For the centre, we have 1 (lx — my — nz) = m (− lx + my — nz) = n (− lx − 'my + nz) ; lx+my - nz - lx+my+nz - lx - my+nz nl ; mn Чт ƒ3 (y + z− x)2 + g3 (z + x − y)3 + h3 (x + y − z)2 − 2gh (z+x − y) (x + y − z) − 2hf (x + y − z) (z + x − y) -2fg (y+z-x)(z+x− y) = 0, a conic touching the three straight lines which join the middle points of the sides of the triangle formed by the three given tangents. Its asymptotes are parallel to those of the curve ƒ2x2 + g3y2 + h3z2 — 2ghyz — 2hfzx — 2fgxy = 0. It will therefore be a rectangular hyperbola, if ƒaa2 + gab2 + h2c2 + (b3 + c2 − a3) gh + (c2 + a2 — b2) hf +(a2+b2 — c3) fg = 0 (Art. 3, Chap. v.), that is, if the point (f, g, h) lie anywhere on the circumference of a certain circle. or, if the point be any one of the points of intersection of the lines drawn from the angular points of the triangle with the points of contact of the four circles which touch the three sides. 15. Let four tangents be given. In this case, but one parabola can be described, and we may anticipate that the locus will be a straight line. This may be proved algebraically as follows. Take the diagonals of the quadrilateral formed by the four given tangents as lines of reference, and let the equations of the four tangents be vwl2+ wnm2 + uvn2 = 0)........... And the centre of (1) is given by the equations (1), (2). SUPPLEMENTARY PROBLEMS. 16. The following theorem is useful in many geometrical investigations. The product of any two determinants is a determinant. βιξ + β,ξ = 0 These equations lead to the following: а1 (α ̧¤ ̧+α ̧¤ ̧) +α ̧ (b ̧ï ̧+b2x) = 0, 2 (α ̧α1 + α ̧b1) x ̧ + (α ̧α, +α2b2) x2 = OL Now, if (2) be satisfied, (3) will be. In the latter case, we have by (1), either or x and x=0. But if x and x, be not =0, then (3) gives Hence (6) is satisfied whenever either (4) or (5) are, and therefore its left-hand member must involve, as factors, the left-hand members of (4) and (5). The only other factor is numerical, and this will be seen, by comparing the coefficients of any term, as for instance a,B,a,b,, to be unity. and so on, for any number of rows and columns. 17. If (f, g, h), (f', g', h'), (f", g", h") be three points which form a conjugate triad with respect to the conic $ (x, y, z) = ux2+vy2+ wz2+2u′yz + 2v′zx + 2w'xy = 0, uf +w'g+v'h, uf' + w'g' + v'h', uf" +w"g"+v'h" where is written for &c. But foƒ+gog+h&n=4 (ƒ, g, h) whatever f, g, h, may be. Similarly, all terms of similar form = 0. Hence the theorem is proved. 18. A triangle is inscribed in the conic, ux2+vy2+ wz2 + 2u′yz+2v′zx+2w'xy = 0, two of its sides passing through the fixed points (f, g, h), (f', g', h'), to find the envelope of the third. Call the fixed points K, K', and the angular points of the triangle PQR, RP passing through K, PQ through K'. Then, by projecting the conic into a circle and the line KK' to infinity, the lines RP, PQ will project into two lines always parallel to themselves, and therefore containing a constant angle, hence QR will project into a line always touching a circle concentric with the given one. Therefore, in the given problem, the envelope of QR will be a conic, having double contact with the given one along the line KK', and will therefore be represented by the equation | f, g, h [2 xp(x, y, z) + f', g, h'=0.... x, y, z (x, y, z) being written, for shortness, instead of ux2 + vy3 + wz2 + 2u'yz + 2v'zx+2w'xy, and λ being a constant to be determined. ..(1), Now we observe, in the first place, that λ must be of two dimensions in f, g, h, of two in f', g', h', and of -1 in u, v, w, u', v', w'. |