Question

The opening under a bridge is in the shape of a semielliptical arch and has a span of 82 feet. The height of the arch, at a distance of 9 feet from the center, is 28 feet. Find the height of the arch at its center. Hint: let the center of the horizontal span be (0, 0).

Answer 1

28.7 feet

Explanation:

The equation of a parabola is in the form:

`(x^2)/(a^2)+ (y^2)/(b^2)=1\\ \\where\ (\pm a,0)=major\ axis,(0,\pm b)=minor\ axis`

The bridge can be represented as:

Since the width of the bridge is 82 feet, therefore 2a = 82, a = 41

`(x^2)/(41^2) +(y^2)/(b^2)=0\\ \\The\ point\ (9,28)\ lie\ on \ the\ ellipse\ hence\ substituting:\\\\(9^2)/(41^2) +(28^2)/(b^2)=1\\\\0.048+(28^2)/(b^2)=1\\\\(28^2)/(b^2)=1-0.048=0.952\\\\b^2=824\\\\b=√(824)\\ \\b=28.7 \\\\`

The arch is 28.7 feet from its center