De Solutione Problematum per MotumIsaac Newtonc.4,143 wordsThe Newton ProjectFalmer2010Newton Project, University of Sussex
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October 1666, c. 4,561 words, 29 pp.29 pp.MS Add. 3958.3, ff. 68r-76v, Cambridge University Library, Cambridge, UKUKCambridgeCambridge University LibraryPortsmouth CollectionMS Add. 3958.3, ff. 68r-76vOctober 1666EnglandLatinUnknown HandScienceMathematicsCatalogue information compiled by Yvonne SantacreuYvonne Santacreu tagged transcriptionCatalogue information updated by Michael HawkinsProofed by Robert IliffeTranscription audited by Michael Hawkins
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De Solutione Problematum per Motum
Ne hujusmodi operationes obscuræ nimis evadant, Lemmata 6 sequēentia brevitèr ideoque non demonstratiapræmittam.
Lemma 1. Si corpus A in circumferentiâ circuli vel sphæræ ADCE moveatur versus ejus centrum B: velocitas ejus ad unaquæque circumferentiæ puncta D, C, E, est ut cordæ AD, AC, AE, ductæ a Corpore A ad ista puncta D, C, E.
Lemma 2. Sit △ ADC similis △AEC, etsi non sunt in eodem plano. Inquam, Si tria Corpora a puncto A, primum ad D, secundum ad E, tertium ad C, uniformiter et in æqdualibꝰbus temporeibus moveantur: Motus tertij componetur ex motibus primi et scdsecondi.
Notetur, quòd hic per corpus intelligitur ejus centrum gravitatis, vel aliquod ejus punctum.
Lemma 3. Omnia puncta corporis parallelismum servantis æqualiter moventur.
Lemma 4. Si corpus solo motur circulari circa axim quemvis rotetur; motus omnium ejus punctorum sunt ut distantiæ ab isto axi. Et hi duo motus simplices vocentur.
Lemma 5. Si motus corporis consideretur ut mixtus e motibus simplicibus: motus omnium ejus punctorum componetur ex motibus eorum simplicibus, eo modo quo motus ab A ad C, in Lem̄mmate 2do componitur ex motibus ab A ad D et E.
Nota. qdquod motus quilibet ad unum horum 3ūum casuum reduci poterit. Et in casu tertio, linea quævis pro axe, (vel si linea aut plana superficies movetatur in plano, quodvis punctum istius plani pro centro) motus assumi potest.
Lemma 6. Sint AE, AH, lineæ motæ et continuò secantes; Ducantur AB, AD, AC, BCCB, CD. Dico quòd, datis quæ requirantur ad proportiones et positiones harum quinque linearum AB, AD, AC, CB, DC, determinandas; illæ designent proportiones et positiones horum quinque motuum, viz: puncti aA in lineâ AE fixi et versus B moventis, puncti A in lineâ AH fixi et versus D moventis, Puncti intersectionis A in plano ABCD est moventis versus C, (lineæ enim 5 istæ semper sunt in eodem plano etsi AE AH non sunt), Puncti intersectionis A in lineâ AE moventis secundum ordinem literarum C, B, et Puncti intersectionis A in lineâ AH moventis secundum ordinem C. D.
Nota. Quod linea recta utpositionemmotûs curviæmotûs curvi eandem designare dicitur
Nota. Quòd cùm linea recta tangitens curvam motu descriptam (ut AB, AD, AC) vel istæ tangenti parallela est (ut CB, CD,), dicitur designare positionem istius motûs in puncto contactûs.
Nota etiam qdquod lineâ AH quiescente (ut in Fig 1, et 4), punctum D et A coincident et puctumpunctum C in lineâ AH, modò sit rectâa (Fig 1,), alitèr in ejus tangente AC (fig 4) reperietur.
Prop 1. Ducere Tangentem ad Ellipsin.
Sit ACB filum per quod ellipsis describi solet, et CE Tangens. Cùm filum AC augetur eâdem velocitate quâ BC diminuitur, i:e: C habet eandem velocitatem versus D et B; eitrit <DCE=ECB.Per Lem 1. Idem de reliquis Conicis intelligatur.
Prop 2. Ducere Tangentem ad Conchoïden
Sint GLC, ALF, GAE, regulæ quibus concha describi solet: fiat GT∥AF⊥CB=∥MN: et NG=CL⊥TN∥RL. Et, Cùm æqualitas proportionalitate simplicior est, ponatur lineam CB=NM esse æqualem velocitati puncti C versus B, vel puncti N versus M. Erit NT=motui circulari puncti N circa G rotantis versus T, (Lem 1): et LR=Motui circulari puncti L in lineâ GL fixi, circa G (Lem 4): et LG = velocitati motui puncti intersectionis L (i:e: velocitati puncti C) a puncto G, sive versus punctum D in lineâ C G moventis, (Lem 6). Iam cum habeatur duplexduplex velocitas puncti C versus viz CB versus B et LG versus D, fiat FD⊥DC=LG: Et motus puncti C erit in lineâ FC, Diametro nempe circuli transientis per puncta LCDBF, (Lem 1): Quæ proinde Conchiam tangit in puncto C.
Prop 3. Invenire punctum C distinguens concavam a convexâ Conchæ portione.
Iis in priori propositione suppositis: Fiat △ustriangulus GFH similis △otriangulo GNT sive LBC: et DF⊥FR∥=HK=2GL; Iungāantur F, K, et fiat KP∥FD. Si Linea DF solum motum Parallelum per CD vel FR directum haberet, (quia CD=GL) motus omnium ejus punctorum esset FR, (Lem 3): Et si solum motum circularem circa centrūumG haberet, motus puncti F, in istâ lineâ DF fixi, esset FH, (Lem 4). At motus istius F ex istis duobus componitur, proinde erit FK, (Lem 5., 2.) et motus puncti intersectionis F per lineas AF, DF, fiacti, et in AF moventis, erit FP (Lem 6). Iam si linea CF Concham tangit in puncto quæsito C, facilè deprehendatur motum puncti intersectionis F esse nullum; proinde P et F coincidere; sive DF et FK in directum jacere; et △GDF, FKH esse similes.
Quæ ut calculo subjiciantur, fiat AG=b. CL=c. CB=y. tum, . . . . Etc. . Quare, . sive . Equatio , priùs inventa, ita resolve, fac . duc of et cum diametro of describe circulum fmao, in quo inscribe fm=b=AG. et cum radio om, fac circulum mv, et et a pūuncto intersectionis v d duc vd⊥ad. erit 3b∶2c∷VD∶Gl. unde datur punctum c.
Vsus autem hujus methodi (ut intelligo) præcipuus est in lineis Mechanicis, ubi deficit Algebraica calculatio. Exempli gra, Tangens Quadratrici ità ducetur.Sint AG, MC regulæ quibus uniformitaèr motis Quadratrix describi intelligitur. Et CB vocetur motus puncti C in lineâ CM fixi et versus B moventis; tum arcus GK erit motus puncti G ciracirca A rotantis, (sup) et arcus CL erit motus puncti C in lineâ AG fixi et circa G rotantis, (lem 4). Quare ( si fiat CL=CD⊥AG∥DF, et BF∥CM.) motus puncti intersectionis C in planò AEK erit CF (Lem 6), quæ proinde Quadrꝰratricem tangit in C.6970 If vce is a Conchoïdes, g its pole, &c: ga=b. ae=lc=vb=c. ma=y & c yethe point betwixt its convexity & concavity, ynthen is . (see pag: 259. lin: 10.). WchWhich Equation hath one affirmative roote (ma) referred to yethe point c. and two negative rootes whereof yethe greater is referred to yethe lower ConchConchoïdes, & yethelesser (I thinke) uselesse to this question.The rootes of this Equation may bee thus found by yethe helfpe of the described ConchConchoïdes & a circle. viz: Suppose ga=b, ae=lc=vb=c. ma=y. (as before) ao=s. oc=r. And thereby may bee found this Equation , to bee compared wthwithyetheformer, (see pag 261 lin 20). But their rootes cannot become equall by reason of their third the termes. Therefore I alter yethe rootes of yethe first equation, as, suppose I make y=z−b. Then is z3∗−3bbz+2b3−2bcc=0. To bee compared wthwithyethe 2d equation: wchwhich cannot yet bee done wthwithout a contradiction, there being but two unknowne quantitys, r & s to bee found by three Equations resulting from yethe comparison of their 2d 3d & 4th termes. But if I make , & substitute this valor into its pallace in yetheprecedent Equation, the result is . The termes of wchwhich being compared with yethe termes of yethe 2d Equation yethe 3d or 4th give . Or 16c6−32bbc4+16b4cc=27b6+27b5 s. & s=16c6−bbc4+6b4. Their second termes give . Or . Therefore from yethe intersection point v (made by yetheConchConchoïdes& a circle whose radius is ov & center o) let fall ved⊥ad; & ar=x=vd, is yethe roote of yethe2d & last equation. Which being found make . Which was to bee done. Prop 1. Suppose ab=x. bd=y⊥ab. And ytthatyethe nature of yethe line addc is such ytthatyethe valor of y is rationall & consists notof no fractions in whose denominator x is, or else wholy of of such fractions in whose denominators x is, but not of divers dimensions: If I then multiply yethe valor of y by x, & divide uni each terme of ytthat valor by so many units as it hath dimensions in ytthatitterme; yethe result shall signifie yethe area abd of yethe afforesaide line addc.As for example. If y=1, or y=x, or y=xx, or x3, or x4 &c ynthenyethe area abd is , or , or , or , or &c: so if , or , or &c: ynthenyethe area abd is , , &c. In like manner if , or , or &c ynthen, or , or =, &c: is yethe area abd (as is by others demontrateddemonstrated). [so if . ynthenis yethe area abd; viz: tis infinite.]. Lastly if &c. yethe area abd is &c. For yethe area abd is compounded of those areas wchwhich are related to & generated by those quantitys of wchwhichyethe valor of y is compounded. & wtwhat those areas are appeare by yethe former example. (nNote ytthat Parabolicall & Hyperbolicall (i:e: negativ &(in respect of bd,) affirmative & negative) areas (thus considered) cannot compound any 3d area, because they are not on yethe same side of yethe line bd.)Prop 2. &c. For these termes are r. aa. aab+aax. aabb+2aabx+aaxx. aab3+3aabbx+3aabxx+aax3, & which termes may bee thus ordered This lest appeares by multiplying both parts by b+xOr by supplying yethe vacant placesNow to reduce yethe first terme to yethe same forme wthwithyethe rest, I consider in what progressions yethe numbers prefixed to these termes proceede, & find ymthem to bee such ytthatyetheany number added to yethe number above it is equall to yethe number following it. Whence any termes may bee found wchwhich are wanting, as in yethe annexed Table. Also any terme, to wchwhichthese numbers are prefixed, being multiplyed by b produceth yethe following litterall terme. Or yetheyethe higher terme multiplyed by , divided by b producceth yethe lower terme. As in yethe following tableProp: 3d. If ab=x. y=db⊥ab⊥ac. & aa (fa=b) & &c. Then (by prop 1), &c: is abde, yethe area of yethe Hyperbola. So if : &c. In like manner if &c. Then (by prop 1) is yethe area abde. (wchwhich may also thus appeare viz: if fb=b+x=z. ynthen. Therefore (prop 1) yethe area . & . so ytthat). And so of yethe rest. As if &c. The area abde is &c.Prop 4th. If . n & m being numbers ytthat signifie yethe dimensions of x & y: Then is, . The area abde, if n is affirmative or mbd iif n is negavenegative of yethe line edm. Prop: 5t. If . ynthen.Prop 4th. To find two or three intermediate termes in yethe above mentioned table of numerallprogressions, I observe ytthat those progressions are of this nature viza.a.aa.a.b.b+c.b+2c.b+3c.b+4c.And ytthatyethe summe of any terme & yethe terme above it is equall tod.d+e.d+2e+f.d+3e+ef.d+4e+6f.yethe terme following it at suchathe distance atofyethe termes in yethe sd numerallg.g+h.g+2h+i.g+3h+3i+k.g+4h+6i+4k.re one from yethe other table. Suppose I would find yethe meanel.l+m.l+2m+n.l+3m+3n+p.l+4m+6n+4p+q.r.r+s.r+2s+t.r+3s+3t+v.r+4s+6t+4v+w.71termes in yethe 3d progression3.*.1.*.0.*.0.*.1.*.3.d−4e+10f.d−3e+6f.d−2e+3f.d−e+fd.d+ed+2e+f.d+3e+fd+4e+6f.d+5e+10f.d+6e+15f.I compare yethe termes of ytthat progression & of yethe correspondent litterall progression & find d=0=2e+f. 4e+6f=1. subduct 4e+2f=0, Or 12e+6f=0 from 4e+6f=1. & yethe rest is 4f=1. Or −8e=1. wchwhich termes being found viz d=0. . . yethe progression must be &c. Hence may be deduced this table viz Note ytthatyethe progression may bee deduced from hence &c & one intermediate terme given yethe rest are easily deduced thence.In like manner if I would find two meanes twixt every terme of ytthat numerall progressressionI compare yethe numerall & correspondent litterall progressions, suppose in yethe 3d progression.1·*·*·0·*·*·0·*·*·1·d−3e+6f.d−2e+3f.d−e+f.d.d+ed+2e+f.d+3e+3f.d+4e+6f.d+5e+10f.d+63+15f.&c.And find ytthatd=0=3e+3f.& 6e+5f= 6f−3e=1. To wchwhich adding 3e+3f=0, or −6f−6e=0. yethe result is 9f=1, or −9e=1. So ytthatyethe progression must bee &c. Hence may bee composed this Note ytthat this progression viz &c gives yethesecond term &c. & this &c gives yethethird terme. Also this progression &c. gives yethe 4th.And in Generall if yethe 2d quantity of any terme is , this progression gives all yethe rest viz &c. And is ever given by supposition for it signifieth yethe distance of yethe terme from 1.0.0.0. In wchwhich x signifieth yethe distance of any terme from yethe first 1.0.0.0.0.0. & y is the quantity of ytthat terme.
Prop 5t.72 If lab is an Hyperbola; cde, ck its Asymptotes, a its vertex, & cag its axis; <ckd=2<kca=<kcd=<cda=<cebif adef is a square & he∥ad& cd=1, &, de=x. ynthen. If also, ef=1. eg=1+x. eh=1+2x+x &c: (athe progression continued is 1+3x+3xx+x3. 1+4x+6x2+4x3+x4. 1+5x+10x2+10x3+5x4+x5 &c). Then, shall the areas of those lines proceedet in this progression. *=adeb. x=adef. . . . &c. As in this table. In wchwhichyethe first area is also inserted. The composition of wchwhich table may be Suppose ytthat adck is a Square abc a circle agc a Parabola. &c. & ytthat de=x. ad∥fe=1=bd. & ytthatyethe progression in wchwhichyethe lines fe, be, ge, he, ie, ne &c proceedes is . . . . . . &c. Then will their areas fade, bade, gade, hade, eiade, &c be in this progression. . &c: as in this table following in wchwhichyethe intermediate termes are inserted. The property of which table is ytthatyethesummme of any figure & yethe figure above it is equall to yethe figure next after it save one. Also yethepronumerall progressions are of these formes.a.a.a.a.b.a+b.2a+b.3a+b.c.b+c.a+2b+c.3a+3b+c.&d.c+d.b+2c+d.a+3b+3c+d.e.d+ec+2d+e.b+3c+3d+e.Where yethe calculation of yethe intermediate termes may bee easily performed. The area abed depends upon yethe 4th Collume &c: (wchwhich progression may bee continued at pleasure by yethe helpe of this rule &c.) Whereby it may appeare ytthat, what ever yethe sine de=x is, yethe area abed is &c. (& yethe area afb is &c.) Whereby also yethe area & angle adb may bee found. The same may bee done thus. yethe areas afd, abd, agd, ahd &c are in this progression . &c As in this following Table By wchwhich it may bee perceived ytthat. &c. And by this meanes haveing yethe area abd, yethe(wchwhichyethe angle adb gives) deyethe sine of yethe angle adb may bee found. Corol: If de=x. & . ynthen abc is an Hyperbola. & its area dabe is . &c.731. Fiat bd∥ad, et ducantur rectæ ab, gd, ag, bd: Conicarum portiones bas, gdt, erunt æquales, item et gasb, bdtg. (fig 1.2) Bisectis enim bg, ad, in punctis r, et e; et ducto cere; erunt brg aed ordinatim applicatæ ad diameterrum cure: proinde portiones reasb=redgtg, et trapezia reab=redg. Ergo eorum differentiæ bas=gdt. Deinde △abg=dgb (per 37.1. Elem), Ergo portiones gasbv=bdtgv.Coroll. Hinc pateat modus ducendi tangentes ad Conicas, ignoratis eorum diametris: DucA dato enim (fig 3) puncto a, duc ac, ab, &et ijs parallelas bd, cf: linea df erit parallela tantgenti ah. Si non, fiat ae∥df, et duc de, ad, af, cg; et erit portio ecd=abft ecd=abf=bac=aecd, qdquod est impos:Porro fiat dn=nf, et duc ank; fiat am=mk, et duc d pmq. Erunt ak, pq diametri.Diametro av describatur circulus abcdv. Et centro a, describantur circuli cr, dm, gn, hq. et a punctis c, d, g, h, duc ce, df, gk, hp, perpendiculares ad diametrum av; & duc abe, acf, adk, agp. In triangulis gqh, pqh &c: angulus hgp=hpq & ∠gqh=∠pqh=recto. Ergo, si ponantur trianguli qgh, qph esse infinitè parvi, ut latera hg, hq sit rectæ; erunt similes et æquales, poindeproinde gq=pq. Hinc omnes lineæ gp, dk, cf, be, & sunt duplices linearum gq, dn cm, br, &c. hoc est longitudo Trochoid: est an est duplex longit: rectæ ag.74De Gravitate Conicarum. Sit λvf Ellipsis, et λwf Parabola. Ita nempe, ut λd vocatâ x; ordinatim applicata d sit , & . Fiat λc=cf=. & dw∥hfz∥kb∥kb∥rp, & vh∥wz∥γg∥βt, tangentes vel sec sese aut curvas in punctis v, l, k, h, r, γ, s, g, c, d, e, f, β, q, t, p, w, a, b, z.Dico quòd, posito λf axe gravitatis, pondus parallelogrammesi vkec est ad pondus portionis vsec, sicut parallelogramgrammiwvbec ad portionem wqec.Ponatur enim pondus lineæ id esse siv dr esse sive , hoc uest dp. Erit pondus omniuum lineæarurum dr in superficie vsec contentarum, hoc est pondus superficiei vsec, æqualei superficiei wqec. Et eâdem ratione pondus ▱miparallelogrami vkec erit ▱mumparallelogramum wbec. Q.E.D.Coroll. Pondus portionis vks est wbq, est portionis vsγ est wqβ, et portionis esf est eqf, & portionis fsg est fqt &c. Quæ omnia dat quadratura Parabolæ, est enim 3βwq=2βwbq, &c.Eâdem ratione gravitas cujuslibet portionis Hyperbolæ cognoscatur, modò axis gravitatis transeat per centrum Hyperbolæ. Et si quævis plana superficies conicis sectionibus ità terminetur, ut omnia conicarum centra sint in eâdem rectâ lineâ: gravitas istius superficiei inveniri potest, positâ rectâ gravitatis axi. Denique, centrum gravitatis cujusvis planæ & finitæ superficiei conicis sectionibus itâa terminatæ inveniri potest, datâ quantitate istius superficiei, & vice versâ, modò centrum gravitatis aixiaxi gravitatis non coïncidat.Sint ac, af Assymptota Hyperbolæ gc, in infinituumversus c continuatæ. Duc de=ab=af. & . & fg∥ac.Dico qdquod. Parallelogramumum ae, & superficies afgc æquiponderant circa axim abc: Etsi superficies aefg afgc versus c sit longitudine & quantitate infinita, & non habet centrum gravitatis.fiat enim ar=ap. & duc pq∥ab. & vr∥ad. et sit pq×ap gravitas lineæ pq, erit vr×vr gravitas lineæ vr (nom ∠vrb=∠apq). sed . æquiponderant ergo lineæ pq, vr. Sed numerus linearum pq in superficie acgf est æqualis numero linearum vr in parallelogrammo ae nam af=ab, ergo ae & ag æquiponderant.Add to the end of the Paragraph. See the mystery. In the first instance from the series &c subduct &c. & then will remain &c &c=1= will remain &c. And so of the rest Or take all the terms but the three first & there will remain &c Or from this series take all the termes b &c take all the termes but the termes but the first & there will remain &c.75Problems of Curves.Probl. 1.To draw Tangents or Perpendiculars to any given point of a given Curve (V B) (Sup. VA=x, AB=y)Put the equation of the Curve=0, multip. by this Progrngression. &c. according to the dimensions of x for a numerator, change the signs and multip. by this Progr.Progression &c for a Denominator, which shall make the Fraction designing AD to be taken forwards from A, if affirmative, otherwise, backwards.Examp. 1. If a+bxy−xxy+y3=0, then (the Subperpendicular.Coroll.Hence for AF (the Subtangent) multiply by the Dimensions of y for a Numerator & by this Progr. &c according to the Dimension of x for a Denominator; thus results .Prob. 2d..From a given point H to draw a Tangent (HB) to any Curve (VB)Sup. VG=p, GH=q and AF=t and it is y−q∶x+p∷y∶ t. Th. , or ty−tq =xy+py. If into this you substitute the value of t (found ꝑper Prob. 1.) you will have the nature of a Line wchwhich described will cut the propounded Curve in the desired Tangt.Tangent points.Exa. Thus substituting for t, there comes bxy−xxy+3y3−bqx+qx2−3qy2=2x2y+2pxy −bx4+bpy=0, the Curve to be described ** But note that this may be reduced to a simpler line by adding or substracting the nature of the given line, viz. ordering this results, it is 3y3−3x2y+2bxy−2pay+qxx−3qyy−bqx+bpy=0. from whence substracting y3−xxy +bxy+athrice there rests −bxy−2pxy+qx2−3qyy−bqx+bqy −3a=0, a Conic section wchwhich described will cut the Curve in the desired tangttangent points.Thus if , then or or & substracting twice (viz so often as the Curve has dimensions) there comes , a strait line wchwhich drawn will cut yethe Curve in the points desired & so of the rest.Where note that this Problem in any Curve is alever solvable by a line of an inferior degree. And also that a line drawn from a given point may touch a Curve of two Dimensions in 1×2 points, of three in 2×3 points of 4, in 3×4 pts.points &c unless it be polar &c.Probl. 3.From a given point (C) to draw a Perpendir.Perpendicular(CB) to any Curve (VB)Make VE=r, EC=s and AD=v so is y−s∶r−sx ∷ y∶v. or put this valor of v in the room of v found ꝑper Prob. 1. And and you have a Curve wchwhichdescribed will cut the propounded Curve in the points to wchwhich the perpendiculrperpendicular may be drawn [but ever try if by means of the nature of the given Curve you can reduce this resulting to any simpler form or degree.Exa. If ax+xx−yy=0, then a+x=v, or ry−xy=y−as+xy−sx, wchwhich appears not reducible to a simpler form.Yet it may be worth while to try if it may be solved by a Circle by assuming d+ex+fy=xx+yy and comparing these 3 equations of the given Conick, found Hyperbola and assum'd Circle to find d, e & f wchwhich if they be found by plane Geometry that Circle described will cut the Curve in the defined points. This, I say, might be tried but it would be found impossible (unless the Conic be a Parabola) because there are 4 given points thró wchwhich the Circle must pass, wchwhich make the Probl. contradicting, since 3 are now to determine a Circle.This might be done by the Parabola, but the Hyperbola falling in so naturally, and being as easily described, 'tis not worth the while.What is said of the Conics may be applied to any other Curves.From hence it appears that this Probl. is ever solvable by a Curve of the same degree and sometimes ꝑhaps by a Curve of an inferior Degree.76Also from a given point may be drawn so many ꝑperpendicrcular, as the square of the Curves Dimensions, unless some ꝑpartpartꝑpartpart extraordrydinary sum out to infinity, or two ꝑpartsparts come together to make it polar.Probl. 4.To find the Points where the Curve has a given inclination to the Basis.Supp. FA∶AB or AB∶AD∷m∶n. Then . Put this eql.equalto the value of t found by Prob. 1. & you have a Curve wchwhich describ'd will cut the given Curve in the desired points.But (if you can) reduce it thsus if , then , or . the like in other cases.Hence Conics have 1×2 points, those of 3 Dimensions 2×3 points, of 4 Dimens. 3×4. &c. will satisfy this Probl, & its ever soluble by a line of an inferior Degree.Probl. 5.From a given point (C) to draw a line CB wchwhich shall cut any given Curve in a given Angle.Supp. BD the ꝑperpendrdicular to yethe desir'd point B & from the point D raise the ꝑperpendr.dicular DH & let fall HF⊥ AE, producing BC & AE to G. Let AD=v, VE=r, EC=s and (yethe ∠ CBD being given) supp. BD∶DH∷m∶n.Now the △striangles BHD & DFH being alike, 'tis m∶n∷AB∶ DF∷AD∶FH. Th. & & . Lastely since VE+EG=VG=VA+AD+DF +FG there results wchwhich reduced is (mr+n put mr+ns =k & nr−ms=l to shorten the terms) this I put equal to the valor of v found as before, and there results a Curve wchwhich described cuts yethe given Curve in the points desired, to wchwhich lines are to be drawn from C that may cut it in the Angles desired.Exa. If dd+xx−yy=0, then is x=v, and so mry −mxy+nsy−nyy=, which is a Conic section passing thrò the Centre of the given Curve. And it it may be made an Hyperbola constantly by substituting dd for yy−2x. For the result will be .Hence it appears that this Problem may ever be solved by a Curve of the same sort, and sometimes ꝑperhaps by one of an inferior.And also that from a given point to any Curve so many lines may be drawn in a given Angle as the sqr.squareof its Dimensions abating those that are imaginary, or coincident to the Pole, or should be drawn where some ꝑpartsparts of the Curve vanishes ad infinitum.