You are watching: What is the minimum eccentricity an ellipse can have

For sufficiently eccentric ellipses, the is possible to perform something similar, however with 4 ellipses surrounding an additional ellipse of the very same size.

What is the minimum eccentricity the ellipses for which this is possible?

I"ll concentrate on the symmetric case, as portrayed here:Numeric results

First, right here are some numeric results:

$a$ and $b$ room the size of the major and young semiaxis. $c$ is the eccentricity, and also $pmvarphi$ is the angle of rotation between the main and one of the neighboring ellipses. $(pm x,pm y)$ is the facility of one of the bordering ellipses. $alpha$ and also $eta$ are two parameters disputed below.

The above is the optimum I uncovered using 150 iterations with 215 bits of numeric precision along the way. I afterwards verified the they are undoubtedly accurate, using the specific algebraic computation explained further below.

## How to achieve them

Here is what ns did: I characterized all my geometry in terms of two parameters, $alphaapprox 0.14778$ and also $etaapprox 0.77656$. The very first defines the conic, together $alpha x^2+y^2=1$. So contrary to mine illustration and the figures above, I"m using $1$ together the length of the boy semiaxis and also $sqrtfrac1alpha$ for the significant internally, and also the figures were scaled afterwards. I then interpret everything projectively, and also use the 2nd parameter to specify $(1,eta,0)^T$ and $(-eta,1,0)^T$, 2 points in ~ infinity which stand for bundles of parallel lines, the 2 bundles orthogonal come one another. From this points I build tangents to the central ellipse, and their allude of intersection. These space the blue lines and red dots in the number above. Ns then compute the change which maps this tangent lines to the name: coordinates lines, and also apply it come the central ellipse to obtain the surrounding one which will definitely touch the name: coordinates axes.

Next ns theoretically take into consideration the two points that intersection which this ellipse could have with the main one. But much more importantly, I take into consideration the line joining these two points. The benefit here is top top the one hand that this heat is in fact easier to compute than the point out of intersection, and also that on the other hand the line will certainly be real even if the ellipses no much longer intersect in any kind of real points.

Now the task have the right to be formulated in terms of the position of this heat wrt. The central ellipse: it need to touch the central ellipse, and any readjust of $eta$ in one of two people direction must reason it to crossing the central ellipse, for this reason it have to be optimal in a certain sense. The adjoint the the procession of the main ellipse will represent a quadratic form, and plugging the line right into that type will result in a solitary number. According to the over conditions, this number have to be both zero and also optimal. In my selection of signs it must be maximal, however sign changes will happen easily.

Now come the numeric refinement via bisection. Each loop adjusts an initial $eta$ then $alpha$. $eta$ is adjusted towards optimality, when $alpha$ is adjusted towards a root, where the an outcome is zero. Therefore in various other words, I pick the angle such that the line is as close come the center of the main ellipse, then select the eccentricity so that for that angle the still just touches the main ellipse. The mediate is a nice stupid bisection, i beg your pardon works because I constantly stay close to the actual solution and also won"t need to worry about other optima. Rate of convergence might be improved, yet I"d rather keep the simplicity of mine approach and also wait a little longer.

I"ve uploaded the sage password for my computations here. If anyone to be to discover symbolic expression because that the minimal eccentricity, then ns guess one could readjust my password to verify that.

Exact solutionAs achille hui pointed out in a comment below, a post by Jim Ferry currently contains specific solution, i.e. One in terms of algebraic numbers. I"ve now been able to reproduce the result. The an essential idea is come work less in terms of building and construction steps, and also instead much more in regards to a characterization of desired properties. This allows one to protect against using square roots, so one can express every little thing (except angles) together polynomials.

## Formulating the emotional conditions

I started with two ellipses, one with horizontal semimajor axis of length $sqrtfrac1alpha$ and also vertical semiminor axis of $1$, the other rotated by $arctaneta$ and also translated by $(x,y)$. Ns formulated both together $3 imes 3$ matrices in the polynomial ring $2175forals.combb Q

$$E_1=eginpmatrixalpha & 0 & 0 \0 & 1 & 0 \0 & 0 & -1endpmatrix\E_3=scriptsizeeginpmatrixeta^2 + alpha & alpha eta - eta & - eta^2 x - alpha eta y - alpha x + eta y \alpha eta - eta & alpha eta^2 + 1 & - alpha eta^2 y - alpha eta x + eta x - y \- eta^2 x - alpha eta y - alpha x + eta y & - alpha eta^2 y - alpha eta x + eta x - y & alpha eta^2 y^2 + eta^2 x^2 + 2 alpha eta x y + alpha x^2 - 2 eta x y - eta^2 + y^2 - 1endpmatrix$$

(Note that i guess I can have rotated my ellipse in the opposite direction, however that"s irrelevant for now.) The truth that a provided line touches a given ellipse have the right to be computed via the double quadric, which deserve to be quickly (i.e. Without divisions or square roots) computed as the adjoint matrix. For this reason the rotated ellipse will certainly touch the axes the the coordinate system iff

$$(1,0,0)cdotoperatornameadj(E_3)cdoteginpmatrix1\0\0endpmatrix=0qquad(0,1,0)cdotoperatornameadj(E_3)cdoteginpmatrix0\1\0endpmatrix=0$$

Next comes the condition that the 2 ellipses touch. As soon as intersecting 2 conics, one strategy starts by looking for roots the $det(mu E_1+E_3)=0$. These worths of $mu$ an outcome in a degenerate conic $mu E_1+E_3$ consists of a pair the lines and also passing with the points of intersection. If 2 points the intersection coincide, i.e. If you have actually a allude of greater multiplicity, then this 3rd degree polynomial in $mu$ must have actually a zero of higher multiplicity as well since few of the degenerate conics will coincide. So to discover out the emotional condition, we look in ~ the discriminant that $det(mu E_1+E_3)$ with respect to $mu$. This is the third condition.

## Computing the solution

Now one deserve to use resultants to get rid of variables, specific to remove $x$ and $y$, in order to combining the three conditions into a solitary polynomial in $a$ and also $b$. For this step, Mr. Lewis offered a technique he occurred himself and called “Dixon-EDF” (for beforehand Detection the Factors). However although his computer system algebra mechanism Fermat comes v a record implementing that method, i haven"t yet worked out how to adjust the various knobs to acquire it to resolve a different set of equations, and the user user interface kept annoying me. So i sticked through sage, and the naive computation took quite some time, about 10 minutes for the computation approximately the final an outcome but greatly for this step here. In the end, the result polynomial had 2069 terms. Ns factorized it and also cheated a little by no considering every potential factors, however only the one which had the smallest absolute value for the almost right numeric values derived above.

We are trying to find the *minimal* worth of $alpha$ satisfying all requirements. This optimality constraint synchronizes to a root of the problem we just found which is a dual root with respect come rotation. ~ all, we want an intersecting situation on both political parties of the systems $eta$. So i computed the discriminant that this polynomial through respect come $eta$, again factorized and used the numerical approximation to find the exactly factor, i beg your pardon is this:

$$151632alpha^12 - 556632alpha^11 - 4029183alpha^10 + 5710568alpha^9 + 2456300alpha^8 - 8614032alpha^7 - 40073338alpha^6 - 8614032alpha^5 + 2456300alpha^4 + 5710568alpha^3 - 4029183alpha^2 - 556632alpha + 151632$$

This is basically the same polynomial Mr. Ferry provided in his strategy as well, other than that because my $alpha$ relates come the *squared* proportion of axes, my exponents are half of his. He describes how one deserve to use the symmetry of the polynomial (he dubbed it a “palindrome”) to usage a level 6 polynomial native which everything else deserve to be computed. However for me this to be enough, therefore I derived real roots, expressed together algebraic numbers in sage, and chose the one closest to the approximate value.

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Plugging the value into the polynomial before, the relevant aspect of the huge polynomial in $alpha$ and also $eta$, I acquired a value for $eta$. That is minimal polynomial is an especially nice and also easy:

$$16eta^12 - 100eta^10 + 615eta^8 - 2468eta^6 + 3186eta^4 - 1728eta^2 + 351$$

(Here you notice that $eta$ is identified only approximately sign, i beg your pardon is the reason why girlfriend don"t have to concern too much about which direction is which when rotating the ellipse.)Having uncovered $alpha$ and also $eta$, I might compute every the rest, and in fact verify the the digits given over were undoubtedly accurate. Ns will upgrade my uploaded password to incorporate the code and output because that this computation.