Answer
(a) See explanations.
(b) See explanations. See graph. $x^{2/3}+y^{2/3}=a^{2/3}$
Work Step by Step
(a) Use the figure given in the Exercise, the coordinates of P can be considered in two parts, the first part is due to the moving center of the circle-C which is $a-b$ away from the origin, thus $x_1=(a-b)cos\theta, y_1=(a-b)sin\theta$. The second part is within circle-C. As circle-C is rolling inside the big circle, the arc length traveled can be expressed in equal two ways $a\theta=b\alpha$ where $\alpha$ is the angle circle-C rotated. Thus we have $\alpha=\frac{a}{b}\theta$ and the angle needed to calculate the second part coordinates is $\alpha-\theta=\frac{a}{b}\theta-\theta=\frac{a-b}{b}\theta$ as shown in the figure (inside circle-C), so that $x_2=b\cdot cos(\frac{a-b}{b}\theta), y_2=b\cdot sin\frac{a-b}{b}\theta)$. We can finally reached the parametric equations as: $x=x_1+x_2=(a-b)cos\theta+b\cdot cos(\frac{a-b}{b}\theta)$, similarly, we have $y=y_1-y_2=(a-b)sin\theta-b\cdot sin(\frac{a-b}{b}\theta)$
(b) In the case of $a=4b$, $x=3b\cdot cos\theta+b\cdot cos(3\theta)$. With $cos3\theta=4cos^3\theta-3cos\theta$ (prove this use the Additional Formula), we have $x=4b(cos^3\theta)=a\cdot cos^3\theta$, similarly we get $y=3b\cdot sin\theta - b\cdot sin(3\theta)$. With $sin3\theta=3sin\theta-4sin^3\theta$, we get $y=4b(sin^3\theta)=a\cdot sin^3\theta$
We can graph the function as shown in the figure. To eliminate $\theta$, we have $cos\theta=(\frac{x}{a})^{1/3}$ and $sin\theta=(\frac{y}{a})^{1/3}$, use $cos^2\theta+sin^2\theta=1$, we get $(\frac{x}{a})^{2/3}+(\frac{y}{a})^{2/3}=1$ or $x^{2/3}+y^{2/3}=a^{2/3}$
