Conference rusure::math

    Nope, take chords equal to the short axis. Proof by drawing:
    Just draw from each of the endpoints of the short axis three chords:
    The short axis, plus one to each side of the ellipse like drawn below.
    The locus bifurcates at the center of the ellipse.
    Note that this is always possible. Chords approximating the length of
    the short axis get the same problem.
    
      xxxxxcxxxxx
     x   c c   c x
    x   c  c  c   x
     x c   c c    x
      xxxxxcxxxxxx

    Re .1:
    
    The problem calls for proof for chords of sufficiently small length. 
    That is, you must prove that, for a given ellipse, there does not exist
    a positive number c such that the locus of midpoints of chords of
    length d, where d < c, is an ellipse.

    
    				-- edp
    
    
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	How about this:

	Project the ellips onto a circle. The property of being
	mid-point of a chord survives this projection. On the
	circle, it's trivial that the mid-points of a fixed-length
	chords describes another circle. Reproject this circle on
	the plane of the original ellips and we have another ellips.

	/Ake

    Re .3:
    
    That's a good idea, but the projection alters the lengths of the
    chords, so a set of chords of a specific length on the circle does not
    project back to a set of chords of all the same length on the ellipse.
    
    
    				-- edp
    
    
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    Combined solutions of Jordi Dou, Barcelona, Spain, and the proposer.
    
    The locus is _not_ another ellipse (unless the given ellipse is a
    circle).  Let the ellipse E satisfy the equation
    
    	x^2 / a^2 + y^2 / b^2 = 1,
    
    and let L be the locus of the midpoints of the chords of E having
    constant length 2k.  Note that the axes of E are axes of symmetry for
    the locus L.  If these axes meet L in C, C' and D, D', then (by
    plugging (c,k) and (k,d) into the equation for E)
    
    	c^2 = a^2 / b^2 * (b^2-k^2),	d^2 = b^2 / a^2 * (a^2-k^2),
    
    where 2c = CC' and 2d = DD'.  By Holditch's theorem [1] the areas E and
    L of the regions bounded by E and L satisfy
    
    	L = E - pi*k^2,
    
    so that
    
    	L^2 = (pi ab - pi k^2)^2 = pi^2 (a^2 b^2 - 2abk^2 + k^4).
    
    An ellipse whose vertices are C, C', D, D' would surround an area of A
    = pi cd so that
    
    	A^2 = pi^2 (a^2 b^2 - (a^2+b^2)k^2 + k^4).
    
    Since L^2 - A^2 = pi^2 (a-b)^2 k^2, the area surrounded by L exceeds
    the area surrounded by this new ellipse whenever a != b and k != 0.
    
    
    [1] Arne Broman, Holditch's Theorem, Math. Mag. 54:3 (1981) 99-108.