Solve the problem.
-An arch in the form of a semiellipse is 52 ft wide at the base and has a height of 20 ft. How wide is the archat a height of 12 ft above the base?
A)41.6 ft
B)20.8 ft
C)35.5 ft
D)17.7 ft

Question

Quizplus · Accepted Answer

The equation of an ellipse centered at the origin with a horizontal major axis is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $2a$ is the length of the major axis and $2b$ is the length of the minor axis. For a semiellipse that is 52 ft wide and 20 ft high, $a = 26$ (half the width) and $b = 20$. To find the width at a height of 12 ft, we set $y = 12$ and solve for $x$: $\frac{x^2}{26^2} + \frac{12^2}{20^2} = 1$. Solving for $x$ gives the half-width at that height, and multiplying by 2 gives the total width, which is approximately 41.6 ft.

Solve the Problem. -An Arch in the Form of a Semiellipse Is 52