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—
| (α) By two pairs of conjugate points. Hence also (β) By one pair of conjugate points and the centre; (γ) By the two foci; (δ) By one focus and one pair of conjugate points; (ε) By one focus and the centre. |
7. The condition that A, B, C and A′, B′, C′ may form an involution may be written in one of the forms—
(AB, CC′) = (A′B′, C′C),
or
(AB, CA′) = (A′B′, C′A),
or