Answer:
The horizontal distance from the center is 49.3883 feet
Step-by-step explanation:
The equation of an ellipse is equal to:
[tex]\frac{x^2}{a^{2} } +\frac{y^2}{b^{2} } =1[/tex]
Where a is the half of the wide, b is the high of the ellipse, x is the horizontal distance from the center and y is the height of the ellipse at that distance.
Then, replacing a by 106/2 and b by 33.9, we get:
[tex]\frac{x^2}{53^{2} } +\frac{y^2}{33.9^{2} } =1\\\frac{x^2}{2809} +\frac{y^2}{1149.21} =1[/tex]
Therefore, the horizontal distances from the center of the arch where the height is equal to 12.3 feet is calculated replacing y by 12.3 and solving for x as:
[tex]\frac{x^2}{2809} +\frac{y^2}{1149.21} =1\\\frac{x^2}{2809} +\frac{12.3^2}{1149.21} =1\\\\\frac{x^2}{2809}=1-\frac{12.3^2}{1149.21}\\\\x^{2} =2809(0.8684)\\x=\sqrt{2809(0.8684)}\\x=49.3883[/tex]
So, the horizontal distance from the center is 49.3883 feet
