Question

A railroad tunnel is shaped like a semiellipse

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

Explanation:

The equation of ellipse is given as:

`(x^2)/(a^2)+(y^2)/(b^2)=1` (1)

Now, from the given information, The ellipse passes through (0, 58), (0, -58), (21, 29), thus equation (1) becomes:

`(0)/(a^2)+((58)^2)/(b^2)=1`

⇒
`b^2=(58)^2`

⇒
`b^2=3364`

Also,
`((21)^2)/(a^2)+((29)^2)/((58)^2)=1`

⇒
`((21)^2)/(a^2)=1-(841)/(3364)`

⇒
`((21)^2)/(a^2)=(3)/(4)`

⇒
`a^2=(4(441))/(3)`

⇒
`a^2=588`

Now, substituting the values of
`a^2 and b^2` in the equation (1), we have

`(x^2)/(588)+(y^2)/(3364)=1`

which is the required equation for ellipse.

Answer 2

y²/58² + x²/b² = 1
(0,58), (-21,29), and (21,29) are points on the ellipse.

29²/58² + 21²/b² = 1
¼ + 21²/b² = 1
21²/b² = ¾
4·21²/3 = b²
b² = 588


y²/3364 + x²/588 = 1