Question

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.

Could someone walk me through how to create an equation for this?

Answer 1

start with the basic equation of an ellipse with an under y of 58

Answer 2

\[\frac{x^2}{a^2}+\frac{y^2}{58}=1\]
let x=21 and y = 29 to solve for a^2

Answer 3

might have to be 58^2 under the y :)

Answer 4

Ahh that makes sense. Thank you! I never understood you could plug them back in.

Answer 5

\[\frac{x^2}{a^2}+\frac{y^2}{58^2}=1\]
\[\frac{21^2}{a^2}+\frac{29^2}{58^2}=1\]
solve for a^2
\[\frac{21^2}{a^2}=1-\frac{29^2}{58^2}\]
\[\Large \frac{21^2}{1-\frac{29^2}{58^2}}=a^2\]