Precalculus (6th Edition) Blitzer

a. $\frac{x^2}{5625}-\frac{y^2}{4375}=1$, right curve b. $58.3\ km$
a. Step 1. Draw a diagram as shown in the figure. Assume the midpoint between A and B to be the origin and the ship at $C(x,y)$ is on a hyperbola with equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with A and B as the foci. Step 2. We have $c=\frac{200}{2}=100\ km$ and $2a=d_1-d_2=(500ms)(0.3 km/ms)=150\ km$; thus $a=75\ km$ and $a^2=5625$ Step 3. We can find $b^2=c^2-a^2=100^2-75^2=4375$ and the possible location of the ship $C(x,y)$ is on the right curve of the hyperbola $\frac{x^2}{5625}-\frac{y^2}{4375}=1$ b. If the ship lies due north of B, we have $x=100$ and $\frac{100^2}{5625}-\frac{y^2}{4375}=1$ which gives $y^2\approx3402.8$ and $y\approx58.3\ km$