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If you dropped two rocks in a pond, the concentric circles emanating from the two spots would osculate $\infty$ times. The locus of osculating points would form a line.

Now imagine that instead of concentric circles, concentric ellipses emanate from the rocks' entry points. Again the concentric ellipses will osculate and again let's consider the locus of points where they do.

What is the shape of the locus?

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"Osculation" is probably not the term you want; the technical use of this is that the first and higher-order derivatives agree (e.g. the circle of curvature of a curve at a certain point has second-order osculation with the curve at said point). Two tangent circles of unequal radius would only agree up to the first derivative at the point of contact. – J. M. Dec 12 '10 at 5:17
Are you assuming anything at all about the ellipses? Eccentricity, or orientation? Also, what is your definition of concentric ellipses? The same ellipse rescaled relative to a fixed centre? Or the family of curves having the same focal points? – Willie Wong Dec 13 '10 at 1:15
@J.M.: but by definition "confocal" ellipses are also concentric. The point of my question is that a circle fixed at a center only has one free parameter, the radius. An ellipse fixed at a center has 3, the minor axis, the major axis, and the tilt. While I think we can assume that the tilt is fixed constant for concentric ellipses, I don't see a preferred scaling law given for the major and minor axis. – Willie Wong Dec 13 '10 at 11:08
"I don't see a preferred scaling law given for the major and minor axis." - I'd agree with you there, @Willie. In any event, water waves certainly do not propagate as ellipses... :) – J. M. Dec 13 '10 at 11:13
@J.M.: to be more precise. Consider the following one-paramter families (parametrized by $\lambda$; $a,b$ are fixed constants. $$ x^2/ (a\lambda)^2 + y^2/(b\lambda)^2 = 1$$ $$x^2/(a\lambda)^2 + y^2/(b\lambda^2)^2 = 1$$ $$ x^2/(a (1 + \lambda)/2)^2 + y^2/(b\lambda)^2 = 1$$ They are all families of ellipses with centre at the origin. And the $\lambda = 1$ representative of the three families all agree. But The three families are, over-all, very different families. – Willie Wong Dec 13 '10 at 11:14
up vote 1 down vote accepted

Just a remark on how to find the general answer to problems of this type. The "ripples" you are describing are level sets of some function $f_1, f_2$. Corresponding to them are their gradient vector fields $\nabla f_1, \nabla f_2$. The set of "osculating points" that you are looking for in your question is the set of points where $\nabla f_1$ and $\nabla f_2$ are parallel. So we can express it using the two-dimensional cross product $v \times w = v_x w_y - v_y w_x$ as

$$ C = \{ p\in\mathbb{R}^2 | \nabla f_1 \times \nabla f_2 = 0 \} $$

For families of confocal ellipses, the level function $f$ is of the form

$$ f(x,y) = \sqrt{(x-x_1)^2 + (y-y_1)^2} + \sqrt{(x-x_2)^2 + (y-y_2)^2} $$

since the family of ellipses are the loci of points where sum of distances to the two foci are constant. So

$$ \nabla f = \frac{(x-x_1, y-y_1)}{\sqrt{(x-x_1)^2 + (y-y_1)^2}} + \frac{(x-x_2, y-y_2)}{\sqrt{(x-x_2)^2 + (y-y_2)^2}} $$

So you see that for generic focal points, the expression defining the algebraic curve $C$ is often very complicated. You can try plotting some examples in Mathematica to see what they will look like.

In your final comment, however, it looks like you are not considering confocal ellipses. Instead, it appears you are considering functions $f$ of the form

$$ f(z) = Q(z-z_0) $$

where $z = (x,y)$ is a point in the plane, and $Q$ is a positive definite quadratic form. $z_0$ is the centre of the ellipse. (In other words, you are considering a family of ellipses that all share a centre and are dilations of each other.) In this case the answer is easier. Remember that a quadratic form can be represented by a matrix: $Q(z-z_0) = (z-z_0)^T A (z-z_0)$. Let $\epsilon$ denote the totally antisymmetric matrix

$$ \epsilon = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} $$

and take $f_1(z) = (z-z_1)^T A_1 (z-z_1)$, $f_2(z) = (z-z_2)^T A_2(z-z_2)$, we have easily that $\nabla f_1 = 2 A_1(z-z1)$ and $\nabla f_2 = 2 A_2(z-z_2)$. So the set $C$ is given by

$$ \nabla f_1 \times \nabla f_2 = 0 \iff (z-z_2)^T A_2^T \epsilon A_1 (z-z_1) = 0 $$

which implies that $C$ is described by a quadratic expression in $z$. That is, $C$ is in general given by a conic section.

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Rather the gradients $\nabla f_1, \nabla f_2$ are anti-parallel, right? – isomorphismes Dec 19 '10 at 10:16
@Willie Wong Is there a free alternative to Mathematica that I could plot them in? – isomorphismes Dec 19 '10 at 10:21
@Willie Wong I guess I don't understand the difference between concentric and confocal ellipses. I got to the point you did in the middle section of this answer. It's tantalizing to think that the problem is reducible to one of quadratic forms.... – isomorphismes Dec 19 '10 at 10:29
Interestingly, I also did this problem with $\mathcal{L}_1$-ellipses (level curves are rhombuses), and the resulting curve $C$ is an L. – isomorphismes Dec 19 '10 at 10:35
@Lao Tzu: (a) I mean parallel (which includes anti-parallel as a sub-case). Go back to the circle case, you see that the gradients can point both in the same direction and in the opposite direction. So restricting to anti-parallel is inaccurate. (b)… (c) Confocal ellipses share the same focal point. So the "smallest one" in the family is just the line segment joining the two foci, and larger ones approach the circle. "Concentric" is ill defined as a notion. – Willie Wong Dec 19 '10 at 14:14

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