Answer
$\frac{4x^2}{8649}-\frac{4y^2}{31351}=1$
Work Step by Step
Step 1. Using the figure given in the exercise, we have $c=100$ and the general form of the hyperbola can be written as $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Step 2. The difference of the distances is$PM_1-PM_2=500(ms)\times 0.186(mi/ms)=93\ mi$ and this distance equals to $2a$ (consider when P is at one of the vertices).
Step 3. Thus we have $a=\frac{93}{2}$ and $b^2=c^2-a^2=100^2-(\frac{93}{2})^2=\frac{31351}{4}$
Step 4. The equation becomes: $\frac{4x^2}{8649}-\frac{4y^2}{31351}=1$
