38. Problem II. The two diarneters m and n being given, and also the angle P which they make with one another; to find the axes and their direction. From the equations mn sin P=ab, and a+b=m* +12°, by a similar process to that in the foregoing problems, we deduce a=iv(m* + n° +2mn sin P) +iv (m? + n°—2mn sin P) and b=} (m? 7. n* +2mn sin P)-(m" + n° -2mn sin P.) The angle C which determines the direction of the two axes, may be found as in the preceding problem. The hgure of which we have now treated, is one of the most useful of all the conic sections. The forms of the orbit in the planatary system are elliptical, and hence a knowledge of its properties is essential in astronomy. The theory of the ellipse is necessary to establish the projection of the sphere, and the representation of a circle whether othographical or schenographical. As walls and other objects are frequently built upon elliptical plans, and arches constructed of this figure, to understand the different methods of describing it, under various circumstances, is as necessary to the architect, as it is to the workman employed under him. It would be impossible to enumerate all the uses to which this beautiful figure can be applied, not only in the research of principles, but in fanciful decoration, which receives much of its elegance from the introduction of the ellipse. OF THE HYPERBOLA. 39. The equation yy sin AB = cx sin B + xx sin (A+B –1809} shews that that the hyperbola meets its axis AP in two the point fja A K2 aa points, one of which is at A, where :=o: the other at a where * c sin B ; therefore supposing that Aa is equal to sin (A + B -180°) c sin B sin (A+B–1809)' M a will be in the opposite hy N perbola M'am'. The points B D A, a, are called the vertices of the hyperbola, the line Aa (2a) is its axis, its middle point C T P C is the centre; lastly, a right line B6 = 2 CB = 26, such 12 la bbsin A sin (A+B-180°) that. ma cos 21 B drawn perpendicular to the axis, and passing through the centre C, is called the second, less, or conjugate axis. 50. These values of a and 6 being substituted in the equation of the hyperbola, gives yy= (2ax +xx). This equation shews that the curve has two branches, AM and Am, equal and infinite in a positive direction. But if x be negative there will be no curve as long as x <- 2a; if x 7 2a the ordinates become real, and the curve will have two other branches indefinitely extended. It is easy to prove that these two branches are equal to those of the positive hyperbola MAm. For since, calling AP (-x), P'm' = a? P'M' (y), we have yy = -2ax +xx, if we make aP' t' we shall 62 have r = 2a+s', and consequently a' y*__2ax' + x'x'; an equa 62 ticn perfectly similar to that of the hyperbola MAm. 66 41. Since yy = (2ax + x.r), we have PM? : AP x Pa :: CB?: CA?. a bb aa aa Tnerefore in the hyperbola, the squares of the ordinates of the transverse acis are to the rectangles of their abscisse, (that is of the distances to the two vertices,) as the square of the second or conjugate axis is to the square of first or transverse axis. If we take the centre Č for the origin of the abscissæ, then cal bb ling CP=z, we shall have yy= (22—aa,) this equation is some aa aa z and y, we shall have MQ*=*(69+CQ“), or yy = aa what more simple than the preceding one. It gives az=(16+yy); 66 ub therefore if we draw MQ perpendicular on the conjugate axis CB, prolonged if necessary; and if we call the co-ordinates CQ, MQ, a? ’ = ?, = (6*+z), 62 66 for the equation to the co-ordinates of the second axis. 42. If a=b, the hyperbola is said to be equilateral, and its equations are yy=2ax+xx, and yy=zz-aa, according as the origin of the abscissæ is at the vertex or at the centre; and the equation to its second axis becomes yy=aa+z2. 43. If, having drawn BA, we take on both sides of the centre C, CF=Cj=BA, the points F, f, are the foci of the hyperbola. The double ordinate Dd passing through one of the foci, is called the parameter, and a line FM, or FM drawn from one of these points to any point of the curve is called the Radius vector. This premised, the distance CF=v(aa+bb). Hence FA X Fa= Conseqnently the semi-conjugate axis of the hyperbola is a mean proportional between the two distance from one of the foci to the two vertices. 6 bb The ordinate DF = therefore the para { v(aa+b)–a} ~{ v(aa+bb)+a} = bb. = . v(a* +6—ao) = a a 266_466 meter p=Dd It is therefore a third proportional 2a to the transverse and conjugate axes. The parameter of the second axes is a line q, a third proportional to the conjugate and transverse 44. If we introduce the expression of the parameter into the 66 66 equations of the hyperbola yy = (2ax +xx); and yy=-(Z2--aa) axes. aa aa we shall have yy= 1 2a (zz-aa) similarly aa the equation yy= (6b+zz) which belongs to the second or cona 66 4 62 Qa jugate axis becomes yy = (6b+zz) (ap+zz). p? р су су a a 45. Call CF=Cf=c, we shall have FM= (yy+zz--2cz+cc) -a and fM +a; hence f M-FM=2a. Therefore in the hyperbola the difference of the lines drawn from the two foci to any point of the curve is every where equal to the transverse aris. From this we may deduce an easy method of describing an hypobola, whose axis are 2a and 26. Take an interval F/=2 vaa +66), and make use of a rule fMo, longer, in proportion as you desire to have a greater portion of the hyperbola : fix one extremity of it at one of the foci, to the point f, for example, so that it may revolve freely about that point. Take then a thread FMO equal in length to f MO-2a. Fasten one of the ends of this thread to the point o of the rule, and the other to the focus F. This done, draw the rule away from the axis as far as the thread FMO will allow, and then bring it again gradually towards the axis, taking care to keep the thread always extended by means of a point or style M which glides along the rule MO. The curve described during this motion by the point M, will be an hyperbolical branch AM, since the difference of the radius-vectors will be every where equal to the transverse axis. 46. This same property will enable us to draw a tangent MT, to any point m of the hyperbola. For if we conceive they arc Mm to be infinitely little, and draw the lines fM, fm, FM, Fm, we may prove, much in the same manner as in the TAE ellipse, that the angles fmM, Mm iz F are equal, and consequently that if we bisect the angle FMF by the line MT, this line will be the required tangent This being premised, in the triangle f MF, we have fM: MF :: fT: FT and therefore by composition fM+FM:JM :: fT+FT 2cz , aa+cz :fT or aa+cz :: 2 : +c, therefore fT-C Х S P aa a aa or CT =9, which gives this proportions CP :CA :: CA : CT, by which it is easy to find the point T, and consequently to draw the tangent MT. 47. We may observe that CT being equal to aa, it is always positive as long as z is so. Therefore every tangent to the hyperbola cuts the axis in some point T, between A and C. But as the ab. scissa increases, the line CT diminishes, so that it is infinitely little or nothing when the abscissa is infinitely great. Hence we see that through the centre C we may draw two right lines cX, C& which will be the limits of the tangents to the hyperbola, these lines, whose position we shall soon determine, are called the asymptotes of the hyperbola. 48. The subtangent PT=% and the tangent bb aa+66 gd-bb aa aa 20+). If we draw the normal MN we shall have the subnore zad rallel to MP, we shall have PT : PM :: AT : AS, or aa :: a ay =bv. Now if we suppose z infinite, the z+a z+a quantity will not differ from unity. Consequently we shalı % ta then have AS-6. Hence it follows that if we draw AD and Ad perpendicular to CA, and each equal to the semiconjugate axis b, the lines CD, Cd, drawn so as to pass through the points D, d, and the centre C, will be the asymptotes to the hyperbola MAM', and if prolonged in a contrary direction, they will be those of the opposite hyperbola. If the hyperbola is equilateral, the angle DCd, made by the asymptotes, is a right angle, for then DA=Ăd=CA. The hyperbola referred to its asymptotes has many properties ; here follow the principle of them : 50. If though any point N of the asymptote we draw the straight line Nn parallel to the line Dd, we shall have CA: DA :: CP: NP B M bz or a : there L : 6::%: P bz fore NM Y, a bas and Mn="+y. Consequently NM x Mn= la a Ff |