# Pencil of ellipses

This is a math problem whose solution would be great for physics, especially space flight planning.

Given two ellipses sharing a focus, could there be a pencil of ellipses sharing the same focus and tangent to the two given ellipses, or is there only one such ellipse? This problem has two versions: 2D (coplanar) and 3D (not coplanar).

Thanks for your answers.

• Can you clarify the 3D version of your problem? It seems to me that if two ellipses share a focus and a point of tangency, then they must be coplanar --- because an ellipse lies in the plane spanned by either of its foci together with any of its tangent lines. Have I misunderstood the question? – Zach Teitler Jun 6 '17 at 21:01
• @ZachTeitler In fact I never mentioned that the two base ellipses were tangent, so the 3D problem is very general, and I don't think really obvious. – Maxence Seymat Jun 7 '17 at 12:24
• Say $B_1$ and $B_2$ are the two base ellipses. If $E$ is an ellipse sharing a focus with $B_1$ and tangent to $B_1$, then $E$ has to lie in the plane of $B_1$. And if $E$ also shares a focus with $B_2$ and lies in the plane of $B_2$, then $E$ has to also lie in the plane of $B_2$. No? – Zach Teitler Jun 7 '17 at 13:23
• You're absolutely right, sorry for not responding. I thought of another problem where the sought ellipse simply crosses the two other, but that's another job. – Maxence Seymat Jun 10 '17 at 15:29

## 2 Answers

It is a pencil (one parametric family). Instead of ellipses with one common focus, consider ellipses with the same center. The reduction is achieved by the following:

Lemma. The map of the complex plane $z\mapsto z^2$ maps ellipses centered at $0$ to ellipses with one focus at zero, bijectively.

Now consider ellipses centered at $0$. Without loss of generality, we may assume that one of the two given ellipses $E_1$ is a circle, another $E_2$ arbitrary. This can be achieved by a linear transformation of the plane $R^2$. Now choose an arbitrary line $L$ through the origin and stretch/compress $E_1$ in the direction $L$. All stretched circles are tangent to the original circle $E_1$ (at the two points of intersection of $E_1$ with the line perpendicular to $L$). Choose the amount of stretch so that the stretched circle is tangent $E_2$. The direction of the line $L_1$ is the free parameter of the pencil.

Proof of the lemma. Every ellipse centered at the origin (other than circle) can be parametrized as $t\mapsto c(re^{it}+r^{-1}e^{-it})$, $r>1$, the image of a circle $|z|=r$ under the Joukowski function. It has foci $2c,-2c$. Squaring gives $t\mapsto c^2(r^2e^{2it}+r^{-2}e^{-2it}+2)$ which has one focus at $0$. Every ellipse with one focus at $0$ (other than circle) has this form, so it is a square of a uniquely defined ellipse with center at zero.

Remark. Of course this fact facilitates space missions to other planets:-) But one of the main things in planning a mission is the energy expense consideration. This makes some times for each given mission better than the other times.

Remark 2. The Lemma I learned from Arnold ("Huygens and Barrow...", Appendix I).

• Nice reduction. I know well the problem of Apollonius, and I thought of a generalisation. I am currently working on a space flight project for an intensive foundation degree, where optimisation of fuel is the main objective. Thanks a lot, I will try to adapt this to the 3D case. – Maxence Seymat Jun 7 '17 at 12:39
• @Maxence Seymat: 1. Which problem of Apollonius? 2. If the two ellipses are not in the same plane, the third one which is tangent to both does not exist, see comments of Zach Teitler to your question. – Alexandre Eremenko Jun 7 '17 at 13:44
• See my response to Zach Teitler's comment. I was referring to the pencil of circles tangent to two given circles, maybe not exactly an Apollonius problem, but it is called Apollonian Circles. – Maxence Seymat Jun 10 '17 at 15:33

Here is another justification of the answer that there is a pencil of ellipses satisfying your conditions, via a naive dimension count. I will assume that you are interested in the plane $\mathbb{R}^2$ (as opposed to a projective plane, complex plane $\mathbb{C}^2$, etc.). There is a $5$-dimensional family of ellipses in the plane (because there is a $6$-dimensional space of quadratic equations, $a_1 x^2 + a_2 xy + a_3 y^2 + a_4 x + a_5 y + a_6 = 0$, and we lose one dimension to scaling, i.e., multiplying both sides of the equation by the same number). (To be a little more precise these are conic curves. The ellipses are an open subset, defined by a discriminant inequality.) To have a focus at a particular point imposes $2$ conditions. To meet a given ellipse at a particular point is $1$ condition; to meet a given ellipse at a particular point, and be tangent at that point, is $2$ conditions; to meet a given ellipse and be tangent to it at some point is $1$ condition. Putting it together, to have a fixed focus and be tangent to two given ellipses is $2+1+1=4$ conditions, on a $5$-dimensional family of ellipses, leaving a $1$-dimensional family of ellipses that meet your conditions.

The only potentially interesting thought I can add is that the fixed focus doesn't have to be related in any way to two given ellipses of tangency. In the above argument it was just an arbitrary point.

Finally, this naive count assumes some "general position." If, for example, the two given ellipses coincide, or if one is degenerate, then some of the $2+1+1$ conditions are redundant, and more ellipses meet your conditions.

Edit: This is NOT a proof. I don't show that there are any real solutions. Even if there are real solutions, I'm just producing a $1$-dimensional family of conic sections; they may not be ellipses.

• Unfortnately, parameter count does not prove the fact, because we need real solutions, rather than complex, and there are also inequalities in the problem conditions. Your conditions have to be explicitly written, and further analysed to prove that the system of equations has real solutions. – Alexandre Eremenko Jun 7 '17 at 7:14