Answer:
64.66 ft
Step-by-step explanation:
The general equation of an ellipse with center (h, k) is:
[tex]\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1[/tex]
Let the center of the bridge at ground level be the center of the ellipse. Place the center of the ellipse (h, k) at the origin (0, 0):
[tex]\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1[/tex]
As the ellipse is horizontal:
Given that the top of the arch of the bridge is 30 feet above the ground, the minor radius is 30 ft, so b = 30:
[tex]\dfrac{x^2}{a^2}+\dfrac{y^2}{30^2}=1\\\\\\\\\dfrac{x^2}{a^2}+\dfrac{y^2}{900}=1[/tex]
Given that the height of the bridge is 15 ft above the ground when it is 28 ft from the center, a point on the ellipse is (28, 15).
Substitute point (28, 15) into the equation to find the value of a:
[tex]\dfrac{28^2}{a^2}+\dfrac{15^2}{900}=1\\\\\\\\\dfrac{784}{a^2}+\dfrac{225}{900}=1\\\\\\\\\dfrac{784}{a^2}+\dfrac{1}{4}=1\\\\\\\\\dfrac{784}{a^2}=\dfrac{3}{4}\\\\\\\\a^2=\dfrac{784\cdot 4}{3}\\\\\\\\a^2=\dfrac{3136}{3}[/tex]
Therefore, the equation of the ellipse is:
[tex]\dfrac{3x^2}{3136}+\dfrac{y^2}{900}=1[/tex]
The span of the bridge is represented by the major axis (2a). Therefore:
[tex]2a=2\cdot \sqrt{\dfrac{3136}{3}}\\\\\\2a=2\cdot \dfrac{56\sqrt{3}}{3}\\\\\\2a= \dfrac{112\sqrt{3}}{3}\\\\\\2a\approx 64.66\; \sf ft \;(2\;d.p.)[/tex]
So, the span of the bridge should be 64.66 ft (rounded to two decimal places).
