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 follows at once from Steiner's theorem, Ex. 3, p. 201. For the four intersections of perpendiculars must lie on the directrix of the parabola, which has the four lines for tangents. It follows in the same way from Cor. 4, p. 196, that the circles circumscribing the four triangles pass through the same point, viz. the focus of the same parabola. If we are 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. 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'¤; bc, a'∞; 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 (p. 144). 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 B E 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 Q 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, 0, 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. Ex. 2. Given five points on a conic, to draw the tangent at any one of them. The point F must then coincide with A, and the line QF drawn through E must therefore take the position q4. The tangent therefore must be pA. Ex. 3. Investigate by trilinear co-ordinates (Art. 62) Mac Laurin's method of generating conics. In other words, find the locus of the vertex of a triangle whose sides pass through fixed points and base angles move on fixed lines. Let a, ẞ, y be the sides of the triangle formed by the fixed points, and let the fixed lines be la + mẞ + ny = 0, l'a + m'ß + n'y = 0. Let the base be auß. Then the line joining to ẞy, the intersection of the base with the first fixed line, is (lu + m) B+ ny = 0. And the line joining to ay, the intersection of the base with the second line, is (l'μ + m') a + n'μy = 0. Eliminating μ from the last two equations, the equation of the locus is found to be Im'aß = (mẞ + ny) (l'a + n'y), a conic passing through the points By, ya, (a, la + mß + ny), (ẞ, l'a + m'ß + n'y). EQUATION REFERRED TO TWO TANGENTS AND THEIR CHORD. 270. It much facilitates computation (Art. 229) when the position of a point on a curve can be expressed by a single variable and this we are able to do in the case of two of the principal forms of equations of conics already given. First, let L, M be any two tangents and R their chord of contact. Then the equation of the conic (Art. 252) is LM = R2; and if μL = R be the equation of the line joining LR to any point on the curve, (which we shall call the point μ), then substituting in the equation of the curve, we get M=μR and μ3L=M_for_the equations of the lines joining the same point to MR and to LM. Any two of these three equations therefore will determine a point on the conic. The equation of the chord joining two points on the curve μ, μ', is μμ'I − (μ + μ') R + M= 0. For it is satisfied by either of the suppositions (μL=R, μR=M), (μ'L=R, μ'R=M). If μ and μ' coincide we get the equation of the tangent, viz. μ3L−2μR + M = 0. Conversely, if the equation of a right line (μL-2μR+M=0) involve an indeterminate μ in the second degree, the line will always touch the conic LM = R2. 271. To find the equation of the polar of any point. The co-ordinates L', M', R' of the point substituted in the equation of either tangent through it, give the result μ3L' — 2μR' + M' = 0. Therefore the co-ordinates of the point of contact satisfy the equation ML' — 2RR' + LM' = 0, which is that of the polar required. If the point had been given as the intersection of the lines aL=R, bR= M, it is found by the same method that the equation of the polar is abL-2aR+M=0. 272. In applying these equations to examples it is useful to take notice that, if we eliminate R between the equations of two tangents μ3L - 2μR+M=0, μ'L-2μ'R+M=0, we get up'LM for the equation of the line joining LM to the intersection of these tangents. Hence, if we are given the product of two u's, μμ' = a, the intersection of the corresponding tangents lies on the fixed line aL = M. In the same case, substituting a for μu' in the equation of the chord joining the points, we see that that chord passes through the fixed point (aL+ M, R). Again, since the equation of the line joining any point μ to LM is μ3L=M, the points +μ, μ lie on a right line passing through LM. - Lastly, if LM = R2, LM = R" be the equations of two conics having L, M for common tangents; then since the equation μ3L= M does not involve R or R', the line joining the point +μ on one conic to either of the points μ on the other, passes through LM the intersection of common tangents. We shall say that the point +μ on the one conic corresponds directly to the point +μ and inversely to the point -μ on the other. And we shall say that the chord joining any two points on one conic corresponds to the chord joining the corresponding points on the other. Ex. 1. Corresponding chords of two conics intersect on one of the chords of intersection of the conics. The conics LM - R2, LM - R'2 have R2 R2 for a pair of common chords. But the chords μμ'L − (μ + μ') R + M = 0, μμ'L − (μ + μ') R' + M = 0, evidently intersect on R - R'. And if we change the signs of μ, μ' in the second equation, they intersect on R + R'. Ex. 2. A triangle is circumscribed to a given conic; two of its vertices move on fixed right lines: to find the locus of the third. Let us take for lines of reference the two tangents through the intersection of the fixed lines, and their chord of contact. Let the equations of the fixed lines be Now we proved (Art. 272) that two tangents which meet on aL - M must have the product of their μ's = a; hence, if one side of the triangle touch at the point μ, α b 9 μ the others will touch at the points and their equations will be μ can easily be eliminated from the last two equations, and the locus of the vertex is found to be LM= 4ab R2, the equation of a conic having double contact with the given one along the line R.* Ex. 3. To find the envelope of the base of a triangle, inscribed in a conic, and whose two sides pass through fixed points. Take the line joining the fixed points for R, let the equation of the conic be LM R2, and those of the lines joining the fixed points to LM be = aL + M = 0, bL+ M = 0. Now, it was proved (Art. 272) that the extremities of any chord passing through (aL+M, R) must have the product of their μ's a. Hence, if the vertex be a b base angles must be and μ and the equation of the base must be The base must, therefore (Art. 270), always touch the conic the a conic having double contact with the given one along the line joining the given points. Ex. 4. To inscribe in a conic section a triangle whose sides pass through three given points. Two of the points being assumed as in the last Example, we saw that the equation of the base must be Now, if this line pass through the point cL - R = 0, dR – M = 0, we must have Now, at the point μ we have μL = R, μ2L = M; hence the co-ordinates of this point must satisfy the equation * This reasoning holds even when the point LM is within the conic, and therefore the tangents L, M imaginary. But it may also be proved by the methods of the next section, that when the equation of the conic is L2+ M2 R2, that of the locus is of the form L2 + M2 = k2 R2. |