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:
- a is the major radius and 2a is the major axis.
- b is the minor radius and 2b is the minor axis.
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).
