Question

question an arch has the shape of a semi - ellipse. the arch has a height of 8 feet and a span of 36 feet. find an equation for the ellipse and use it to find the height, in feet, of the arch at a distance of 2 feet from the center. (round your answer to the nearest hundredth if necessary.) provide your answer below: ft

Explanation:

Step1: Determine the semi - major and semi - minor axes

The span of the semi - ellipse is 36 feet, so the length of the major axis $2a = 36$, then $a=18$. The height of the semi - ellipse is 8 feet, so $b = 8$. The standard form of the equation of an ellipse centered at the origin is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1$. Substituting $a = 18$ and $b = 8$, we get $\frac{x^{2}}{18^{2}}+\frac{y^{2}}{8^{2}}=1$, or $\frac{x^{2}}{324}+\frac{y^{2}}{64}=1$.

Step2: Find the height at $x = 2$

We substitute $x = 2$ into the equation $\frac{x^{2}}{324}+\frac{y^{2}}{64}=1$. So $\frac{2^{2}}{324}+\frac{y^{2}}{64}=1$. First, simplify $\frac{4}{324}=\frac{1}{81}$. Then the equation becomes $\frac{1}{81}+\frac{y^{2}}{64}=1$. Rearrange to solve for $y^{2}$: $\frac{y^{2}}{64}=1-\frac{1}{81}=\frac{81 - 1}{81}=\frac{80}{81}$. Then $y^{2}=\frac{80\times64}{81}$. Take the square root of both sides: $y=\sqrt{\frac{80\times64}{81}}=\frac{\sqrt{80\times64}}{9}=\frac{\sqrt{5120}}{9}\approx\frac{71.55}{9}\approx7.95$.

Answer:

$7.95$