Question

A railroad tunnel is shaped like a semi-eclipse. The height of the tunnel is 28ft and the total width is 50ft. Find the vertical clearance at a point 13ft to the right from the center. Please show all work that leads to your answer. Round your final answer to the nearest tenth.

Answer 1

Let the ellipse be modeled by

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

Since the base has width 50, and the height is 28, we have

b = 28
a = 25

and the ellipse is

x^2/25^2 + y^2/28^2 = 1

When x=13, y=23.9

Answer 2

Standard form equation of an ellipse:

x² / a² + y² / b² = 1

where:

a is the semimajor axis

b is the semiminor axis


In this case:

a = total width / 2 = 50ft / 2 = 25ft

b = The height = 28ft


x² / a² + y² / b² = 1

x² / 25² + y² / 28² = 1


The vertical clearance at a point 13ft mean:

x = 13


x² / 25² + y² / 28² = 1

13² / 25² + y² / 28² = 1

Subtract 13² / 25² to both sides

13² / 25² + y² / 28² *- 13² / 25² = 1 - 13² / 25²

y² / 28² = 1 - 13² / 25²

y² / 28² = 25² / 25² - 13² / 25²

y² / 28² = ( 25² - 13² ) / 25²

Multiply both sides by 28²

y² = 28² ∙ ( 25² - 13² ) / 25²

Take the square root of both sides

y = ± √ [ 28² ∙ ( 25² - 13² ) / 25² ]

y = ± √ 28² ∙ √ ( 25² - 13² ) / √ 25²

y = ± 28 ∙ √ ( 25² - 13² ) / 25

y = ± 28 ∙ √ ( 625 - 169 ) / 25

y = ± 28 ∙ √ 456 / 25

y = ± 28 ∙ √ ( 4 ∙ 114 ) / 25

y = ± 28 ∙ √ 4 ∙ √ 114 / 25

y = ± 28 ∙ 2 ∙ √ 114 / 25

y = ± 56 ∙ √ 114 / 25

y = ± 56 ∙ 10.677078252 / 25

y = ± 597.916382112 / 25

y = ± 23.91665528448


y = ± 23.9 ft

rounded to the nearest tenth