Solve the problem.
-An arch for a bridge over a highway is in the form of a semiellipse. The top of the arch is 35 feet aboveground (the major axis). What should the span of the bridge be (the length of its minor axis)if the height26 feet from the center is to be 12 feet above ground?
A)55.36 ft
B)27.68 ft
C)151.67 ft
D)35.85 ft

Question

Quizplus · Accepted Answer

We can use the equation for a semiellipse:
$\frac{(y-k)^2}{b^2}+\frac{(x-h)^2}{a^2}=1$

where $(h,k)$ is the center of the ellipse, $a$ is the distance from the center to the end of the major axis, $b$ is the distance from the center to the end of the minor axis.

We are told that the top of the arch is 35 feet above ground, so the center of the ellipse is $(0,35)$. We are also told that the height 26 feet from the center is to be 12 feet above ground, so we know that the point $(x,35-12)=(x,23)$ is on the ellipse.

Substituting these values into the equation for a semiellipse, we get:

$\frac{(23-35)^2}{b^2}+\frac{(x-0)^2}{a^2}=1$

Simplifying:

$\frac{12^2}{b^2}+\frac{x^2}{a^2}=1$

We want to solve for $a$ (the length of the minor axis) when $b=26$. We know that the top of the arch is 35 feet above ground, so we also know that $a^2=b^2+(35-0)^2=26^2+35^2=1505$.

Plugging in $b=26$, we get:

$\frac{12^2}{26^2}+\frac{x^2}{1505}=1$

Simplifying:

$x^2=\frac{1505\cdot12^2}{26^2}-1505=\frac{12^2}{26^2}(1505-26^2)=\frac{12^2}{26^2}\cdot411$

Taking the square root:

$x=\sqrt{\frac{12^2}{26^2}\cdot411}=\frac{12}{26}\sqrt{411} \approx 55.36$

So the span of the bridge should be approximately 55.36 feet, or choice A.

Solve the Problem. -An Arch for a Bridge Over a Highway Is in Highway