Answer
The coordinates of $P$ is $(R \cos \theta,r \sin \theta)$ for $0\leq \theta\leq 2\pi$.
The curve traced by $P$ is the ellipse $(x/R)^2+(y/r)^2=1$.
Work Step by Step
Let the origin be O. So, we have the vector position of OP as OP=OB+BP.
From Figure 19, we get
OB=$R \cos \theta \ \textbf{i}+R \sin \theta \ \textbf{j}$.
BP=$-(R-r)\sin \theta \ \textbf{j}$
So,
OP=OB+BP
OP=$R \cos \theta \ \textbf{i}+R \sin \theta \ \textbf{j}-(R-r)\sin \theta \ \textbf{j}$
OP=$R \cos \theta \ \textbf{i}+R \sin \theta \ \textbf{j}-R \sin \theta \ \textbf{j}+r \sin \theta \ \textbf{j}$
OP=$R \cos \theta \ \textbf{i}+r \sin \theta \ \textbf{j}$.
Thus, the coordinates of $P$ is $(R \cos \theta,r \sin \theta)$ for $0\leq \theta\leq 2\pi$. So we have the parametric equations $x=R \cos \theta$ and $y=r \sin \theta$.
This is the parametrization of an ellipse (see Example 5) that satisfies the equation $(x/R)^2+(y/r)^2=1$.
