a).
[ 2a = 50 implies a = 25 , text{ft} ]
The maximum height of the arch is 20 ft, which corresponds to the semi-minor axis (b).
[ b = 20 , text{ft} ] So, the equation of the elliptical arch is:
[ frac{x^2}{25^2} + frac{y^2}{20^2} = 1 ]
2. Find the height of the arch at the given point:
We need to find the height (y) of the arch at a point that is 15 ft away from the center horizontally. This means we set (x = 15).
[ frac{15^2}{25^2} + frac{y^2}{20^2} = 1 ]
[ frac{225}{625} + frac{y^2}{400} = 1 ]
Simplifying the fraction: [ frac{9 times 25}{25 times 25} = frac{9}{25} ]
So the equation becomes:
[ frac{9}{25} + frac{y^2}{400} = 1 ]
Now, solve for (y^2): [ frac{y^2}{400} = 1 - frac{9}{25} = frac{25 - 9}{25} = frac{16}{25} ]
[ y^2 = 400 times frac{16}{25} = (16 times 25) times frac{16}{25} = 16 times 16 = 256 ]
[ y = sqrt{256} = 16 , text{ft} ]
This is the vertical height of the arch 15 ft from the center.
3. Determine the height of the bridge road:
The highest point of the arch is its maximum height, which is (b = 20) ft.
The bridge road is constructed 4 ft above this highest point.
[ text{Road Height} = 20 , text{ft} + 4 , text{ft} = 24 , text{ft} ]
4. Calculate the required vertical distance:
The vertical distance of the point on the arch from the bridge road is the difference between the road's height and the arch's height at (x = 15).
[ text{Distance} = text{Road Height} - y = 24 , text{ft} - 16 , text{ft} = 8 , text{ft} ]

Step 4: Final Answer:
The vertical distance of the point from the bridge road is 8 ft.