The locus of focal points of two tangent concentric ellipses is a rectangular hyperbola passing through the common tangency points of the two ellipses and having one asymptote identical to the middle line of the two parallel tangents.
The proof follows from a standard argument determining the foci from the general equation of the conic, as this is discussed in [Loney, p. 366]. According to this, if the equation of the conic is given in the form (1) F(x,y) = ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, the coordinates (x,y) of the focal points of the conic are determined by solving the system of equaitons: (2) [(ax+hy+g)2 -(hx+by+f)2]/[a-b] = F(x,y), (3) [(ax+hy+g)(hx+by+f)]/h = F(x,y). To apply these equations in simplified mode assume that the equation of one of the ellipses is given in the coordinate system centered at the center of the conic, whereas the line of the tangency points coincides with the y-axis: x = 0. Thus the general member-conic of the family under consideration is given by an equation of the form: (4) (a+k)x2 + 2hxy + by2 + c = 0, where parameter k determines the member conic of the family. For convenience I set a+k = a'. Applying (2)+(3) in this case (in which the linear term coefficients vanish) we obtain: (5) [(a'x+hy)2-(hx+by)2]/[a'-b] = [(a'x+hy)(hx+by)]/h = F(x,y). These, after some simplification reduce to
(6) x2 - y2 = [(a'-b)c]/[a'b-h2], xy = [hc]/[a'b-h2]. From this we deduce z = [x2-y2]/[xy] = [a'-b]/h = (a-b+k)/h => k = hz + (b-a) => x2-y2 = z(xy) = hz/(b(hz+b)-h2) => bh(x2-y2) + (b2-h2)xy - h = 0, which is obviously a rectangular hyperbola (see section-11 of Conic_Equation.html ). To verify the other claim about the asymptotic direction it suffices to see (ibid section-3) that the direction of tangent (-b, h) of its intersection point C with the y-axis satisfies the equation
Remark This property leads to the solution of a problem proposed by Nikolaos Dergiades (Hyacinthos Message18698) asking for the construction of an ellipse c' tangent and concentric to a given one c and with given focal points {A,B}. As Dergiades notes (ibid Message 18708) the hyperbola is easily constructible as a conic passing through five points {A,B,C,D,H}, last point being the orthocenter of triangle BCD. The property of the orthocenter H to belong to a rectangular hyperbola that passes through the vertices of a triangle is proved in OrthoRectangular.html .