hyperbola. But in such a quadrilateral the intersections of the
diagonals and the points of contact of opposite sides lie in a line
(§ 54). If therefore DEFG
(fig. 28) is such a quadrilateral,
then the diagonals
DF and GE will meet on
the line which joins the
points of contact of the
asymptotes, that is, on the
line at infinity; hence they
are parallel. From this
the following theorem is
a simple deduction:
All triangles formed by a tangent and the asymptotes of an hyperbola are equal in area.
If we draw at a point P (fig. 28) on an hyperbola a tangent, the part HK between the asymptotes is bisected at P. The parallelogram PQOQ′ formed by the asymptotes and lines parallel to them through P will be half the triangle OHK, and will therefore be constant. If we now take the asymptotes OX and OY as oblique axes of co-ordinates, the lines OQ and QP will be the co-ordinates of P, and will satisfy the equation xy = const. = a2.
 |
| Fig. 28. |
For the asymptotes as axes of co-ordinates the equation of the hyperbola is xy = const.
Involution
 |
| Fig. 29. |
§ 76. If we have two projective rows, ABC on u and A′B′C′ on u′, and place their bases on the same line, then each point in this line counts twice, once as a point in the row u and once as a point in the row u ′. In fig. 29 we denote the points as points in the one row by letters above the line A, B, C ..., and as points in the second row by A′, B′, C′ ... below the line. Let now A and B′ be the same point, then to A will correspond a point A′ in the second, and to B′ a point B in the first row. In general these points A′ and B will be different. It may, however, happen that they coincide. Then the correspondence is a peculiar one, as the following theorem shows:
If two projective rows lie on the same base, and if it happens that to one point in the base the same point corresponds, whether we consider the point as belonging to the first or to the second row, then the same will happen for every point in the base—that is to say, to every point in the line corresponds the same point in the first as in the second row.
 |
| Fig. 30. |
In order to determine the correspondence, we may assume three pairs of corresponding points in two projective rows. Let then A′, B′, C′, in fig. 30, correspond to A, B, C, so that A and B′, and also B and A′, denote the same point. Let us further denote the point C′ when considered as a point in the first row by D; then it is to be proved that the point D′, which corresponds to D, is the same point as C. We know that the cross-ratio of four points is equal to that of the corresponding row. Hence
but replacing the dashed letters by those undashed ones which denote the same points, the second cross-ratio equals (BA, DD′), which, according to § 15, equals (AB, D′D); so that the equation becomes
This requires that C and D′ coincide.
§ 77. Two projective rows on the same base, which have the above property, that to every point, whether it be considered as a point in the one or in the other row, corresponds the same point, are said to be in involution, or to form an involution of points on the line.
We mention, but without proving it, that any two projective rows may be placed so as to form an involution.
An involution may be said to consist of a row of pairs of points, to every point A corresponding a point A′, and to A′ again the point A. These points are said to be conjugate, or, better, one point is termed the “mate” of the other.
From the definition, according to which an involution may be considered as made up of two projective rows, follow at once the following important properties:
1. The cross-ratio of four points equals that of the four conjugate points.
2. If we call a point which coincides with its mate a “focus” or “double point” of the involution, we may say: An involution has either two foci, or one, or none, and is called respectively a hyperbolic, parabolic or elliptic involution (§ 34).
3. In an hyperbolic involution any two conjugate points are harmonic conjugates with regard to the two foci.
For if A, A′ be two conjugate points, F1, F2 the two foci, then to the points F1, F2, A, A′ in the one row correspond the points F1, F2, A′, A in the other, each focus corresponding to itself. Hence (F1F2, AA′) = (F1F2, A′A)—that is, we may interchange the two points AA′ without altering the value of the cross-ratio, which is the characteristic property of harmonic conjugates (§ 18).
4. The point conjugate to the point at infinity is called the “centre” of the involution. Every involution has a centre, unless the point at infinity be a focus, in which case we may say that the centre is at infinity.
In an hyperbolic involution the centre is the middle point between the foci.
5. The product of the distances of two conjugate points A, A′ from the centre O is constant: OA · OA′ = c.
For let A, A′ and B, B′ be two pairs of conjugate points, the centre, I the point at infinity, then
or
In order to determine the distances of the foci from the centre, we write F for A and A′ and get
Hence if c is positive OF is real, and has two values, equal and opposite. The involution is hyperbolic.
If c = 0, OF = 0, and the two foci both coincide with the centre. If c is negative, √c becomes imaginary, and there are no foci. Hence we may write—
| In an hyperbolic involution, | OA · OA′ = k2, |
| In a parabolic involution, | OA · OA′ = 0, |
| In an elliptic involution, | OA · OA′ = −k2. |
From these expressions it follows that conjugate points A, A′ in an hyperbolic involution lie on the same side of the centre, and in an elliptic involution on opposite sides of the centre, and that in a parabolic involution one coincides with the centre.
In the first case, for instance, OA · OA′ is positive; hence OA and OA′ have the same sign.
It also follows that two segments, AA′ and BB′, between pairs of conjugate points have the following positions: in an hyperbolic involution they lie either one altogether within or altogether without each other; in a parabolic involution they have one point in common; and in an elliptic involution they overlap, each being partly within and partly without the other.
Proof.—We have OA . OA′ = OB · OB′ = k2 in case of an hyperbolic involution. Let A and B be the points in each pair which are nearer to the centre O. If now A, A′ and B, B′ lie on the same side of O, and if B is nearer to O than A, so that OB < OA, then OB′ > OA′; hence B′ lies farther away from O than A′, or the segment AA′ lies within BB′. And so on for the other cases.
6. An involution is determined—
| |
7. The condition that A, B, C and A′, B′, C′ may form an involution may be written in one of the forms—
or
or
for each expresses that in the two projective rows in which A, B, C
