Hello, NKS!
A bridge is built in the shape of a semielliptical arch.
It has a span of 106 feet.
The height of the arch 25 feet from the center is 5 feet.
Find the height of the arch at its center. Code:

* o *
*  o
*  : *
* b : *
 :5
*  : *
*      +    o  o  
53  25 53
We have: .$\displaystyle \frac{x^2}{53^2} + \frac{y^2}{b^2} \:=\:1 \quad\Rightarrow\quad b^2 \:=\:\frac{53^2y^2}{2809  x^2}$
Since $\displaystyle (25,5)$ is on the ellipse: .$\displaystyle b^2 \:=\:\frac{53^25^2}{2809  25^2} \;=\;\frac{53^25^2}{2184}$
. . $\displaystyle b \;=\;\frac{53\cdot5}{\sqrt{2184}} \:=\:\frac{265}{2\sqrt{546}} \:=\:\frac{265\sqrt{546}}{1092} $
The height is about 5.67 feet.