The second equation must therefore be identical with the first, or can only differ from it by a constant multiplier. Hence t1t - Atzt, must be identical with c2 - Ac2. Now e2- Ac2 = 0 represents a pair of right lines passing through the intersection of C1, C2, and harmonically conjugate with them; and the equivalent form shows that these right lines join the points t1tз, tął, and t ̧t, tąłз. For t ̧tą - λtst, = 0 must denote a locus passing through these points.

Ex. 3. If 2a, 28, 27, 28 be the eccentric angles of four points on a central conic, form the equation of the diagonals of the quadrilateral formed by their tangents. Here we have

Hence reasoning, as in the last example, we find for the equations of the diagonals

264. If three conics have each double contact with a fourth, six of their chords of intersection will pass three by three through the same points, thus forming the sides and diagonals of a quadrilateral.

Let the conics be

S+L=0, S+ M2 = 0, S+N2 = 0.

By the last Article the chords will be

L-M=0, M-N=0, N-L=0;

L+M=0, M+N=0, N-L=0;
L+M=0, M-N=0, N+L=0;

L-M=0, M+N=0, N+L=0.

As in the last Article, we may deduce hence many particular theorems, by supposing one or more of the conics to break up into right lines. Thus, for example, if S break up into right lines, it represents two common tangents to S+M2, S+N2; and if I denote any right line through the intersection of those common tangents, then S+L also breaks up into right lines, and represents any two right lines passing through the intersection of the common tangents. Hence, if through the intersection of the common tangents of two conics we draw any pair of right lines, the chords of each conic joining the extremities of those lines will meet on one of the common chords of the conics.

This is the

extension of Art 116. Or, again, tangents at the extremities of either of these right lines will meet on one of the common chords.

265. If S+L, S+ M3, S+N", all break up into pairs of right lines, they will form a hexagon circumscribing S, the chords of intersection will be diagonals of that hexagon, and we get Brianchon's theorem: "The three opposite diagonals of every hexagon circumscribing a conic intersect in a point." By the opposite diagonals we mean (if the sides of the hexagon be numbered 1, 2, 3, 4, 5, 6) the lines joining (1, 2) to (4, 5), (2, 3) to (5, 6), and (3, 4) to (6, 1); and by changing the order in which we take the sides we may consider the same lines as forming a number (sixty) of different hexagons, for each of which the present theorem is true. The proof may also be stated as in Ex. 2, Art. 263. If

tt-c=0, ttc=0, tt-c2 = 0,

be equivalent forms of the equation of S, then c, = c, = c, represents three intersecting diagonals.*

266. If three conic sections have one chord common to all, their three other chords will pass through the same point.

Let the equation of one be S = 0, and of the common chord L=0, then the equations of the other two are of the form

S+ LM = 0, S+LN=0,

which must have, for their intersection with each other, L(M-N) = 0;

but M - N is a line passing through the point (MN).

According to the remark in Art. 257, this is only an extension of the theorem (Art. 108), that the radical axes of three circles meet in a point. For three circles have one chord (the line at infinity) common to all, and the radical axes are their other common chords.

* Mr. Todhunter has with justice objected to this proof, that since no rule is given which of the diagonals of tttts is c1 = + c2, all that is in strictness proved is that the lines joining (1, 2) to (4, 5) and (2, 3) to (5, 6) intersect either on the line joining (3, 4) to (6, 1), or on that joining (1, 3) to (4, 6). But if the latter were the case the triangles 123, 456 would be homologous (see Ex. 3, p. 59), and therefore the interrections 14, 25, 36 on a right line; and if we suppose five of these tangents fixed, the sixth instead of touching a conic would pass through a fixed point.

The theorem of Art. 264 may be considered as a still further extension of the same theorem, and three conics which have each double contact with a fourth may be considered as having four radical centres, through each of which pass three of their common chords.

The theorem of this Article may, as in Art. 108, be otherwise enunciated: Given four points on a conic section, its chord of intersection with a fixed conic passing through two of these points will pass through a fixed point.

Ex. 1. If through one of the points of intersection of two conics we draw any line meeting the conics in the points P, p, and through any other point of intersection B a line meeting the conics in the points Q, q, then the lines PQ, pq will meet on CD, the other chord of intersection. This is got by supposing one of the conics to reduce to the pair of lines OA, OB.

Ex. 2. If two right lines, drawn through the point of contact of two conics, meet the curves in points P, p, Q, q, then the chords PQ, pq will meet on the chord of intersection of the conics.

This is also a particular case of a theorem given in Art. 264, since one intersection of common tangents to two conics which touch reduces to the point of contact (Cor., Art. 117).

267. The equation of a conic circumscribing a quadrilateral (ay=kß8) furnishes us with a proof of "Pascal's theorem," that the three intersections of the opposite sides of any hexagon inscribed in a conic section are in one right line.

Let the vertices be abcdef, and let ab = 0 denote the equation of the line joining the points a, b; then, since the conic circumscribes the quadrilateral abcd, its equation must be capable of being put into the form

ab.cd-bc.ad= 0.

But since it also circumscribes the quadrilateral defa, the same equation must be capable of being expressed in the form

de.faef.ad= 0.

From the identity of these expressions, we have

ab.cd-de.fa= (bc — ef) ad.

Hence, we learn that the left-hand side of this equation (which from its form represents a figure circumscribing the quadrilateral formed by the lines ab, de, cd, af) is resolvable into two factors, which must therefore represent the diagonals of that quadrilateral. But ad is evidently the diagonal which joins the vertices

a and d, therefore bc- ef must be the other, and must join the points (ab, de), (cd, af); and since from its form it denotes a line through the point (bc, ef), it follows that these three points are in one right line.

268. We may, as in the case of Brianchon's theorem, obtain a number of different theorems concerning the same six points, according to the different orders in which we take them. Thus, since the conic circumscribes the quadrilateral beef, its equation can be expressed in the form.

be.cf-bc.ef=0.

Now, from identifying this with the first form given in the last Article, we have

ab.cd-be.cf=(ad - ef) bc;

whence, as before, we learn that the three points (ab, cf), (cd, be), (ad, ef) lie in one right line, viz. ad — ef = 0.

In like manner, from identifying the second and third forms. of the equation of the conic, we learn that the three points (de, cf), (fa, be), (ad, bc) lie in one right line, viz. bc - ad=0. But the three right lines

bc-ef=0, ef- ad = 0, ad - bc = 0,

meet in a point (Art. 41). Hence we have Steiner's theorem, that "the three Pascal's lines which are obtained by taking the vertices in the orders respectively, abcdef, adcfeb, afcbed, meet in a point." For some further developments on this subject we refer the reader to the note at the end of the volume.

Ex. 1. If a, b, c be three points on a right line; a', b', c' three points on another line, then the intersections (bc', b'c), (ca', c'a), (ab', a'b) lie in a right line. This is a particular case of Pascal's theorem. It remains true if the second line be at infinity and the lines ba', ca' be parallel to a given line, and similarly for cb', ab'; ac', bc'. Ex. 2. From four lines can be made four triangles, by leaving out in turn one line: the four intersections of perpendiculars of these triangles lie in a right line. Let a, b, c, d be the right lines; a', b', c', d' lines perpendicular to them; then the theorem follows by applying the last example to the three points of intersection of a, b, c with d, and the three points at infinity on a', b', c'.*

* This proof was given me independently by Prof. De Morgan and by Mr. Burnside. The theorem itself, of which another proof has been given p. 217, may also be deduced from Steiner's theorem, Ex. 3, p. 212. For the four intersections of perpendiculars must lie on the directrix of the parabola, which has the four lines for tangents. The line joining the middle points of diagonals is parallel to the axis (see Ex. 1, p. 212). It follows in the same way from Cor. 4, p. 207, that the circles circumscribing the four triangles pass through the same point, viz. the focus of the same parabola. If we are

Ex. 3. Steiner's theorem, that the perpendiculars of the triangle formed by three tangents to a parabola intersect on the directrix is a particular case of Brianchon's theorem. For let the three tangents be a, b, c; let three tangents perpendicular to them be a', b', c', and let the line at infinity, which is also a tangent (Art. 254) be. Then consider the six tangents a, b, c, c', ∞, a'; and the lines joining ab, c'∞o; bc, a'o; cc', aa' meet in a point. The first two are perpendiculars of the triangle, and the last is the directrix on which intersect every pair of rectangular tangents (Art. 221). This proof is by Mr. John C. Moore.

Ex. 4. Given five tangents to a conic, to find the point of contact of any. Let ABCDE be the pentagon formed by the tangents; then, if AC and BE intersect in O, DO passes through the point of contact of AB. This is derived from Brianchon's theorem by supposing two sides of the hexagon to be indefinitely near, since any tangent is intersected by a consecutive tangent at its point of contact (Art. 147).

269. Pascal's theorem enables us, given five points A, B, C, D, E, to construct a conic; for if we draw any line AP through

F

B

D P P

one of the given points, we can find the point F in which that line meets the conic again, and can so determine as many points on the conic as we please. For, by Pascal's theorem, the points. of intersection (AB, DE), (BC, EF), (CD, AF) are in one right line. But the points (AB, DE), (CD, AF) are by hypothesis known. If then we join these points O, P, and join to E the point in which OP meets BC, the intersection of QE with AP determines F. In other words, F is the vertex of a triangle FPQ whose sides pass through the fixed points A, E, O, and whose base angles P, Q move along the fixed lines CD, CB (see Ex. 3, p. 42). The theorem was stated in this form by MacLaurin.

Ex. 1. Given five points on a conic, to find its centre. Draw AP parallel to BC and determine the point F. Then AF and BC are two parallel chords and the line

joining their middle points is a diameter. In like manner, by drawing QE parallel

to CD we can find another diameter, and thus the centre.

given five lines, M. Auguste Miquel has proved (see Catalan's Théorèmes et Problèmes de Géométrie Elémentaire, p. 93) that the foci of the five parabolas which have four of the given lines for tangents lie on a circle (see Higher Plane Curves, Art. 146).