Consider an ellipse fixed at the origin of the xy-plane having semi-major axis 4 and semi-minor axis 3. A congruent ellipse rolls over the first ellipse without slipping. Initially, they share a common major axis.
Find the locus of focus (the one nearer to the origin) of the rolling ellipse, and calculate the distance traveled by the focus in one revolution (around the fixed ellipse).
The distance from the focus to the center of the ellipse is sqrt(7).
This means when the rolling ellipse begins, its focus is at (8-sqrt(7),0) and after rotating 360 degrees it is on the opposite side at (-8-sqrt(7),0)
If it rotates 180 degrees it ends up above or below the fixed ellipse, and the focus is (sqrt(7), +/- 6).
What's interesting about these easy to find points is they are on a circle with radius 8 centered on the focus of the fixed ellipse (-sqrt(7),0)
The next easiest to find points are the rotations of 90 degrees. The foci of the two ellipses are always reflections across their common tangent line. In this case, the line is x+y=5 and the point of tangency is (3.2,1.8). The focus becomes (5,5-sqrt(7)). Yes, this is 8 away from that other focus.
At 270 degrees we similarly get the point (-5,5+5sqrt(7)). Also 8 away.
So I have evidence from these eight points that the locus is a circle of radius 8.
Note: the problem is termed by revolution. The ellipse rotates 720 degrees in one revolution.
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Posted by Jer
on 2018-12-07 17:57:12 |
Evidence
