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The Eccentric CircleGiven a line l, a point F not on l, and ε > 0, we can construct the conic with directrix l, focus F, and eccentricity ε by using an associated circle, called an eccentric circle of the conic. Let O be any point not on l and not equal to F. Let s be its perpendicular distance from l. Construct the circle with radius ε ⋅ s centered at O. Let E be a point on the circle. If EF is not parallel to l, let R be the point where it intersects l. Let P be the point where the line through F parallel to EO intersects RO, and let p be the distance from P to l. Triangle similarity implies that EO/PF = OR/PR and OP/PR = s/p. So s/p = EO/PF. Now, EO = ε ⋅ s, so PF/p = ε. So P is a point on the conic. On the other hand, if EF is parallel to l, let P and p be as before. Note that p = s. Quadrilateral EFPO is a parallelogram, so PF = EO. It follows that PF/p = EO/s = ε. So P is on the conic. The idea is that, as E travels around the circle, P sweeps out the conic. Click here to see an animation of the ellipse being swept out; click here to see an animation of the hyperbola. We discover the following:
The beauty of the eccentric circle construction is that it allows us to translate well-known properties of the circle into properties of conics. For instance, from any point in the exterior of a circle, there exist exactly two lines tangent to the circle. This allows us to prove that, from any point in the exterior of a conic (where exterior is defined as in the previous section), there exist exactly two lines tangent to the conic. To see this, let O be any point in the exterior of the conic. Construct the eccentric circle centered at O. There are exactly two tangents m and n from F to the eccentric circle. Use them to obtain points P and Q on the conic. It remains to show that OP and OQ are tangent to the conic. Note that PF is perpendicular to m. Let T be any point on OP other than P. Let t be the distance from T to the directrix, and s the distance from T to m. Then s/t = ε. On the other hand, TF > s since the shortest distance from T to m is along the perpendicular, so TF/t > ε. It follows that T is in the exterior of the conic. So OP is tangent to the conic. Similarly for OQ. This establishes that there are at least two tangents from O to the conic. On the other hand, each tangent from O to the conic produces a tangent from F to the eccentric circle, and there are only two such, so the tangents just constructed are the only tangents possible. If O happens to be on l, then it is not possible to construct the eccentric circle. The points of tangency are the points where the line perpendicular to OF at F intersects the conic. This is not difficult to prove. The fact that there is exactly one tangent to any point on a conic follows similarly. Let P be a point on the conic. The eccentric circle centered at P is the circle passing through F. There is exactly one tangent to the circle at F. Use this to produce a tangent to the conic. This establishes existence. On the other hand, given a tangent at P, the reverse of this operation produces a tangent to the circle at F, of which there is exactly one. This establishes uniqueness. |
Michael Ortiz, Ph.D.
Assistant Professor of Mathematics
Department of Natural and Behavioral Sciences
Sul Ross State University Rio Grande College
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Copyright © 2013 Michael Luis Ortiz. All rights reserved.