B'E as punctuated lines. Let the remaining two tangents intersect BE and B'E in A, C and A', C' respectively.

Project A and C from S, A' and C' from S' by rays meeting in a and c, through which two points draw the axis of

projection xy. Let d be any third point on xy, and project d from S and S' by rays SD, S'D' intersecting BE in D and B'E in D'; then DD' will be a sixth tangent. The point of contact K is found by joining S'E meeting xy in e' and Se' meeting BE in K. The point K' is found in a similar way. The point of contact m is found by joining C'Z and CZ' intersecting in z, and by then drawing a line through the centre O and z meeting CC' in the required point of contact (see Solution II.).

PROOF.-AS will be proved in Solution II., the system of tangents enveloping the conic divide any pair of their number, such as EZ and EZ', homologously into two projective punctuated lines. Moreover, the two pencils S and S' are in perspective; that is, are in perspective of the same punctuated line; for the ray SB, passing through S and S' and forming part of the pencil whose centre is S, is correlative of the ray S'B' passing through S and S' and forming part of the pencil whose centre is S'. Wherefore, according to Lemma II., Art. 7, the pencils S and S'are perspective of the same line xy. It will be readily perceived that K, the point of contact on the tangent BE, is correlative of E, regarded as a point on B'E; and that similarly K' is the correlative of E, regarded as a point on BE.

SOLUTION II.-Newton has solved this same problem by the aid of the following two interesting Lemmas :LEMMA III.—With the system of tangents (Fig. 30) we have

AF. BG=CD2,

where CD is the half-diameter parallel to the tangents AF and BG, and EFG a third tangent intercepting AF and BG in F and G. For, by a well-known law,

From this Lemma may be deduced a very important proposition of modern projective geometry.

Thus, if a series of tangents to the conic intersect the tangent at A in a series of points CDEF, etc., and the tangent at B in the homologous series C'D'E'F', etc., we shall have

But the right-hand members of equations (2) and (3) are equal (eq. 1); hence (§ 3, Def. I.) the lines AP and BQ (Fig. 30) are homologously divided by the other tangents to the conic, and form, therefore, two projective punctuated lines. The system of enveloping tangents is termed a pencil of the second order, which may be more rigorously defined as the series of rays joining the several homologous points of two projectively punctuated lines. Therefore the lines joining the homologous points of two projectively punctuated lines situate in one plane envelop a conic.

LEMMA IV. With the system of tangents shewn in Fig. 31 we have, from Lemma III.,

By similar reasoning with respect to the point K instead of M

Thus, if we are given a system of five tangents consisting of four forming a parallelogram MILK and a fifth EH

traversing this system, we can find a sixth tangent eq

by simply choosing any point e on MI and setting off q on KI so as to fulfil this condition. Again, since a parallelogram can be circumscribed to the conic, having any pair of tangents meeting in a point for its non-parallel sides, it follows from the constancy of the product ME. KQ,