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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1848.0. "Math Magazine 1442" by RUSURE::EDP (Always mount a scratch monkey.) Tue Mar 01 1994 13:40

    Proposed by W. O. Egerland and C. E. Hansen, Aberdeen Proving Ground,
    Maryland.
    
    Prove that two ellipses with exactly one focus in common intersect in
    at most two points.
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1848.1Intersections are on a hyperbolaWIBBIN::NOYCEDEC 21064-200DX5 : 130 SPECint @ $36KFri Mar 11 1994 15:5621
An ellipse with foci A and B is the set of points X such that
	len(AX)+len(XB) = K
where K defines the "size" of the ellipse.

For our two ellipses with common focus A, we have at each intersection X
	len(AX)+len(XB) = K1
	len(AX)+len(XC) = K2
so	  len(XB) - len(XC) = K1-K2

Thus there is a hyperbola with foci B and C that passes through each
intersection.

handwaving starts here...

If the hyperbola is a straight line (K1-K2), it obviously intersects an
ellipse at most twice.

Otherwise the hyperbola curves toward one focus, and away from the other.
Suppose WLOG that it curves away from B.  It seems obvious that the
hyperbola can intersect the (A,B) ellipse in at most two points, since
it's curving away from a point in the interior of the ellipse...