Answer
The exact location of the ship in the first quadrant is $\left( 1,1 \right)$
Work Step by Step
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 $\left( 1,1 \right)$.
