Answer
The answers:
(a) The path followed by A is $(0,10\sin \theta)$ for $\theta$-values between $0$ and $\pi/2$.
(b) The path followed by B is $(10 \cos \theta,0)$ for $\theta$-values between $0$ and $\pi/2$.
(c) The path followed by P is $(4\cos \theta,6\sin \theta)$ for $\theta$-values between $0$ and $\pi/2$.
This is the parametrization of the ellipse $(x/4)^2+(y/6)^2=1$.
Work Step by Step
(a) The position vector of OA is
OA$=(0,10\sin \theta)$.
So, the parametrization of the path followed by A is $(0,10\sin \theta)$ for $\theta$-values between $0$ and $\pi/2$.
(b) The position vector of OB is
OB$=(10 \cos \theta,0)$.
So, the parametrization of the path followed by B is $(10 \cos \theta,0)$ for $\theta$-values between $0$ and $\pi/2$.
(c) Let Q be the intersection of the vertical line through P and the $x$-axis. So, the position vector of OP is
OP=OQ+QP.
But the vector OQ is OQ=$(4\cos \theta,0)$ and the vector QP is QP=$(0,6\sin \theta)$. Therefore,
OP=OQ+QP=$(4\cos \theta,0)+(0,6\sin \theta)$,
OP=$(4\cos \theta,6\sin \theta)$.
So, the parametrization of the path followed by P is $(4\cos \theta,6\sin \theta)$ for $\theta$-values between $0$ and $\pi/2$.
This is the parametrization of an ellipse (see Example 5) that satisfies the equation $(x/4)^2+(y/6)^2=1$.
