first; and if we make k=0, we find that the point of meeting of the first line with a consecutive line of the system is determined by the equations μ3L− 2μR+M=0, μL-R=0; or, what comes to the same thing, by the equations μL-R=0, μR – M=0. Now since any point on a curve may be considered as the intersection of two of its consecutive tangents (Art. 147), the point where any line meets its envelope is the same as that where it meets a consecutive tangent to the envelope; and therefore the two equations last written determine the point on the envelope which has the line 'L-2μR+ M for its tangent. And by eliminating μ between the equations we get the equation of the locus of all the points on the envelope, namely LM = R'. A similar argument will prove, even if L, M, R do not represent right lines, that the curve represented by u3L−2μR+ M always touches the curve LM = R". The envelope of L cosø + M sino-R, where is indeterminate, may be either investigated directly in like manner, or may be reduced to the preceding by assuming tan10=μ, when on substituting and clearing of fractions, we get an equation in which μ only enters in the second degree. 284. We might also proceed as follows: The line μ*L-2μR+M is obviously a tangent to a curve of the second class (see note, p. 147); for only two lines of the system can be drawn through a given point: namely, those answering to the values of μ determined by the equation where L', R', M' are the results of substituting the coordinates of the given point in L, R, M. Now these values of μ will vidently coincide, or the point will be the intersection of two consecutive tangents, if its coordinates satisfy the equation LM-R. And, generally, if the indeterminate μ enter algebraically and in the nth degree, into the equation of a line, the line will touch a curve of the nth class, whose equation is found by expressing the condition that the equation in μ shall have equal roots. Ex. 1. The vertices of a triangle move along the three fixed lines a, ß, y, and two of the sides pass through two fixed points a'ß'y', a"ß"y", find the envelope of the third side. Let a + uß be the line joining to aß the vertex which moves along y, then the equations of the sides through the fixed points are V = 0. y' (a + μß) − (a' + μß′) y = 0, y" (a + μß) – (a” + μß") y = And the equation of the base is for it can be easily verified that this passes through the intersection of the first line with a, and of the second line with B. Arranging according to the powers of μ, we find for the envelope (aß'y" + By'a” — ya'′ß” — ya′′ß')2 = 4a′ß" (ay" — a′′z) (By' – B'y). This example may also be solved by arranging according to the powers of a, the equation in Ex. 3, p. 49. Ex. 2. Find the envelope of a line such that the product of the perpendiculars, on it from two fixed points may be constant. Take for axes the line joining the fixed points and a perpendicular through its middle point, so that the coordinates of the fixed points may be y = 0, x = c; then if the variable line be y - mx + n = 0, we have by the condition of the question Ex. 3. Find the envelope of a line such that the sum of the squares of the perpendiculars on it from two fixed points may be constant. Ans. 2x2 2y2 + = 1. b2-2c2 62 Ex. 4. Find the envelope if the difference of squares of perpendiculars be given. Ans. A parabola. Ex. 5. Through a fixed point O any line OP is drawn to meet a fixed line; to find the envelope of PQ drawn so as to make the angle OPQ constant. Let OP make the angle with the perpendicular on the fixed line, and its length is p sec; but the perpendicular from 0 on PQ makes a fixed angle ß with OP, therefore its length is p sec cosẞ; and since this perpendicular makes an angle =0+ẞ with the perpendicular on the fixed line, if we assume the latter for the axis of x, the equation of PQ is x cos (0+ẞ) + y sin (0 + ẞ) = p sec 8 cos ẞ, or ☛ cos (20+ ẞ) + y sin (20 + B) = 2p cos ẞ- x cosẞ — y sin ß, an equation of the form whose envelope, therefore, is x2 + y2 = (x cos ẞ + y sin ß - 2p cos ẞ)", the equation of a parabola having the point O for its focus. Ex. 6. Find the envelope of the line connected by the relation μ + μ' = C. A B +=1, where the indeterminates are μ μ We may substitute for ', C-μ, and clear of fractions; the envelope is thus found to be A2 + B2 + C2 — 2AB — 2AC – 2BC = 0, an equation to which the following form will be found to be equivalent, ±√A ± JB± JC = 0. Thus, for example,-Given vertical angle and sum of sides of a triangle to find the envelope of base. In like manner,-Given in position two conjugate diameters of an ellipse, and the sum of their squares, to find its envelope. The ellipse, therefore, must always touch four fixed right lines. 285. If the coefficients in the equation of any right line λa+μß + vy be connected by any relation of the second order in λ, H, V, Aλ* + Bμ* + Cv2 + 2 Fμv + 2 Gvλ + 2Hλμ = 0, the envelope of the line is a conic section. Eliminating v between the equation of the right line and the given relation, we have (Ay* — 2 Gya + Ca3) λ2 + 2 (Hy3 – Fya - Gyß + Caß) λμ and the envelope is +(By* −2Fyß + Cß3) μ3 = 0, (Ay3—2 (¥ya+ Ca3) (By3—2Fyß+ Cß3) = (Hy3 −Fya—Gyß+Caß)". Expanding this equation, and dividing by y3, we get (BC − F”) x2 + (CA — G2) ẞ2 + (AB – H3) Ÿ3 2 + (GH – AF) By + 2 (HF − BG) ya + 2 (FG − CH) aß = 0. The result of this article may be stated thus: Any tangential equation of the second order in λ, μ, v represents a conic, whose trilinear equation is found from the tangential by exactly the same process that the tangential is found from the trilinear. For it is proved (as in Art. 151) that the condition that λα + β + my shall touch aa2 + bß* + cy2+2ƒBy+2gya+2haß = 0, or, in other words, the tangential equation of that conic is (bc — ƒ3) λ2 + (ca − g3) μ3 + (ab — h2) v2 ν + 2 (gh− af ) μv + 2 (hƒ − bg) vλ + 2 (ƒg — ch) Xμ = 0. Conversely, the envelope of a line whose coefficients λ, μ, v fulfil the condition last written, is the conic aa+&c. = 0; and this may be verified by the equation of this article. For, if we write for A, B, &c., bc-ƒ", ca- g3, &c., the equation (BC - F) a+ &c. = 0 becomes (abc+2fgh—aƒ”—bg”—ch2) (aa2+bB*+cy3 +2ƒBy+2gya+ 2haß)=0. Ex. 1. We may deduce, as particular cases of the above, the results of Arts. 127, F G H 130, namely, that the envelope of a line which fulfils the condition +-+ =0 is (Fa) + (GB) + √(Hy) = 0; and of one which fulfils the condition F G H (FA) + (Gu) + (Hz) == 0 is + + = 0. a B Ex. 2. What is the condition that λa + μß + vy should meet the conic given by the general equation in real points? Ans. The line meets in real points when the quantity (bc −ƒ ̃2) λ2 + &c. is negative; in imaginary points when this quantity is positive; and touches when it vanishes. Ex. 3. What is the condition that the tangents drawn through a point a'ß'y' should be real? Ans. The tangents are real when the quantity (BC — F2) a22 + &c. is negative; or, in other words, when the quantities abc + 2ƒgh + &c. and aa"2 + bß" + &c. have opposite signs. The point will be inside the conic and the tangents imaginary when these quantities have like signs. 286. It is proved, as at Art. 76, that if the condition be fulfilled, ABC+2FGH-AF - BG - CH2 = 0, then the equation Ax2 + Bμ* + Cv3 +2 Fμv + 2 Gvλ + 2Hλμ = 0 may be resolved into two factors, and is equivalent to one of the form (aλ + B'μ + ý'v) (a′′λ + ß′′μ + y′′v) = 0. And since the equation is satisfied if either factor vanish, it denotes (Art. 51) that the line λa+μẞ+vy passes through one or other of two fixed points. If, as in the last article, we write for A, bc-f2, &c., it will be found that the quantity ABC+2FGH+ &c. is the square of abc + 2fgh+ &c. Ex. If a conic pass through two given points and have double contact with a fixed conic, the chord of contact passes through one or other of two fixed points. For let S be the fixed conic, and let the equation of the other be S = (λa + μß + vy). Then substituting the coordinates of the two given points, we have whence S' = (λa' + μß' + vy′)2; _S" = (λa” + μß" + vy”)2; (λa' + μß' + vy) √(S′′) = ± (\a′′ + μß" + vy") √(S'), showing that a + μẞ+vy passes through one or other of two fixed points, since S', S" are known constants. 287. To find the equation of a conic having double contact with two given conics, S and S'. Let E and F be a pair of their chords of intersection, so that S- S'EF; then represents a conic having double contact with S and S'; for it may be written μ (μE+F)2=1μS, or (μE - F)* = 4μ S'. Since is of the second degree, we see that through any point can be drawn two conics of this system; and there are three such systems, since there are three pairs of chords E, F. If S' break up into right lines, there are only two pairs of chords distinct from S', and but two systems of touching conics. And when both S and S' break up into right lines there is but one such system. Ex. Find the equation of a conic touching four given lines. Ans. μ2E2 - 2μ (AC + BD) + F2 = 0, where A, B, C, D are the sides; E, F the diagonals, and AC - BD = EF. Or more symmetrically if L, M, N be the diagonals, L + M±N the sides, μ2 L2 — μ (L2 + M2 − N2) + M2 = 0. For this always touches 4L2M — (L2 + M2 — N2)2 = (L + M + N) (M + N − L) (L + N − M) (M + L – M). |