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I would like to know if there is any proof to this: Prove that an arc segment of an ellipse cannot be similar to an arc segment of an ellipse with different eccentricity.

I specifically exclude circles, which can be considered ellipses with coincident focal points.

Motivation: this seems obvious to me, and maybe to many others, I'm curious if there's any formal proof of it mathematically. I should add: I may be wrong and this may not actually be the case. I would like a formal proof of the *negation, if that's the case.

Mike
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  • Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read the title. – José Carlos Santos Oct 01 '21 at 06:15
  • It would also be helpful to indicate at what level of geometrical argument you are looking for a proof. Euclidean geometry, analytic geometry, differential geometry? Also, what is meant by the "converse" of the statement in the title? –  Oct 01 '21 at 06:45
  • boojum, by "converse" I mean a proof of the opposite: that a part of an ellipse can be part of another ellipse of different eccentricity. – Mike Oct 01 '21 at 06:48
  • José, I have edited the question – Mike Oct 01 '21 at 06:49
  • boojum, I am comfortable with a proof at any level. – Mike Oct 01 '21 at 06:49
  • That's why I asked -- that's not the converse of your proposition. A proposition $ \ p \Rightarrow q \ $ has as its converse $ \ q \Rightarrow p \ \ . $ You seem to be asking whether there is a proof of the negation, that is, whether there is a proof by contradiction? –  Oct 01 '21 at 07:32
  • Understood, boojum. I'll edit the question. – Mike Oct 01 '21 at 07:48
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    The curvature function for an ellipse with semi-axes $ \ a \ $ and $ \ b \ $ (and so $ \ e \ = \ \sqrt{1 \ - \ \frac{b^2}{a^2}} \ \ $ ) is derived, for example, in this post: https://math.stackexchange.com/questions/1451959/curvature-of-ellipse . Varying the eccentricity changes the ratio of $ \ a \ $ to $ \ b \ $ so the curvatures for two ellipses of different eccentricity will differ everywhere over the curves. –  Oct 01 '21 at 07:55
  • Thanks boojum. Your comment is actually the right answer to my question. You should post it as an answer. I don't think I have enough reputation points to accept answers but I'm hopeful those with the points agree. – Mike Oct 01 '21 at 08:24
  • @boojum The curvature of ellipse is a continuous function that changes between the smallest value at the ends of the minor axis and the largest value at the ends of the major axis. So there will be points on ellipses of different eccentricities that have the same curvature. – Conifold Oct 01 '21 at 09:17
  • @Conifold, I understand the reasoning. However my understanding of the word "arc" is something with non-zero length, hence not a point. – Mike Oct 01 '21 at 10:27
  • The problem is how exactly are you going to compare curvatures over an arc? If they have equal curvatures at a point then, by continuity, they will have points of equal curvature in its vicinity as well. You need a stronger invariant here. For example, consider the natural equation (curvature as a function of arclength). For ellipse it is analytic, so the entire ellipse is determined (up to congruence) by its arbitrarily small arc. In particular, the eccentricity is so determined. And similar ellipses have equal eccentricities. – Conifold Oct 01 '21 at 10:37
  • @Conifold: you are right. The question of "...how exactly are you going to compare curvatures over an arc.." may be what triggered the question in my mind. – Mike Oct 01 '21 at 11:33

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This is true of any conic section, but let this proceed with two similar ellipse segments, $S_1$ and $S_2$, segments of ellipses having corresponding eccentricity $e_1$ and $e_2$.

Let $S_2$ be dilated by the ratio of similitude, so that its image is $S_3$, congruent to $S_1$. The dilation does not change the eccentricity, so $S_3$ is still a segment of an ellipse having eccentricity $e_2$.

Ellipse segments $S_1$ and $S_3$ are congruent. Therefore their corresponding eccentricities, $e_1$ and $e_2$, are equal.

See the Conica of Apollonius, where Book 6, Proposition 6 states that if any segment of a conic section can be fitted to a segment of another, then the entire sections are congruent. That goes for congruent segments. Extending it to similar segments is only a matter of scaling.

Pope
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  • Thanks Pope. Please explain further the two eccentricities in your first paragraph, $e_{1}$ and $e_{2}$.. are they equal? What kind of transformation do you mean by 'dilation'? My understanding is that anything other than translation (or reflection) would change the eccentricity.. so you may start off with ellipses of different eccentricity, but the dilation makes them equal eccentricity.. I don't want to compare ellipses of the same eccentricity, but of different eccentricity. – Mike Oct 01 '21 at 07:21
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    Your third paragraph restates the claim, without any proof. Referring to an historical book does not help. –  Oct 01 '21 at 08:29
  • @Mike: Each of the two similar ellipse segments has an eccentricity. They are equal, but not as a given condition. They were proved to be equal. A dilation is a scaling in all directions with respect to some fixed point. It conserves angles and ratios of distances in any direction. In this case, it would conserve the ratio of the focus distance to the semi-major, and that ratio is the eccentricity. – Pope Oct 01 '21 at 10:06
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    @Mike: I see now what you mean by beginning with unequal eccentricities. Your proposition is that if the eccentricities are not equal, then the ellipse segments cannot be similar. What I proved is that if the segments are similar, then the eccentricities are equal. That is the contrapositive of your proposition, hence logically equivalent, so the proof does stand up. – Pope Oct 01 '21 at 10:22
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All parabolas are similar. An arc of one parabola can be made to coincide with another parabola arc by three transformations/ geometric operations viz., Zoom,translation and rotation in the plane.

This happens because only a single constant is involved in its $(x,y)$ description equation, for example in its polar form

$$ p/r = (1- \cos \theta) \tag 1$$

For an ellipse two constants occur. Zoom,translation and rotation cannot make a rigid arc to be placed inside another arc where eccentricity $e$ is now introduced:

$$ p/r = (1- e \cos \theta) \tag2$$

For such a match to take effect for an ellipse a fourth transformation/operation is necessary and that is.. changing of Aspect Ratio after zoom magnification/ uniform reduction or dilation in the plane.

In other words for a matching between two given elliptic arcs as mentioned in the question to take place, these four unique operations have be determined and applied.

Geometric similarity occurs when $p/r$ has the same value at a given $\theta$.

This can occur if and only if the value of eccentricity $e$ as a constant is same for either ellipse (or conic section)... and that completes the proof for ellipse and any curve described by two parameters.

Narasimham
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Experimental aspect using radius of curvature

For circle,

$$\rho=a \implies \frac{d\rho}{ds}=0$$

For logarithmic spiral $r=ae^{\theta \cot \beta}$ which is self-similar,

$$\rho=a\csc \beta+s\cot \beta \implies \frac{d\rho}{ds}=\cot \beta$$

For general conics

\begin{align} 0 &= F(x,y) \\ &\equiv \begin{pmatrix} x & y & 1 \end{pmatrix} \begin{pmatrix} a & h & g \\ h & b & f \\ g & f & c \end{pmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} \\ \begin{pmatrix} u \\ v \end{pmatrix} &= \begin{pmatrix} a & h \\ h & b \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}+ \begin{pmatrix} g \\ f \end{pmatrix} \\ \Delta &= \det \begin{pmatrix} a & h & g \\ h & b & f \\ g & f & c \end{pmatrix} \\ \kappa &= \frac{\Delta}{(u^2+v^2)^{3/2}} \\ \kappa' &= \frac{3\Delta}{(u^2+v^2)^3} [hu^2+(b-a)uv-hv^2] \\ \frac{d\rho}{ds} &= -\frac{\kappa'}{\kappa^2} \\ &= -\frac{3}{\Delta} [hu^2+(b-a)uv-hv^2] \end{align}

which is closely related to the evolute.

Observations without proofs.

  • The conics are similar by varying $c$ only.

  • Testing with CAS, $\dfrac{d\rho}{ds}$ is invariant for the same relative position of similar conics.

  • Similar conics implying similar evolutes.

  • Up to scale, evolute varies as eccentricity.

  • Conics with different eccentricities $e_i>0$ are locally similar only at vertices where $\kappa'=0$.

  • Osculating conics should have the same $\kappa$, $\kappa'$ and so on, but they are usually not (locally) similar.

Ng Chung Tak
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