Centers of the osculating circles along an ellipse Consider an ellipse on the plane $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. We will use the usual parametrization: $P(t)=(x(t),y(t))=(a\cos t,b\sin t)$.
Then the tangent vector is $T(t)=(-a\sin t, b\cos t)$,
and the (inward) normal vector is $N(t)=(-b\cos t,-a\sin t)$.
Also we know the radius of curvature at $P(t)$ is $\displaystyle  r(t)=\frac{(a^2\sin^2t+b^2\cos^2t)^{3/2}}{ab}$. So the center of the osculating circle at $P(t)$, or center of curvature, is given by 
$$C(t)=P(t)+r(t)\frac{N(t)}{|N(t)|}=(a\cos t,b\sin t)-\frac{a^2\sin^2t+b^2\cos^2t}{ab}(b\cos t,a\sin t).$$
My question is, does these centers also trace out an ellipse? If not, could we describe the coordinates by some quadratic/algebraic equation?
Thank you!
 A: In general, the Evolute of a curve is the locus of all its centers of curvature. 
So your question is: how does the evolute of an ellipse looks like?
The answer is that it is not an ellipse.  The coordinates $x,y$ of $C(t)$ satisfy a polynomial equation of total degree $6$:
$$  {a}^{6}{x}^{6}+3\,{a}^{4}{b}^{2}{x}^{4}{y}^{2}+3\,{a}^{2}{b}^{4}{x}^{2
}{y}^{4}+{b}^{6}{y}^{6}\\+ \left( -3\,{a}^{8}+6\,{a}^{6}{b}^{2}-3\,{a}^{
4}{b}^{4} \right) {x}^{4}+ \left( 21\,{a}^{6}{b}^{2}-42\,{a}^{4}{b}^{4
}+21\,{a}^{2}{b}^{6} \right) {x}^{2}{y}^{2}+ \left( -3\,{a}^{4}{b}^{4}
+6\,{a}^{2}{b}^{6}-3\,{b}^{8} \right) {y}^{4}\\+ \left( 3\,{a}^{10}-12\,
{a}^{8}{b}^{2}+18\,{a}^{6}{b}^{4}-12\,{a}^{4}{b}^{6}+3\,{a}^{2}{b}^{8}
 \right) {x}^{2}+ \left( 3\,{a}^{8}{b}^{2}-12\,{a}^{6}{b}^{4}+18\,{a}^
{4}{b}^{6}-12\,{a}^{2}{b}^{8}+3\,{b}^{10} \right) {y}^{2}\\-{a}^{12}+6\,
{a}^{10}{b}^{2}-15\,{a}^{8}{b}^{4}+20\,{a}^{6}{b}^{6}-15\,{a}^{4}{b}^{
8}+6\,{a}^{2}{b}^{10}-{b}^{12} = 0 $$
Here are the ellipse (blue) and $C(t)$ (red) in the case $a=2,b=1$.

EDIT: Here's an animation showing the osculating circle moving.

