Chapter 10.3, Problem 76AYU

To determine

To find: The span of the bridge if the height 28 feet from the center is to be 13 feet.

Answer to Problem 76AYU

Span of the bridge is $\frac{x^2}{400}+\frac{y^2}{400}=1$ .

Explanation of Solution

Given:A bridge over a highway is in the form of half an ellipse.

It is given that the arch of abridge over a highway is in the form of ellipse and top of the arch is 20 feet above from the ground level. The highway has four lanes, each 12 feet wide; a center safety strip 8 feet wide; and two side strips, each four feet wide.

Now the span of bridge is

  $\begin{array}{l}h=0\\ k=0\\ a=20\\ b=20\end{array}$

Now equation of the ellipse is

  $\begin{array}{l}\frac{x^2}{{20}^2}+\frac{y^2}{{20}^2}=1\\ \frac{x^2}{400}+\frac{y^2}{400}=1\end{array}$

Hence, span of the bridge is $\frac{x^2}{400}+\frac{y^2}{400}=1$ .