Locus of osculation of concentric ellipses (elliptic pond ripples) 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?
 A: 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. 
