Monday Math 120

Given a (non-degenerate) conic section, other than a parabola, draw two parallel lines, each of which intersects the conic at two points. Next, find the midpoints of these two parallel chords. Then the center of the conic section lies on the line connecting these two midpoints. This can be used to find the center (and thus the axis and focus/foci) of any conic with only straightedge and compass.

Here, we will give an analytic proof. Let our conic be written as . If p and q are both positive, we have an ellipse (or a circle, if p=q); if they differ in sign, we have a hyperbola. In all cases, the center is at the origin. Given the line , we substitute to find the intersections:

;
If the line has two intersections with the conic, then the discriminant of the above quadratic is positive. The x-coordinate of the midpoint is the average of the solutions to the above quadratic; considering the quadratic formula, this is
;
since the midpoint lies on the line, the y-coordinate is
.
The line connecting this to the origin has slope
.
Note that this is independent of the intercept b of the intersecting line. Thus, a parallel line will give a midpoint that gives the same slope, and thus lies on the same line through the center.

For the parabola, we note that it can be considered the limiting case of an ellipse, or hyperbola, as the eccentricity approaches 1 while the center goes to infinity; hence, a line through the midpoints of paralell chords, and thus through the center, will in the limiting case of the parabola will be parallel to the axis of the parabola (the path of the center as it moves toward infinity).

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