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Harmonic Conjugates
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Suppose you are given a line segment [AB]. Let P and Q be points that divide the line segment internally and externally in the same ratio. That is:
Then P and Q are said to be harmonic conjugates with respect to the segment [AB].
In the diagram on the right, Q has been constructed to be the harmonic conjugate of P. You can drag P and observe Q. (You can't drag Q.) Some things that you might observe include:
P and Q are always on the same side of the midpoint;
If P is on the segment, then Q is outside it, and vice versa;
As P approaches the midpoint, Q goes infinitely far away.
It can be shown that if M is the midpoint of [AB], then P and Q will be harmonic conjugates if |MP|.|MQ| = |MB|2.
Exercise: Prove this.
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Harmonic conjugates with respect to a segment.
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Locus of harmonic conjugates; pole & polar
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Suppose you have a circle, and a point p that is not the centre of the circle. Draw a line through p, cutting the circle at two points a and b. Let q be the harmonic conjugate of p with respect to the segment [ab]. Now, there are many different lines that we could have drawn through p so as to generate the chord [ab]. Each different chord [ab] gives a different location for q. We are interested in the locus of q as the line moves through being all such possible chords, (or, equivalently, the locus of q as the endpoint a moves around the circle.)
The situation is illustrated in the diagram on the right. Here, as you drag a around the circle, q is tracing a path.
If you want to try a different location for p, don't forget to clear the trace before you start dragging a again.
The path traced out by q is the locus of harmonic conjugates of p with respect to all chords of the circle whose chordal lines contain p. This locus has a particular name: it is called the polar of the point p with respect to the circle. Also, we refer to p as the pole corresponding to this polar.
It appears from the diagram that when p is inside the circle, the polar is a straight line outside the circle, and when p is outside the circle, the polar is a chord of the circle.
This is indeed the case, (addressed below by way of an exercise). Also, in the case where p is outside the circle, imagine drawing two tangents from p to the circle. The polar is actually the line segment joining the two points of contact of these tangents.
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You might also have noticed that the polar is perpendicular to a line joining p to the centre of the circle. (That can be seen from the diagram if you drag a to trace out the polar, and then bring it into a position that makes [ab] a diameter.)
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Anyway, here's another picture for you to play with. Needless to say, you can drag p.
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Polar of the pole p with respect to the circle
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Exercise:
(This is a tough one, unless you're really handy at your oul' geometry. It strikes me as too difficult for an exam question, unless there were some pretty extensive "hand-holding".)
(i) Let p be a point inside the circle. let r be the harmonic conjugate of p with respect to the diameter through p. Prove that the line through r perpendicular to pr is the polar of p with respect to the circle. (That is, prove that if q is any point on this line, and if qp cuts the circle at a and b, then p and q are harmonic conjugates with respect to the chord ab.)
(ii) Let p be a point outside the circle. let r be the harmonic conjugate of p with respect to the diameter which, when extended, passes through p. Let s and t be the points at which tangents from p touch the circle. Prove that the chord st passes through r, and that it is the polar of p with respect to the circle. (That is, prove that if q is any point on this line segment, and if qp cuts the circle at a and b, then p and q are harmonic conjugates with respect to the chord ab.)
Exercise:
(Not so tough)
Use (i) above to show that the equation of the polar of (x1, y1) with respect to the circle x2+y2=r2 is xx1+yy1=r2. (In the case where the pole is outside, the polar is of course only part of this line.)
And finally...
Notice that the definition of harmonic conjugates involved only ratios along a single line. Harmonic conjugates are therefore affine invariant. That is, if p and q are harmonic conjugates with respect to a segment [ab], then p' and q' are harmonic conjugates with respect to the image segment [a'b']. Clearly, if a locus is defined by a property that is affine invvariant, then "the locus of the image is the image of the locus".
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Accordingly, if we define the polar of a point with respect to an ellipse in the same way that we defined it for a circle, then an affine transformation that maps the unit circle to the ellipse will map pole/polar pairs to pole/polar pairs.
In the case of the ellipse, the polar is not in general perpendicular to diameter conatining the pole. Hence, we cannot use such a fact to find the equation of the polar with respect to an ellipse. However, we can use our usual trick of mapping from an ellipse back to the circle in order to derive the equation. Guess what: do it as an exercise!
The diagram on the right shows the polar of a point with respect to an ellipse. (The button to show the tangents only has an effect when the pole is outside the ellipse.)
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polar with respect to an ellipse
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Sin a bhfuil!
Now, toddle off with yourself and do all the past papers!
If you've found these pages interesting and/or helpful, please do e-mail me to let me know.
Likewise, if you have any comments or suggestions, please let me know those too!
In case you're interested, the interactive Java diagrams on this page were created using JavaSketchpad, a World-Wide-Web component of The Geometer's Sketchpad. The equations were created using MathType, which is a souped up version of the Equation Editor in Word. It's capable of exporting in lots of different formats, but I just got it to generate "gifs", as this doesn't place any great demands on browsers by way of plug-ins, etc. Everything else is hand-coded html (i.e. typed up, tags and all, in "notepad"! Am I a masochist or what?)
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