SUPPLEMENTARY PROBLEMS. 173 Next, let V be the point of intersection of two consecutive positions of QR. Then, if a triangle inscribed in the conic so that two of its sides always pass through K', V, the envelope of the third side will pass through K. Hence (1) must be satisfied when we exchange x, y, z, with f, g, h. λ, being what λ becomes when x, y, z are written for f, g, Hence λp (x, y, z) = λ,p (f, g, h) identically, h. whence, involves (f, g, h) as a factor. Similarly it involves (f', g, h) as a factor. Hence we may write being a function of u, v, w, u', v', w', of three dimensions, since is of - 1. The equation then becomes f, g, h 2 $ (f, g, h) $ (f', g', h') & (a, B, y) + μ | ƒ', g', h' α, ß, y = 0, To determine μ, we may suppose, since it is independent of the co-ordinates of K, K', that each of these points lies on the polar of the other.. Then, the envelope of QR must pass through the pole of KK', as may be seen by projecting KK' to infinity, for then QR will always pass through the centre of the conic. Hence, if (f", g", h") be the polar of KK' f, g, h 12 $ (f, g, h) o (f', g', h') o (f.", g′′, h') + μ ƒ', g', h' = 0. Therefore the required envelope is u, w v', v' f, g, h f', g', h' f, g, h) (f', g', h') + (x, y, z) = w', v, u' 19. A triangle is described about a conic, x2+y2+z2 = 0, two of its vertices move on fixed straight lines, lx+my+nz = 0, l'x+my+n'z = 0; to prove that the locus of the third vertex will be given by the equation 1, m, n (ll'+mm' + nn') (x2 + y2 + z2) + l′, m', n' x, y, z = = 0. It may be shewn, by reciprocating the theorem in the last article, that the locus will have double contact with the given conic along the pole of the intersection of the two given straight lines; hence its equation will be of the form For this purpose let the straight line lx+my+nz=0 cut the given conic in P, P'. Let T be the pole of PP'. Now, suppose one side of the triangle to become the tangent at P, then the other tangent through P will coincide with it, hence the required locus passes through the point of intersection of l'x + m'y + n'z = 0, with the tangent at P, and also with the tangent at P'. Now, the co-ordinates of T are l, m, n, hence these two tangents are represented by the equation (l2 + m2 + n2) (x2 + y2 + z2) − (lx + my + nz)2 = 0........(2), Hence (1) must be satisfied by the values of x, y, z, which satisfy (2), and also make = TRILINEAR CO-ORDINATES OF FOCI. 12 12 12 + m2 +n2, ll' + mm' + nn', lx +my+nz which if l'x+my+n'z = 0, becomes = == = (12 + m2 +n2) (l'2 + m22 + n'2) (x2 + y2+z2) · (ll' + mm' + nn')2 (x2 + y2 + z2) − (712 + m2 + n'2) (lx+my+nz)2 − (ll' + mm' + nn')2 (x2 + y2+ z2), if (2) be satisfied. Hence, by (1) λ (ll' + mm' + nn')2 = 0, identically, therefore, the equation of the required locus becomes l, m, n (l' + mm' + nn')2 (x2 + y2 + z3) + | l', m', n' 2 = = 0. x, y, z 175 TRILINEAR CO-ORDINATES OF THE FOCI OF A CONIC. 20. The following investigation of the trilinear co-ordinates of the foci does not introduce the conception of the imaginary circular points at infinity, or of imaginary tangents. The trilinear co-ordinates of the focus of the conic ua2 + vß2+wy2+2u' By + 2v'ya + 2w'aß=0, may be investigated in the following manner. Draw two tangents to the conic parallel to a=0, respective distances from that line. co-ordinates of a focus, we have and let f, f be their Then, if f, g, h be the (f-f) (f-ƒ) = the square on the semi-axis minor. If the equation of either tangent be a' (bB+cy) = (2A — aa') a, which represents a line parallel to, and at a distance a' from a=0, the two values of a' obtained by introducing the condition of tangency, will be f1, f. Now, the condition of tangency is ♦. +2U'ba' .ca' + 2 V 'ca' (aa' — 2A) + 2 W ' (aa' — 2A) ba' = 0, which may be written 12 (Ua2 + Vb2+Wc2 + 2 U'bc + 2 V 'ca +2 W'ab) a'2 − 4▲ (Ua + W'b + V'c) a' + 4A2U= 0. Hence, the left-hand member of this equation is identically equal to (Ua2+ Vb2 + Wc+2V'bc + 2 V'ca +2 W'ab) (a' -f) (a' —ƒ»), and therefore the square on the semi-axis minor (Ua + W'b + V'c) ƒ— AU Ua+ Vb*+ Wc+2 U'bc+2 V'ca + 2 W'ab Two similar expressions being obtained, we get for the determination of the foci, the equations = (Ua2 + Vb2 + Wc2+2U'bc + 2 V'ca +2 W'ab) f2 - 4A (Ua+ W'b+V'c)ƒ+4A3U = (Ua2 + Vb2 + We2 + 2U'bc+2 V'ca + 2 W'ab) g2 -4A (Vb+Uc+W'a) g+4A3V = (Ua3 + Vb2 + Wc2 + 2U'bc + 2 V'ca +2 W'ab) h2 − 4A (Wc + V'a + U'b) h + 4A3W, the same as those obtained in Chap. VI. Art. 33. coincides with the centre of gravity of the triangle of reference. 2. Prove that | 0, 1, 1, 1|=(x+y+z) (x−y−z) (y−z−x) (≈—x−y). 3. Prove that the square on the radius of the circle, described about the triangle of which the angular points are a, b, c, Investigate a similar expression for the square on the radius of the sphere, described about the tetrahedron of which the angular points are a, b, c, d. 4. S is a focus of a conic, PQ a chord subtending a constant angle at S; SR, ST are drawn meeting the tangents at P and Q in R, T respectively, so that the angles PSR, QST are constant ; prove that RT always touches a conic having S for a focus, and a directrix in common with the given conic. |