## Precalculus (6th Edition) Blitzer

a. $\frac{x^2}{2304}+\frac{y^2}{529}=1, y\geq0$ b. $42\ ft$.
a. Using the figure given in the exercise, we have $a=48, b=23$ and the equation of the archway can be written as $\frac{x^2}{48^2}+\frac{y^2}{23^2}=1$ or $\frac{x^2}{2304}+\frac{y^2}{529}=1, y\geq0$ b. Using the above results, we have $c=\sqrt {a^2-b^2}=\sqrt {1775}\approx42\ ft$. Thus he should place his desk about 42 feet from the center.