How to find ye axes vertices Diamiters, \Centers,/ or Asymptotes of any Crooked Line supposeing it have them.

[1] Definitions. If kgach is a crooked line, & ef touch it at ye point a . & if ye line lines cg hk & all other lines wch being par{illeg}|a|lle{illeg}|l|l to one another & extended from one si{illeg}de of ye \{illeg}/ croked line to ye other and also bisected by ye streight line ai . Then is ai a Diameter, & cbg one of those lines wch are ordinately applyed to it, & if ye angles {illeg}, cbd , &c are right ones Then is abd t{illeg}|he| axis, & cbg one of those lines wch are ordinately applyed to ye axis.

[2] 3 The Vertex of a crooked line is yt point where ye crooked line intersect the diameter or axis as at a

[3] 4 An {illeg} The \The/ Asymptote{illeg} of {illeg} crooked lines are such lines wch being produced both ways infinitely have noe least distance twixt ym & ye crooked line & yet {doe} noe where intersect it. \or touch it/ as dδ , dθ .

[4] 5 Those lines wch are limited on all sides \& have axes but/ as acxkλ are Ellipses if ye of ye first, 2d, 3d, 4th kind &c

6 Those wch are not ellipses & have \axes but/ noe Asyptotes are {illeg} Parabolas of ye first, 2d, 3d, 4th kind {illeg}|&|c. as zkah .

7 Those wch have {illeg} Asymptotes, are Hyperbolas of ye 1st, 2d, 3d 4th kind, &c as (upon) whose asymptotes are βd , γd .

8 There are some lines of a middle nature twixt {illeg}|a| Parab: & hyperb: have an asymto{te} haveing an Asymptote for one of its s{illeg}|i|d{illeg}|e|s wch are but none for ye other as βαγe , one side αγ haveing ye asymptoe {sic} δe , ye other side αβ haveing none.

10 If an Ellipsis have 2 axes (a{illeg}|s| am & xλ ) ye longer is ye transverse axis {illeg}|(|as am ) ye shorter is ye right axis (as xλ ).

If all those lines wch are parallell to one of ye diameters of ye crooked line & are terminated by ye crooked line be bisected

9 If two diameters of ye same Ellipsis be ordinately applyed {illeg}|ye| one to ye other ye shortest of them is called ye right diameter, ye longest ye transverse one. ({illeg} {illeg}) (as am & xλ ).

If {illeg} all ye parallells wch are terminated by \ye same or by/ 2 divers figures, are bisected by ye same streight line, yn is yt line \called/ a \right/ diameter. But if That yt diameter wch is ordin{illeg}|a|tely applied to it (i.e. wch is par{illeg}|a|llell to these parallells) is a transverse diameter.

1 If all ye parallell lines wch are terminated by {illeg}|t|he same or by 2 divers figures, bee bisected by a streight line; yt bisecting line is a diameter, & those paralle{l} lines, are lines ordinately a{illeg}|p|plied to yt diameter.

2 If those parallell lines intersect ye diameter at right angles yt|e| diameter is {illeg} an axis

A line is said to bee of ye first kind wh

In an Ellipsis ye point where 2 diameters intersect, is {illeg} \ye/ center.

The center of an{illeg} figure \Ellipsis/ is yt point where {illeg} two \of {illeg} its/ diameters intersect.

{illeg} The center of two di{illeg}ers figure{illeg}|s| of t \opposite/ /Hyperbol{illeg}|a|s\ is yt point where two of their diameters intersect one another or else where there {sic} Asymptotes intersect.


Propositions. The lines ordinately ap{illeg}|p|lie{illeg}|d| to ye axis of a crooked line are parallell to ye tangent of ye crooked line at its vertex.

[5] Demonstr. Suppose \ chad a Parab &/ dc (being ordinately applied to ye axis ab ) not parallell to ye tan{illeg}|g|ent an but to some oth{illeg}|e|r line like ah. If dc bee understood to move {illeg}|to|wards a db continually decreas|i||ng| \{illeg}/ un{illeg}|t|ill it vanish into nothing at ye conjucti{illeg}|o|n of ye points a & d . but cb sinc & since cb mus{illeg}|t| always be equall to ah at ye conju{illeg}|n|tion of ye point {sic} a & d . it followeth yt cb cannot decrease so as to vanish into nothing at ye same time wch bd doth & therefore cannot allways be = to bd .

[6] Otherwise. if dc is not parallell to ye tangent an but to some other line as ah . Then ab doth not bisec{illeg}|t| all ye parallell li{illeg}|n|es (as oe ) wch are terminated by ye crooked line cad . & therefore cannot bee its diam:

[7] |2dly.| If ad is ye axis of a crooked line & cb=y, is ordinately a{illeg}|p|plied to ad . yt is if bc=ce=y . Then y {illeg} must be found noe where of odd dimensions in ye Equation \expressing ye nature of ye line cod/. For (suppose{in}g y=bc=ce to be ye unknowne quantity) y hath 2 valors bc & cd equall {illeg}|to| one anot{illeg}|h|er excepting yt ye one bc is affirmative, ye other ce is negative. wch two valors canot {sic} bee exprest by an equation in wch y is of od {sic} dimensions for suppose yy=aa . yn i{illeg}|s| is {sic} y=+a, or y=a y= aa=a, since a×a=aa. & aa=a, since a×a=aa. & y=aa therefore is y=+a, or y=a. soe if y4=a4. yn is yy=aa , & y=a or y=a but if y3=a3. yn y=c:a3=a . but not y=c:a3=a. soe i{illeg}|f| y5=a5. yn y=a=qc:a5 but not y=5=qc:a5. The same reason is cogent in compound equations. as if yy2xy+xx=ax. Then, y=xOax. where though ye root a+ax is affirmative & ye roote aax may bee negative yet they can never be equall in length, & though ye 2 roots of an equation wch differ in signes should bee equally long y{illeg}|e|t yt is w{illeg}|h|en there is but one quantity considere ye Equation is fully determined.

[8] Prop: 4th. If x is of more dimensions in a quantity not multiplied by x yn in one multiplied by it (as in y2={illeg}xy+aa) yn y is not parallel to one of ye lines Asymptotes. & e contra. Otherwise x & y are parallel to ye Asymptotes of ye line. et e contra.

Prop 3d. If ag is ye Asymptote of ye crooked line dcf , & ab=x is coincident wth it, then & bc=y. then in ye Equation (expressing ye relation twixt x & y ,) x must be of more dimensions in some \one/ quantity \onely/ where it is multiplyed by y or yy: &c in any other quan{illeg}|t|ity in ye Equation \& y of as many/. & if ae is an Asymptote to ye same line dcf , & bc=y bee parallell to it then y must noe where be of soe many dimensions as in if \one/ quantity \onely/ in wch {illeg}|t|is drawne into x , xx , \or/ x3 &c. Demonstraco. If this be false yn suppose ye equation expressing ye nature relation twixt x & y to be xyy+ayy {illeg} bxyd3=0 . first if x is coincident wth ye Asymptote ag yn, if y | x | =eeo (yt is if x being infinite in length) i{illeg}|t| is y=0, if x=eeo (yt {illeg}|i|s y vanisheth into nothing if x be infinite. {illeg}|)| therefore tansposeing {sic} eeo into ye place of x , I find eeyyo+ayyobeeyod3 o=0. Or eeyy+o×ayybeeyo×d3=0 that is eeyy=beey /+||{ee}\, or y=b. Now since y vanishe{illeg}|th| not it followeth y{illeg}|t| x is not coin
cident wth ye Asymptote. But if I blot out {one} of ye terme{illeg} {illeg} as bxy So: if ye equation be xx+xy=aa . yn by substituteing eeo into ye place of x I have e4oo+eeyo=aa, or y=o2aae4oee=eeo soe yt y not vanishing but being infinite x cannot be asymptote. But if ye equation {illeg} was ayybxy=d3 . yn by writeing eeo for x , there results, eeyy0+ayy=d3. or yy=od3oa+ee=0 . & therefore in this case x is coincident ye asymptote ag . Soe if ayybxyd3=0 , by making eeo=x . it is ayybeeyo {illeg} d3=0 . Or by extracting ye roote bee2aoOe4bb4d3ooa4aaoo . Soe yt y hath 2 roots ye one infinitly \grate {sic}/ wch is bee2ao+e4bb4d3aoo2ao=beeao=y . ye other is infinitely little wch is bee2aoe4bbo4aaoo=beebee2ao=o=y .
bee multiplied by y wherever i{illeg}|t| is of its greatest dimensions. & if ae is an asymptote to ye line dcf , & bc=y be parallel to it, ye & ab=x terminated by it at y{illeg}|e| point a , {illeg}|th|en must y be multiplied by x wherever it is of its greatest dimensions. Example: Suppose axx+yxx=b3. because in these 2 termes axx+yxx x is of it {sic} greatest dimensions; but in one of ym (viz: axx ) it is not multiplyed by y therefore x is not coincident wth ag ye asymptote. If yyxx+ayxx + /\ a3xa4=0 . then since x is of its greate{illeg}|s|t dimens: in yyxx & ayxx onely & is drawne into y in both of ym therefore x is coincident wth ye Asymptote{;} I Also since y is of its greatest dimension{illeg}|s| in xxyy onely, {illeg}d (wch {illeg} {b}{l} multiplied {illeg} by x ) therefore y is parallel to {illeg} x {illeg} {terminated} by an Asymptote. &c.

Demonstr: If x is coincident wth ye Asymptote yn { eeo=x } when o=y . i:e: x is infinite when {illeg} {wisheth}. Now suppose yyxx+ayxx a3x =a4. yn i{illeg}|f| y=o is x=aaoo+ao . i:e: x is {illeg} but if yyxx+aaxx=a4. yn if y=o it is {x=aaoo+aa=Oa }. soe {yt} x is finite & therefor{e}{illeg} coincident wth ye {illeg} de{illeg}{illeg} of {illeg} is like{illeg}


[9] Haveing {an} the nature of a crooked line expresed in al{illeg}|g|ebraicall termes to find its axi|e|s if {{sic}}{i} {shave} { anq }

[10] Draw a line infinitely both ways fix upon some point (as b ) for ye begining of one of ye unknowne quantitys (wch I call x . Then reduce ye Equation to such an order (if it bee not already so) yt x may be always found in ye line bc . wth one end fixed at b , & having y moving making right angles wth it at ye other end: yt end of y wch is remote from x , describing ye crooked line. wch may bee {illeg}|al|ways done wthout any great difficulty. As may be percei{illeg}|v|ed by these examples.

[11] Suppose ye given Equation was y3=axx . x & y {illeg} soe yt bg=x . gd=y . dc being perpendicular to bc & ye angle dgc being given, ye proportion twixt dg & dc is given, wch I supose as d to e . yn is deyeyd {illeg}c {illeg} gd2gc2=dc2 yyeeyydd=w2 . dwddee=y. bg+gcx+eyd deydc=eyd=w . & dwe=y . gc2=gd2dc2 =yyww =ddwweewwee . or gcddee . & bc=v=wddeee . Or evwddeee=x .[12] Therefore I write d3w3e3 for y3 , {illeg}|&| eevv2evwddee+ddwweewwee for xx in ye equation y3=axx , & soe I have this equation d3w3e3=aeevv2aevwddee+addw2 aeew2ee . wch expresseth ye relation twixt w & v , yt is twixt dc & bc , {illeg}|w|riteing therefore y for dc , & x for bc : I have this equation d3y3aeeexx= \{o}{ 0 }/ d3y3+ae3yyaeddyy+2aeexyddeeae3xx

[13] Soe if x \ =bd / turned about ye pole b & y \=dg/ about ye pole g j{illeg} describing ye croo{illeg}|k|ed line ad by ye conjunction at ye extremitys. & ye equation expresing ye relation wch they bea{illeg}|r|e to one another is xx {illeg}xy xx=ay. ye distance of ye poles is given wch I call b=bg .[14] perpendic: to bg I draw dc=w & make bc=v . Then, is dc2+bc2=bd2. w2+v2=xx. dc2+cg2=dg2. w2+bb2bv+vv=yy. soe yt for xx=ay {illeg} I write w2+v2=aw2+bb2bv+vv. Or w4+2w2v2+w4=aaw2+aabb2aabv+aavv . wch expresseth ye relation wch bc beareth to dc , & by makeing bc=x , dc=y , it is, y4+2xxyy+x4=0 aayy+2aabx aaxx aabb .

Examp 3d If bg=x {illeg}|b|e always in ye line bc . & fd turning about ye pole f {illeg}|&| passing by ye end of bg=x wth its other end d describes ye crooked line bdh , soe yt calling gd y x=y . yn drawing ef & dc perpendicular to bh . be=a , & ef=b are given. & I make bc=v {.} dc=w . yn is eg=xa . gc=vx . efeggccd. bxavxw. bw=vxav+axxx. or by extracting ye roote x=+a+v2Oaa+2av+vv4bw4 . againe gc2+dc2=gd2 vv2vx+xx+w2=yy=xx. . Or vv+w22w=x=a+v2Oaa+2av+vv4bw4 . & by transposeing a+v2 {illeg}|to| ye other side & so squareing both pts w42avw2=2aw3+v44bwv2 . wch equation expresseth ye relation twixt w=dc , & v {illeg} =bc . & so by {illeg} calling dc y , & bc x , it is, y42axyy+4bxxyx42ax3=0 .

Example 4th. if bd=x turnes about ye pole b , & gd (a given line =a) slides upon bg wth one end & intersecting bd at right angles at ye other end describes ye crooked line \ bde / by its intersection wth bd . then makeing bc=v , dc=w . bc2+dc2=bd2. v2+w2=x2. bcbddcdg. vxwa. & avw=x therefore vv+ww=aavvww or w4+vvww=aavv . & so by writing x for v & y for w , I have ye relation twixt x=bc & y=dc ex{illeg}|p|rest in this equation y4+xxyyaaxx=0 .

Or if ye relation twixt bd & dg {illeg} was exprest in this equation ({illeg}|m|a{illeg}|k|ing dg=y . bd=x ) xxy+ayy = a3 . then {illeg} as before bc2+dc2=bd2. v 2+w2=x2. bcbddcdg. vvv+w2wy. therefore y=wvv+w2v . first therefore I take away xx by making a3ayyy=xx=vv+w2 . or by ordering it a3aayvvywwy=0 . Then I take away y by substituteing it {sic} valor wvv+w2v into its r{illeg}|o|ome & it will be a3aw2v2aw4vv=vvwvv+w2+w3vv+w2v . & by □ing both pts. a6v42a4v4w22a4vvw4+aav4w42aavvw6+aaw8v8ww3v6w4=0 3v4w6vvw8 . & by writeing x for v & y for w ye equation will be aay8+2aaxxy6+aax4y4x8yy+a6x4=0 xxy83x4y63x6y42a4x4yy 2a 4x x y4 .

The like may as easily be performed in any other case.

After ye equation is brought to this order observe yt if y is noe where of odd dimensions yn ye line bc in (wch {illeg} is coincident wth x ) is ꝑte of an axis of y{illeg} e crooked line, as in ye 2d Example. And if {illeg} x is noe \where/ of odd dimensions (as in this, a4+yyaa=aax2 {illeg}|)| Then I draw bk from ye point b at ye begine|i|ng of x . I draw {illeg} \ bk / perpendicular to {illeg} \to {sic} bc / wch is coincident wth ye axis of ye crooked line. And if neither x nor y bee of unequall dimensions in any terme of ye equation then both bk & bc may bee taken for axes of ye crooked line or lin{e}s \whose nature are/ expressed by ye equation. As in ye 4th Example.

But if y is of odd dimension{s} in ye Equation then ordering ye Equation according to y see if y is of eaven dimensions in ye first te{r}me {illeg} { x } not found in ye 2nd. if so take away ye 2nd terme of ye equation & if there result an Equation in wch y is noe where of {odd} dim{en}sions. Then I draw ce perpendicular to bc & equall to yt quantity wch added or substracted from y yt m{illeg}|i|ght take away ye {illeg}|2|d terme; through ye point {illeg} P{illeg} f{illeg}{illeg}{illeg}bee {illeg}.


[15] As in this Example, yy+\2/ay x3a=0. Then to take away ye {illeg}|2|d terme I make za=y . & soe I have, zzaax3a=0 . in wch z is not of odd dimensions. Then drawing bc for x , dc for y , {illeg} & de for z : or wch is ye same (since za=y ) I make {illeg} ce=a that is I draw dc & ce on 2 contrary sides of ye line bc . & through ye point e I draw ae parallell to bc & make it ye axis of ye line dag

[16] Example ye 2d. y48ayyy+24aay23axy232a3y \ +16a4 / +12aaxy=0 . yt ye first terme may bee of eaven dimensions I multiply ye {illeg} Then by making z+2a=y I take away the 2d terme. & ye Equation z43axzz+12a3x=0 . in wch z is onely of eaven dimensions. Then I draw bc for x . dc for y de for, or wch is ye same (sinc {sic} z+ \ 2/ a=y ) I make ec=2a , yt is I draw ec & dc on ye same side of bc then through ye point e para{illeg}|l|lell to bc I draw ea for ye axis of ye lines dac d khg .

[17] In like manner, if x is of odd dimensions in some terme of ye Equation, ye Axis bk perpendicular to x=bc may bee found. As for Example. xx+ {{illeg}}{{illeg}| 2 |} axa3a=0 . by makei{ng} za2=x , I take away ye 2d terme and soe have {illeg}|this| equation zzy3aaa4=0 . therefore I draw bc=x from ye fixed point b , & ce=z, or wch is ye same (since za2=x ) I draw eb=12a , yt is I draw eb & bc on two contrary sides of ye line k b yn throug{h} ye point e , parallell to bk I draw ea ye axis of ye line

[18] Example 2d. x44ax3+4aaxx4a3xaayyaaby=0. by makeing x=z+a I have this Equat x42aazzaayyaaby+a4=0. In wch z is noe where of odd dimensions. therefore assumeing b for ye begining of x & making b c=x, & ce={illeg} z , or wch is ye same if {illeg} bc=+x I I make be=+a, since x=z+a; that is if bc is affirmative I take be & bc on ye same side of ye line kb . otherwise I describe ym on contrary sides of it. then through ye point e parallell to bk I draw eg {{illeg}|a|n} axis of ye lines dbmd & nhr . Againe I order ye Equation according to y & it is a2yy+aabyz4=0 +2aazz a4 . {illeg}|&| soe \since x is not in ye 2d terme/ makeing vb2=y . I take away ye 2d terme, & it is aavvaabb4z4=0 +2aazz a4 . Therefore I draw ec=z , dc=y , & df=v . or wch differs not {illeg} (since vb2=y ) I make cf=b2 & through ye point f parallell to ce I draw { fg } for another axis of ye lines dbmd , & ndhdr .

[19] But if {illeg} but if {sic} {illeg} {illeg} ye unknowne quantity ( x or y ) is of odd dimensions in ye first terme or if {illeg}|b|oth ye unknowne quantitys are in ye 2d terme, or {illeg}|if| by this {sic} meanes ye equation is irreducible to such a forme yt x , or, y , or both of ym bee of odd dimensions noewhere in ye Equ{illeg}|a|t: Then try to find ye axes by ye following method. Observing by ye way yt

If +x begins at ye {illeg} point b {illeg}|&| extends to{illeg}|w|ards c in ye line sr yn x is taken ye contrary way towards s , & all ye \affirmative/ lines parallell to sr are drawne ye same way wch +x is but ye negative lines parallell t{illeg}|o| sr are drawn ye same way wth x as if from ye point m I must draw a line =a, I draw it towards n but if from ye sme point m I must draw a line =a I draw it towards l . soe if from ye point {illeg}| d | I must draw dλ=+b , yn I draw it towards θ , but if dλ=b then I draw it towards γ . A{illeg}|g|aine if +y is drawne toward p from ye line sr , yn y is drawne from ye same line sr ye contrary way towards o , & those lines wch are affected wth an affirmative signe {illeg}|&| are paralell {sic} to y they are {illeg}|d|rawne ye same way wch y is but those lines wch are negative are drawn ye contrary way. as if a= then I draw a{illeg} aπ towards e but if a= yn I draw it towards t . soe if =+d yn I draw it from v towards δ, if =d , I draw it towards w.

A generall rule to find ye axes of any line.

[20] [21] Suppose bc=x. cd=y. & kg to be ye axis. yn parallel to y from ye point b to ye axis kg draw bf=c . from d ye end of y , d{illeg}d perpendicular to kg draw dh=ϩ . & make fh=ϱ & suppose fefgde . yn is fg=exd . & dedh=ϩeϩd=dg . dg2dh2=gh2 eeϩϩddϩϩ=gh2 . gh=ϩeedd d gh=fh+fgϱ+exd=ϩeeddd . therefore dϱ+ϩeedde=x . Againe ge+ecdg=dc , that is eexxdxx+ceϩd=y. or for x writeing its valor, y=cedϩ+ϱeedde . Now assumeing any quantity for e , yt I may find ye valors of c & d . I substitute these valors of x & y into theire roome in ye Equation. as if ye equation be x22xy+ay+yy=0 . by making e=a . ye valor of x is dϱ+ϩaadda & ye valor of y is acdϩ+ϱaadda . wch 2 valors substituting into their roome in ye equation, {illeg} their {sic} results { aaϩϩ+2dϩϩa2d2+4ddϱϩ2dϱϱaadd 2aaϱϩ+2acϱaadd a2dϩ+aaϱaadd 2acdϩ+aaϱϱ 2acϩa2d22dacϱ +a3c +aacc }=0 } Now yt ϩ I may have an equation in wch {illeg}| ϩ | is of \{illeg} {2}/ eaven dimensions o{ne}ly I suppose ye 2d terme =0 & soe have this equation 4ddϱϩ2aaϩϱ {illeg} aadϩ2acdϩ2acϩaadd=0 & yt ye teres in this {feigned} equation may destiny one another I order it according to { ϱ } & soe suppose each terme =0 . & so I have these equations 4ddϱϩ2aaϱϩ=0 & aadϩ2acdϩ2acϩaadd=0 . by ye first I find 2dd=aa, or d=aa2. by ye 2d aad2acd2acaadd=0 . & by substituteing ye valor of d into its {roome} I find a34aac2=0 . or c=a4 therefore from { b } perpendic: to bc I draw bf=a4 . through ye point f paralell {sic} to bc I draw fe=a2 & since f{illeg} =a therefore I draw ge=aaaa2=a2 . & lastly {illeg}{illeg}ye points f & g I draw fg ye axis of ye crooked line bah .


But since there is noe use of those termes in wch ϩ is of eaven dimensions ye Calculation will bee much abre{illeg}|v|iated by this following table.

x= ϩeedd e . y=dϩe . yy=2cdϩe . y3=3ccdϩe . y4=4c3dϩe . y5=5c4dϩe . &c xy=cϩeedd e . xyy=ccϩeedd e . xy3=c3ϩeedd e . xy4=c4ϩeedd e . xy5=c5ϩeedd e . &c.

[22] {illeg} x=eedd . y=d . yy=2cd . y3=3ccd . y4=4c3d . y5=5c4d . y6=6c5d &c xy=ceedd . xyy=cceedd . xy3=c3eedd . xy4=c4eedd . xy5=c5eedd . &c.

[23] xx=2deedd . yy=21×2}deedd . y3=62×3}cdeedd . y4=123×4}ccdeedd . y5=204×5}c3deedd . y6=305×6}c4deedd . xy=2dd+ee . xyy=4cdd+2cee . xy3=6ccdd+3ccee . xy4=8c3dd+4c3ee . xy5=5c4ee104dd . &c. xxy=2cdeedd. xxyy=2ccdeedd. xxy3=2c3deedd. xxy4=2c4deedd. xxy5=2c5deedd. &c.

[24] x3=3ddeedd. y3=3dee+3d3 . y4=2×6cd32×6cdee. y5=3×10ccd33×10ccdee. y6=4×15c3d34×15c3dee. xyy=3ddeedd+eeeedd. xy3=3ceeeedd9cddeedd. xy4 =6cceeeedd18ccddeedd. xy5=10c3eeeedd30c3ddeedd. &c xxy=2dee3d3 . xxyy=4cdee6cd3 . xxy3=6ccdee9ccd3 . xxy4=8c3dee12c3d3 . x3y=3cddeedd. x3yy=3ccddeedd. x3y3=3c3d2eedd. x3y4=3c4ddeedd.

[25] x4=4d3eedd. y4=1×4de63e4dd+3eed4d6. y 5=2×10cde63e4dd+3eed4d6. y 6=3×20ccde6&c: y 7=4×35c3de6&c. xy3=e42ddee+1d43ddee+3d4 . xy4=4ce4812cddee+4+12cd4 . xy5=10cce42030ccddee+10+30ccd4 . xy6=20c3e44060c3ddee+20+60c3d4 &c. xxyy= 1×2dee 4d3 eedd . xxy3= 2×3cdee 12cd3 eedd . xxy4= +12ccdee 24ccd3 eedd . xxy5= +20c3dee 40c3d3 eedd . x3y=3ddee4d4 . x3yy=6cddee8cd4 . x3y3=9ccddee12ccd4 . x3y4=12c3ddee16c3d4 . x4y=4 cd3eedd. x4yy=4ccd3eedd. x4y3=4c3d3eedd. x4y4=4c4d3eedd.

[26] x5=5d4eedd . y5=1×5de4+10d3ee5d5 . y6=2×15cde4+60cd3ee30cd5 . y7=3×35ccde4+210ccd3ee105ccd5 . y8=4×70c3de4 \+560c3d3ee280c3d5 / &c xy4=e4 2ddee+1d4 4ddee+4d4 ineedd . xy5=5ece4 10cddee+5cd4 20cddee+20cd4 ineedd . xy6=15cce4 30ccddee+15ccd4 60ccddee+60ccd4 ineedd . xy7=35c3e4 70c3ddee+35c3d4 140c3ddee+140c3d4 ineedd . xxy3=2de4 4d3ee+2d5 3d3ee+3d5 . xxy4=2×4cde4 2×8cd3ee+8cd5 12cd3ee+12cd5 . xxy5=20ccde4 40ccd3ee+2×10ccd5 30ccd3ee+30ccd5 . xxy6=40c3de4 80c3d3ee+2×20c3d5 60c3d3ee+60c3d5 x3yy= 1×3ddee 5d4 eedd . x3y3= 3×3ddcee 15cd4 eedd . x3y4= 6×3ccddee 6×5ccd4 eedd . x3y5= 10×3c3ddee 50c3d4 eedd . x3y6= 15×3c4ddee 75c4d4 eedd . x 4y=1×4d3ee5d5 . / 5d5 \ x 4yy=2×4cd3ee10cd5 . x 4y 3=3×4ccd3ee15ccd5 . x4y4=4×4c3d3ee20c3d5 . x4y5=5×4c4d3ee25c4d5 . x5y=5cd4eedd . x5yy=5ccd4eedd . x5y3=5c3d4eedd .

[27] x6=6d5eedd. y6= 6×1de105e8dd+10e6d410e4d6+5eed8d10. y7=6×7dce10&c: y8=6×28ccde10&c: y9=6×84c3de10&c: xy5=1×1e6 3dde4+3d4eed6 5dde4+3d4ee5d6 . xy6=1×6ce6 18cdde4+18cd4ee6cd6 6×5cdde4+60cd4ee30cd6 . xy7=1×21cce6 63cce4dd+63cceed421ccd6 21×5cce4dd+210cceed4105ccd6 . xy8=1×53c3e6&c xxy4=2×1de4 4d3ee+2d5 4d3ee+4d5 eedd . xxy5=2×5cde4 20cd3ee+10cd5 5×4d3ee+20cd5 ineedd . xxy6=2×15ccde4 60ccd3ee+30ccd5 4×15ccd3ee+60ccd5 ×eedd . x3y3=3×1dde4 6d4ee+3d6 3d4ee+3d6 . x3y4=3×4cdde4 24cd4ee+12cd6 3×4cd4ee+12cd6 . x3y5=3×10ccdde4 60ccd4ee+30ccd6 30ccd4ee+30ccd6 . x4yy=4×1dddee 4d5 2d5 ineedd . x4y3=4×3d3cee 12cd5 6cd5 ineedd . x4y4=4×6ccd3ee 24ccd5 12ccd5 ineedd . x4y5=4×10c3d3ee&c . x5y=1×5d4ee 5d6 1d6 . x5yy=5×2 cd4ee 10cd6 2cd6 . x5y3=5×3ccd4ee 15ccd6 3ccd6 . x5y4=5×4c3d4ee 20ccd6 4ccd6 . x6=1×6d5eedd . x6y=6cd5eedd . x6yy=6ccd5eedd . x6y3 −  {sic}6c3d5eedd. &c.

[28] y3{illeg}d3. y4=4cd3 . y5=10ccd3 . y6=20c3d3. y7=35c4d3. y8=56c5d3. xyy=ddeedd. xy3=cddeedd. xy4=ccddeedd. xy5=c3ddeedd. xxy=d3eed. xxyy=2cd32ceed. xxy3=3ccd33cceed. xxy4=4c3d34cceed. x3=eeddineedd . x3y=ceecddeedd . x3yy=cceeccddineedd . x3y3=c3e63e4dd+3eed4d6.

[29] y4={illeg} 1×4d3eedd . y5=5×4cd3eedd . y6=15×4ccd3eedd . y7=35×4c3d3eedd . {illeg} xy3=1×3ddee 3d4 1d4 . xy4=4×3cddee 12cd4 4×1d4 . xy5=10×3ccddee 30ccd4 10ccd4 . xy6=20×3c3ddee 60c3d4 20c3d4 . xxyy=1×2dee +2d3 +2d3 ineedd. xxy3=3×2cdee +6cd3 +6cd3 . xxy4=6×2ccdee +12ccd3 +12ccd3. xxy5=10×2c3dee +20c3d3 +20c3d3. x3y=1×1e4 2ddee+d4 3ddee+3d4. x3yy=1×2ce4 4ceedd+2d4c 6ceedd+6cd4 . x3y3=3cce4 6cceedd+3ccd4 9cceedd+9ccd4 . x3yy=4c3e4&c. x4=4eed4d3ineedd. x4y=4ceed4cd3ineedd. x4yy=4ccde63e4dd+3eed4d6. x4y3=4c3de6&c .

[30] y5=1×10eed3+10d5 . y6=6×10ceed3+60cd5 . y7=21×10cceed3+210ccd5 . y8=56×10c3eed3+560c3d5 . xy4=1×6eedd 6d4 4d4 ineedd. xy5=5×6ceedd 30cd4 20cd4 ineedd. xy6=15×6cceedd 90ccd4 60ccd4 ineedd. xxy3=1×3e4d+6eed33d5 +6eed36d5 1d5 . xxy4=4×3ce4d+24ceed312cd5 +24ceed324cd5 4cd5 . xxy5=10×3cce4d+60cceed330ccd5 +60cceed360ccd5 10ccd5 . x3yy=e42ddee+d4 6ddee+6d4 +3d4 ineedd. x3y3=3×1ce46cddee+3cd4 18cddee+18cd4 +9cd4 ineedd. x3y4=6cce412ccddee+6ccd4 36ccddee+30ccd4 +18ccd4 ineedd. x4y=4de49d3ee+d5 6d3ee+d5 . x4yy=2×4cde416cd3ee+8cd5 12cd3ee+12cd5 . x4y3=3×4ccde424ccd3ee+12ccd5 18ccd3ee+18ccd5 . x5=10 {illeg}dd10d4 {illeg}{eedd}. 10c {illeg}dd10cd4 {illeg}eedd=x5y . x5yy=10 {illeg} cdde63e4dd+3eed4d6


T{illeg}|h|e use of ye precedent table in fin{illeg}ding ye Axes of crooked Lines, declared by Examples.

[31] Suppose I had this Equation given, xx2xy+ay+yy. That I may find ye axis of ye line signified {illeg}|b|y it, first I observe of how many dimensions one of ye unknowne quanties {sic} \or ye rectang of ym both/ is found at most in ye Equation, (as in this Example they have noe more yn 2) then I take every quantiy {sic} in wch one of ye unknowne quantitys is of or ye rectangle of ym both is of soe many dimensions (wch in this case are xx2xy+yy.) Then lookeing in ye Table, (either amongst ye rules of ye first or 2d sort &c) for such a rule in wch ye \f{illeg}|i|rst/ quantity {illeg} is {illeg} of soe many dimensions {illeg} I substitute ye valors of ye unknowne quantitys, found by yt rule, into their place in ye Equation \selected quantitys/ & supposeing ye product =0, I find ye proportion of d to e thereby, that is I {illeg} find ye angle wch ye axis makes wth ye unknowne quantity called x . As in this case I take ye 2d Rule of ye first sort, & by it I find ye valors of xx {illeg} xx=2deedd. xy=ee2dd. yy=2deedd. wch valors substituting into ye roome of ye unknown quantitys {illeg}in these selected termes xx2xy+yy. I have this equation. 2deedd2ee+4dd2deedd=0. or, 2dd=ee. & e=d2 so yt by assuming any quantity for ee as a I have ye valor of d , for {illeg} d=a2 . therefore dd2dea2aaa2 &c. In ye next place yt I may find ye length of ye line bf=c . I take another rule whose first quantity is {illeg} not of soe many nor of fewer dimensions yn one of ye unknowne quantitys or ye rectangle ym both is any \some{illeg}/ where in ye Equa{illeg}|t|ion. Then select every quantity out of ye Equation, ye valor of whose unknowne quantity may be found by this rule, & substituting their valors, found thereby, into yr pl{illeg}|a|ces in this /these\ selected termes make ye p{illeg}|r|oduct =0. & find ye valor of c thereby. {illeg}|A|s in this example I must take ye first rule of ye 1st sort. By wch I find xy=ceedd, y=d, yy=2dc: but ye valor of xx ci|a|{illeg}|n|not be found by it. therefore I onely take ye termes 2xy+ay+yy, & by substituting ye valors of ye unknown quantitys into their roomes I have 2ceeddad2dc=0. Then by {illeg}|s|ubstituting ye \about found/ valors of d=a & e=2aa into yr places, it is O2ac+aa+2ac. Or {illeg} +2ac+aa+2ac. or c=a4. Soe yt if I make b ye beginning of x , & +x to tend towards c \in ye line bc /, & +y towards k perpendicularly to bc . then must I draw bf=a4 from ye point b perpendicular to bc ; & fe=a , & parallell to bc ; yn eg=gf2fe2 eg=a , & parallell to bf . Lastly through ye points f & g draw gf ye axis of ye line sought. Otherwise suppose eedd : it may be done thus {illeg} egefbfbh . therefore I take bh=cdeedd=a4 , & through ye points f & h eedddccdeedd . I draw af ye axis sought.

[32] Example ye 2d. If ye Equation bee x3axy+y3=0. ye Rule whose first quantity is of as many dimensions as either of ye {illeg} unknowne quantitys {illeg}|i|n {illeg} \this/ Equation, is ye 3d of ye first sort or ye first of ye 2d sort. Selecting therefore \onely/ x3+y3 out of ye Equation (since {illeg} \in/ neither of the{illeg}|se| \rules/ ye valor{illeg} of xy {illeg}|i|s found) by ye 3d rule of ye first sort I find x3=3ddeedd , y3=3d33dee . therefore ye selected termes x3+y3=3ddeedd+3d33dee=0 . & deedd {illeg}eedd. Or, ee=2dd . In like manner by ye first rule of ye 2d sort tis found y3=d3 . x3=eeddineedd . & therefore x3+y3= e63e4dd+3eed4d6d3=0 . & c:e63e4dd+3eed4d6=dd. Or ee=2dd as before. Soe yt egfeddeeddd . therefore eg=fe . Now yt I may find bf=c I take ye 2d Rule of ye first sort (whose first quantity yy is of fewer dimensions yn x3 or y3 but not of fewer xy ,) The quatitys {sic} in ye Equation whose valors are expressed in this rule are xy , & y3 for xy=ee2dd . y3=6cdeedd . {illeg}|S|oe yt I write 6cdeeddaee+2add instead of y3axy. soe yt c=2addaee6ddeedd . Or since 2ddee=0 , it is, c=0×a6dd=0 . Had I taken ye first rule of ye fir{illeg}st sort I had found xy=ceedd. & y3=3cdd . therefore y3axy=3ccdaceedd=0 wch is right since c=0 . but by this equation c hath other valors for 3cdaeedd=0 or 3cOa=0 , & c=Oa3 . &c. Whence obse{illeg}|r|ve yt {illeg}|fo|r ye most {pt} it will bee most convenient to find ye {illeg} c by yt rule whose 1st quantity hath one dimension lesse yn ye first quantity of yt rule by wch ye proportion twixt d & e were found.


Example y{illeg}|e| 3d. If ye Equation be xxyy+4bxyy+4bbyy=0 2axxy8abxy8abby a4 . xxyy being of 4 dimensions I take ye 4th rule of ye first sort, or ye 2d rule of ye 2d sort. By ye 4th rule of ye 1st sort I find xxyy=2deeeedd4e3eedd & since by that rule I can find ye valor of noe other quantity in ye Equation I {illeg} {illeg}|m|ake xxyy= 2dee 4d3 eedd=0. Or {illeg} e65dde4+8d4ee4d6=0. Whose rootes are eedd , ee2dd , & ee2dd . therefore either ee=dd , or ee=2dd Which is divisible by d , & ee2dd , & by eedd . therefore either d=0 ; or, ee2dd=0 ; or, eedd=0 . The operation is ye same if I make use of ye 2d rule of ye 2d sort. Againe I take ye 3d rule of ye 1st sort & by it I find, xyy=eeeedd3ddeedd. xxy=2dee3d3 . therefore 4bxyy2axxy 4ee 12dd eedd+6d34dee=0 . & if d=0 . then & xxyy=4cdee6cd3 . therefore xxyy 2axxy +4bxyy= 4cdee4adee+4bee 6cd3+6ad312bdd eedd=0 . Or c=a +2beeeeddbbddeedd 3d32dee. & if d=0, then c=a+2beeeedd0 . or c is infinitely long. but if ee2dd=0 . then c=aO4bO6b . & if eedd=0 , then c=a. Againe I take ye fir{illeg}|s|t rule of ye 2d sort & by it I find xxyy=2cd32cdee. xxy=d3eed. xyy=ddeedd . therefore xxyy2axxy+4bxyy=2cd32cdee2ad3+2aedd+4bddeedd=0 And c=a+2bdeedd Now if d=0. then Or c=ad3+aedd Now if d=0. or if eedd=0, then ye termes \of this Equat/ destroy one another soe yt ye valor of c may not be found thereby. but if ee2dd=0, then I find 2cd34cd32ad3+4ad3+4bdddd=0. Or 2cd3=2ad3O4bd3. Or c=aO2b. Againe I take ye 2d rule of ye 1st sort & by it I find xxyy2axxy+4bxyy +4bbyy8abxy =2cddeedd8bbdeedd4acdeedd+16ddab8abee16cddb+8ceeb=0. & if d=0 then 8abee+8ceeb=0, or /cc=2acee+ceeb0×ee\ If d=0 then cc=2ac4ceeb0×ee+4bb+4abee0ee. or c {illeg} aO2eb0O4eebb0×0+4aeb O4aeb0+aa+4bb. that is c is infinitely long as was found before. If eedd=0 also i{illeg}|t| may bee found to bee 8ceeb8aeeb=0 , or c=a but upon this supposition d=0 it was not before found c=a & therefore c=a is false, when d=0 . If eedd=0 . then I find 8abdd8cbdd=0. or c=a . &c. If ee2dd=0 . then ccdd4bbdd2acdd=0. c=aOaa4bb. Which valor not being found before I conclude ee2dd=0 to bee false. Lastly by useing ye first eq rule of ye first sort I find, 4bcceedd8abceedd8cdbb+8abbd=4bxyy+4bbyy8abxy8abby=0 . & by supposeing d=0 I have {{illeg} c= {illeg} 2a}. & c=a . & if ee=dd , then c=a wch being always found upon ye supposition dd=ee. I conclude ye valor of dd to be ee & of c {illeg} to be a {illeg}. & so draw ye axis gf parallell to x & distant from it ye length of a . But here of|b|serve yt this might have beene better performed by taking away ye 2d terme of ye Equation xxyy+4bxyy+4bbyy=0 2axxy8abxy8abby a4 . Or xx+4bx+4bba4yy2ay=0 as was {illeg} observed before.


[33]To find ye Diameter or axis of any crooked line which hath it

[34] Suppose ye \crooked/ line to bee lgc , ye \diameter or/ Axis kd , ye unde{illeg}|t|ermined quantitys describing ye line to be ab=x, bc=y. \to/ from ye begining point a (ye begining of x), perpendicular to kb draw ah=pb=c, {illeg} cutting ye axis kd in h . paralell {sic} to kb draw hp=x. & produce cb {illeg} soe yt it intersect ye axis in ye point d . & suppose yt hp is to hd . as d to e : or yt hd=exd . & therefore dp=xeeddd. le{illeg}|t| ec bee one =el=ϩ be one of those lines wch are ordinately applied to ye diameter he=ϩ . & lastly suppose that {illeg} ec is to ef , {illeg}|as| {illeg}| d | to f : yn is ef=fϩd . & fc=ϩ eefff. then hppdfcfd{illeg}|.| /xx eedddϩ ddffffd.\ therefore fd=ϩ eeddeeffdddd+ ddffdf . & hd=exd=dfϱ+ϩ e4eeffeedd+ddffdf draw cf perpendicular to hd ye Axis &
{sic} by ordering ye Equation it will bee & hd=exd= {illeg} ϱfϩd+ϩ eeddeeffd4+ ddffdf & by ordering ye equation
lastly suppose yt ec is to ef as e is to f : yn is ef=fϩe ; & fc= {illeg} ϩ eeffe then hppdfcfd.
xx eedddϩ eefffϩ e4eeffeedd+ddffde=fd & hd=ϱfϩe+ϩ e4eeffeedd+ddffde=exd. & by ordering ye Equation it will be, x= dd {illeg} edϱdfϩ+ϩ e4eeffeedd+ddffee . Againe, defccd=ϩ eeffd . & dp=x eeddd, or by substituteing {illeg}|t|he valor of x into its ro dp+pbdc=y=x eedd+dcϩ eeffd or by substituteing ye valor of x into its place it is y=eϱ eedd+fϩ eedddϩ eeff+eecee . {illeg}|A|nd yt I may abbreviate ye termes I make eedd=t ; & +f eedd+d eeffe=v ; & so ye Equation is y=tϱvϩ+ece . Also by supposeing s= e4eeff+ddffeedddfe, I lesson {sic} ye termes of ye Equation x=edϱdfϩ+ϩ e4eeffeedd+ddffee , by writeing instead dϱ+sϩe=x.

Now therefore by meanes of t substituting these valors of x & y into yr stead I take ym out of ye Equation expressing ye {illeg} reltion {sic} twixt ym soe yt yn I have an equation expressing ye relation twixt ϱ & ϩ . And to that end it will bee convenient to have a table of ye squares, cube, squaresquares, square=cubs, rectangles \&c/ of ye valors of x & y , {illeg} After ye manner of yt wch follows.

x=dϱ+sϩe . xx=ssϩϩ+2dsϱϩ+ddϱϱee . x3=s3ϩ3+3dϱssϩϩ+3ddϱϱsϩ+d3ϱ3e3. x4=s4ϩ4+4dϱs3ϩ3+6ddϱϱssϩϩ+4d3ϱ3sϩ+d4ϱ4e4. x5=s5ϩ5+5dϱs4ϩ4+10ddϱϱs3ϩ3+10d3ϱ3ssϩϩ+5d4ϱ4sϩ+d5ϱ5e5. \&c./ y=tϱvϩ+ece. {illeg} yy= vvϩϩ2tϱvϩ+ttϱϱ 2ecvϩ+2ectϱ +eecc ee. y3= v3ϩ3+3tϱvvϩϩ 3ttϱϱvϩ+t3ϱ3 +3ecvv6ectϱv+3ttϱϱec 3eeccv+3tϱeecc +e3c3 e3. y4=v4ϩ44tϱv3ϩ3 +6ttϱϱvvϩϩ4t3ϱ3vϩ+t4ϱ4 4ec +12cetϱ 12ttϱϱcev+4cet3ϱ3 +6ccee 12cceetϱv+4c3e3tϱ 4c3e3v+c4e4 . y5=v5ϩ5+5tϱv4ϩ4 10ttϱϱv3ϩ3 + 10t3ϱ3vvϩϩ5t4ϱ4vϩ+t5ϱ5 +5ce20cetϱ+ 30ttϱϱce20cet3ϱ3+5cet4ϱ4 10ccee+ 30cceetϱ30cceettϱϱ+10cceet3ϱ3 + 10c3e320c3e3tϱ+10c3e3t2ϱ2 5c4ϱ4+5c4e4tϱ +c5e5 . \&c/ xy= svϩϩ+eϱsϩ+dtϱϱ +ecs+decϱ dvϱ ee . xxy= ssvϩ3+sstϱϩϩ+2ecdsϱ+ecddϱϱ +ssec+2tdsϱϱ+tddϱ3 2dsvϱddv eee . x3y= vs3ϩ4+2tϱs3ϩ3+3ecdssϱϩ2+3ecddsϱϱϩ+ecd3ϱ3 +ecs3+3tdsϱ2ϩ2+3tddsϱ3 +td3ϱ4 3dϱssv3ddsvϱϱ d3v ϱ3 e4 . &c. xyy= vvsϩ32tϱvsϩϩ+sttϱϱϩ+ttdϩ3 2ecvs+2stceϱ+2ectdϱ2 +dvvϱ+sccee+eeccdϱ 2dtvϱϱ 2dϱecv e3 . xxyy= ssvvϩ42sstvϱϩ3+ssttϱϱϩ2+2ttdsϱ3ϩ+ddttϱ4 2ssecv+2ssectϱ+4ectdsϱ2+2ddectϱ3 +2dsϱvv+sseecc+2eeccdsϱ+ddeeccϱϱ 4dsϱϱtv2ddtvϱ3 4dsϱecv2ddecvϱϱ +ddvv e4 . xy3
xy3= sv3ϩ4+3stϱvvϩ33sttϱϱvϩ2+st3ϱ3ϩ+dt3ϱ4 +3secvv6sectϱv+3sttϱϱec+3dttϱ3ce dϱv33seeccv+3stϱeecc+3dtϱϱccee 3dtϱϱvv3dttϱ3v +3decvv6dectϱϱv 3deeccv e4 . &c If there bee occasions to doe these operaco_ns in Equations of 5 or 6 or more dimensions this table may be easily enlarged.

As for example. If ye relation twixt x & y bee exprest in this Equation, xx+ax2xy+yy=0 . then into ye place of xx , x , xy , yy , I substitute their valors found by this table, & there results +ssϩϩ+2dsϱϩ+ddϱϱ=0 +2sv+aes+adeϱ +vv2tsϱ2dtϱϱ 2eds2decϱ +2dvϱ+ttϱϱ 2tϱv+2ectϱ 2ecv+eec . Which Equation expresseth ye relation twixt ϱ & ϩ . yt is twixt ge & le or ec . Now yt ge=ϱ be ye diameter of ye line & {illeg} & le=ec=Oϩ be ordinately applied to it, it is required (by Prop 2d) yt in this Equation ϩ be not of odd dimensions. & that may bee soe {illeg} ye quan\ti/tys in th{illeg}|e| 2d terme \wch ϩ is {illeg} of {illeg} / must {illeg} one another wch canot {sic} be unlesse thes{illeg}|e| {illeg} quantitys destroy one another in wch ye unknowne quantitys ϱ & ϩ destroy one /{illeg}are of ye {same}\ <21v> dimensions. Which things being considered it will appeare yt I must divide ye 2d terme into two pts, makeing, 2dsϱϩ2tsϱϩ+2dvϱϩ=0 2tvϱϩ ; &, aesϩ2ecsϩ2evcϩ=0 . & divideing ye first by 2ϱϩ & ye 2d by eϩ they will be, dststv+dv=0 . {illeg} Hitherto useing useing ye letters s , t , & v for b{illeg}|r|evitys sake, I must now write their valors in theire stead (yt I may \find/ ye length of c , & ye proportion of d to e wch determine ye position of ye axis, & also ye proportion of e to f wch determines ye position of ye lines applyed to ye axis.) & soe {instead} of ye Equation dststv+dv=0 ; there results, +de e4eeff+ddffddee+ddee+deeeff+2dfeeddeef=0 wch by squareing /both pts is\ e62de5+ddffdde4+2dff+2d3e3ddffd4e2 Or eeff=eefddfdfeedd ddee+de+deeedd . Or +de e4eeff+ddffddee+ddee+deeeff=eef+2dfeedd . And by squareing both pts {illeg} {illeg} it is /& ordering ye product it is.\ { 2e64de5+2dde43ffe43de4+4dffe3+4d3ddfee+d4ee+ddffeed3ffed4ff+2d3ee2d3ff+de4+deeff+2e52e3ffeedd=2d3ff2deeffeeff Or 2e64de53ffe4dde4+4dffe3+4d3+2ddeeff4d3ff=2d3ee+4d3ff+4d56deeff+2e3ff2ef2e5ineedd .} dde4ffe4+3ddffee2d4ee2d4ff=+2e4f+2e3df6eeddf4ed3f+4d4fineeff . Which b{illeg}|ei|ng divided by ee2dd the product is ddeeffee+ddff=2eef+2edf2ddfineeff . Wherefore I conclude yt ee=2dd . or yt dededd. Againe by inserting \ye valors/ of s & v into ye Equation as2cs2cv there results a e4eeff+ddffddeeadf=2c e4eeff+dffddee+2cfeedd+2ceeeff2cdf . or by 2dfeedd=eefeeeeff2ddeeff . & by squareing both pts & ordering ye product it is, e44ddee+4d4=4ddfeeff2eefeeff . Which is divisible by 2ddee=0 , for ye quote will bee 2ddee=2feeff . & therefore 2dd=ee . Or, 2dd=ee+2feeff . Againe by inserting \ye valors/ {illeg}|of| s & v into ye Equation, as2cs2cv=0 , there resulteth, +a2c e4eeddeeff+ddff2cfeedd2cdeeddadf+2cdf=0 . & by writing 2dd instead of ee & d{illeg}|i|videing it by d there resulteth +a4c2ddffaf=0 Or a4 af 42ddff =c . Thus haveing found ye proportion of d to e , & ye valor of c since theire {sic} remaines noe more equations by wch I may find ye proportion of e to f I concluded {illeg} it to be undetermined, soe yt I may assume any proportion betwixt ym. As if I make f=0 . Then ye angle ceh is a right one & eh ye axis of ye line, & c=a4=ah . & d 2dddehphd . or d2ddhpdp . that is hp=dp ; As in ye 1st figure.[35] Or if I make eeff21 . that is ee=2ff . or f=d , then I find yt c=0 . that is that ye diameter ed intersects ye line ap at ye point a ye begining of x . & yt ye lines ec are parallell to ma as in ye 2d figure[36] &c. Soe yt by assumeing any proportion twixt ec & ef , that is, {illeg}|s|upposeing ye angles fec of any bigness, ye position of ye dia{illeg}|mi|ter fa , {illeg} may be found i|a|fter ye same manner. As If I would have ye angle fec to be an anangle of 60 degrees. yn must ec=ϩ be double to fe=12ϩ , & fc=34yy . i.e. ef21. & 2dd=ee, therefore dd2=ff. I found before yt a4 af 42ddff =c . or writeing ye valor of f in its roome, tis a4 a12dd 43dd2=c that is a4Oa43=c . Or a4 since c must be lesse yn a4 it must a4a43=c=ah. & ph=dp since ee=2dd . As in ye 3d figure.[37] But if I would make ye angle ceh of 60 degr: th{illeg}|e|n as before e=2f , & dd2=ff, & a4Oa43=c , or since c must be greater yn a4 tis a4+a43=c , as in ye 4th fig. &c.[38]

Example 2d. If ye Equation expressing ye nature of ye line be x33xyy+3 x33xxy+3xyyy3=0 ayy .


[39] To find ye Axis or Diame{illeg}|t|er of any crooked Line supposeing it have ym.

[40] [41] [42] Suppose bc=x ; cd=y ; {illeg} nad ye line whose axis \or Diameter/ is sought; pk its axis or Diameter; a its vertex; hd=hn=ϩ= lines ordinately applied to its Diameter; bm a perpendicular to pc drawne from ye point b , i.e. from ye begining of x ; bf=c= pte of ye line bm intercepted twixt ye diameter & pc ; mh=ϱ= a line parallell to bc & drawne from ye bm to ye intersection of fk & nd ; & he=mf & parallell to mf ; & dgh a right angled triangle. dhhgdf &, mh=fehe=mfde.

Then since {illeg} deϱhe therefore eϱd=he {illeg}

Then deϩfϩd=hg=eq . &, eq+fe=dϱ+fϩd=x . Againe gd=ddϩ2ffϩ2dd deϱhe=eϱd=gq . &, gq+qc+dg=cd ; or, eϱ+dc+ϩddff d =y .

Now therefore {illeg} by substituteing dϱ+fϩd into ye place of x , & eϱ+dcϩddff d into ye place of y , & theire □s & cubes &c: into ye place of x2 , x3 , y2y3 &c. I take x & { y } out of ye Equation expressing ye relation twixt ym & {illeg}|S|oe have an Equation expressing ye relation twixt ϱ & ϩ . And to yt end it will be convenient to have a table of ye squares, cubes, & rectangles &c: of ye valors of x & y , like yt wch follows.

d4xxyy= dx=dϱ+fϩ ddxx=ddρϱ+2dϱfϩ+ffϩ2 d3x3=f3ϩ3+3ffϩϩdϱ+3ddϱϱfϩ+d3ϱ3 d4x4=f4ϩ4+4dϱf3ϩ3 +6ddϱ2f2ϩ2 +4d3ϱ3fϩ+d4ϱ4 0 dy= +ϩddff+eϱ +dc 0 ddyy= +ddϩϩ+2eϱϩddff+eeϱϱ +2dc+2edcϱ +ddcc ddxy= +fϩ2ddff+feϱϩ+deϱϱ +fdcϩ+ddcϱ +dϱϩddff 0 d3x2y= +ffϩ3ddff+ffeϱϩ2+2efϱ2ϩ+ddeϱ3 +ffdc+2ddcfϱ+d3cϱϱ +2dfϱddff+ddϱϱddff 0 d4x3y= +f3ϩ4ddff+ef3ϱϩ3+3deffϱϱϩϩ+3ddefϱϱϩϩ+d3eϱ4 +dcf3+3ddcffϱ+3cd3fϱϱ+d4cϱ3 +3ffdϱddff+3ddfϱϱddff+d3ϱ3ddff 0 0 d3y3= ddffϩ3ddff+3ddeϱϩϩ+3eeϱϱϩddff+e3ϱ3 +3dddc+6edcϱ+3eedcϱ2 3ffeϱ+3ddcc+3ddcceeϱ 3ffdc+d3c3 0 d4y4= d4ϩ4+4eϱddϩ3ddff+6ddeeϱϱϩϩ+4e3ϱ3ϩddff+e4ϱ4 2ddff+4d3c6ffeeϱϱ+12dceeϱϱ+4dce3ϱ3 +f44eϱff+12d3ecϱ+12ddecϱ+6ddcceeϱϱ 4dcff12dffeeϱ+4d3c3+4d3c3eϱ +6d4cc+d4c4 6ddffcc 0 d3xyy= fddϩ3+d3ϱϩϩ+eeϱϱϩ+deeϱ3 f3dffϱ+2cdefϱ+2cddeϱϱ +2efϱddff+ccddf+ccd3ϱ +2cdfddff+2cϱ2ddff +2cddϱddff 0 d4xxyy= 0 0 d4xy3=

As for example if ye relation twixt x & y bee exprest by, xx+ax2xy+yy=0 then in stead of xx , { x }, xy , yy , writeing their valors found by this table there resulteth { ddϩϩ+2dfϱϩ+ddϱϱ=0 +2fϩϩddff+adf+addϱ 2efϱ2deϱϱ 2dcf2ddcϱ 2dϱddff+eeϱϱ +2ϱdd+2edcϱ 2dc+dd .} Which equation espresseth ye relation twixt ϱ & ϩ when any valors are assumed for c , d , e , & f . And if ye valors of c , d , e , & f bee such that ye 2d terme in ϩ is not of odd dimensions {illeg} in any terme ye Equation (that is yt ye 2d terme of this Equation be wa{illeg}) then (by Prop: ye 2d) Oy=hn=hd is ord{illeg} <24r> is ordinately applyed to ye Diameter pk .Now yt ye 2d terme of this Equation vanish it is necessary yt those termes destroy one another in wch ye unknowne quan{illeg}|t|ys {sic} are ϱ & ϩ are not diverse nor differ in dimensions. Whence it appeares yt I must divide {illeg}|t|he 2d terme into 2 pts making 2dfϱϩ2efϱϩ2dϱϩddff+2eϱϩddff=0. & adfϩ2dcfϩ+2dcϩddff=0 . Or by divideing the first of these by 2ϱϩ , & ye 2d by dϩ. they are, dfefd+eddff=0 , & af2cf+2cddff=0 . The first being divided by de=0 . there results, f+ddff=0 . Therefore one or both these propositions d=e ; dd=2ff , is {illeg} trew. by ye 2d tis found yt af2f2ddff =c . Now since by assumeing some quantitys for ye valors of d , c , or f I cannot find ye valor of e unless by ye Equation d=e. therefore I conclude d=e . whence it is not necessary yt dd=2ff , \or ye proportion of d to f bee limited/ soe yt by assuming ye angle ahd of any bigness I may find ye position of ye axis ahd . As if I suppose ye angle fhd to be a right one (i.e. yt ah is ye axis of ye line) then are ye △s feh & hgd alike, & therefore fhfedhghdd+eee & gh=eϩdd+ee df. & eddd+ee=f . Or because d=e . therefore +f=d2 . Soe yt I draw fq parallell to pb & qh equall & c=a4 . There Soe yt I draw bf=a4. & fq parallell to pb qk=fq & parallell bf & through ye points f & k I draw kh ye axis of ye line nad , wch is a Parab. /as in figure 1st\[43] So if I would have hd {illeg} paralell {sic} to qk i.e. ye an{illeg}|g|leg dhf of 45 degre{illeg}|e|s. then this evident yt hg=0=f . & c=af2fO2ddff . Threfore through ye point b I draw ye axis kh , so yt bq=kq , as before. {illeg} /&c.\ {illeg}|&| note yt since kh t{illeg}|h|e axis is always paralell {sic} to it selfe ye line dbn is a parabola.[44]

[45] Example ye 2d, x3+y3=a3. B{illeg}|e|ing first to write ye valors of x3 & y3 (found by ye precedent table) in{illeg}|to| their roome, since I have {illeg} noe neede of those termes in wch ϩ is of eaven dimensions I leave ym out, & soe for x3+y3a3=0 I write onely f3ϩ3+ddddff+3ddfϱϱϩ+6edcϱϩddff+3ddccϩddff=0 ff+3eeϱϱϩddff . Then sorting these quantitys together in wch ye unknowne quantitys are ye sam{illeg}|e| there these 4 Equations (ye {illeg} 1st being divided by ϩ3 , y{illeg}|e| 2d by 3ϱϩ2 , y{illeg}|e| 3d by 6ϩϱ , ye 4th by 3ϩ ) viz: f3+ddffddff=0 ; ddf+eeddff=0 ; +cdeddff=0 ; +ddccddff=0 . In ye first Equa
tion f3=dd+ffddff, I extract ye cube roote & tis f=ddff. or dd=2ff. In ye 2d ddf=eeddff , ddf=eef , or d=e. By the 3d, +cdeddff=0 , or c=+0de f=0 . & so by ye fourth.[46] Now therefore since \ c=0 / d={illeg} f d={illeg} e . In ye line bq from some point as q perpendicular to bq I \draw/ kq=+e , & ql=e , both of ym =bq=d . then from ye points k & l through b I draw ye two lines ak & gl both wch (since they it cuts one another \ye lines hnd applied to them/ at right angles) is are axe|i|s of ye lines ndr & {illeg} wch appeares al{illeg} in {illeg} yt dd=2ff, for therefore nt2=2st2=st2+ns2 , soe yt ns=st & nt perpen
dicular to bk .

[47] Example 3d If ye nature of ye given line bee e{illeg}|x|pressed in these termes x33xxy+2xyy2ayy=0 . Then by supplanting ye {illeg}|v|alors of x & y into theire roome & working as before, there will bee, f3+2fdd3ffddff=0 . & &|2|dly 3ddf6def3dd+4edddff=0 . & 3dly 6cddf+4cdef+4cddddff4adeddff=0 . & 4tly, 2ccddf4acddddff=0 . The first of these divided by f=0 . is ff+2dd=3fddff . Or □ing tis {illeg} & ordering ye product tis 10f413ffdd+4d4=0. Which being 2ff dd=0 . there results 5ff 4dd=0 . Wherefore I conclude one of these 3 to be ye valors of f viz: f=0=dd2=4dd5. Now yt I may know wch of those is ye right valor of f I try y singly, {illeg}|&| fi\r/s{illeg}|t| suppose f=0 ; If so yn by ye 4th Equation 4acddddff=0, therefore c=0. If {illeg} c=0=f , yn in ye 3d Eq{illeg}/u\ation all ye termes vanish except 4adedd=0 : therefore e=04addd=0 . & since 0=e=f=0 , all ye termes in ye 2d Equat vanish except except 3dddd , therefore also d=0 , wch since it ought not to bee I conclude yt f=0 is false. Therefore I passe ye 2d valor of ff=dd2 , or, Of=ddff. & soe divideing ye 4th Equat by 2ddf {i{t}} results cc=O2ac . wch is divisible by c=0 & by cO2a=0 , Now yt I may know wch is ye right valor of c first I suppose c=0 : & soe all ye termes in ye Equation vanish except, 4adeddff=0 . {illeg} or, { e=04adf=0 . } & since e=0 , by ye 2d Equation tis 3ddf3ddddff=0 or f=Oddff=ff=+ddff . {Which things since} they agree I conclude yt {illeg} f=ddff , or dd=2ff; c=0 ; e=0 . Since e=0 {illeg} {illeg} must be parallell to x & {illeg} c=0 {illeg} must bee {coincident} wth it. then {illeg} ye axis { bc I take some {illeg} {illeg} } & fro{illeg} perpendicular {illeg} {illeg} {illeg}ye {illeg}


[48] Example ye 4th. If ye Equation bee bx3+ayxx=a4 . by takeing \onely/ those termes (of ye valors of x3 & yxx found by ye precedent table) in wch ϩ is of odd dimensions, & sorting those together in {illeg}|w|ch are multiplied by ye same unkowne {sic} quantit{illeg}|y|s are ye same & of ye same dimensions \as before/. there will result these Equations. first bf3+affddff=0 . 2dly 3ddbf+2ddbf+addddff=0. & 3dly, 2addfc=0. ye 1st is divisible by f=0, fb+addff=0 . To know wch of these 2 are ye valors of f first I suppose f=0 to be trew, & yn all ye termes in ye 2d Equation vanish except adddd , or dd=0add=0 . by ye 3d Equation 2addfc=0 vanisheth since f=0 & there|fore| ye valor of c cannot bee found soe yt if I assume some valor for it as c=0 now since both d & f should never bee =0 therefore I conclude yt f=0 is false & so pass to its other valor f=addffb . or bfa=ddff . & soe by ye 2d Equation tis {illeg} ddb=ade. wch is divisible by d=0 , db+ae=0 . If db=ae tis baedddfff. & soe ye lines ordinat diameter will bee parallel to ye lines ordinately applied to it wch cannot bee therefore I try ye other valor of d=0 . And if d=0 , yn ye 3d Equation 2addfc=0 vanisheth & soe c cannot bee found & is therefore unlimited. Now since I find noe repugnancys in these Equations f=addffb , & d=0 , to be trew I conclude ym trew. & since d=0 . I draw bh perpendicular wc from b ye begining of x , wch shall bee ye Diameters of ye lines enm & dpl . then in yt diameter I tak{illeg}|e| some point as b or h & from yt point dra{illeg}|w| {illeg} \ gh=+f or/ th=f , i.e. of any length, & paralell {sic} to bc . then from ye pointe t or g \perpendicular to tg / I draw ts=addffb , or gr=addffb . that{illeg} is, abthtsghgr. & so through ye points s & h or h & r I draw sr wch shall be ordinate parallel to ye lines ordinately applied to ye Diameter bh .

Example ye {illeg} 5t. Suppose x3=aay . Then by selecting those termes out ye valors of x3 & y in wch ϩ is of {illeg} o{illeg}|d| dimensions , & sorting them together in wch ye unknowne quantitys differ not, I have, f3ϩ3=0 ; 3ddfϱϱϩ=0 ; & 3dly aaddϩddff=0 . by ye first f=0 , & therefore ye 2d vanisheth{illeg}; & ye 3d divided by aaϩ is, ddddff=0 ; or {illeg} 0=ddd . wch may not bee since f=0 . Now since d=f=0 , & ye proprortion {sic} of d to e {illeg} {illeg}|&| ye length of c cannot bee found tis e{illeg}|v|ident ye line hath n{illeg}|o|e axis or diameter.


November 1664

[49] [50] Observe ye Axes, Diameters & position of ye lines ordinately applied to ye|m| may bee for ye most pte easlier obteined {illeg} by making bc=x. cd=y. bf=c=cq . mh=ϱ=fe. hd=ϩ=hn. hddgdfϩfϩd=dg. ye angles bcd , mgd , fqd , {illeg} mbc , feh , right ones. fehedeϱeϱd=he=gq. qc+gq+gd=cd=fϩ+eϱ+cdd=y. hg=ϩϩffϩϩdd=eq . fe+eq=bc=dϱ+ϩddffd=x. Then for readiness in these operations make a table of ye □s, cubes, rectangles, &c of these valors of x & y . As was done before dx=ϩddff+dϱ. 0 ddxx=ddϩϩ+2dϱϩddff+ddϱϱ. ff 0 d3x3=ddϩ3ddff+3d3ϱϩϩ+3ddϱϱϩddff+d3ϱ3. ff3dffϱϩϩ 0 dy=fϩ+eϱ+cd. 0 ddyy=ffϩϩ+2efϱϩ+eeϱϱ. +2cdf+2cdeϱ +ccdd 0 d3y3=f3ϩ3+3effϱϩϩ+3eefϱϱϩ+e3ϱ3. +3cdff+6cdefϱ+3eecdϱϱ +3ccddf+3ccddeϱ +c3d3 ddxy=fϩϩddff+dfϱϩ+deϱϱ. +eϱϩddff+cddϱ +cdϩddff 0 d3xyy=ffϩ3ddff+dffϱϩϩ+2defϱϱϩ+deeϱ3. +2cddfϱ+2cddeϱϱ +2efϱddff+eeϱϱeeff+ccd3ϱ +2cdfddff+2cdeϱeeff +ccddeeff 0 d3xxy=ddfϩ3+ddeϱϩϩ+ddeϱ3. f3ffeϱ+cd3ϱϱ +2dfϱddff+ddfϱϱϩ +cd3+2deϱϱϩddff&c. dcff+2cddϱϩddff 0

[51] Example If ye relation twixt bc & cd be expressed by ayybxy+bbx=0 aby . then by inserting those quant{illeg}|y|{sic} (of ye valors of x & y found by this table) in wch ϩ is of odd dimensions, into place of yy , xy , y , x in this Equation, & supposeing those to destroy one another wch are multiplied by ye same unknowne quantitys there will bee these 2 Equations 2aefbdfbeddff=0, & ∼ 2acdf+bcdddff+bbdddffabdf=0. The 2d is divisible by d=0 & there ∼ results 2acf+bcddff+bbddffabf=0. Now to try wch of these {illeg}|t|wo are true fir{illeg}|s|t I suppose d=0[52], & soe ye first Equation will bee 2aefbeff=0. wch is imposible {sic} unlesse f=0, & yn ye valors of e & c canot {sic} bee found, Therefore d=0 is false. And therefore by ye 2d Equat c=bbddff+abf 2af+bddff . & by ye first e=bdf 2afbddff . 2afObddffbfde. & c= Obbddff+abf 2afObddff . Whence ye proportion twixt d & f {illeg} yt is ye angle fhn is undetermined, & e & c have double valors viz: when e=bdf 2afbddff , then c= abfbbddff 2af+bddff . And when e=bdf 2af+bddff , then c= afb+bbddff 2afbddff . wherefore ye lin{illeg}|e| hath 2 axes.

[53] For avoyding mistakes \ (wch might have happened in ye 4th Example where I found d=0. & f= addff b )/ it will no{illeg}|t| be amisse to make hddgfgϩgϩf=dg. & fehedeϱeϱd=he. & soe it will be gϩf+eϱd+c=y. & ϱ+ϩffgg f =x . Or, dgϩ+feϱ+dfc df =y. & fϱ+ϩffgg f =x. And then observe yt it can never happen yt f=0. or d+e=0 . observe alsoe yt if deg+ffgg. yn ye line fh is ye axis, otherwise ye diameter of ye crooked line. when d=0 \{illeg}/ ye axis is perpendicular to x from ye point b as also if c=a0 : And yn it will be convenient to doe ye worke over againe changing {illeg}{illeg}|y|e names of x & y {yt} is writeing y instead of x & x instead {illeg}|o|f y .


[54] Haveing ye Diameter to find ye Vertex of ye line.

[55] Suppose bc=x, or tr=x. cd or ra=y. bf=c . fqkqdefs=br=xas=yc. yt is ex=dydc. soe yt into ye given equation I insert this valor of x=dydce or of y=ex +dcd into ye place of x or (wch may more readily bee done). {illeg}

[56] As in ye first example I found d{illeg} =e & ye proportion twixt d & f to bee unlimited so yt if I would fk to bee ye diameter Axis I make defddff. (vide {illeg} C) or {illeg} f=ddff. & there I found c=af2f2ddff . or sinc {sic} f=ddff it is c=a4 . As may {illeg}|b|ee seene in yt example. Now yt I may find ye vertex of ye line was there exprest in these termes. xx+ax2xy+yy=0. I suppose dexyc. {illeg}|tha|t is x=yc . or x=ya4 . or y=x+a4 . & writeing this valor of y into its r{illeg}|o|ome in ye Equation xx+ax2xy+yy=0; there results ax+aa16=0 . or x=a16 . Therefore from ye point b I draw br=a16 . & from ye point r I draw ye perpendicular rd until it {cu{illeg}t}{cu{illeg}|t|t} ye axis hd , yt is, soe yt rd=hd. & ye point d shall bee ye vertex of ye Parab: {sic} mdo .

[57] Soe in ye 2d Example of ye line x3+y3=a3 , it was found d=e . & c=0. & therefore y=ex+dcd . or y=x . therefore I write x3 for y3 in ye Equation x3+y3=a3. & it is 2x3=a3 . or x=ac:2 . therefore I take br=ac:2 & soe draw ye perpendicular ar , which shall intersect ye axis ab at ye vertex of ye crooked line. [58] & yn (calling br=h ) {illeg} it shall be ar=c+ehd. Soe yt in this case ar=a2.

[59] In ye 3d Example ye Equation being x33xxy+2xyy2ayy=0, It was found, c=0. deq0. yt is e=0 . therefore y=ex+dcd=0 . Therefore by {illeg} writeing 0 instead of y in ye Equation all ye termes vanish except ∼ x3=0 , or x=0=br . & ar=c+ehd=0 . soe yt ye vertex of ye line bdn must bee at ye point b .

[60] But in ye 4th Example, bx3+axxy=a4 . It was found d=0 de0{1 }{i}. \or d=0./ & c {illeg} was unlimited, I make therefore c=0 . & s{illeg}|in|ce for y I make y=ex+dcd=x0 . Or bx3+ax30=a4 . 0bx3+ax3=0a4 , Or ax3=0 , & x=0 ye axis is perpendicular to x therefore I insert ye valor of x into ye equ{illeg}|a|tion x=dydce=0 & there results b000+a00y=a4. or y=a4b00000a=a300 . Wherefore I conclude ye vertex of ye line to be infinitely distant from b towards m.

[61] ☞ If ye position of any line (as ts ) be given ye point where it intersects ye given crooked line dsa may be found by ye same manner; for suppose ∼ ar {illeg}|o|r ac=x . cd or rs=y . & {illeg} rx=yy . ta=a. tq=b. pq=d. angles tqp , srt , dct right ones; yn, tqqptrrs to find ye point s where ye crooked line dsa is intersected by ye line fp , I suppose tqpqtrrs. yt is, {illeg} y=da+dxb . or bydad=x . & since by ye nature of ye line rx=yy . it follows yt {illeg} yy=brydard . {illeg}|&| ddxx+2ddax+ddaabb=rx . & by extracting ye rootes of ym. both, y= br 2d O +bbrr4ddar 4dd . & x=abbrddO x=bbr2ddaO b4rr4d4 abbrdd . therefore I take ar=bbr2ddaObd bbrr4dd ar . & rs=br2dO bbrr4dd aa .

By ye same manner ye intersec{illeg}|t|ion by 2 crooked lines may be found.


[62] Having ye nature of any lines expressed in Algebraicall termes, to find its Asymptotes if have any

[63] Suppose \rh , & rs ye asymptotes of ye line dtn . & hd parallel to rs drawne from ye Asymptote to ye line \ dtn // bc=x. cd=y. bf=c. fehede. hddgfg. fe=mh=ϱ . hd=ϩ . ye angles dcb , {illeg} cbm , dgm , bfq , hef , {illeg} c to bee right ones. yn is, eh=eϱd. dg=gϩf. dg+gq hg= ϩffggf . dg+gq+qc= dgϩ+efϱ+dfc df =y. & fe+eq=fϱ+ ϩffggf =x. Now for readiness in operation it bee convenient to have in readinesse a table of the{illeg}|se|{illeg} valors of x & y wch will bee ye same wth yt by wch ye diameters of crooked lines are determined. viz.
fx=fϱ+ϩffgg. 0 ffxx=ffϩϩ+2fϱϩffgg+ffϱϱ. gg 0 f3x3=ffffggϩ3+3fffϱϩ2+3ffϱ2ϩffgg+f3ϱ3. gg3ggfϱ 0 dfy=dgϩ+efϱ+dfc. 0 ddffyy=ddggϩϩ+2defgϱϩ+eeffϱϱ +2ddcfg+2cdeffϱ +ccddff 0 d3f3y3=d3g3ϩ3+3ddefggϱϩϩ+3deeffgϱϱϩ+e3f3ϱ3. +3cd3fgg+6cddeffgϱ+3cdeeffϱϱ +3ccd3ffg+3ccddefϱ +c3d3f3 dffxy=dgffggϩϩ+dfgϱϩ+effϱϱ. +efϱffgg+dcffϱ +cdfffgg 0 df3xxy=dffgϩ3+ef3ϱϩϩ+dgffϱϱϩ+ef3ϱ3. gggdefggϱ+2effϱϱffgg+cdf3ϱϱ +cdf3+2cdffϱffgg cdfgg +2dfgϱffgg 0 ddf3xyy=ddggϩ3ffgg+ddfggϱϩϩ+2deffgϱϱϩ+eef3ϱ3. +2defgϱffgg+2cddffgϱ+2cdef3ϱϱ +2cddfgffgg+eeffϱϱffgg+ccddf3ϱ +2cdeffϱffgg +ccddffffgg This table may be continued {illeg}|w|hen ye nature of ye lines are expre{illeg}|s|sed by Equations of 4 or more dimensions. This like ye former rules will be {illeg} be{illeg}|s|te{illeg} perceived by Examples yn precepts. As

Example ye 1st. To find ye asymptotes of ye line \whose nature is/ exprest by rx+rxxq=yy . first I write ye valors of x , xx & yy (found by this table) into theire places in ye Equation rx+rxxqyy=0. & there results rffϩϩrggϩϩ+2rfϱϩffgg+rffϱϱ qff ddggϩϩ2defgϱϩeeffϱϱ ccddff2ddcfgϩ2cdeffϱ ddff +rfϱ+rϩffggf =0 . \or {illeg} by ordering it,/ +rq rggqff ggff ϩϩ +2rϱffggqf 2egϱdf 2cgff +rffggf ϩ +rϱϱq eeϱϱdd 2ceϱd +rϱcc =0 . Or, +ddffrϩϩ+2ddfrϱffggϩ+ddffrϱϱ ddggr2defgqϱeeffqϱϱ ddggq2cddfgq+ddffqrϱ +ddfqrffgg2cdq effϱ ccddffq =0 . Now by assumeing any valors for c ; d , e ; f , g . I have, by this equation, ye relation wch ϱ beares to ϩ , yt is wch mh {illeg}|b|eares to hd . But yt ye valors of c , d , e , f & g , may be such {illeg} yt hr (to wch hd is applyed) may be one asymptote of ye {illeg} line \&/ hd paralell {sic} to ye other, it is necessary (by Prop: 3d) yt neither ϱ nor ϩ bee any where of soe many or {illeg}|o|f mor{illeg}|e| dimensi{illeg}ons, yn in those {illeg} termes in wch {illeg}|t|hey multiply one another. Therefore I consider of how many dimensions \ ϱ is/ at ye most in any terme multiplied by ϩ, & how {illeg}|ma|ny ϩ is in any terme multiplied by ϱ ; & find ym but of one. & therefore conclude yt ϱ & ϩ {illeg} \ou/ght to be found in noe terme in this equation ∼ unlesse wher{illeg}|e| they multiply one another. [64][65] Moreover tis manifest yt ye Equation (expresing ye nature of ye {illeg} given line) will ever be of one \& but of one/ dimension more yn ϱ or ϩ in some termes in wch they multiply one another: & therefore this may bee put for a Ge{illeg}|ne|rall Rule viz. All those termes must destroy one another in wch there is not ϱϩ & wch are of as many, {illeg} or want but one dimension of being of as many dimensions as ye Equation is. Now that the{illeg}|se| termes destroy one another, tis necessary yt those be =0 in wch ye unknowne quantitys ϱ & ϩ {illeg} are ye same. Upon wch considerations it will appeare yt in this example I must make, ddffrϩϩddggrϩϩddggqϩϩ=0 . 2dly, +ddfqrffggϩ2cddfgqϩ=0 . {illeg}|t|hi{rdly} {ddffrϱϱeeffqϱϱ=0 }. 4thly ddffqrϱ2cdqeffϱ=0 . Or by dividing ym by those quantitys wch {illeg} /{neede}\ {illeg} =0 . they are ddrffddggrddggq=0 . 2dly ddrffgg2cddg=0 . thirdly, ddreeq=0 . 4thly, ddqr2cdqe=0 . by ye 3d, d=eqr . & since tis not d=0 , by ye first ff=gg+qggr . by ye 2d c= rffgg 2g . or { c=grq2 }. by ye 4th dr2e=c=rq2 . Therefore from ye point b I draw bf & bk=c=Orq2 . from f I draw fq parallell to bc . from q I draw qv , soe yt fqqvdeqr. & through ye points f & v I draw fv wch shall be {illeg} \{one}/ Asymptote yn <27v> or which is ye same I make (since tis not c=0) I make. edcbl=dce. Or, bl= cq r . & soe draw ye asyptote {sic} passing through ye line points l & f . Againe if from some point in ye Asymptote lh , as h I draw {illeg} hg & from g I draw gd= Then from ye point k I draw pk paralell {sic} to {illeg} bl & pv paralell {sic} to bk soe yt (assumeing some other proportion twixt d & e \yn before/ if there be any other) depkpvqr. & soe through ye points k & v I draw ye other Asymptote. Or since {illeg} it is not c=0; I make ed bkdce=bl=+cqr. & soe t{illeg}|h|rough ye points l & k I draw ye other asymptote, wch shall be paralle{illeg}|l|l to hd.

Example ye 2d. Supose {sic} ye Asymptotes of xxyx+ay=0 were to bee determined, Since I have noe use of ye termes in wch is ϱϩ I onely select those termes out of ye valors of xx , | y | & yx in wch ϱϩ is not & sorting them as was before {illeg} taught I have these equations, 1st ffϩϩggϩϩgϩϩffgg=0 2dly agϩcϩffgg=0 . 3dly, dϱϱeϱϱ=0 . 4thly aeϱ d cϱ=0 . by ye {illeg}|th|ir{illeg}|d| d=e. by ye {{illeg}|4|{illeg}|th|} a=c. by the 1st g=+ffgg. by the 2d c=a

Example ye 3d, yy=bx2c+yb . by working as before I have these Equations {illeg} gggffgg=0 . 2cgcffggbffgg=0. +eϱϱdϱϱ=0 . {+xae}
gg{illeg} bff+bgg c =0. 2cgbg=0 . eedd=0.
Suppose xxax+byyy=0. then by workeing as before I have these Equations, ff2gg=0. affgg+bg2cg=0. ddee=0. { bead2ce=0 .} by ye 2d c= bOa 2 . & soe by ye 4th by ye 4th d=Oe. by ye 1st f=Og2. or { g=Offgg .} by ye 2d (by suposeing {sic} g=+ffgg) tis ba2=c : & by ye 4th (by supposeing d=+e) tis ba2=c . {illeg}|B|ut by ye 2d (by supposing g=ffgg) tis a+b2=c . & by ye 4th (by supposeing d=e ) tis a+b2=c . Whence I conclude yt when c=ba2 yn is d=e, & g=ffgg; & when c=b+a2 yn d=e & g=ffgg.


[66] To find ye Quantity of crookednesse in lines.

[67] Suppose ab=x. be=y. bc=o=gh. bg=c. \ ed & df tangents \secants/ to ye crooked line intersecting at d ./ ye angles abe , acf , egd , right ones. & let rx=yy, be ye relation twixt x & y . soe yt aef is a {illeg}|P|arab. Then be=rx. bn=v=r2. {illeg} \ebbneggd/ rxr2c+rxcr+rrx2rx=gd=r2+ cr 2rx . cf=rx+ro. {illeg} \cfcmfhhd/ rx+ror2c+rx+ro cr+rrx+ro 2rx+ro =hd=r2o+cr2rx. That is 2crrx+2rrrxx+rrox=2rrrxx+rrox4orrxx+rrox+2crrx+ro. Or, Sqareing {sic} both sides 4ccr3x=16oorrxx+16o3rrx16crrxo16crrxorx+4ccr3x+4ccr3o that is ({illeg} by blotting out 4ccr3x on both sides, divideing ye rest by o, & then supposeing o=bc to vanish) 16crrxrx+4ccr3=0. Or c=4xrx r therefore makeing ab=x . bg=4xrx r . gd=bn ×egeb=12r+2x. & describing a circle with ye Rad de=16x3r+12xx+3rx+rr4, ye circle shall have ye same quantity of crookednesse wch ye Parabola hath at ye point e.

[68] Or thus. If ab=x. cb=y. bd=v. cd \& em/ perpendiculars to ye crooked line cma wch intersect at ye point e . af=c. fe=d. ye angle{illeg}|s| abc , baf , afe , \ mna / right ones.

Supose {sic} rx=yy, expresseth ye relation twixt {illeg} ab & bc . First I find ye length of bd=v ({illeg}|se|d|e| fol: 8th hujus, or Des=Cartes his Geom: pag 40) wch is v=r2. cbbdgcge. /yvc+ycv+vyy.\ {illeg} ab+ge=x+v+cvy=d. Or dy+cv+vy+xy=0. Out of these termes first I {illeg}|t|ake away v by w{illeg}|r|iteing its valor in its rome {sic} wch in this case is 12r & there remaines results, cr2+yr2+xydy=0. Then I take away either x or y (wch may bee{illeg} e{illeg}|a|sliest {sic} done) by ye helpe of ye Equation expressing ye nature of ye li{ne wch} is now rx=yy. or yyr=x . And there results cr2+yr2+y3rdy=0. Now tis Evident yt when ye lines cm & ce are coincident yt ce is {illeg}|y|e radius of a circle wch hath ye same quantity of crookednesse wch ye Parabola mca hath at ye point c . Wherefore I suppose eb & nm \2 of/ ye {illeg} rootes of ye Equation 2y3+rry2dry+crr=0, to be equall to one another. & {illeg}|s|o by Huddenius his method I multiply it 2y3+rry2dry+crr=03110. & there results, 6yyrr 2r=d. againe otherwise 2y3+rry2dry+crr=02001, & there results 4y3rr=c . Soe yt if cb=qa=y . yn ab=yyr. af=4y3rr. fe= 6yyrr 2r {illeg}. ce=00000 then ye circle described by y{illeg}|e| radius ec shall bee as crooked as ye Parabola \at ye point c/

Or better thus. All th Make ab=x. bc=y. cg=c. fe=d. bd=v. Then yvccvy=ge. cv+xydy=0.

Or thus. Make ab=x. bc=y. bd=v. bg=e. ge=f. & fy=vy+ev. Thus in ye former Example rx=yy. v=r2. fy=12ry12re Or if out of ye Equa

[69][70] The crookednesse of \equall portions of/ circles are as their diameters, reciprocally.

Demonstr. The crookednesse of {illeg}th any whole circle ( bfd , gcme ) {illeg}|a|mounts to 4 right angles, therefore there is as much crookednesse in ye circle bfd as in cmeg . Now supposing ye perimeter fbdf is equall to ye arch cme , Then as ye arch emc=fdbf is to ye circumference cmegc , soe is ye crookednese {sic} of y{illeg}|e| arch cme to ye crookednesse of ye perimeter cmegc , or of bdfb . so is ab to ac .


To find ye Quantity of crookednesse in lines[71]

[72] Suppose ndf & efm perpendiculars to ye c{illeg}|r|ooked line adeo , wch intersect one another at f . ac=x. ce=y. cm=v. ag=ch=c. gf=d. & ye angles abd , ace , mag, agf right ones. Then, ec=ycm=veh=ychf=vyvcy. gf=gh+hf=x+vyvcy=d. Or dyvy+vcxy=0.

Haveing therefore ye relation twixt x & y (as if it be rxrqxx=yy) first I find ye valor of v ({illeg} see Cartes Geom: pag 40th. or fol: 8th of this) {illeg} (as in this Example tis 12rrxq=v) by wch I take v out of ye Eq{illeg}|u|atio dyxyvy+vc=0, (& in this case {illeg}|t|here results dyxy12ry+rxyq+12rcrxcq=0.) then by meanes of ye Equation expressing ye relation twixt x & y I take out either x or y , wch may easliest bee done (as in this example I take out y by writeing rxrxxq in its stead & there results
+d x r2 +rxq rxrxxq=rcxq12rc. Or squareing bi|o|th ptes {ddrxdrr2drx+2drrqddrqdqxqr } Or 2dq2xqrq+2rxrxqrxx=2rcxqrcqq . & by squareing {illeg} both pts {illeg} 10122104ddq3rx4ddqqrxx+8dqqrx34qqrx4=q3rrcc4qqrrccx+4rrqccxx 4dq3rr8dq3r8dqrr+8qrr +q3r3+4dqqrr+4q3r4r3 +8dqqrr4qqrr +4q3rr8qqrr qqr3+4qr3 4qqr3+4qr3 ) Then if I assume any valors for c {illeg}|&| d yt is {illeg} if I determine ye point f , I have an Equation by wch I can find all ye perpendiculars to ye crooked line, drawne from ye point f . for if I tooke x out of ye Equation, ye rootes of ye Equation will bee (db=y {illeg}|&| ec=y &c:) all such lines as are drawn from ye points of interse{illeg}|ct|ion d , e , k , h , to ye line {illeg} ao (as db , ec , &c) but if I tooke y out of ye Equation yn ye roots of ye Equation will bee those lines drawne from a to ye perpendiculars (as ab , ac , &c. Now by how much ye nigher ye lines points d & f are to one another, soe much ye lesse difference there will bee twixt ye crookednesse of ye pte of ye line de , & a circle described by ye radius df or ef . And should ye line df be understood to m{illeg}|o|ve untill it bee coincident wth ef , taking f for ye last point where they ceased to intersect at theire coincidence, ye circle described by ye radius {illeg} ef , would have ye & ye given crooked line at ye point e , would bee alike crooked. And when {illeg}|y|e 2 lines df & ef are c{illeg}|o|incident |2 of| ye rootes of ye Equation (viz db & ec , if ye Equation ab {illeg} & ac ) shall bee equall to one another; Wherfor {sic} to find ye crookednesse of ye line at ye point e I {illeg} supose {sic} ye e{illeg}|q|uation to have 2 equall rootes & so ordering it Ac{illeg}|c|ording D: Cartes or Huddenius his Method, ye valor of any of these {illeg} \3/ { xy} } c d being giv{illeg}|e|n, ye valor of ye other 2 may be found. [73] (as in this Example ye valor of x being given I multiply ye Equation according to Huddenius method & it is 004ddq3x+8dqqx38qqx4+4drq38dqr+16qrrrq3+4q38rr+8qrr 4qqrx2q3x =cc= 4qqx4+8dqqx34ddqqxx+4ddq3x +8qr8dqr8dq34dq3r4rr+4q3+12dqqr+q3rr12qqr+4q3r+8qrr5qqrr 4qrxx4qqrx+q3r [74] |Then by divideing both ye numerators by x & ye denominators| by 2qrxqqr, & so multiplying ym {in crucem} & ordering ye product it is. { 4ddq44q4rd+q4rr=0 16q4x+8q4rx +16q3rx8q3rrx +24q3xx+12q4xx 24qqrxx36q3rxx 16qqx3+24qqrrxx +16qrx324q3x3 +56qqrx3 32qrrx3 +16qqx4 32qrx4 +16rrd4 }.

Now considering yt if {illeg} q , r , & x bee known, yt is, if ye Ellipsis eak be determined, & ye line ac given{,} there are onely two points in ye line (viz: e & k ) to be considered. And ye roots of this equation \valors of d / are ( gf , am , sq ) {illeg} {such} lines as are drawne from ye line gas to ye points where ye perpendiculars efm kqm intersect (as m ) or to such points as of where two perpendiculars (as ef & df ) {ceased} to intersect at theire coincidence into one (as f & y ). Therefore \{illeg}of ye first {illeg} roots/ I get ye valor of ye {line} cm=x+r2rxq=d. as {illeg} this Equation by {illeg} yt is dx+r2rxq=0 {illeg} +2dq2qx+qrrx=0; {illeg} there results <31v> +2dq3q3r6q3x+6qqrx+12qqxx12qrxx8qx3+8rx3=0.[75] That is dividi{ng} it by 2y3; d=r2+3x3rx6xxq+6rxx+4x3qq4rx3q3. Which Equation \expresseth/ ye length of ye lines (qs=d, & gf=d) wch are drawne from ye line sag to ye points q & f at wch ye coincident perpendiculars last intersected \one another/ before theire coincidence. Now haveing ye length of gf or sq it will not be difficult to find, c=ag=ch , or, as=cl=c; for it was found before yt dyvyxy+vc=0 Or c=vy+xydyv . Likewise it will not bee difficult to find ef or kq , for (supposeing lq=dx ; hf=dx ; lc=+c ; hc=c ; ec=+y ; ck=y ). ef=e. or kq=e) it is, yy2cy+cc+xx2dx+dd=ee={kq×kqef×ef, {illeg} Lastly ye circle described wth ye radius ef shall have ye same quantity of crookedness wch ye Ellipsis hath at ye point e .

[76] Example ye 2d. Were I to find ye quantity of crookedness a{illeg}|t| some given point of ye line exprest by rx+rxxq=yy ; I might consider yt it differs from ye former Example {illeg}|o|nely in yt there I have rxxq here or rxxq , hi|e|re rxxq , yt is in ye former q was negative in this is affirmative. Soe yt this operation will bee ye same wth ye former ye signe of q being changed soe yt it will be found
{illeg} qf or hq=d=r2+3x+3rx+6xxq+6rxx+4x3qq+4rx3q3 . &c as before.

[77] Example ye 3d. Had I ye Par In ye Parabola, rx=yy. & v=r2. In ye above mentioned Equation dyxyvy+vc=0 I take out v by write{ing} r2 in its roome & it is dyxyry+rc2=0. yn I take out y by writeing rx in its stead /{illeg}\ 2drx+2xrx+rrx=rc. & by □ing both sides, 4ddx8dxx4drx+4x3+4rxx+rrx=rcc 1213210 . Which is an equation |haveing 2 equal{illeg}|l| roots| & therfore multiplied accordind|g| Huddenius his method soe yt rcc be blotted out, & therere result{illeg} {illeg}|d|ivided by x it is, 4dd16dx+12xx=0 4dr+8rx +rr . Now tis evident |yt {illeg} x=ac| being determind {sic} {illeg} there are 2 points (viz: e & k ) from wch perpendiculars being drawne they intersect one another in ye axis at {illeg}| m |, wherefore ∼ am=x+r2 is one of ye rootes of ye Equation & therefore it being divided by dxr2=0, or by 2d2xr=0 there results 2d6xr=0 Or d=12r+3x=sq=gf. Then into ye above found Equation ∼ ∼ 2xrx2drx+rrxr=c, I substitud|t| th{illeg}|i|s valor of d & there results 4xrxr=c=ag=ch=pf. |ef=r+4x2rrr+4rx .| Soe yt I have eh=rx+4xrxr. And hf=12r+2x . & therefore ef=14rr+3rx+12xx+16x3r shall be ye Rad of a circle wch is as crooked as ye Parabola at ye point e .

[78] Or it might have be{illeg}|e|ne done thus, haveing ye Equation dyxyry+rc2=0, I might have pr{illeg}t /writ {sic}\ yyr in stead of x , & soe have had dyy3rry2+rc2=0 1310 wch must have 2 equall \roots/ & therefore by ye Method de max: & min: I {bott {sic}} out rc2 & there results, dy3y3rry2=0. Or, d=r2+ryyr. makeing ad=y, de=x. cm=v. gf=d. ag=fp=c. now if ad=y bee determined it is manifest yt there is but one point of ye Parab: (viz: e) to bee considered from wch ye perpendiculars wch are drawne doe noe where intersect one another & therefore this equation hath not superfluous rootes like ye former.

[79] Example ye 4th. If it bee supposed yt rx{illeg}xx=ry{illeg}rx+xx=0 /ye nature of ye line is contained in\ ryyyrx=0. {illeg} & if tis ad=x . y={edek . v={dtdy. {illeg} fe & jk 2 perpendiculars to ye crooked line, f, & q two points where ye coincident perpendiculars last intersected d={ap=qsfg . c={pqpf. Then is v=ryr2y . by wch I take v out of ye above named Equation \ dyxyvy+vc=0 /, & ye result being divided by y , it is, ry+2dy2xydr+rxcr Or ({illeg} y=drrx+crr+2d2x: Then I substitute this valor of y into its place in Equation rx+yyry & there results rx + ddrr2drrx+rrxx+2dcrr2crrx+ccrr rr4rx+4xx+4dd+4dr3dx rrxdrrdrr r+2d2x =0. or by ordering it {it} will bee {{illeg} d3+ddxx+4ddxddr 5r+6dr+2cr +2rrdrr crr 3210 = 0 } Which Equation |must have {illeg}|


two equall roots & therefore by ordering it according to Huddenius Method de Maximis & Minimis, I blot out ye last terme & ye result is 6xx8dx+2dd 5rx+3dr +rr =0 . Or 2dd+3dr+6xx 8dx5rx +rr =0 . By what was said before tis evident yt the perpendicular rm drawne from ye line asg to ye point where ye two perpendiculars intersect, is one of ye rootes of this Equation.

[80] And yt I may have a general rule to find ye line{illeg} rm (or had there beene 3 or more perpendiculars, to find all those lines wch are drawne from ye line acrw to each /every\ intersection of ye perpendiculars) I consider yt if ac=λv=θb=c be not drawne from ye line at to ye point of intersection m; yn d hath two valors as vc&bc but if they bee dr{illeg}|a|wne to ye point m, yt is, if they be ∼ coincident wth nm ; yn d hath c{illeg}t yn ye two roots of d are equall to one ∼ another, being ye same wth ye line rm . Likewise if =cv=wz=d be drawne from ye line aw to ye perpendiculars fe , qk {illeg} but not from ye point where they ∼ intersect; then hath c two root{illeg}s (as λv , λz ) wch will also be equall to one ∼ another & coincident wth ye line mn , when d is ye same wth rm . This being considered; if I would ye valor of nm , I must order ye affore found Equation (in wch x was supposed to have 2 equall roots) according to c & it will bee rccrrcdrr+6drx8dxx ddr+4ddx+4x3 +2rrx5rxx 21000 =0 . wch must have 2 equall roots & therefore by Hudenꝰ {sic} Meth: de Max: & Min: I take away ye last terme & Soe I have, 2rccrrc=0; or, c=r2=nm: {illeg}|B|ut if I would have ye valor of rm I order ye Equation according to ye letter d & it is 4ddrdrr+ccrddr+6drxcrr8dxx+2rrx+4x3=0. wch Equation must likewise have two equall roots & therefore takeing away ye last terme there re by Hud: meth: de Max: & Min: there resulteth this, 8ddxdrr 2ddr+6drx 8dxx=0. Or d=xr2=rm & this x12r is \one o of/ ye rootes of ye Equation 2dd+3dr+6xx 8dx5rx +rr=0, wch was required, therefore I must divide this equation by dx+r2=0. yt is by 2d2x+r=0 , & there will result, d3x+r=0 . That is d=3xr=fg=qs. Whence i{illeg}|t| will not be difficult to find ye points q & f & ∼ consequently ye lines qk , fe wch shall be ye radij of circles wch have ye same quantity of crookednesse ye line aek hath at ye points e & k . Makeing { c={asag }.

Note yt these Equations have not rm or ar as one of th{illeg}|e|ire rootes unlesse when ye axis of ye axis line is either paralell {sic} to x {illeg}|(| for yn \onely/ a circle whose center is at ye intersection m can touch ye crooked line in both k & e together) & then perhaps they may easlyer bee found {y{illeg}|n|} by ye foregoeing rule.

May 1665

[81] 1. Note yt ye crooked line φfγqπ (described by ye points q & f ) is always touched by the (perpendicular) kq ; & that in such sort as to bee measured by it{,} they applying themselves the one to the other, point by point; soe yt if = the shortest of all ye lines qk be substracted from qk the{illeg}re remaines = . By this meanes ye length of as many crooked lines may bee found as is desired

2. Also if ye line qk is applyed to ye crooked line qγ point by point, every point of ye line qk (as k ) shall describe lines to wch (as akw) to wch qξk is perpendicular.

[82] 3. The line {qπxfy } is ye same (if wka be a Parab:) {Heura{illeg}} found. & {illeg} perpendicular to th{e} line efb , & abd a tangent & ye position & {illeg} point {illeg} wth ye tangent (as if they were inherent in ye same {body}) while ye tangent gl{illeg} ye crooked crooked {sic} line ebf, soe yt ye point {af{illeg}} describe {illeg} dag{illeg} then from {illeg} draw perpendiculars to ye line dag (or {illeg} then shall the \{illeg}/ point c , {illeg} perpendiculars intersect {illeg} ye po{illeg} of ye {illeg} {illeg} {illeg} {least} {illeg}


[83] The Crookednesse in lines may bee otherwise found as in the following Examples

[84] In the Parabola aeg suppose e ye point where ye crookednesss {sic} is sought for, & yt f is the center & fe ye Radius of a Circle equally crooked wth ye Parabola at e . Then naming ye quantitys {illeg} ce=y. ap=d. pf=c. ef=s. Bye {sic} nature of ye line ac=yyr. ce+pf=eh={illeg}yc. {cp=yrd=hf}. eh2+hf2=ef2, That is, y4rr2dyyr+yycy+cc+dd=ss 422100 wch Equation must have 2 \equall/ rootes that ef may be ⊥ to ye Parab: & therefore multiplyed according to Hudden {sic}'s Method {illeg} it produceth 2y3rr2dyr+yc=0 3110 . Which Equation hath soe many rootes as there ca{illeg}|n| be drawn perpendiculars {illeg}|t|o ye Parab: from {illeg}|t|he determined point f . And two of ther{illeg}|s|{sic} rootes must be{illeg}|c|ome equall, yt f may bee the center of ye required Circle, therefore this equation \is to bee/ multiplyed againe, & it will produce 6y2rr2dr+1=0 that is 3yyr+r2=d. Or 3x+r2=d : As was found in ye 3d precedent example.

[85] Here observe yt in ye 1st of these 3 Equations y hath 4 valors gl , ec , hs & kv . \see fig 2d/ when d ; c , & s are det{illeg}ermined. But d , c , & y=ec being determined s hath but one valor =ef. And if d , s , & y=ec bee determined yn c hath 2 valors pf & pm. And c, s , & y=ec being determined d hath 2 valors an & ap as that {illeg} first equation denotes by{illeg} ye dimensions of ye quantitys in it. Figure By the 2d of these Equations 2 of ye valors of y are united but not by ye increasing or diminishing ye Rad: of {illeg} ye ci valor of d=fe &c. But 2 rootes one as {illeg} {illeg} first suppose ye circle soe little as noe where to intersect ye Parabola, it being increased gradually will fir{illeg}|st| touch ye Parab: at r (fig 3d) then ceasing to touch it it intersects it in 2 points g & k (fig 2d) wch two points growing more distant at last it untill it touch ye Parab: in t (fig 3d) wch being divided into two intersection points e & h (fig 2d) the points g & e draw neered|r| {sic} untill they conj{illeg}|o|yne in ye touch point w & soe ye circle ceaseth (by still increasing) to touch {o is} intersect ye Parab: or intersect it unlesse in h & k . Whence from one point f may be drawne 3 perpendiculars fr , fw , ft , to ye Parabola twar . And therefore in this 2d Equation y must have 3 valors wc , tl , & vr , when ap=d, & pf=c, are determined then also hath s three valors fr , fw , & ft .

By ye 3d Equation Two of the {illeg} valors of y in ye 2d Equation are united by incresing or diminishing ye length of pf=c. For begining at ye point at the point p (from wch ye {illeg} \3 perpendiculars/ fall upon γ , a & β ) if ye point f doth gradually move from p , the perpendicular ftfwfr moves from γaβ towards aγλ Soe yt ye two perpendiculars = wf & tf will at last conjoyne into one {illeg}f EF , Which shall be ye Rad: of a Circle as crooked as ye Parab: at E .

This 3d operation might have beene done by making pf determined & by = increasing or diminishing ap=d . That is by destroying ye term 2dyr in stead of c in ye 2d Equation. And so might ye 2d Operacon beene done otherwise by {illeg} determining ye circle egh , &|O|r taking c or d out of ye 1st Equation instead of ss .

[86] There is another way of finding ye crookednesse in lines & yt is not by supposing {illeg}|t|wo perpendiculars ( wf & ft , or wf & fr ). but 3 intersections of a circle wth ye figure, (fig 2d h , e , g : or e , g & h ). And then shall xy} have 3 equall valors al,ac,as.lg,fc,eh Or ac,al,avce,lg,vk . As if (in ye last example I had this equation y42dryy+rryy2cyrr+ccrr+ddrrssrr=03110111 . Supposing it to have 3 equall rootes by Huddenius his method tis 3y42dryy+rryyccrrddrr+ssrr=0422000 . (Which equation doth not determine ye perpendiculars to eag ) as 2y5rr2dyr+y=c {illeg} doth for by { 12y42dryy+rryy=0 }. this I can find ye valor of c ( y being determined) {illeg} but by it I {illeg} {illeg} neither find ye valor of { c } nor {illeg}till one of ym is {illeg}|t|aken out of ye Equation). That Equation multiplyed {illeg} ye dimensions of y produceth 6y22dr+rr Or {3yyr+r2 {illeg}3x{illeg}{illeg}=d }.

[87] The same may be done thus. If a circle touch a crooked line at one point & intersect it {illeg}er when two points come together yt circle {illeg} to {illeg} or {illeg}

As if bc=y. {ac=yyr an=mt=d. pv=o{illeg} c{illeg}={illeg}=r2. {illeg} {illeg} yyr . r2ydryyr2dryy2rr=bq}. {r2{illeg} +y {illeg} =dyy2oyr 2dry+2ro2y3oyy rr =pg. {illeg} may ever bee {illeg}). Then {illeg}} {bq2+qt2{illeg} 6y48dryyo+rryyo+2ddrrodr3o=0 . Or dd=8dryy2rr {illeg}.} Or d= {illeg} Or, {illeg} d= {illeg} wch cannot {illeg} T{illeg} by ye {illeg} perpendiculars {illeg} by supposeing ye circle described by ye Rad: {illeg} pt & {illeg} to {illeg}


[88] Haveing found an Equa (by ye former rule) {illeg}|a|n Equation expressing \by which/ ye quantity of crookednesse in any line may bee found to find ye greatest or least crookednes of any that line.

[89] In ye 3|4|d|t|h Example I had found gf=qs=d=3xr . And by a rule there shewed viz {asag}=c= xy+vydy v }: or writeing 3xr in stead of d It was there found 4x38dxx+4ddxddr 5r+6dr+ccr +2rrdrr crr = 0 . {illeg}|N|ow by writeing {illeg} 3xr in stead of d & ordering ye product according to ye letter c it is ccrcrr+16x3 12rxx 3rrx = 0 . Or extracting ye roote i{illeg}|t| is c=r2O12xx3rx+rr416x3r={as.ag. Also by ye nature of ye line, kded}=y=r2Orr4rx. Therefore kh=gl=kdsagaed=rr4rx+12xx3rx+rr416x3r . Also qh=lefg=r2x . And Since {he2+gl2kh2+ he2+gl2=qk2 Therefore
3rr28rx+16xx16x3r+212xx3rx+rr416x3r×rr4rx=qkqk=zz; Supposing qk=z The roote of ye Surde quantity extracted the Equation is 16x3r+24xx12rx+2rr=zz. Or 16x324rxx+12rrx2r3+rzz=0. In wch Equation ye least valor of z=fe is to {illeg} bee found & yt should happen when x hath 2 equall {illeg}|v|alors or rootes. But because fe=z {illeg} being determined x can have but one valor =ad ye other 2 {illeg} rootes being imaginary tis impossible yt it should have 2 equall rootes: Therefore I substitute {y}{ y } take away x out of ye Equation &|b|y substituting its valor ryyyr in its stead & there results 16y648ry5+24rry4+32r3y336r4yy+12r5y2r6+r4zz=0 65432100 . In wch equation z {illeg} or ef=qk being {illeg}|d|etermined y hath 2 valors de & dk ye other foure being imaginary & when ef is {illeg} the longest or shortest that may bee then these two valors become one & then is ye line aek {illeg} more or least crooked. If therefore (yt y 's valors become equall) this Equation is multiplyed according to its dimensions there will result 8y520y4+8rry3+8r3yy6r4y+r5=0 . wch is divisible by yr2=0 , or by 2yr=0 (for there results 4y48ry3+4r3yr4=0 ). And if y=r2 , yn is x=r4 . Therefore I take ad=r4 & de=r2 & |at| ye point e shall bee ye least crookednesse.

Here may bee noted Huddenius his mistake, yt if i{illeg} some quantity in an equation designe a maximum or minimu yt Equation hath two{illeg} equall rootes & {illeg} wch {illeg}|is| false in ye equation 16x324rxx+12rrx2r3+rzz=0. & in all other equation's {sic} which have but one roote.

Or because

Another way. May 1665

[90] Or because ye lines fn & qn described by ye points f & q doe touch one another point's {sic} n from wch points onely {illeg} lines drawne perpendicular to ye croked {sic} line kea will bee perpendicular to ye \point of/ greatest or least point crookednesse: And also since all those are points of greatest or least crookedness f|t|o wch such {illeg} \perpendiculars/ are drawne: The difficulty will be to find ye point n . Now suppose yt am=d be determined yn c hath two valors for mfmq}=c . And alsoe y hath two valors for dkde}=y . A{illeg}|l|soe (when am is not parallell to ye axis of ye line) x hath {illeg} two (or more) valors adad}=x . wch valors of c , x , or y become equall if am=an : {illeg}|b|y wch meanes ye point n may bee found: Excepting onely {illeg}|w|hen fm , mq , are parallel to ye crooked line at n \({ un }{)}/ [91] yt is, parallell \perpendicular/ to ye streightest or most crooked ptes of ye line aek . But if as=c be de{illeg}|t|ermined, then d={st.sf. x={av.ad. y={rv.ed. (but if as is parallel to ye axis of ye line ye two valors of y are equall & soe not usefull). Which valors of d , x, & y become equall if as=bun : excepting onely when as is perpendicular to ye most streight or crooke{illeg}|d| pts of ye line ake .

[92] As for example. In ye precedent example it was found d+r3x=0 . But b{illeg}|e|cause an or x is parallell to ye axis of ye line, in yt {illeg}|E|quation x hath but one dimension. Therefore substitute either ye valors of d or of x into their stead. As if I substitute ye valor of x=ryyyr into its place it will bee d+r3y+3yyr=0 or 3yy3ry+rr+dr=0 2100 . wch must have 2 equall roots & therefore multiplyed according to y 's dimensions tis 6yy3ry=0 . Or y=r2 as before. But if I had substitu{illeg}|t|ed d 's valor into its stead it would have beene 16x312rxx+3rrxcrr+ccr=0 32100 {illeg}|w|hich {illeg}ving 2 equall roots being rightly ordered is 48x324rx2+3rrx=0 . Or 16x28rx+rr=0 . Or 4xr=0 . Or x=r4=0 , as before.

{In ye first} Example of finding ye quantity of crookednesse in lines ye {illeg} found { 4rx30 {illeg} 00xx+3rxq3xr2=0 110 }. wch {must} have 2 equall rootes & therefore by Hudde{nius} method it is {{illeg}xx4qxx4qrx+4qqx+qqrq3=0}. {illeg} \wch being divided by { 2xq } it is/ 2rx2qxqr . Or, 4xx4qx+qq=0 .{illeg} That is 2x=y . or x= {illeg} =y Or {illeg} = {illeg} =y {illeg} yt if I take { us=r2 } & {illeg} }= {illeg} =y {illeg} ye points {illeg} ye greatest or least crookednesse {illeg} ye line {illeg} {illeg} crookednesse wch {illeg} found

{These Equations} {illeg} superfluous rootes {illeg} often as {illeg} ye perpendiculars {illeg} {illeg}


The points of greatest or least crookednesse may bee yet otherwise found by an equation of 4 equall rootes. As in ye example of ye 2d way of finding ye quantity of crookedness in lines it was found y42dryy+rryy2crry+ccrr+ddrrssrr=0 . wch being compared wth an equation like it y44ey3+beeyy4e3y+e4=0 . by ye 2d terme tis 4ey3=0 , or y=0 . In like manner by & yyr=00r=0=x . Soe yt ye Parab at ye {illeg}|b|egining is most crooked (at a ).


[93]If ye body{illeg} b {illeg} {illeg}|m|ove {fro} ye line bd & from ye point {illeg}| d | two lines da , dc bee drawne ye motion of ye body b from ad is to its m{illeg}|ot|ion from dc as abdc is to cbad.

[94] Coroll: 1. The body b receiving two divers forces from a & c & ye force from ba is to ye forc{e} from { bc } as ba to bc , yn draw adbc & cdab , ye body b shall bee moved in ye line {illeg} / bd \.

[95] 2d Or if y{illeg}|e| body { b }{ d } is suspended by ye {thred} bd & is forced from a to {illeg} & from { c } towards f , yn draw dcab & da{illeg}, & make daab force from c to f ∷ force from {illeg} ye body {illeg} {illeg}nd in Equilibrio is b .

{illeg}Coroll: 3d. ye force of ye body b from d i{illeg}|s| to its force from a as bd to ba .


vide pag: 15. But here observe yt unlesse ye reflecting line adn bee drawne from ye through ye point w the center of motion in ye whole body aiwn ye determinacon of ye motion of adn will not be ye same wth ye determinacon of ye motion of g before reflection (as in ye first figure[96]) but verge from it (as in ye 2d fig[97]) yt is wl & gdi will not bee parrallell {sic}. For since ye cheife {sic} resistan{ce} is in ye line of ye body adn is from its center of motion (prop 32) from d w towards d , & not from i towards d , the body g will find more opposition on yt side towards ye center w , yn on ye other side towards a & therefore at its reflection it{illeg} must incline toward v {illeg} from ye \(ax 120)/ &{illeg} not returne in ye line dg . But if ye body awn presse g towards w yn g presseth ye body awn towards ye contrary pte \as/ from w towards l (ax 119) & not from w towards m, if wmdg. But if adn {illeg}|t|he line adn pass through ye point w (as in fig: 1st[98]) yn

38 If ye superficies abr (fig 3d[99]) circulate all its points in ye line cd move wth equall velocity from c towards d. For make sfr=tsr=recto . & srf=crt & draw tadc than is ye motion of ye point e from c to ye motion of ye point f from c as ae to sf. but ae=sf (for △ rsf similis △ ret therefore er×sffr=et. also aet similis △ erf therefore erfretae. or, er×aefr=et=er×sffr & ae=sf ) therefore ye motion of e from c is equall to ye motion of f from c.

39 If ye body g move reflect on ye immoveable surface \dv / a{illeg}|t| its corner {illeg}| o | (fig 4th[100]) its parallell motion (viz from d to v ) shall not bee hindered {illeg} by ye surface dv , (viz: if ye center of g 's motion were distant from ye perpendicular dm an inch at one minute before reflection it shall bee s{illeg}|o| farr {sic} distant from it one minute after reflection). For dv i{illeg}|s| noe ways opposed to motion parallell to it, & a body might slidemove upon it wthout l{illeg}|o|oseing any motion, & if at ye first moment of contact ye body g should loose its ⊥ & onely keepe its ∥ motion it would (perhaps) continue to slide upon it & not reflect.

40 The body g reflecting on ye plaine vd at its cor{illeg}ner {illeg} o all its points in ye ⊥ line opvd shall move from ye plaine vd wth ye same velocity wch before reflection {illeg} \they/ moved fro to it. For ye point o (prop 9) moves wth yt velocity bacwards {sic} wch it before did forwards (vi{illeg}|z| to vd ) & all ye other points (prop 38) move wth ye same velocity from it.


[101] A Method for finding theorems concerning Quæstions de Maximis e{illeg}|t| minimis. And 1st Concerning ye invention of Tangents to crooked lines.

[102] Suppose ab=x. eb=y. bd=v. bc=o. cf=z. & ed=df. ye nature of ye line ax+xx=yy. Then is ac=x+o. ax+ao+xx+2ox+oo=zz. vv+yy=ed2=fd2=zz+vv2ov+oo. Or yy=oo2ov+zz Or yy=oo2ov+ax+ao+xx+2ox+oo. & since ax=yyxx. Therefore 0=2oo2vo+ao+2xo. Or 2×o2v+a+2x=0. Now yt ed may bee perpendicular to ye line tis required yt ye points e , & f conjoyne, wch will hapen {sic} when bc=o {illeg}|v|ani{illeg}sheth into nothing. Therefore in the equation 2×o2v+a+2x=0. Or \{illeg}/ v=o+a2+x, those termes in wch o is must be blotted out, & there remaines v=x+a2=bd. wch determines ye perpendicular ed .

[103] Hence it appeares yt in such like operations those termes may be ever blotted out in wch o=b is of more yn one dimension.

As if ye nature of ye line was x3+xxy+xyy=ayy . Then is {illeg} since ac=o+x it is x3+3x2o+3xoo+o3+xxz+2xoz+ooz+xzz+ozz=azz. That is x3+3x2o+xxz+2xoz+ooz+xz2+oz2=az2 . Also vv+yy=vv2vo+oo+zz . or yy+2vo=zz . Therefore {illeg}
ayyxyyxxy=x3+3xxo+xxz+2xoz+ozz+xzzazz=+xyy+2voxayy2voa =0. That is xxy+3xxo xxy3xxo2xozozz+ 2voa2vox=xxyy+2vo. That is (both pts □ed {illeg}|&| those terms left out in wch o is of more yn one dimension) x+yy+2xxyin2voa 2vox3xxo2xozozz _ =x4yy+2vox4 Or yin2voa 2vox3xxo2xozozz=voxx. That is 3xxy2xzyzzy=vxx+2vxy2vay. Now if bc=o vanisheth yn is z=y. And consequently 3xxy2xyyy3 xx+2xy2ay=v= 3xxy+2xyy+y3 2ay2xyxx.

[104] Hence I observe yt if {illeg} in ye valor of y there be divers termes in wch x is then in ye valor of z there are those same termes & also those termes \each of ym/ multiplyed according to ye by so many units as x hath dimensions in yt terme & againe multiplyed by o & divided by x . As if ∼ ∼ ∼ x3+xxy+xyyayy=0 . Then, x3+xxz+xzzazz+ 3x3o+2xxoz+xozz x =0. Which operacon may bee conveniently symbolized by (ordering ye equation according to ye dimensions of y ) &) making some letter \(as a . e . m . n . p )/ to signifie a terme, & ye same letter wth some marke (as ä, c̈, ë, g̈, m̈, n̈, p̈ &c), to signifie ye same terme multiplyed according to ye dimensions of x in it as in ye former example (suposing {sic} xa=m. xx=n. x3=p .) The nature of ye line is { in lettersxyyayy+xxy+x3=0xzzaz2+x2z+x3+ xozz+2xxoz+3x3o x =0. & in their symbolsmyy+ny+p=0mzz+nz+p+ m̈zzo+n̈zo+p̈o x =0. Soe if a4+ax3+bbx2abbx=y4.y4=m=y4. Then 3aox2+2bboxabbo+ ax3+bbx2 abbx+a4=z4. m̈ox+ m =z4. m̈ox+m=z4 .

A{illeg} And as any particular Equation may be thus symbolized so divers equations may bee represented by ye same caracters as 0=a+cy+yye may represent all equations in wch y is of one & two dimensions

Now if a generall Theoreme be required fo{illeg}|r| drawing tangents to such lines it may bee thus found. eb=y, bd=v, ab=x, bc=o, fc=z, by supposition, a+cy+eyy=0 . Then {illeg}/x\ \by observation ye 2d/, a+cz+ezz+ äo+c̈oz+ëozz x =0. Or, cyxeyyx=xa+czx+ezzx+äo+c̈oz+ëozz=0. Againe eb2+bd2=cf2+cd2 that is. yy+2ov=zz. Which valor {illeg} of zz put into its stead in ye termes ezzx \& czx/ in ye former Equation the result is +cyxäoc̈ozëozz2eovx=cxyy+2ov. And both pts squared it is (by ye first Observacon) ccyyx22cyxäo2cxyc̈oz2cxyëozz4cxyeovx=c2xxyy+2ccxxov. Wch rightly ordered is äyc̈yzëyzz=2exyv+cxv. And since ye points e & f conjoin{illeg}|e| to make ed a perpendicular therefore is z=y & consequently äyc̈yyëy3 cx+2exy =v . Wch is ye Theorem sought for. As for example were it required to draw a perpendicular to ye line whose nature is x3+xxy+xayy=0. a+cy+eyy=0. Then is äyc̈yyëy3 cx+2exy = 3x3y2xxyyxy3 x3+2xxy2axy =v or v= 3xxy+2xyy+y3 2ay2xyxx .

In like manner to draw tangents to those lines in wch y is of 1, 2 & 3 dimensions suppose a+cy+eyy+gy3=0. Then is {illeg} yy+gy3=ax \by 2d observacon/ cyxeyyxgy3x+äo+czx+ {illeg}/{illeg}\ =0 & by writeing ye valor of z=yy+2vo) in its stead in th{illeg}|o|se te{illeg}rmes in wch {illeg} not (viz {illeg} +gz3x there results { cyx+gy3xäoc̈zoez2ogz3o2exvo=cx+gyyx2vogxyy+2vo }. {illeg} {illeg} by {illeg} it is
{illeg}{2goz3=4eovx=cx+gyyxincx+gyy }{illeg}inyy+2vo. Or {illeg}{+gyyovx+4vogxyy }. That is { äyc̈yyëy3g̈y4 cx+2exy+2gxyy =v }.

By ye s{illeg}|a|me proceeding {illeg} of 1, 2, 3 {illeg} dimensions in {a+cy+{illeg} gy3 {illeg}} it would be found {illeg} =v . &c {illeg} {illeg}


[105] Having ye nature of a crooked line expressed in Algebraicall termes wch are not put one pte equall to another but all of ym equall to nothing, if each of ye {illeg} termes be multiplyed by soe many units as x hath dimensions in them. & then multiplyed by y & divided by x they shall be a numerator: Also if the \signes be changed &/ each terme be multiplyed by soe many units as y hath dimensions in yt terme{illeg} & yn divided by y they shall bee a denominator in ye valor of v .

Example 1st. If rx+ rxxq yy=0. Then 120 rx+rq xxyyinyx rxrq xx+yyin1y 002 = ry+2rq xy 2y =r2+rqx=v . Example 2d. If x3+xxy+xyyayy=0. Then 3x3+2xxy+xyyinyx xxy2xyy+2ay2in1y =v= 3xxy+2xyy+y3 2ay2xyxx . Exam: 3d. If x3+bxxcdx+bcd+dxy=0 . Then 3xxy2bxy cdy+dyy dx = 2byd 3xyd+ cyx yyx=v. And by taking y out of ye valor of v yn, v=2x3dd3bxxdd+bbxdd2cxd+2bcd+bccxxbbccx3 .[106]

Note. That haveing xy given, it will be often {illeg} \more/ convenient to find yx by ye equation expressing ye nature of ye line & yn having x & y to find v by them both, Then to take yx out of v 's valor & soe to find it by xy alone.

The Perpendiculars {illeg}|t|o crooked lines & also ye Theorems ∼ for finding them may otherwis{illeg} more conveniently \be/ found thus

[107] Supposing ab=x; cb=o, db=v, eb=y, cf=z. {illeg} And if ye fc bee supp distance twixt fc & fb , bee imagined to bee infinitely little, yt is if ye triangle efr is supposed to bee infinitely little then bebdbgberefryvozy. That is yzyy=vo. Or z=y+voy.

Now suppose ye nature of ye line bee rxrxxqyy=0. Then is rx+rorxx2roxrroz2{=0}

In wch equation instead of rx=rxxq+yy & zz=yy+2vo+ vvoo yy write theire valos {sic} & ye result is ro 2rox2voqrooq vvoo yy =0. Or r 2rxro q 2v vvoyy =0. but these two termes ro , vvoyy are infinitely little, yt is if compared to finite termes they vanish therefore I blot ym out & there rests r2rxq=v=db .

Suppose ye nature of ye line be p+qy+ryy=0 Then (by observation ye 2d) it is {illeg} qyxryyx=px+p̈o+qzx+q̈zo+rzzx+r̈zzo=0. Then writeing ye valor of z=y+voy in its stead in these termes qzx+rzzx, There results p̈o+qxvoy +q̈zo+2rxvo+r̈zzo=0. Or because ye difference twixt z & y is infinitely little it is p̈y+q̈yy+r̈y3 qx2rxy =v.

[108] And though ye angle ebg made by intersection of x & y is not determined {illeg}|w|hither {sic} it acute obtuse or a right one, yet may ye line bg bee found as easily after ye same manner wch determines ye position of ye tangnt eg . For suppose {illeg}=t . cb {illeg} cb=y, fc=z, ab=x, & yt zy. Then (supposing ye distance of fc & eb to be {infinit}ely little) it is, tyt+oy+oyt=z. Now if ye nature of ye line is p+qy{illeg}y3={0 } Then is qyxrq2xsy3x=px+p̈o+qxz+q̈oz+rxz2+r̈oz2+sxz3{illeg} And by putting ye valor of z into its stead in those terms in wch {illeg} results p̈o+qxoyt+q̈oz+2rxoy2t+r̈oz2+3sxoy3t+s̈oz3=0. Or t= {illeg}

Soe yt ye variation of ye angle ebg makes no variation.

Note yt th{illeg}|e| foundacon of this operacon of yt {illeg} {illeg} pag 131) {illeg} tangents {illeg} But since {illeg} Equation is the sa{illeg} {illeg}stons yt it would bee if {illeg} {illeg}


To draw perpendiculars to crooked lines in all other cases.

Although ye unknone {sic} quantitys x & y are not related to one another as in the precedent rules (yt{illeg} that is soe yt y move upon x in a given angle), y{illeg}|et| may there be drawne tangents to them by ye same metho{illeg}|d|.

[109] As if efm is an Elipsis {sic} described by ye thred aen (as is usuall) Then make gb=v. en=y. an=a. x+y={illeg} And let ye relation twixt x & y be x+y=b. ab=z . Then is bn=az. eb2=xxzz=yyaa+2azzz . And consequently xxyy+aa 2a =z=ab And. And xx x4+2xxyy2xxaay4+2aayya4 4aa =eb2= 2xxyy+2xxaa+2aayyx4y4a4 4aa =ae2ab2 . Againe suppose af=s, fn=t, bc=o. Then is ac=x+o=o+ xxyy+aa 2a = sstt+aa 2a . That is 2ao+xxyy+tt=ss. Alsoe vgv+og+ogv=fc. And fc2=gg+ 2g2o v = 2sstt+2ssaa+2ttaas4t4a4 4aa {illeg} Making eb=g . And consequently 2xxyy+2xxaa+2aayyx4y4a4 4aa + 2g2o v = 2sstt+2ssaa+2ttaas4t4a4 4aa Or 2xxyy+2xxaa+2aayyx4y4a4+ 8aag2o v =4ttao+2ttxx2ttxx2ttyy+t4+4a3o3+2aaxx2aayy+4aatta4s4 . [110] Or 4aoxx4aoyy+8aaggov+4aayyt44a3o4aatt=0 . Lastly by ye nature of ye line bt=s. And ba2bt+tt=ss=2ao+xxyy+tt. Or 2aoxx+yy+bb 2b =t. And { 4tt= 4aoxx4aoy24abbo+Q:xx+yy+bb: 2bb } wch valor substituted into ye Equation A, the result is {t48x3o+6axyyx4y4a4+2axyy2xxaa=0} 8aaggovx4y4a4+2xxyy2xxaa+2aayy=0.
{8aaggov +4a3bbo4a3bbox4a2+2xxyyaa+2 aabbxxaay4+2aabbyyb4a bb }.

[111] As if

[Editorial Note 2]

[1] Figure

[2] Figure

[3] Figure

[4] Figure

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[6] Figure

[7] Figure

[8] Figure

[9] October 1664

[10] Figure

[11] Figure

[12] Figure

[13] Figure

[14] Figure

[15] Figure

[16] Figure

[17] Figure

[18] Figure

[19] Figure

[20] Figure

[21] Figure

[22] For ye first equation of ye first sort

[23] For ye 2d

[24] For ye 3d

[25] For ye 4th

[26] For ye 5t

[27] For ye 6t &c

[28] For ye first Equation of ye seacond Sort

[29] For ye seacond

[30] For ye 3d.

[31] Figure

[32] Figure

[33] November 1664

[34] Figure

[35] Figure Figure

[36] Figure

[37] Figure

[38] Figure

[39] November 1664

[40] A

[41] Figure

[42] Figure

[43] Figure

[44] Figure

[45] Figure

[46] Figure

[47] Figure

[48] Figure

[49] B

[50] Figure

[51] This line is a streight one ye equation being divisible by b=y=0

[52] Endeavor not to find ye quantity d in these cases, but suppose it given[Editorial Note 1]

[Editorial Note 1] There is a line connecting the end of this note to the following one

[53] Or else C

[54] December

[55] Figure

[56] Figure

[57] Figure


[59] Figure

[60] Figure

[61] Figure

[62] F

[63] Figure

[64] G

[65] Figure

[66] December 1664

[67] Figure

[68] Figure

[69] Theorema

[70] Figure

[71] December 1664.

[72] Figure

[73] Figure

[74] Figure

[75] Figure

[76] Figure

[77] Figure

[78] Figure

[79] Figure

[80] Figure

[81] Figure

[82] Figure

[83] Feb 1664

[84] Figure

[85] Figure

[86] Another way.

[87] Figure

[88] December 1664

[89] Figure

[90] Figure

[91] Figure

[92] Figure

[93] Figure

[94] Of compound force.

[95] Figure

[96] Figure

[97] Figure

[98] Figure

[99] Figure

[100] Figure

[101] March \May/ 20th 1665

[102] Figure

[103] Theoreme 1st Observation 1st

[104] Observacon 2d

[105] An universall theorem for tangents to crooked lines, when yx.

[106] See Des Cartes his Geometry. booke 2d, pag 42, 46, 47. Or thus, x3bxxcdx+dyx+bcd 21001 { 2xxybxy dx +bcdy dxx =v }. And bcy xx +byd2xyd=v .

[107] Figure

[108] An universall theo{illeg}|r|em for drawing tangents to crooked lines when x & y intersect at any determined angle

[109] Figure

[110] A=

[111] Figure

[Editorial Note 2] The rest of the page is damaged.

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Professor Rob Iliffe
Director, AHRC Newton Papers Project

Scott Mandelbrote,
Fellow & Perne librarian, Peterhouse, Cambridge

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