China Mathematical Competition

Suppose that there is a point $P(4\cos \alpha ,3\sin \alpha )$ on the ellipse. When $P$ and the origin $O$ are not on the same side of $AB$, the distance from $P$ to $AB$ is $$\begin{array}{rl}& \frac{3(4\cos \alpha )+4(3\sin \alpha )-12}{5}\\ & =\frac{12}{5}(\cos \alpha +\sin \alpha -1)\\ & \le \frac{12}{5}(\sqrt{2}-1)\\ & <\frac{6}{5}.\end{array}$$ But $AB=5$, so $△PAB<\frac{1}{2}\times 5\times \frac{6}{5}=3$. Therefore, when the area of $△PAB$ is equal to $3$, points $P$ and $O$ are on the same side of $AB$. There are two such points $P$. Answer: B.