DE, TG, MPK, perpendicular to this axis, and ML, K parallel to it, the triangles MTL, MK, CDE, will be equal and similar. If we call CP=u, PM=%, CE=K=ML=r, MK=DE=TL=s, and as before CM=m, TM=u, CA=a, AB = b, we shall have TGz+s, CG = u+r, and % +5:u+r :: 6 : a, and consequently as tas=bu + br; besides TL : ML:; MP: PS, or s:r::%: and aazz 67 1 aasz PS= Therefore r = ; substituting this value in the equation az +as=bu+br we have (bu-as) (bu-az)= 0. Now bu-az cannot be equal to zero; therefore bu-as=o. Hence bu=as, and consequently az=br, Therefore we have CP: DE::a: 6::CE: MP. 58. Consequently to the triangles CED, CMP, are equal in surface II°. If we draw DM, we shall have DMC, or ; CDTM=to the :-98-12 su - 72 trapezium DEMP=(3+2) "") = $u+uz , 2 a? 62 -= ab; therefore the parallelo2a 26 2ab 2ab gram TT constructed on the conjugale diameters is equal to the rectangle of the axes. 62 = a = III. DE=CP Hence DE= CP'=6*+PM', and DE-PM'=6'. La IV°. CE = MP, and CE’= MP-=CP-a. Consequently 62 CP_CEʻ=a'. Voa_b=CP+PM_DE_CE:=CM_CD". Therefore the difference of the squares of any two conjugate diameters is equal to the difference of the squares of the two ares. Hence also in the equilateral hyperbola the conjugate diameters are equal. 59. Call P the angle DCM comprized between the two conjugate diameters; we shall have the two equations mn sin P=ab, and m_no=a_62; by means of these two expressions it will be easy to resolve the two following problems. Problem 1. Given the two axes a and b of an hyperbola, to find two conjugate diameters making with each other a given angle P. The preceding equations give a_6")* a' b + sin :P -6°)? a' b* * = + sin 'P There now only remains to find the direction of one these diameters, or the angle MCP which I shall call C. In the triangle CMP m=v{ } (a?—64) + [ 1+ v[ }} am am sin C we have 1 :m:: sin C: PM=m sin C. Therefore CE= 6 and in the triangle DCE we have 1; n:: cos (P+C): am sin c 6 Consequently sin C = cos P cos C-sin P sin C, and tang C = bn bn cos P 62 mn sin P am + bn sin P-6 + m2 6 Problem 11. Given two conjugate diameters m and n of a hyperbola, and the angle P which they make with each other, to find the two axes and their direction? We have first v { }(m_—-*) + } v (m* +2mno +n*_4mo no cos "p)} =v { 1 (mo—n") + } ~ (cm* +n*)_4 m n cos *p)} and then Ş=={ 1 v { }(n—m") + įv (m* +n") s mono cos -P *P)} b =v{t(n—m") + } v ((mon") + 4 mono sin P ) Lastly, we find the direction of the axes, or the angle C, as in the preceding problems. But it is better to employ the asymptotes. Through the extremity M of the principal diameter CM, draw TMt making with MQ the angle TMQ=P, and having taken TM=M=n, draw CT, Ct. Then if we bisect the angle TCt by the line CA, we shall have the direction of the transverse axis. OF SEVERAL OTHER CURVES. Among the curves most frequently used in geometry the conic sections undoubtedly occupy the first rank : but there are several other curves of which it will be necessary to give some account. I. THE CONCHOID OF NICOMEDES. 60. If through a point B taken at pleasure without a straight line GH, we draw the lines BQM, BAD, &c. such that their parts QM, AD, &c. be equal, the curve MDM which passes through the points M, D, &c. is called a conchoid. The point B is its pole, the line M D M G through the points m, d, &c. determined in this manner, will be the inferior conchoid, or rather the d lower part of the same conchoid. 61. From its construction it follows, I', that GH is the asymp B tote of the curve; II° that dD measure its greatest breadth, when BA is perpendicular to GH. But as BA may be greater, less, or equal to dA, let us investigate the figure of the curve in these three In the first case it will be such as represented in the preceding figure. M In the second case, it will have a node Bndn'. H A G m cases. In the third case the node disappears í M and there remains only a cusp, or point of reflexion at B. G 62. To find whether the conchoid is of the number of algebraic curves, draw PM perpendicular on AP, (fig. 1, 60) and call AD, or QM=a, AB =b, I CM = 111 112 PM=Y, AP=i. We shall then have PY : PM :: AQ : AB, or ✓(aa-yy):y :: 1-v(aa-yy) : 6. Consequently xy = 6+yv (aa-yy); this is the equation to the co-ordinates of the upper or superior conchoid. The same calculus gives xy=b-y v (aa-yy) for the inferior conchoid. The equation is still the same for the nodated conchoid. This curve is therefore algebraic, and disembarrassing its equation from the radical quantity we shall find that it is a line of the fourth order or a curve of the third, having for its equation yt+2by' + (64-a2 +) ya 2a by=a* 6* It may be described by the continual intersection of a rule BCM, moveable about the point B, with a circle G described with the radius :a, made to move along GH, so that the centre C may always be on that line. For this purpose it B is only necessary that the rule should constantly pass through the centre of the circle. 63. We may even construct an indefinite number of different conchoids. For if instead of a circle we D cause any curve CM to move along GH, its intersection with a rule BM moveable about the point B, and subjected to pass through a fixed point Q in the axis ΔΑ 1 P C G of the curve CM, will describe a conchoid of which the equation will be easily found. For if we draw MP and B В AB perpendicular on the directrix, and suppose AP=x, PM=y, CP=2, CQ=a, AB=b, we shall have PQ : PM::AQ : AB, or s-ary::+a_%:6; hence z=a+ ху Then substituting this value in the equation of the curve CM, we shall have that of the conchoid MD. For example if the curve CM is a circle, whose centre is Q, we 2az-zz, which gives xy=(6+y) v(aa-yy) for the equation of the ordinary conchoid. 64. But if the moveable curve is a parabola whose equation y'= pz, then ys + by — apy-apb=pxy becomes the equation of the parabolic conchoid which Descartes employed to resolve a general equation of the sixth degree. M b+y have yy II. THE CISSOID OF DIOCLES. TGC c 12 65. Let ANBn be a circle whose diame. ter is AB, and to which QBq is a tangent at the point B. If after having drawn from the point A the right lines AQ to different M м points of the tangent, we take QM = AN, the curve MAm which passes through the N the points M, m, thus determined is called a cissoid. A л P B We easily perceive that it is composed of two similar and equal parts AM, Am forming at A, a cusp, or point of reflexion, and which after having cut the circumference at the points C, c, equally distant from A and B, recede indefinitely from each other without ever reaching the tangent QBq, which therefore is their asymptote. 66. To find the equation of the cissoid, draw OM parallel to AP, and MP, NG perpendicular to d. Call AP=x, PM=y, and AB or the diameter of the generating circle=a. Since AN = MQ we have AG=PB, and AG : GN :: AP : PM that is, -a: (xx-ax):: . : y = Na-x) 23 consequently y = the equation required. 67. This equation shews l'. That the cissoid is an algebraic curve of the second order ; II'. That to each abscissa AP correspond two equal ordinates PM, Pm, one positive, the other negative; and therefore that the curve has too perfectly equal and similar branches ; III°. That when x = 0, y is also =0; consequently the curve passes through the origin of the abscissæ ; IV°. That when x = { a, then y = + ļa; that is, the two branches of the cissoid cut the circumference in two points C, c, equally distant from A and B; Vo. That if x=a, y is infinite, and therefore the tangent BQ is the asymptote of this curve, as we had already concluded from its description. The conchoid and cissoid were employed by their inventors Nicomedes and Diocles to find the duplication of the cube, a problem celebrated among ancient geometers, but which is no longer considered as either difficult or interesting by modern analisis. |