Answer to Question #259682 in Analytic Geometry for josel

Question

The arch of a bridge is in the shape of semi-ellipse, with its major axis at the water level. Suppose the arch is 20ft high in the middle, and 120ft across its major axis. How high above the water level is the arch, at a point 20ft from the center(horizontally)?

Answer 1

The equation of the ellipse is $\frac{x^2}{60^2}+\frac{y^2}{20^2}=1$ for $y>-0$ and we want the values of y when $x=-20\ and \ x=20$

By substituting those values of x into, we get $\frac{20^2}{60^2}+\frac{y^2}{20^2}=\frac{1}{9}+\frac{y^2}{20^2}$

$\frac{y^2}{60^2}=\frac{8}{9}$

$\frac{y}{20}=\sqrt{\frac{8}{9}}=18.86$

so, at point 20 ft. from the center (horizontally) the arch is 18.86ft above the water