## Precalculus (6th Edition) Blitzer

The exact location of the ship in the first quadrant is $\left( 1,1 \right)$
The provided equations of the paths are, \begin{align} & 2{{y}^{2}}-{{x}^{2}}=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\ & 2{{x}^{2}}-{{y}^{2}}=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \end{align} Multiply equation (I) by 2 and add equation (II). Therefore, \begin{align} & -2{{x}^{2}}+4{{y}^{2}}=2 \\ & 2{{x}^{2}}-{{y}^{2}}=1 \\ & \overline{\begin{align} & 3{{y}^{2}}=3\,\,\,\,\,\,\,\,\, \\ & y=\pm 1 \\ \end{align}} \\ \end{align} Now, If $y=1$ then, \begin{align} & \,2{{x}^{2}}-1=1 \\ & 2{{x}^{2}}=2 \\ & {{x}^{2}}=1 \\ & x=\pm 1 \end{align} If $y=-1$ Then, \begin{align} & 2{{x}^{2}}-\left( -1 \right)=1 \\ & 2{{x}^{2}}+1=1 \\ & {{x}^{2}}=0 \\ & x=0 \\ \end{align} Here, the ship is located in the first quadrant, so take only $x=1$ and $y=1$. Thus, the ship’s coordinate in the first quadrant is $\left( 1,1 \right)$.