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Consider an ellipse $x^2 + 4y^2 = 4$ given in parametrised form $(2 \cos t, \sin t)$. At a given point $p_0 = (2 \cos t_0, \sin t_0)$ we want to measure how round the ellipse is (i.e. how similar to a circle it is). To do this, let $C(x,y) = (x-a)^2 + (y-b)^2 - \lambda$ be a circle with centre $(a,b)$. If this circle goes through the point $p_0$ then

$$ (2\cos t_0 - a)^2 + (\sin t_0 - b)^2 -\lambda = 0 = g(t_0)$$

A point of $k$ contact between this circle and ellipse is any point for which the first $k-1$ derivatives $g^{(i)}$ vanish and $g^{(k)} \neq 0$.

For this example there are only four points for which we can have $4$-point contact. The centres of the corresponding circles are: $(\pm {3 \over 2}, 0), (0, \pm 3)$.

My question is: Why are these the only points where $4$-point contact is possible?

Ideally, I would like to gain geometric insight from any answer, if possible. It is clear to me that $3$-point contact is possible at every point on the ellipse.

student
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    Those four points don't lie on your ellipse. – TonyK May 05 '15 at 08:14
  • @TonyK That's correct: They are the centres of the osculating circles. – student May 06 '15 at 03:31
  • I don't believe that $(0,\pm 3)$ are anything of the sort. The semiminor axis is $1$, and the osculating circle must be centered inside the ellipse. Something's seriously wrong there. – Ted Shifrin May 06 '15 at 17:07

1 Answers1

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You're correct that you can find a circle with $3$-point contact at any point $P$ on the ellipse, namely, the osculating circle at $P$. You will have $4$-point contact at the critical points $P$ of the curvature. These are the so-called vertices of the ellipse. There are, in fact, precisely 4: $(\pm 2,0)$ and $(0,\pm1)$.

EDIT: So here's the argument. Start with an arclength-parametrized plane curve $\alpha=\alpha(s)$. Recall that $T(s)=\alpha'(s)$ is the unit tangent vector and curvature $\kappa$ is defined by $T'(s)=\kappa(s)N(s)$ ($N$ is the principal normal). We're going to work with the point $\alpha(0)$ on our curve. Now consider $$f(s)=\frac12\left(\|\alpha(s)-P\|^2-r^2\right),$$ where $P=(a,b)$ is the center of the circle and $r=\|\alpha(0)-P\|$. Recall, also, that because $T(s)\cdot N(s)=0$ for all $s$, we'll have $N'(s)=-\kappa(s)T(s)$. OK, now we calculate: \begin{align} f'(s) &= (\alpha(s)-P)\cdot\alpha'(s) = (\alpha(s)-P)\cdot T(s); \\ f''(s) &= \kappa(s)(\alpha(s)-P)\cdot N(s) + T(s)\cdot T(s) = \kappa(s)(\alpha(s)-P)\cdot N(s)+1; \\ f'''(s) &= \kappa'(s)(\alpha(s)-P)\cdot N(s) - \kappa(s)^2 T(s)\cdot(\alpha(s)-P) + \kappa(s)N(s)\cdot T(s) \\ &= \kappa'(s)(\alpha(s)-P)\cdot N(s) - \kappa(s)^2 T(s)\cdot(\alpha(s)-P) \,. \end{align} Now, let's use $f'(0)=f''(0) = 0$ to conclude that $\alpha(0)-P$ is parallel to $N(0)$ and $\kappa(0) = 1/r$. This gives us the osculating circle at $P$. We then see that $f'''(0)=0$ if and only if $\kappa'(0)=0$. That is, the osculating circle will have $4$-point contact with the curve at $P$ precisely when $P$ is a critical point of $\kappa$.

Ted Shifrin
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  • But my question is: why there are only 4 points with 4-contact possible on the ellipse? Does it have to do with what a critical point of curvature is? – student May 06 '15 at 03:17
  • A point where its derivative (say, with respect to arclength on the curve) vanishes. In this case, it's points of max/min curvature. I'm grading final exams for the next day, but will try to remember to give you a proof soon. – Ted Shifrin May 06 '15 at 03:21
  • That is very kind of you, thank you. But a proof of what? I am more looking for a geometric explanation or intuition of why the 3rd derivative can only vanish in 4 points. It's not obvious to me since the ellipse seems to be curved the exact same way all the way around... do you understand what I'm trying to ask? – student May 06 '15 at 03:25
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    I refer to a proof that 4th order contact occurs only at max/min points of curvature. 2-pt contact means the circle is tangent to your curve; 3-pt contact means it has the same curvature as your curve; what's next? – Ted Shifrin May 06 '15 at 03:31
  • So cool, that is exactly what I am asking! I am looking forward to seeing the proof! This is exciting! ^o^ I don't know what's next. Maybe something like "rate of change of curvature"? But then it seems to me that the two curves would be equal so that's probably wrong... – student May 06 '15 at 03:35
  • If it was "rate of change of curvature" then it would make sense to me though: it seems plausible that the ellipse looks like a circle in a neighbourhood for only four points and that for other points it has the wrong rate of change... come to think about it: the circle has constant curvature so the rate of change has to be zero. But the ellipse has varying curvature. Except at those for points. Did I get it? – student May 06 '15 at 03:42
  • (Even if I do get it I would still be very excited to see a proof of this tomorrow!) – student May 06 '15 at 03:43
  • Many thanks for the proof, I am starting to understand it (I read it a few times). I'm still struggling to get the idea: you start with some arbitrary point on $\alpha$ (here we chose $\alpha (0)$). It looks to me like there is a circle centred at $P=(a,b)$ with radius chosen so that the circle intersects $\alpha$ at $\alpha (0)$. But it's not clear to me why this circle is useful. Have I understood so far what's going on or am I misunderstanding something? – student May 12 '15 at 13:08
  • Michael, that's the same circle you were denoting $C(x,y)$ in your question. My $f(s)$ is the equivalent of your $g(t)$. – Ted Shifrin May 12 '15 at 13:36
  • Okay, I understand it slightly better now (I also deleted my other question). What I still don't understand now is why you choose $\alpha (0)$. Could you not have done the same argument for arbitrary $\alpha (t_0)$? I guess I'm still a bit confused.... – student May 16 '15 at 00:16
  • Michael: Of course. I just used $0$ for ease of typing. You can always adjust a parametrization to make any given point where you "start," anyway :) – Ted Shifrin May 16 '15 at 02:32