Question

A railroad tunnel is shaped like a semiellipse as shown below. The height of the tunnel at the center is 58 ft and the vertical clearance must be 29 ft at a point 21 ft from the center. Find an equation for the ellipse.

Answer 1

x^2/588 + y^2/3364 = 1

Explanation:

The equation of ellipse is:

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

The ellipse passes through (0, 58) and (21, 29),then

0^2/a^2 + 58^2/b^2 =1

b^2 = 3364

Also, we have the point (21, 29)

21^2/a^2 + 29^2/3364=1

a^2 = 588

∴Thus, the equation of the ellipse is:

x^2/588 + y^2/3364 = 1

Answer 2

Equation of an ellipse is given by x^2/a^2 + y^2/b^2 = 1

The ellipse passes through (0, 58), (0, -58), (21, 29)

0^2/a^2 + (58)^2/b^2 = 1
b^2 = (58)^2 = 3,364

(21)^2/a^2 + (29)^2/(58)^2 = 1
(21)^2/a^2 = 1 - 841/3364 = 3/4
a^2 = 4(441)/3 = 588

Required equation is x^2/588 + y^2/3,364 = 1