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

Answer:

28.7 feet

Step-by-step explanation:

The equation of a parabola is in the form:

$\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1\\ \\where\ (\pm a,0)=major\ axis,(0,\pm b)=minor\ axis$

The bridge can be represented as:

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

$\frac{x^2}{41^2} +\frac{y^2}{b^2}=0\\ \\The\ point\ (9,28)\ lie\ on \ the\ ellipse\ hence\ substituting:\\\\\frac{9^2}{41^2} +\frac{28^2}{b^2}=1\\\\0.048+\frac{28^2}{b^2}=1\\\\\frac{28^2}{b^2}=1-0.048=0.952\\\\b^2=824\\\\b=\sqrt{824}\\ \\b=28.7$

The arch is 28.7 feet from its center