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

Center is (0,0)

Answer 2

\[\frac{ x^2 }{ a^2 } + \frac{ y^2 }{ b^2 } = 1\]

Where x,y are the coordinates of any point on the ellipse and
a, b are the radius on the x and y axes respectively.

Answer 3

But it doesn't give the width of the tunnel.

Answer 4

If the ellipse is centered on the origin, ( its center at 0,0 ) the equation is the same as I mentioned before ^
Where a is the radius along the x-axis and b is the radius along the y-axis.

Answer 5

But what's the width of the tunnel?

Answer 6

\[\frac{ x^2 }{ a^2 } + \frac{ y^2 }{ 3364 }\]
okaii we got what "b" is.
I got that since 58^2 = 3364 and "b" is the radius along the y-axis.

Answer 7

Are you good till now?

Answer 8

Okay.

Answer 9

Can you find "a" that is the radius along the x-axis?

Answer 10

I don't know how.

Answer 11

@some_someone

Answer 12

yeah, okay well the visual (picture) I attached might help you get a better picture of the situation, but i will be working on this also. But we got to work together and you got to gve it a try also, (I know you dont really know how but you got to try :) ) I cant just do all the work and give you the answer. I basically did half of the problem in finding "b" now you got to find "a" and square it.

Answer 13

b = 58, "a" is unknown, and the point (21,29) is on it
let the equation be
\[\frac{ x^2 }{ a^2 } + \frac{ y^2 }{ 58^2 } = 1\]
\[\frac{ 21^2 }{ a^2 } + \frac{ 29^2 }{ 58^2 } = 1\]
\[\frac{ 441 }{ a^2 } = 1 - \frac{ 841 }{ 3364 }\]
\[\frac{ 441 }{ a^2 } = \frac{ 3 }{ 4 }\]
\[a^2 = 588\]
\[\frac{ x^2 }{ 588 } +\frac{ y^2 }{ 3364 } = 1 \]

Answer 14

@jim_thompson5910
@Mertsj
@Callisto
@hartnn
@zepdrix
IS THIS CORRECT, RIGHT?

Answer 15

yes