自然哲学的数学原理-14

PS X PT under the other two PS and PT, the conic section will becomea circle. And the same thing will happen if the four lines are drawn inany angles, and the rectangle PQ X PR, under one pair of the lines drawn,is to the rectangle PS X PT under the other pair as the rectangle underthe sines of the angles S, T, in which the two last lines PS, PT are drawnto the rectangle under the sines of the angles Q, R, in which the first tw134 THE MATHEMATICAL PRINCIPLES [BOOK 1.PQ, PR are drawn. In all other cases the locus of the point P will beone of the three figures which pass commonly by the name of the conicsections. But in room of the trapezium A BCD, we may substitute aquadrilateral figure whose two opposite sides cross one another like diagonals. And one or two of the four points A, B, C, D may be supposed tobe removed to an infinite distance, by which means the sides of the figurewhich converge to those points, will become parallel ; and in this case theconic section will pass through the other points, and will go the same wayas the parallels in, infinitum.LEMMA XIX.To find a point P from which if four right lines PQ, PR, PS, PT andrawn to as many other right lines AB, CD, AC, BD, given by position, each to each, at given angles, the rectangle PQ X PR, under anytwo of the lines drawn, shall be to the rectangle PS X PT, under theother tivo. in a given ratio.Suppose the lines AB, CD, to which the tworight lines PQ, PR, containing one of the rectangles, are drawn to meet two other lines, givenby position, in the points A, B, C, D. From oneof those, as A, draw any right line AH, in whichyou would find the point P. Let this cut theopposite lines BD, CD, in H and I; and, becauseall the angles of the figure are given, the ratio ofPQ to PA, and PA to PS, and therefore of PQto PS, will be also given. Subducting this ratio from the given ratio oiPQ X PR to PS X PT, the ratio of PR to PT will be given ; and adding the given ratios of PI to PR, and PT to PH, the ratio of PI to PH.and therefore the point P will be given. Q.E.I.COR. 1. Hence also a tangent may be drawn to any point D of thelocus of all the points P. For the chord PD, where the points P and Dmeet, that is, where AH is drawn through the point D, becomes a tangent.In which case the ultimate ratio of the evanescent lines IP and PH willbe found as above. Therefore draw CF parallel to AD, meeting BD inF, and cut it in E in the same ultimate ratio, then DE will be the tangent ;because CF and the evanescent IH are parallel, and similarly cut inE and P.COR. 2. Hence also the locus of all the points P may be determined.Through any of the points A, B, C, D, as A, draw AE touching the locus,and through any other point B parallel to the tangent, draw BF meetingthe locus in F ; and find the point F by this Lemma. Bisect BF in G,and, drawing the indefinite line AG, this will be the position of the diameter to which BG and FG are ordinates. Let this AG meet the locusSEC. V.J OF NATURAL PHILOSOPHY.in H, and AH will be its diameter or latus transversum.to which the latus rectum will be as BG2to AG X GH. If AG nowhere meets the locus,the line AH being infinite, the locus will be a parabola ; and its latus rectum corresponding to thediameter AG will be -.-7^ AC*But if it does meet itanywhere, the locus will be an hyperbola, whenthe points A and H are placed on the same side the point G ; and anellipsis, if the point G falls between the points A and H ; unless, perhaps,the angle AGB is a right angle, and at the same time BG2equal to therectangle AGH, in which case the locus will be a circle.And so we have given in this Corollary a solution of that famous Problem of the ancients concerning four lines, begun by Euclid, and carried onby Apollonius ; and this not an analytical calculus, but a geometrical composition, such as the ancients required.LEMMA XX.If the two opposite angularpoints A and P of any parallelogram ASPQtouch any conic section in the points A and P ; and the sides AQ, ASof one of those angles, indefinitely produced, meet the same conic sectionin B and C ; and from the points of concourse, B and C to any fifthpoint D of the conic section, two right lines BD, CD are drawn meetingtlie two other sides PS, PQ of the parallelogram, indefinitely produced in T and R ; the parts PR and PT, cut off from the sides, willalways be one to the other in a given ratio. And vice versa, if thoseparts cut off are one to the other in a given ratio, the locus of the pointD will be a conic section passing through the four points A, B, C, FCASE 1. Join BP, CP, and from the pointD draw the two right lines DG, DE, of whichthe first DG shall be parallel to AB, andmeet PB, PQ, CA in H, I, G ; and the otherDE shall be parallel to AC, and meet PC,PS, AB, in F, K, E ; and (by Lem. XVII)the rectangle DE X DF will be to the rectangle DG X DH in a given ratio. ButPQ is to DE (or IQ) as PB to HB, and consequently as PT to DH ; and by permutation PQ, is to PT as DE toDH. Likewise PR is to DF as RC to DC, and therefore as (IG or) PSto DG ; and by permutation PR is to PS as DF to DG ; and, by compounding those ratios, the rectangle PQ X PR will be to the rectanglePS X PT as the rectangle DE X DF is to the rectangle DG X DH.and consequently in "a given ratio. But PQ and PS are given, and therefore the ratio of PR to PT is given. Q.E.D.136 THE MATHEMATICAL PRINCIPLESCASE 2. But if PR and PT are supposed to be in a given ratio one tothe other, then by going back again, by a like reasoning, it will followthat the rectangle DE X DF is to the rectangle DG X DH in a givenrati) ; and so the point D (by Lem. XVIII) will lie in a conic section passing through the points A., B, C, P, as its locus. Q.E.I).COR. 1. Hence if we draw BC cutting PQ in r and in PT take Pt toPr in the same ratio which PT has to PR ; then Et will touch the conicsection in the point B. For suppose the point D to coalesce with the pointB, so that the chord BD vanishing, BT shall become a tangent, and CDand BT will coincide with CB and Bt.COR. 2. And, vice versa, if Bt is a tangent, and the lines BD, CD meetin any point D of a conic section, PR will be to PT as Pr to Pt. And,on the contrary, if PR is to PT as Pr to Pt, then BD and CD will meetin some point D of a conic section.COR. 3. One conic section cannot cut another conic section in more thanfour points. For, if it is possible, let two conic sections pass through thehve points A, B, C, P, O ; and let the right line BD cut them in thepoints D, d, and the right line Cd cut the right line PQ, in q. ThereforePR is to PT as Pq to PT : whence PR and Pq are equal one to the other,against the supposition.LEMMA XXI.If two moveable and indefinite right lines BM, CM drawn through givenpoints B, C, as poles, do by their point of concourse M describe a thirdright line MN given by position ; and other two indefinite right linesBD,CD are drawn, making with the former two at those given pointsB, C, given angles, MBD, MCD : I say, that those two right lines BD,CD will by their point of concourse D describe a conic section passingthrough the points B, C. And, vice versa, if the right lints BD, CDdo by their point of concourse D describe a conic section passingthrough the given points B, C, A, and the angle DBM is alwaysequal to the giren angle ABC, as well as the angle DCM alwaysequal to the given angle ACB, the point M will lie in a right linegiven by position, as its locus.For in the right line MN let a pointN be given, and when the moveable pointM falls on the immoveable point N. letthe moveable point D fall on an immovable point P. Join ON, BN, CP, BP,and from the point P draw the right linesPT, PR meeting BD, CD in T and R, Cand making the angle BPT c jual to thegiven angle BNM, and the angle CPRSEC. V.J OF NATURAL PHILOSOPHY. 137equal to the given angle CNM. Wherefore since (by supposition) the angles MBD, NBP are equal, as also the angles MOD, NCP, take away theangles NBD and NOD that are common, and there will remain the anglesNBM and PBT, NCM and PCR equal; and therefore the triangles NBM,PBT are similar, as also the triangles NCM, PCR. Wherefore PT is toNM as PB to NB ; and PR to NM as PC to NC. But the points, B, C,N, P are immovable: wheiefore PT and PR have a given ratio to NM,and consequently a given ratio between themselves; and therefore, (byLemma XX) the point D wherein the moveable right lines BT and CRperpetually concur, will be placed in a conic section passing through thepoints B. C, P. Q.E.D.And, vice versa, if the moveable pointD lies in a conic section passing throughthe given points B, C, A ; and the angleDBM is always equal to the given angle ABC, and the angle DCM alwaysequal to the given angle ACB, and whenthe point D falls successively on anytwo immovable points p, P, of the conicsection, the moveable point M falls successively on two immovable points /?, N.Through these points ??, N, draw the right line nN : this line nN will bethe perpetual locus of that moveable point M. For, if possible, let thepoint M be placed in any curve line. Therefore the point D will be placedin a conic section passing through the five points B, C, A, p, P, when thepoint M is perpetually placed in a curve line. But from what was demonstrated before, the point D will be also placed in a conic section passing through the same five points B, C, A, p, P, when the point M is perpetually placed in a right line. Wherefore the two conic sections will bothpass through the same five points, against Corol. 3, Lem. XX. It istherefore absurd to suppose that the point M is placed in a curve line.QE.D.PROPOSITION XXII. PROBLEM XIV.To describe a trajectory that shall pass through Jive given points.Let the five given points be A, B, C, P, D. cFrom any one of them, as A, to any other svtwo as B, C, which may be called the poles,draw the right lines AB, AC, and parallel tothose the lines TPS, PRO, through the fourthpoint P. Then from the two poles B, C,draw through the fifth point D two indefinitelines BDT, CRD, meeting with the last drawn lines TPS, PRQ (the138 THE MATHEMATICAL PRINCIPLES IBOOK Lformer with the former, and the latter with the latter) in T and R. Thendrawing the right line tr parallel to TR, cutting off from the right linesPT, PR, any segments Pt, Pr, proportional to PT, PR ; and if throughtheir extremities, t, r, and the poles B, C, the right lines lit, Cr are drawn,meeting in d, that point d will be placed in the trajectory required. For(by Lena. XX) that point d is placed in a conic section passing throughthe four points A, B, C, P ; and the lines R/ , TV vanishing, the point dcomes to coincide with the point D. Wherefore the conic section passesthrough the five points A, B, C, P, D. Q.E.D.The same otherwise.Of the given points join any three, as A, B,C ; and about two of them 15, C, as poles,making the angles ABC, ACB of a givenmagnitude to revolve, apply the legs BA,CA, first to the point D, then to the point P,and mark the points M, N, in which the otherlegs BL, CL intersect each other in both cases. CDraw the indefinite right line MN, and letthose moveable angles revolve about theirpoles B, C, in such manner that the intersection, which is now supposed tobe ???, of the legs BL, CL; or BM7 CM, may always fall in that indefiniteright line MN ; and the intersection, which is now supposed to be d, of thelegs BA ^A, or BD; CD, will describe the trajectory required, PADc/B.For (by Lem. XXI) the point d will be placed in a conic section passingthroughthe points B, C ; and when the point m comes to coincide withthe points L, M, N, the point d will (by construction) come to coincide with the points A, D, P. Wherefore a conic section will be describedthat shall pass through the five points A, B. C, P, D. Q,.E.F.COR. 1. Hence a right line may be readily drawn which shall be a tangent to the trajectory in any given point B. Let the point d come to coincide with the point B, arid the right line Bt/ Avill become the tangentrequired.COR. 2. Hence also may be found the centres, diameters, and latera rectaof the trajectories, as in Cor. 2, Lem. XIX.SCHOLIUM.The former of these constructions will be- ccome something more simple by joining ,and in that line, produced, if need be, akingBp to BP as PR is to PT ; and t rough pdraw the indefinite right inc j0e parallel to SPT, and in that line pe taking always peequal to Pi , and draw the right lines Be, CrSEC. Y.J OF NATURAL PHILOSOPHY. 139to meet in d. For since Pr to Pt, PR to PT, pB to PB, pe to Pt, are all inthe same ratio, pe and Pr will be always equal. After this manner thepoints of the trajectory are most readily found, unless you would ratherdescribe the curve mechanically, as in the second construction.PROPOSITION XXIII. PROBLEM XV.To describe a trajectory that shall pass through four given points, andtouch a right line given by position.CASE 1. Suppose that HB is thegiven tangent, B the point of contact,and C, 1., P, the three other givenpoints. Jo n BC. and draw IS parallel to BH, and PQ parallel to BC ;complete the parallelogram BSPQ.Draw BD cutting SP in T, and CDcutting PQ, in R. Lastly, draw anyline tr parallel to TR, cutting offfrom PQ, PS, the segments Pr, Pt proportional to PR, PT respectively ;and draw Cr, Bt their point of concourse d will (by Lem. XX) always fallon the trajectory to be described.The same otherwise.1 et tl e angle CBH of a given magnitude revolve about the pole B; as also the rectilinear rad: us 1C, both ways produced, about the pole C.Mark the points M, N, on which the leg BC ofthe angle cuts that radius when BH; the otherleg thereof, meets the same radius in the pointsP and D. Then drawing the indefinite line MN,let that radius CP or CD and the leg BC of theangle perpetually meet in this Ikie; and thepoint of concourse of the other leg BH with theradius will delineate the trajectory required.For if in the constructions of the preceding Problem the point A comesto a coincidence with the point B, the lines CA and CB will coincide, andthe line AB, in its last situation, will become the tangent BH ; and therefore the constructions there set down will become the same with the constructions here described. Wherefore the concourse of the leg BH withthe radius will describe a conic section passing through the points C, D,P, and touching the line BH in the point B. Q.E.F.CASE 2. Suppose the four points B, C, D, P, given, being situated withontthe tangent HI. Join each two by the lines BD, CP meeting in G,and cutting the tangent in H and I. Cut the tangent in A in such mannr:140 THE MATHEMATICAL PRINCIPLES [BOOK IX ITthat HA may be to IA as the rectangle under a mean proportional between CG andGP, and a mean proportional between BHand HD is to a rectangle under a mean proportional between GD and GB, and a meanproportional betweeen PI and 1C, and A willbe the point of contact. For if HX, a parallel to the right line PI, cuts the trajectoryin any points X and Y, the point A (by theproperties of the conic sections) will come to be so placed, that HA2 willbecome to AP in a ratio that is compounded out of the ratio of the rectangle XHY to the rectangle BHD, or of the rectangle CGP to the rectangle DGB; and the ratio of the rectangle BHD to the rectangle PIC.But after the point of contac.t A is found, the trajectory will be described asin the first Case. Q.E.F. But the point A may be taken either betweenor without the points H and I, upon which account a twofold trajectorymay be described.PROPOSITION XXIV. PROBLEM XVI.To describe a trajectory that shall pass through three given points, andtouch two right lines given by position.Suppose HI, KL to be the given tangentsand B, C, D, the given points. Through anytwo of those points, as B, D, draw the indefinite right line BD meeting the tangents inthe points H, K. Then likewise throughany other two of these points, as C, D, drawthe indefinite right line CD meeting the tangents in the points I, L. Cut the lines drawnin R and S, so that HR may be to KR asthe mean proportional between BH and HD is to the mean proportionalbetween BK and KD ; and IS to LS as the mean pioportional betweenCI and ID is to the mean proportional between CL and LD. But youmay cut, at pleasure, either within or between the points K and H, I andL, or without them ; then draw RS cutting the tangents in A and P, andA and P will be the points of contact. For if A and P are supposed tobe the points of contact, situated anywhere else in the tangents, and throughany of the points H, I, K, L, as I, situated in either tangent HI, a rightline IY is drawn parallel to the other tangent KL, and meeting the curvein X and Y, and in that right line there be taken IZ equal to a mean proportional between IX and IY, the rectangle XIY or IZ2, will (by the properties of the conic sections) be to LP2 as the rectangle CID is to the rectangle CLD, that is (by the construction), as SI is to SL2; and thereforeSEC. V.] OF NATUKAL PHILOSOPHY. 141IZ is to LP as SI to SL. Wherefore the points S, P, Z. are in one rightline. Moreover, since the tangents meet in G, the rectangle XIY or IZ2will (by the properties of the conic sections) be to IA2 as GP2is to GA2,and consequently IZ will be to IA as GP to GA. Wherefore the pointsP, Z, A, lie in one right line, and therefore the points S, P, and A are inone right line. And the same argument will prove that the points R, P,and A are in one right line. Wherefore the points of contact A and P liein the right line RS. But after these points are found, the trajectory maybe described, as in the first Case of the preceding Problem. Q,.E.F.In this Proposition, and Case 2 of the foregoing, the constructions arethe same, whether the right line XY cut the trajectory in X and Y, ornot ; neither do they depend upon that section. But the constructionsbeing demonstrated where that right line does cut the trajectory, the constructions where it does not are also known ; and therefore, for brevity ssake, I omit any farther demonstration of them.LEMMA XXII.To transform figures into other figures of the same kind.Suppose that any figure HGI is to betransformed. Draw, at pleasure, two parallel lines AO, BL, cutting any third lineAB, given by position, in A and B, and fromany point G of the figure, draw out anyright line GD, parallel to OA, till it meetthe right line AB. Then from any givenpoint in the line OA, draw to the pointD the right line OD, meeting BL in d ; andfrom the point of concourse raise the rightline dg containing any given angle with the right line BL, and havingsuch ratio to Qd as DG has to OD ; and g will be the point in the newfigure hgi, corresponding to the point G. And in like manner the severalpoints of the first figure will give as many correspondent points of the newfigure. If we therefore conceive the point G to be carried along by a continual motion through all the points of the first figure, the point g willbe likewise carried along by a continual motion through all the points ofthe new figure, and describe the same. For distinction s sake, let us callDG the first ordinate, dg the new ordinate, AD the first abscissa, ad thenew abscissa ; O the pole. OD the abscinding radius, OA the first ordinateradius, and Oa (by which the parallelogram OABa is completed) the newordinate radius.I say, then, that if the point G is placed in a right line given by position, the point g will be also placed in a right line given by position. Ifthe point G is placed in a conic section, the point g will be likewise placedJ42 THE MATHEMATICAL PRINCIPLES [BOOK 1.in a conic section. And here I understand the circle as one of the conicsections. But farther, if the point G is placed in a line of the third analytical order, the point g will also be placed in a line of the third order,and so on in curve lines of higher orders. The two lines in which thepoints G, g, are placed, will be always of the same analytical order. Foras ad is to OA, so are Od to OD, dg to DG, and AB to AD ; and there-OA X AB OA X dg fore AD is equal to , , and DG equal to 7 . Now if the ad adpoint G is placed in a right line, and therefore, in any equation by whichthe relation between the abscissa AD and the ordinate GD is expressed,those indetermined lines AD and DG rise no higher than to one dimenvv xu- ,. OA X AB . OA X dgsion, by writing this equation . m place of AD, and -. -in place of DG, a new equation will be produced, in which the new abscissa ad and new ordinate dg rise only to one dimension ; and whichtherefore must denote a right line. But if AD and DG (or either ofthem) had risen to two dimensions in the first equation, ad and dg wouldlikewise have risen to tAvo dimensions in the second equation. And so onin three or more dimensions. The indetermined lines, ad} dg in thesecond equation, and AD, DG, in the first, will always rise to the samenumber of dimensions ; and therefore the lines in which the points G, g,are placed are of the same analytical order.I say farther, that if any right line touches the curve line in the firstfigure, the same right line transferred the same way with the curve intothe new figure will touch that curve line in the new figure, and vice versa.For if any two points of the curve in the first figure are supposed to approach one the other till they come to coincide, the same points transferredwill approach one the other till they come to coincide in the new figure ;and therefore the right lines with which those points are joined will become together tangents of the curves in both figures. I might have givendemonstrations of these assertions in a more geometrical form ; but I studyto be brief.Wherefore if one rectilinear figure is to be transformed into another, we