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!

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.

• I intend no disrespect, in any way whatsoever, and was first to up vote your answer. But how did you answer this so quickly, including entering all the notation correctly in LaTeX? That is amazing! Sorry, rhetorical question. Mar 25, 2013 at 1:41
• Actually, now I'm confused. The question is titled, "Center of the osculating circle of an ellipse". I see the ellipse. I see your answer, which says that C(t) is not an ellipse, but rather, is described by coordinates x,y that are the solution to a 6th degree polynomial equation. Where is the circle that is referenced in the title of the equation? Mar 25, 2013 at 1:46
• @FeralOink I will revise the title, to match my question better ^-^ Mar 25, 2013 at 2:43
• Thank you Pengfei. As for Robert Israel, that is just delightful! I am so happy! I am smiling and giggled slightly. Thank you so much. That is wonderful. Mar 25, 2013 at 2:52
• @FeralOink haha that is cute Mar 25, 2013 at 16:10