Joachimsthal's Notations

Ferdinand Joachimsthal (1818-1861) was a German mathematician and educator famous for the high quality of his lectures and the books he wrote. The notations named after him and discussed below serve one of the examples where the language of mathematics is especially auspicious for derivation and memorization of properties of mathematical objects. Joachimsthal's notations have had extended influence beyond the study of second order equations and conic sections, compare for example the work of F. Morley.

A general second degree equation

(1) Ax2 + 2Bxy + Cy2 + 2Fx + 2Gy + H = 0

represents a plane conic, or a conic section, i.e., the intersection of a circular two-sided cone with a plane. The equations for ellipses, parabolas, and hyperbolas all can be written in this form. These curves are said to be non-degenerate conics. Non-degenerate conics are obtained when the plane cutting a cone does not pass through its vertex. If the plane does go through the cone's vertex, the intersection may be either two crossing straight lines, a single straight line and even a point. These point sets are said to be degenerate conics. In the following, we shall be only concerned with a non-degenerate case.

The left-hand side in (1) will be conveniently denoted as s:

(2) s = Ax2 + 2Bxy + Cy2 + 2Fx + 2Gy + H

so that the second degree equation (1) acquires a very short form:

(3) s = 0.

A point P(x1, y1) may or may not lie on the conic defined by (1) or (3). If it does, we get an identity by substituting x = x1 and y = y1 into (1):

(4) Ax12 + 2Bx1y1 + Cy12 + 2Fx1 + 2Gy1 + H = 0,

which has a convenient Joachimsthal's equivalent

(5) s11 = 0.

For another point P(x2, y2) we similarly define s22 and, in general, for points P(xi, yi) or P(xj, yj) we define sii and sjj, where, for example,

(6) sii = Axi2 + 2Bxiyi + Cyi2 + 2Fxi + 2Gyi + H.

Thus, sii = 0 means that P(xi, yi) lies on the conic (3), sii ≠ 0 that it does not.

There is also a mixed notation. For two points P(xi, yi) and P(xj, yj), we define

(7) sij = Axixj + B(xiyj + xjyi) + Cyiyj + F(xi + xj) + G(yi + yj) + H.

Clearly for P(xi, yi) = P(xj, yj), (7) reduces to (6). An important observation is that sij is symmetrical in its indices:

(8) sij = sji.

The last of Joachimsthal's conventions brings the first whiff of an indication as to how useful the notations may be. In sij both P(xi, yi) and P(xj, yj) are quite generic. The indices are only needed to distinguish between two points. But if we omit the indices from one of them, the points will be as distinct as before. One additional convention accommodates this case: for points P(xi, yi) and P(x, y) we write (7) with one index only,

(9) si = Axix + B(xiy + xyi) + Cyiy + F(xi + x) + G(yi + y) + H.

The curious thing about (9) is that, although sij was probably perceived as a number, si appears to dependent on "variable" x and y and thus is mostly perceived as a function of these variables. As a function of x and y, (9) is linear, i.e. of first degree, so that si = 0 is an equation of a straight line. What straight line is it? How does it relate to the conic (1)? The beauty of Joachimsthal's notations is that the relation between s = 0 and si = 0 is quite transparent.

Theorem

Let point P(xi, yi) lie on the conic s = 0. In other words, assume that sii = 0. Then si = 0 is an equation of the line tangent to s = 0 at P(xi, yi).

Proof

Any point P(x, y) on the line through two distinct points P(x1, y1) and P(x2, y2) is a linear combination of the two points:

(10) P(x, y) = t·P(x1, x1) + (1 - t)·P(x2, x2),

which is just a parametric equation of the straight line. Substitute (10) into (2). The exercise may be a little tedious but is quite straightforward. The result is a quadratic expression in t:

(11) s(t) = t2·(s11 + s22 - 2s12) + 2t·(s12 - s22) + s22.

Line (10) and conic (1) will have 0, 1, or 2 common points depending on the number of roots of the quadratic equation s(t) = 0, which is determined by the value of the discriminant

(12)
D= (s12 - s22)2 - (s11 + s22 - 2s12)·s22
 = s122 - s11·s22.

The line is tangent to the conic iff the quadratic equation has two equal roots, i.e. when D = 0, or

(13) s122 = s11·s22.

This is an interesting identity valid for any line tangent to the conic, with P(x1, y1) and P(x2, y2) chosen arbitrarily on the line. We can use this arbitrariness to our advantage. Indeed, what could be more natural in these circumstances than picking up the point of tangency. Let's P(x1, y1) be such a point. This in particular means that the point lies on the conic so that, according to (5), s11 = 0. But then (12) implies s12 = 0. Let's say this again: If P(x1, y1) is the point of tangency of a conic and a line through another arbitrary point P(x2, y2) then

(14) s12 = 0.

Now, since this is true for any point P(x2, y2) on the tangent at P(x1, y1) we may as well drop the index. The conclusion just drops into our lap: the tangent to a conic s = 0 at point P(x1, y1) on the conic is given by

(15) s1 = 0,

which proves the theorem.

Tangent Pair

If two points P(x1, y1) and P(x2, y2) are such that the line joining them is tangent to a conic s = 0, then as in (13), s122 = s11·s22. Now, fixing P(x1, y1) outside the conic and making P(x2, y2) an arbitrary point on the tangent from P(x1, y1), we can remove the second index:

(16) s12 = s11·s.

The latter is a quadratic equation which may be factorized into the product of two linear equations each representing a tangent to the conic through P(x1, y1).

Example

Let s = x2 + 4y2 - 25, so that s = 0 is an ellipse x2 + 4y2 = 25. What are the tangents from P(0, 0) to the ellipse? Let's see that there are none. First,

  s11 = -25 and s1 = -25.

So that (16) becomes

  s = -25, or
x2 + 4y2 = 0.

Obviously the equation has no real roots (besides x = y = 0), nor linear factors. We conclude that there are no tangents from (0, 0) to the ellipse. Naturally. But let's now take a different point, say P(5, 5/2). In this case,

  s11 = 25 and s1 = 5x + 10y - 25.

(16) then becomes

  (5x + 10y - 25)2 = 25·(x2 + 4y2 - 25).

First, let's simplify this to

  (x + 2y - 5)2 = x2 + 4y2 - 25.

Second, let's multiply out and simplify by collecting the like terms:

  2xy - 5x - 10y + 25 = 0,

which is factorized into

  (x - 5)·(2y - 5) = 0.

Conclusion: here are two tangents from (5, 5/2) to the ellipse: x = 5 and y = 5/2.

Poles and Polars With Respect To a Conic

Let P(x1, y1) be a point outside a conic s = 0 and P(x2, y2) and P(x3, y3) be the points where the tangents from P(x1, y1) meet the conic.

Then the tangents have the equations (15)

(17) s2 = 0 and s3 = 0

and also meet at P(x1, y1):

(18) s21 = 0 and s31 = 0.

Because of the symmetry of the notations, we have

(19) s12 = 0 and s13 = 0,

which says that points P(x2, y2) and P(x3, y3) lie on the straight line

(20) s1 = 0.

The latter is uniquely determined by P(x1, y1), which, too, can be retrieved from (20). We define s1 = 0 as the polar of P(x1, y1) with respect to the conic s = 0. P(x1, y1) is said to be the pole of its polar. Obviously, for a point on the conic, the polar is exactly the tangent at this point.

Thus we see that the pole/polar definitions generalize naturally from the circle to other non-degenerate conics. We now prove La Hire's

Theorem

If point P(x1, y1) lies on the polar of P(x2, y2) with respect to a conic s = 0, then P(x2, y2) lies on the polar of P(x1, y1) with respect to the same conic.

Proof

Ineed, P(x1, y1) lies on the polar s2 = 0 if and only if s21 = 0. Because of the symmetry of the notations, this is the same as s12 = 0, which says that P(x2, y2) lies on s1 = 0.

Beautiful.

References

  1. D. A. Brannan et al, Geometry, Cambridge University Press, 2002

Poles and Polars

Conics

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