Given the equation of an elipse
[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex]
from the question,
[tex]\begin{gathered} \text{major axis}\Rightarrow2a \\ \therefore2a=52 \\ a=\frac{52}{2}=26ft \\ b=13ft \end{gathered}[/tex]
Given that
[tex]x=14ft[/tex]
Substitute, for a,b, and x in the elipse formula to find y
[tex]\begin{gathered} \frac{14^2}{26^2}+\frac{y^2}{13^2}=1 \\ \frac{196}{676}+\frac{y^2}{169}=1 \end{gathered}[/tex]
Multiply through by 169
[tex]\begin{gathered} 49+y^2=169 \\ y^2=169-49 \\ y^2=120 \\ y=\sqrt[]{120}=10.95ft \end{gathered}[/tex]
Hence, it clear the arch because the height of the archway of the bridge 7 feet from the center is approximatelyfeet
