Conic Diameters
In a circle, a chord is called a diameter if it passes through the center of the circle. One way of characterizing the diameter is this. Given a particular chord, we can consider the family of parallel chords as shown in orange. The diameter for this family is the red chord that bisects every member of the family.
The notion of a diameter being a chord that passes through the center doesn't apply to conics, because at this point we haven't even established that conics have a center. But we can still use the characterization described above. Namely, to each family of parallel chords, we associate the diameter, which is the locus of the midpoints of the chords in the family. It seems reasonable to conclude that the diameter is itself a chord. We must be careful not to let the analogy mislead us into making an unspoken assumption, however. It is far from obvious that the locus of midpoints actually sits on one line. This has to be proved.
Consider the conic with focus F and directrix l. Let PQ and RS be parallel chords, and let M and N be their midpoints. Assume first that they are not perpendicular to the directrix. Let T be the point of intersection between l and
the perpendicular from F to the chords. Let X and Y be the points where
the perpendicular intersects the chords, and let U and V be the points where
the secants intersect l. Then Then PF/PU = QF/QU, so
QF^{2}/QU^{2} = (QF^{2}– PF^{2})/(QU^{2}– PU^{2})
The Pythagorean Theorem implies that PF^{2} = PX^{2} + XF^{2} and QF^{2} = QX^{2} + XF^{2}, so
QF^{2}/QU^{2} = (QX^{2}– PX^{2})/(QU^{2}– PU^{2}) = (2MX ⋅ PQ)/(2MU ⋅ PU) = MX/MU
On the other hand,
SF^{2}/SV^{2} = NY/NV
by a similar argument. Now, QF/QU = SF/SV , so MX/MU = NY/NV.
It follows that the triangle TXM is similar to the triangle TYN, so the angles MTX and NTY are congruent, and M and N are on the same line from T.
Now suppose that PQ and RS are perpendicular to the directrix. Consider the line through F parallel to the directrix. Define U, V , X, and Y as before. In
this case, MX/MU = ε^{2} = NY/NV. It follows that MX = NY. So MN is parallel to the directrix.
Notice that we showed more than we set out to prove. Not only do the points of the diameter lie on a single line, but that line is concurrent with the directrix and the line from the focus perpendicular to the family of parallel chords. So, to determine the diameter, we can first construct the perpendicular from F to any chord in the family, and extend it through the directrix. Then we construct the line from the directrix through the midpoint of the chord. This line contains the diameter. Click here for an animation.
The diameter of an ellipse consists of a single line segment passing through the center of the ellipse (which we have yet to establish). On a hyperbola there are two types of chords — those that connect the two branches and those that lie between two points on the same branch — and the diameters come in two types accordingly. If the family of chords connects the two branches, then the diameter consists of the entire line passing through the center of the hyperbola (shown here). If each chord lies in a single branch, then the diameter consists of two rays on the same line passing through the center, each pointing in an opposite direction.
The case of the parabola will be considered next.
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