bridge arch, ellipse question

Question

A certain bridge arch is in the shape of half an ellipse 142 feet wide and 29.6 feet high. At what horizontal distance from the center of the arch is the height equal to 14.2 feet?

Answer 1

We can start by finding the equation of the ellipse in standard form, which is:

x^2 / a^2 + y^2 / b^2 = 1

where a is half the width of the ellipse and b is half the height of the ellipse.

Given that the width is 142 feet and the height is 29.6 feet, we have:

a = 142 / 2 = 71 b = 29.6 / 2 = 14.8

So the equation of the ellipse is:

x^2 / 71^2 + y^2 / 14.8^2 = 1

To find the horizontal distance from the center of the arch where the height is equal to 14.2 feet, we need to substitute y = 14.2 into the equation and solve for x.

x^2 / 71^2 + 14.2^2 / 14.8^2 = 1

Simplifying the equation, we get:

x^2 / 71^2 = 1 - 14.2^2 / 14.8^2

x^2 / 71^2 = 0.1083

Taking the square root of both sides, we get:

x / 71 = ±√0.1083

x = ±√0.1083 x 71

x ≈ ±2.58

Since we are only interested in the positive value of x, we have:

x ≈ 2.58 feet



Therefore, the horizontal distance from the center of the arch where the height is equal to 14.2 feet is approximately 2.58 feet.

Answer 2

The equation for the half ellipse is

(x²/71²) + (y²/29.6²) = 1

Set y=14.2 and solve for x.