Answer:
[tex]\frac{x^2}{100}+\frac{y^2}{64}=1[/tex]
At a distance of 5 feet from the center, the height is 6.93 ft
Step-by-step explanation:
Assume that the center of the ellipse is at the origin. The general equation for an ellipse is:
[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex]
Where "a" is half of the span of the ellipse, and "b" is the height of the arch (half of the amplitude of the ellipse). Thus, the equation is:
[tex]\frac{x^2}{10^2}+\frac{y^2}{8^2}=1\\\frac{x^2}{100}+\frac{y^2}{64}=1[/tex]
For x = 5 feet, the height of the arch is:
[tex]\frac{5^2}{100}+\frac{y^2}{64}=1\\y^2=\frac{100-25}{100}*64\\y=6.93\ ft[/tex]
At a distance of 5 feet from the center, the height is 6.93 ft
The height of the arch at a distance of 5 feet from the center is approximately 6.93 feet.
The standard equation of the ellipse centered at origin behind the shape of arch is presented below:
[tex]\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1[/tex] (1)
Where:
If we know that [tex]x = \pm 5\,ft[/tex], [tex]a = 10\,ft[/tex] and [tex]b = 8\,ft[/tex], then the height of the arch at this location is:
[tex]y = b\cdot \sqrt{1-\frac{x^{2}}{a^{2}} }[/tex]
[tex]y = 8\cdot \sqrt{1-\left(\pm \frac{5}{10} \right)^{2}}[/tex]
[tex]y \approx 6.93\,ft[/tex]
The height of the arch at a distance of 5 feet from the center is approximately 6.93 feet.
We kindly invite to check this question on ellipses: https://brainly.com/question/14281133
