自然哲学的数学原理-13

difference between the latus rectum and 4DS.COR. 3. Hence also if a body move in any conic section, and is forcedout of its orbit by any impulse, you may discover the orbit in which it willafterwards pursue its Bourse. For bv compounding the proper motion oiSEC. IV.] OF NATURAL PHILOSOPHY. 125the body with that motion, which the impulse alone would generate, youwill have the motion with which the body will go off from a given placeof impulse in the direction of a right line given in position.COR. 4. And if that body is continually disturbed by the action of someforeign force, we may nearly know its course, by collecting the changeswhich that force introduces in some points, and estimating the continualchanges it will undergo in the intermediate places, from the analogy thatappears in the progress of the series.SCHOLIUM.If a body P, by means of a centripetalforce tending to any given point R, movein the perimeter of any given conic section whose centre is C ; and the law ofthe centripetal force is required : drawCG parallel to the radius RP, and meeting the tangent PG of the orbit in G ;and the force required (by Cor. 1, andCG3Schol. Prop. X., and Cor. 3, Prop. VII.) will be as -SECTION IV.Of the finding of elliptic, parabolic, and hyperbolic orbits, from ttu.focus given.LEMMA XV.Iffrom the two foci S, II, of any ellipsis or hyberbola, we draw to anythird point V the right lines SV, HV, whereof one HV is equal to theprincipal axis of the figure, thai is, to the axis in which the foci aresituated, the other, SV, is bisected in T by t/ie perpendicular TR letfall upon it ; that perpendicular TR will somewhere touch the conicsection : and, vice versa, if it does touch it, HV will be equal to theprincipal axis of the figure.For, let the perpendicular TR cut the right lineHV, produced, if need be, in R ; and join SR. Because TS, TV are equal, therefore the right lines SR,VR, as well as the angles TRS, TRV, will be alsoequal. Whence the point R will be in the conic section, and the perpendicular TR will touch the same ; and the contrary. Q.E.D.126 THE MATHEMATICAL PBINCIP, -ES [BOOK 1PROPOSITION XVIII. PROBLEM X.From a focus and the principal axes given, to describe elliptic and hyperbolic trajectories, which shall pass through given points, and touchright lines given by position.Let S be the common focus of the figures ; AB A 33the length of the principal axis of any trajectory ; r p T~*P a point through which the trajectory should 。 /Rpass ; and TR a right line which it should touch. / 。About the centre P, with the interval AB SP, 。 S~~yfif the orbit is an ellipsis, or AB {- SP, if they>G ^orbit is an hyperbola, describe the circle HG. On the tangent TR let fallthe perpendicular ST, and produce the same to V, so that TV may beequal to ST; and about V as a centre with the interval AB describe thecircle FH. In this manner, whether two points P, p, are given, or twotangents TR, tr, or a point P and a tangent TR, we are to describe twocircles. Let H be their common intersection, and from the foci S, H, withthe given axis describe the trajectory : I say, the thing is done. For (because PH -f- SP in the ellipsis, and PH SP in the hyperbola, is equalto the axis) the described trajectory will pass through the point P, and (bythe preceding Lemma) will touch the right line TR. And by the sameargument it will either pass through the two points P, p, or touch the tworight lines TR, tr. Q.E.F.PROPOSITION XIX. PROBLEM XI.About a given focus, to describe a parabolic trajectory, which shall passthrough given points, and touch right lines given by position.Let S be the focus, P a point, and TR a tangent ofthe trajectory to be described. About P as a centre,with the interval PS, describe the circle FG. Fromthe focus let fall ST perpendicular on the tangent, andproduce the same to V, so as TV may be equal to ST.After the same manner another circle fg is to be described, if another point p is given ;or another point vis to be found, if another tangent tr is given; then drawthe right line IF, which shall touch the two circles YG,fg, if two pointsP, p are given ; or pass through the two points V, v, if two tangents TR,tr, are given : or touch the circle FG, and pass through the point V, if thepoint P and the tangent TR are given. On FI let fall the perpendicularSI, and bisect the same in K ; and with the axis SK and principal vertex Kdescribe a parabola : I say the thing is done. For this parabola (becauseSK is equal to IK, and SP to FP) will pass through the point P ; and/KSSEC. IV.] OF NATURAL PHILOSOPHY. 127(by Cor. 3, Lem. XIV) because ST is equal to TV. and STR a light angle, it will touch the right line TR. Q.E.F.PROPOSITION XX. PROBLEM XII.About a given focus to describe any trajectory given in specie which shahpass through given points, and touch right lines given by position.CASE 1. About the focus S it is reuiredto describe a trajectory ABC, passing through two points B, C. Because thetrajectory is given in specie, the ratio of theprincipal axis to the distance of the foci GAS Hwill be given. In that ratio take KB to BS, and LC to CS. About thecentres B, C, with the intervals BK, CL, describe two circles ; and on theright line KL, that touches the same in K and L, let fall the perpendicular SG ; which cut in A and a, so that GA may be to AS, and Ga to aS,as KB to BS ; and with the axis A., and vertices A, a, describe a trajectory :I say the thing is done. For let H be the other focus of the describedfigure, and seeing GA is to AS as Ga to aS, then by division we shallhave Ga GA, or Aa to S AS, or SH in the same ratio; and thereforein the ratio which the principal axis of the figure to be described has tothe distance of its foci; and therefore the described figure is of the samespecies with the figure which was to be described. And since KB to BS,and LC to CS, are in the same ratio, this figure will pass through thtpointsB, C, as is manifest from the conic sections.CASE 2. About the focus S it is required todescribe a trajectory which shall somewheretouch two right lines TR, tr. From the focuson those tangents let fall the perpendicularsST, St, which produce to V, v, so that TV, tvmay be equal to TS, tS. Bisect Vv in O, and jerect the indefinite perpendicular OH, and cut I.the right line VS infinitely produced in K and Vk, so that VK be to KS, and VA* to A~S, as the principal axis of the trajectory to be described is to the distance of its foci. On the diameterK/J describe a circle cutting OH in H ; and with the foci S, H, andprincipal axis equal to VH, describe a trajectory : I say, the thing is done.For bisecting Kk in X, and joining HX, HS, HV, Hv, because VK is toKS as VA- to A*S; and by composition, as VK -f- V/c to KS + kS ; andby division, as VA* VK to kS KS, that is, as 2VX to 2KX, and2KX to 2SX, and therefore as VX to HX and HX to SX, the trianglesVXH, HXS will be similar; therefore VH will be to SH as VX to XH ;and therefore as VK to KS. Wherefore VH, the principal axis of thedescribed trajectory, has the same ratio to SH, the distance of the foci, as12S THE MATHEMATICAL PRINCIPLES [BOOK 1.K Sthe principal axis of the trajectory which was to be described has to thedistance of its foci; and is therefore of the same species. Arid seeing VH,vH are equal to the principal axis, and VS, vS are perpendicularly bisectedby the right lines TR, tr, it is evident (by Lem. XV) that those rightlines touch the described trajectory. Q,.E.F.CASE. 3. About the focus S it is required to describe a trajectory, whichshall touch a right line TR in a given Point R. On the right line TRJet fall the perpendicular ST, which produce to V, so that TV may beequal to ST ; join VR, and cut the right line VS indefinitely producedin K and k, so. that VK may be to SK, and V& to SAr, as the principalaxis of the ellipsis to be described to the distance of its foci ; and on thediameter KA: describing a circle, cut the Hright line VR produced in H ; then withthe foci S, H, and principal axis equal to RVH, describe a trajectory : I say, the thing .---is done. For VH is to SH as VK to SK, V" "1and therefore as the principal axis of the trajectory which was to be described to the distance of its foci (as appears from what we have demonstrated in Case 2) ; and therefore the described trajectory is of the samespecies with that which was to be described ; but that the right line TR,by which the angle VRS is bisected, touches the trajectory in the point R,is certain from the properties of the conic sections. Q.E.F.CASE 4. About the focus S it is rrequired to describe a trajectoryAPB that shall touch a right lineTR, and pass through any givenpoint P without the tangent, andshall be similar to the figure apb,described with the principal axis ab,and foci s, h. On the tangent TRlet fall the perpendicular ST, which / .. ,.---""produce to V, so that TV may beequal to ST ; and making the angles hsq, shq, equal to the angles VSP, SVP, about q as a centre, andwith an interval which shall be to ab as SP to VS, describe a circle cutting the figure apb in p : join sp, and drawSH such that it may be to sh as SP is to sp,and may make the angle PSH equal to theangle psh, and the angle VSH equal to theangle pyq. Then with the foci S, H, and Bprincipal axis AB, equal to the distance VH,describe a conic section : I say, the thing isdone ; for if sv is drawn so that it shall be toSEC. IV.] OF NATURAL PHILOSOPHY. 129sp as sh is to sq, and shall make the angle vsp equal to the angle hsq, andthe angle vsh equal to the angle psq, the triangles svh, spq, will be similar,and therefore vh will be to pq as sh is to sq ; that is (because of the similar triangles VSP, hsq), as VS is to SP? or as ab to pq. Whereforevh and ab are equal. But, because of the similar triangles VSH, vsh, VHis to SH as vh to sh ; that is, the axis of the conic section now describedis to the distance of its foci as the axis ab to the distance of the foci sh ;and therefore the figure now described is similar to the figure aph. But,because the triangle PSH is similar to the triangle psh, this figure passesthrough the point P ; and because VH is equal to its axis, and VS is perpendicularly bisected by the rght line TR, the said figure touches theright line TR. Q.E.F.LEMMA XVI.From three given points to draw to afonrth point that is not given threeright lines whose differences shall be either given, or none at all.CASE 1. Let the given points be A, B, C, and Z the fourth point whichwe are to find ; because of the given difference of the lines AZ, BZ, thelocus of the point Z will be an hyperbolawhose foci are A and B, and whose principal axis is the given difference. Let thataxis be MN. Taking PM to MA as MNis to AB, erect PR perpendicular to AB,and let fall ZR perpendicular to PR ; thenfrom the nature of the hyperbola, ZR willbe to AZ as MN is to AB. And by thelike argument, the locus of the point Z willbe another hyperbola, whose foci are A, C, and whose principal axis is thedifference between AZ and CZ ; and QS a perpendicular on AC may bedrawn, to which (QS) if from any point Z of this hyperbola a perpendicularZS is let fall (this ZS), shall be to AZ as the difference between AZ andCZ is to AC. Wherefore the ratios of ZR and ZS to AZ are given, andconsequently the ratio of ZR to ZS one to the other ; and therefore if theright lines RP, SQ, meet in T, and TZ and TA are drawn, the figureTRZS will be given in specie, and the right line TZ, in which the pointZ is somewhere placed, will be given in position. There will be givenalso the right line TA, and the angle ATZ ; and because the ratios of AZand TZ to ZS are given, their ratio to each other is given also ; andthence will be given likewise the triangle ATZ, whose vertex is the pointZ. Q.E.I.CASE 2. If two of the three lines, for example AZ and BZ, are equal,draw the right line TZ so as to bisect the right line AB ; then find thetriangle ATZ as above. Q.E.I.130 THE MATHEMATICAL PRINCIPLES [BOOK I.CASE 3. If all the three are equal, the point Z will be placed in thecentre of a circle that passes through the points A, B, C. Q.E.I.This problematic Lemma is likewise solved in Apollonius s Book oiTactions restored by Vieta.PROPOSITION XXL PROBLEM XIII.About a given focus to describe a trajectory that shall pass throughgiven points and touch right Hues given by position.Let the focus S, the point P, and the tangent TR be given, and supposethat the other focus H is to be found.On the tangent let fall the perpendicularST, which produce to Y, so that TY maybe equal to ST, and YH will be equalto the principal axis. Join SP, HP, andSP will be the difference between HP andthe principal axis. After this manner,if more tangents TR are given, or morepoints P. we shall always determine asmany lines YH, or PH, drawn from the said points Y or P, to the focusH, which either shall be equal to the axes, or differ from the axes by givenlengths SP ; and therefore which shall either be equal among themselves,or shall have given differences ; from whence (by the preceding Lemma).that other focus H is given. But having the foci and the length of theaxis (which is either YH, or, if the trajectory be an ellipsis, PH -f SP ;or PH SP, if it be an hyperbola), the trajectory is given. Q.E.I.SCHOLIUM.When the trajectory is an hyperbola, I do not comprehend its conjugatehyperbola under the name of tins trajectory. For a body going on with acontinued motion can never pass out of one hyperbola into its conjugatehyperbola.The case when three points are givenis more readily solved thus. Let B, C,I), be the given points. Join BC, CD,and produce them to E, F, so as EB maybe to EC as SB to SC ; and FC to FDas SC to SD. On EF drawn and produced let fall the perpendiculars SG,BH, and in GS produced indefinitely Etake GA to AS, and Ga to aS, as HBis to BS ; then A will be the vertex, and Aa the principal axis of the trajectory ; which, according as GA is greater than, equal to, or less thanSEC. V.] OF NATURAL PHILOSOPHY. 131AS. will be either an ellipsis, a parabola, or an hyperbola ; the point a inthe first case falling on the same side of the line GP as the point A ;inthe second, going oft* to an infinite distance ;in the third, falling on theother side of the line GP. For if on GF the perpendiculars CI, DK arelet fall, TC will be to HB as EC to EB ; that is, as SO to SB ; and bypermutation, 1C to SC as HB to SB, or as GA to SA. And, by the likeargument, we may prove that KD is to SD in the same ratio. Wherefore the points B, C, D lie in a conic section described about the focus S,in such manner that all the right lines drawn from the focus S to theseveral points of the section, and the perpendiculars let fall from the samepoints on the right line GF, are in that given ratio.That excellent geometer M. De la Hire has solved this Problem muchafter the same way, in his Conies, Prop. XXV., Lib. VIII.SECTION V.How the orbits are to be found when neither focus is given.LEMMA XVII.Iffrom any point P of a given conic section, to the four produced sidesAB, CD, AC, DB, of any trapezium ABDC inscribed in that section,as many right lines PQ, PR, PS, PT are drawn in given ang 7ei,each line to each side ; the rectangle PQ, X PR of those on the oppositesides AB, CD, will be to the rectangle PS X PT of those on tie othertwo opposite sides AC, BD, in a given ratio.CASE 1. Let us suppose, first, that the lines drawnto one pair of opposite sides are parallel to either of I ^^ p ; Tthe other sides ; as PQ and PR to the side AC, and s|PS and PT to the side AB. And farther, that onepair of the opposite sides, as AC and BD, are parallelbetwixt themselves; then the right line which bisects^ IQ I3those parallel sides will be one of the diameters of the 1Lconic section, and will likewise bisect RQ. Let O be the point in whichRQ is bisected, and PO will be an ordinate to that diameter. ProducePO to K, so that OK may be equal to PO, and OK will be an ordinateon the other side of that diameter. Since, therefore, the points A, B; Pand K are placed in the conic section, and PK cuts AB in a given angle,the rectangle PQK (by Prop. XVII., XIX., XXI. and XXI1L, Book III.,of Apollonius s Conies) will be to the rectangle AQB in a given ratio.But QK and PR are equal, as being the differences of the equal lines OK,OP, and OQ, OR ; whence the rectangles PQK and PQ X PR are equal ;and therefore the rectangle PQ X PR is to the rectangle A^ B, that Is, tothe rectangle PS X PT in a given ratio. Q.E.D132 THE MATHEMATICAL PRINCIPLES [BOOK ICASE 2. Let us next suppose that the opposite sides AC and BD of the trapezium are notparallel. Draw Be/ parallel to AC, and meetingas well the right line ST in /, as the conic sectionin d. Join Cd cutting PQ in r, and draw DMparallel to PQ, cutting Cd in M, and AB in N.Then (because of the similar triangles BTt,DBN), Et or PQ is to Tt as DN to NB. And ^^ Q Nso Rr is to AQ or PS as DM to AN. Wherefore, by multiplying the antecedentsby the antecedents, and the consequents by the consequents, as therectangle PQ X Rr is to the rectangle PS X Tt, so will the rectangleN i)M be to the rectangle ANB ; and (by Case 1) so is the rectanglePQ X Pr to the rectangle PS X Pt : and by division, so is the rectanglePQ X PR to the rectangle PS X PT. Q.E.D.CASE 3. Let us suppose, lastly, the four lines?Q, PR, PS, PT, not to be parallel to the sidesAC, AB, but any way inclined to them. In theirplace draw Pq, Pr, parallel to AC ; and Ps, Ptparallel to AB ; and because the angles of thetriangles PQ</, PRr, PSs, PTt are given, the ratiosof IQ to Pq, PR to Pr, PS to P*, PT to Ptwill b? also given; and therefore the compounded ratios Pk X PR to P? X Pr, and PS X PT to Ps X Pt aregiven. But from what we have demonstrated before, the ratio of Pq X Pito Ps X Pt is given ; and therefore also the ratio of PQ X PR to PS XPT. Q.E.D.LEMMA XVIII.The s niL things supposed, if the rectangle PQ X PR of the lines drawnto the two opposite sides of the trapezium is to the rectangle PS X PTof those drawn to the other two sides in a given ratio, the point P,from whence those lines are drawn, will be placed in a conic sectiondescribed about the trapezium.Conceive a conic section to be described passing through the points A, B, C, D, and anyone of the infinite number of points P, as forexample p ;I say, the point P will be always c1placed in this section. If you deny the thing,join AP cutting this conic section somewhereelse, if possible, than in P, as in b. Thereforeif from those points p and b, in the given angles ^ Bto the sides of the trapezium, we draw the rightlines pq, pr, ps, pt, and bk, bn, bf, bd, we shall have, as bk X bn to bf X bd,SEC. V.] OF NATURAL PHILOSOPHY 133so (by Lem. XVII) pq X pr to ps X pt ; and so (by supposition) PQ xPR to PS X PT. And because of the similar trapezia bkAf, PQAS, asbk to bf, so PQ to PS. Wherefore by dividing the terms of the precedingproportion by the correspondent terms of this, we shall have bn to bd asPR to PT. And therefore the equiangular trapezia ~Dnbd, DRPT, aresimilar, and consequently their diagonals D6, DP do coincide. Whereforeb falls in the intersection of the right lines AP, DP, and consequentlycoincides with the point P. And therefore the point P, wherever it istaken, falls to be in the assigned conic section. Q.E.D.COR. Hence if three right lines PQ, PR, PS, are drawn from a common point P, to as many other right lines given in position, AB, CD, AC,each to each, in as many angles respectively given, and the rectangle PQX PR under any two of the lines drawn be to the square of the third PSin a given ratio ; the point P, from which the right lines are drawn, willbe placed in a conic section that touches the lines AB; CD in A and Cand the contrary. For the position of the three right lines AB, CD, ACremaining the same, let the line BD approach to and coincide with theline AC ; then let the line PT come likewise to coincide with the line PS ;and the rectangle PS X PT will become PS2, and the right lines AB, CD,which before did cut the curve in the points A and B, C and D, can no(onger cut, but only touch, the curve in those coinciding points.SCHOLIUM.In this Lemma, the name of conic section is to be understood in a largesense, comprehending as well the rectilinear section through the vertex ofthe cone, as the circular one parallel to the base. For if the point p happens to be in a right line, by which the points A and D, or C and B arejoined, the conic section will be changed into two right lines, one of whichis that right line upon which the point p falls,and the other is a right line that joins the othertwo of *he four points. If the two opposite angles of the trapezium taken together are equal cto two right angles, and if the four lines PQ,PR, PS, PT, are drawn to the sides thereof atright angles, or any other equal angles, and therectangle PQ X PR under two of the linesdrawn PQ and PR, is equal to the rectangle