Answer
(a) $\quad\left(0, \frac{-5}{2}\right), \left(\frac{5}{3}, 0\right) $
(b) Clockwise
(c) Point
Work Step by Step
Since
\[
\begin{array}{ll}
-3 \sin t=x^{\prime}, & -3 \cos t=x^{\prime \prime} \\
2 \cos t=y^{\prime}, & -2 \sin t=y^{\prime \prime}
\end{array}
\]
We use the formulas to calculate the coordinates of the center of curvature $(\zeta, \eta)$ :
\[
\begin{aligned}
\ &-y^{\prime} \frac{\left(y^{\prime}\right)^{2}+\left(x^{\prime}\right)^{2}}{x^{\prime} y^{\prime \prime}-x^{\prime \prime} y^{\prime}}+x=zeta \\
&-2 \cos t \frac{4 \cos ^{2} t+9 \sin ^{2} t}{6}+3 \cos t = \zeta \\
\frac{5}{3}=\zeta(0) &
\end{aligned}
\]
So
\[
x^{\prime} \frac{\left(x^{\prime}\right)^{2}+\left(y^{\prime}\right)^{2}}{x^{\prime} y^{\prime \prime}-x^{\prime \prime} y^{\prime}}+y=\eta
\]
\[
\begin{aligned}
\eta &=2 \sin t-3 \sin t \frac{9 \sin ^{2} t+4 \cos ^{2} t}{6} \\
\eta(0) &=0
\end{aligned}
\]
The center of curvature is $\left(\frac{5}{3}, 0\right)$
\[
\begin{array}{l}
0=\zeta(\pi / 2) \\
\frac{-5}{2}=\eta(\pi / 2)
\end{array}
\]
And then $\left(0, \frac{-5}{2}\right)$
The direction of the evolute that is tracked while increasing $t$ from 0 to $2\pi$ is tracked clockwise.
The evolution of a circle is a point, which is the center of the circle.
