The height of the arch is about 6.71 feet when measured from its base at a distance of one foot.
This is further explained below.
Generally, An ellipse is a collection of points whose distances from a fixed point (focus) or fixed line (directrix) is constant and less than 1.
Depending on whether the transverse axis is horizontal or vertical, we may construct an equation for the ellipse. The main axis is located along the transverse axis. The ellipse may be represented using the following equations:
[tex]\begin{aligned}&\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \\&\frac{(y-k)^2}{a^2}+\frac{(x-h)^2}{b^2}=1\end{aligned}[/tex]
An elliptical arch is described as having a center height of 9 feet and a width of 6 feet. The precise height of the arch from the base has to be determined.
[tex]\frac{(y-k)^2}{a^2}+\frac{(x-h)^2}{b^2}=1[/tex]
The center (h, k) is where the graph is since we centered it at the origin (0,0). We only need to enter the numbers into the equation at this point.
[tex]\begin{aligned}\frac{(y-0)^2}{9^2}+\frac{(x-0)^2}{3^2} &=1 \\\frac{y^2}{81}+\frac{x^2}{9} &=1\end{aligned}[/tex]
Since we divided the arch in half, 0 [tex]\leq[/tex] 9 in this instance. All that is left to do is calculate the arch's height in feet from the base.
The height of 1 foot from each end of the arch must thus be determined. The locations of the arch's ends are x=-3 and x=3. Here, we have a choice between the two. We may choose x=3-1=2 for this. Therefore, all that is left to do is enter x=2 in our usual form and work out the value of y that results.
[tex]\frac{y^2}{81}+\frac{(2)^2}{9}=1[/tex]
y^2+36=81
y^2+36-36=81-36
y=3√5
Therefore, the height of the arch at 1ft from the base is 3 √5ft or approximately 6.71ft
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