Mathesis enucleata, or, The elements of the mathematicks by J. Christ. Sturmius ; made English by J.R. and R.S.S.

SECTION II.

CHAP. I. Of the chief Properties of the Conick Sections.

Proposition I.

IN the Parabola (GKEH Fig. 114) the(α) 1.1 square of the semi-ordinate (IK) is equal to the rectangle IL made by the Latus Rectum EL and the abscissa EI.

Demonstration.

MAke the sides of the cone that is supposed to be cut, AB =a, BC=b, and moreover EB=oa, and EI=eb, and AC=c; therefore NI will be = ec, by reason of the si∣militude of the ▵ ▵ BCA and EIN; and EP or IO=oc, by reason of the similitude of the ▵ ▵ ABC and EBP. There∣fore ▭ NIO=oecc=□ IK by the Schol. of Prop. 34 (n. 3) and Prop 17. Lib. 1. Now if a line be sought which with the abscissa EI shall make the ▭ IL=□ IK you will

have it by dividing the said square by the Abscissa EI. viz. 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = EL. And this is called the Latus Rectum, viz. in relation to the Abscissa EI with which it makes that rectangle, which, it's evident, is = □ IK, and from this equality the section has the name of Parabola, in Apollo∣nius.

CONSECTARYS.

I. THis Latus Rectum, expressed by the quantity 〈 math 〉〈 math 〉, may be found out after a shorter way, if you make as b to c (the side of the cone parallel to the section BC at the Diame∣ter of the base AC) so oc (the side EP called by some the Latus Primarium) to a fourth.

II. But if any one, with Apollonius, had rather express this by meer data in the cone it self as cut (because oc or that La∣tus Primarium EP is not a line belonging to the cone it self) he may easily perceive, if the quantity of the Latus Rectum found above, be multiplyed by the other side of the cone a, there will be produc'd the equivalent 〈 math 〉〈 math 〉 which instead of the proportion above will furnish us with this other,

• as ab− to cc− so oa
• □ of AB into BC−□ AC−EB

to a fourth; which is the very proportion of Apollonius in Prop. 9. Lib. 1. and confirms our former.

SCHOLIƲM I.

HEnce you have an easie and plain way of describing a Parabola, having the top of the ax and the Latus Re∣ctum given, viz. by drawing several semiordinates whose extreme points connected together will exhibit the curvity of the Parabola. But you may find as many semiordinates as you please, if having cut off as many parts of the Ax as you please, you find as many mean proportionals between the

Latus Rectum and each of those parts or Abscissa's. See n. 2. and 3. Fig. 47. Introduct. to Specious Analysis.

SCHOLIƲM II.

HEnce also we have a new genesis of the parabola in Plan•• from the spculations of De Witte, viz. if the rectiline∣ar angle HBG (Fig. 115.) conceived to be moveable about the fixed point B be conceived so to move out of its first situ∣ation with its other leg BH along the immoveable rule EF, that it may at the same time move also the ruler HG, from its first situation DK, all along parallel to it self, and with the other leg BG let it all along cut the said ruler HG, and with this point of its intersection continually moving from B to∣wards G it will describe a curve. That this curve will be the parabola of the antients is hence manifest, because it will have this same first property of the parabola. For, 1. if the an∣gle HBG (n. 1.) be supposed to be a right one, and BD or HI=a, BI or KG=b (viz. in that station of the angle and rule HG by which they denote the point G in the inter∣section) you'l have by reason of the right angle at B, BI, i. e. b a mean proportional Between HI i. e. a, and IG or BK, and so this as an abscissa = 〈 math 〉〈 math 〉. Wherefore if BK i. e. 〈 math 〉〈 math 〉 be multiplyed by BD=a, the rectangle DBK will be = bb =□KG; which is the first property of the parabola: So that it follows, since the same inference may be made of any other point in this curve, that this curve will be the parabo∣la, BD or HI its Latus Rectum, KG a semiordinate, and BK its axis, &c. 2. If the angle HBG be an oblique one (num. 2) it may be easily shewn from what we have supposed that the ▵ ▵ DBH and BKG will be equiangular: Therefore as BD (i. e. a) to DH sc. BI (i. e. b) so KG sc BI (i. e. b) to BK (i. e. 〈 math 〉〈 math 〉) Therefore again the □ DBK=bb□KG. QED.

Consect. 3. It is also evident in this second case, that BK drawn parallel to the ax, but not thro' the middle of the pa∣rabola,

will be a diameter which will have for its vertex B, its Latus Rectum BD, and semiordinate GK, &c.

Consect. 4. Therefore you may find the Latus Rectum in a given parabola geometrically, if you draw any semiordinate whatsoever IK (Fig. 116.) and make the abscissa EF equal to it, and from F draw a parallel to the semiordinate IK, and from E draw the right line EK thro' K cutting off FH the Latus Rectum sought; since as EI to IK so is EF (i. e. IK) to FH by Prop 34. lib. 1. wherefore having the ab∣scissa and semiordinate given arithmetically, the Latus Re∣ctum will be a third proportional.

Consect. 5. Since therefore the Latus Rectum found above is 〈 math 〉〈 math 〉, if you conceive it to be applyed to the parabola in LM, so that N shall be that point which is called the Focus, LN will be 〈 math 〉〈 math 〉 and its square 〈 math 〉〈 math 〉 and this divided by the Latus Rectum 〈 math 〉〈 math 〉 will give occ〈 math 〉〈 math 〉 for the abscissa EN: So that the distance of the Focus from the Vertex will be ¼ of the Latus Re∣ctum.

Consect. 6. Since therefore EN is = 〈 math 〉〈 math 〉 if for EF you put ib, NF will be = 〈 math 〉〈 math 〉, whose square will be found to be 〈 math 〉〈 math 〉, to which if there be added □ GF=oicc, by Prop. 1. the square of NG will be = 〈 math 〉〈 math 〉 whose root (as the extraction of it and, without that, the ana∣logy of the square NF with the square NG manifestly shews) will be 〈 math 〉〈 math 〉; so that a right line drawn from the Focus to the end of the ordinate, will always be equal to the abscissa EF

+EN i. e. (if EO be made equal to EN) to the compounded line FO.

SCHOLIƲM III.

HEnce you have an easier way of describing the parabola in Plano from the given Focus and Vertex, viz. (Fig: 117.) the axis being prolonged thro' the vertex E to O, so that EO shall = EN, if a ruler HI be so moved by the hand G, according to PQ, from OF to HI, that putting in a style or pin, it shall always keep the part of the Thred NGI (which must be of the same length with the rule HI) as fast as if it were glued to it (which perhaps might also be done with the Compasses by an artifice which we will hereafter al∣so accommodate to the hyperbola) and at the same time it will describe in Plano the part of the line EGR. That this will be a parabola is evident from the foregoing Consect. because as the whole thred is always = to the ruler IH; so the part GN is always necessarily equal to the part GH, i. e. to the line FO. Moreover from the same sixth Consect. and Fig. 116. may be drawn another easie way of describing the para∣bola in Plano from the Focus and Vertex given thro' innume∣rable points G to be found after the same way: viz. If from any assumed point in the ax F you draw to the ax a perpendi∣cular, and at the interval FO from the Focus N you make an intersection in G. Which innumerable points G will be de∣termined with the same facility, having given only, or assu∣med the axis and Latus Rectum, by vertue of the present Proposition. For if, having assumed at pleasure the point F in the axis, you find a mean proportional between the Latus Rectum and the abscissa EF, a semiordinate FG made e∣qual to it, will denote or mark the point G in the parabola sought.

Proposition II.

IN the hyperbola (GKEH Fig 118)(α) 1.2 the square of the semiordinate (IK) is equal to the rectangle (IL) made of the Latus Rectum (EL) and the abscissa (EI) together with ano∣ther rectangle LS of the said abscissa (EI or LR) and RS a fourth proportional to DE the Latus Transver∣sum, (EL) the Latus Rectum, and EI the Abscissa.

Demonstration.

Suppose the side of the cone AB here also = a, and BM parallel to the section = b, and the intercepted line AM=c, and EI=eb; all according to the analogy we have observed in the parabola; and NI will be as there = ec. Making moreover MC=d and the Latus Transversum DE=ob, so that DI shall be = ob+eb; then will (by reason of the simi∣litude of the ▵ ▵ BMC, DEP, and DIO) EP be = od, and IO=od+ed, and so QO=ed. Therefore □ NIO will be = oecd+eecd=□IK But this square divided by the Abscissa EI=eb gives 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 for the line IS which with the abscissa would make the rectangle ES = to the said square ••K. Now therefore, if here also we call a Line the Latus Rectum found after the same way as in the parabola, viz by making as b− to c− so od to a fourth 〈 math 〉〈 math 〉 (as a line parallel to the section — to the intercepted diam. so the Latus Primarium, but that the other part 〈 math 〉〈 math 〉 will be a fourth proportional to bc and ed or to eb, ec▪ and ed, or (to speak with Apollonius as we have done in the Prop.) to ob, 〈 math 〉〈 math 〉 and eb (for in these three cases you I have the same fourth 〈 math 〉〈 math 〉) Wherefore now it is evident that the square of the se∣miordinate

oecd+eecd is equal to the rectangle IL (made by the Latus Rectum 〈 math 〉〈 math 〉 into the abscissa eb=oecd) together with ▭ LS of this fourth proportional 〈 math 〉〈 math 〉 into the same ab∣scissa Eb, which is = eecd. Which was to be found and de∣monstrated.

CONSECTARYS.

I. HEnce you have in the first place the reason why Apol∣lonius called this Section an Hyperbola; viz. because the square of the ordinate IK exceeds or is greater than the rectangle of the Latus Rectum and the Abscissa.

II. Since therefore the Latus Rectum here also as well as in the parabola is found by making as b to c so od to 〈 math 〉〈 math 〉 (i. e. as the parallel to the section BM is to the intercepted Diam. AM so is the Latus Primarium EP to a fourth EL.) If any one had rather express this Latus Rectum after Apollonius's way, he will easily perceive, this quantity being found and multi∣plyed both Numerator and Denominator by b the parallel to the section, there will come out the equivalent quantity 〈 math 〉〈 math 〉 which gives us instead of the former proportion this other,

• as bb− to cd− so ob to a fourth;
• □BM−▭AMC—Latus Transversum to a fourth;

which is that of Apollonius in Prop. 12. Lib. 1. and consequent∣ly herein confirms our former.

III. You may also have this Latus Rectum geometrically, by finding a third proportional (as we have done in the pa∣rabola Consect. 4. Prop. 1.) to the abscissa EI (Fig. 119.) and the semiordinate IK (= EF;) and then find a fourth pro∣portional EL to DI (the sum of the Latus Transversum and abscissa) and FH already found, or IS equal to it, and DE (the Latus Transversum) and that will be the Latus Rectum sought.

[illustration]

Pag. 172.

119

120

121

122

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124

125

126

126

127

128

SHCOLIƲM.

FRom this third Consectary, we may reciprocally from the Latus Rectum and transverse given, find out and apply as many semiordinates to the ax as you please, and so describe the hyperbola thro' their (ends or) infinite points: viz. if assuming any part of the abscissa EI, you make as DE to EL so DI to IS; and then find a mean proportional IK between IS and the abscissa EI, and that will be the semiordinate sought: And both this praxis and the Consect. may be abun∣dantly proved by setting it down in, and making use of, the literal Calculus.

Proposition III.

IN the Ellipsis (KDEK, Fig. 120.) the(α) 1.3 square of the semiordinate (IK) is equal to the rectangle (IL) of the Latus Rectum (EL) and the abscissa (EI) (less or) taking first out another rectangle (LS) of the same abscissa (EI or LR) and RS a fourth proportional to (DE) the Latus Transversum (EL) the Latus Rectum and (EI) the abscissa.

Demonstration.

Suppose the side of the cone to be AB here also = a and BM parallel to the section = b and the intercepted AM=c, and EI=eb; and NI will be again = ec, all as in the hyperbola. And mak••ng also here as in the hyper∣bola MC=d, and the Latus Transversum DE=ob, so that DI will be ob−eb; then will (by reason of the simili∣tude of the ▵ ▵ BMC, DEP and DIO) EP be=od, and IO=od−ed. Therefore ▭ of NIO will be = oecd−eecd=□IK. But this square divided by the abscissa EI=eb gives 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 for that line IS which with the abscissa would make the rectangle ES= to the said square IK. Now therefore if we call the Latus Rectum a right line found after the same way as in the parabola, by making ac∣cording to Cons. 1. Prop. 1.

as b to c− so od− to a fourth 〈 math 〉〈 math 〉 i. e. as the line pa∣rallel to the section — to the intercepted diameter — so the Latus Primarium, &c. It is manifest that the Latus Re∣ctum is one part of the line just now found; and the other part 〈 math 〉〈 math 〉 is a fourth proportional to b, c and ed, or (to speak with Apollonius as we have done in the Prop.) to 〈 math 〉〈 math 〉 and eb (for there will come out the same quantity 〈 math 〉〈 math 〉) where∣fore now it is evident that the □ of the semiordinate IK is e∣qual to the ▭ IL (of the Latus Rectum 〈 math 〉〈 math 〉 and the ab∣scissa eb=oecd) having first taken out thence the ▭ LS, or eecd out of that fourth proportional 〈 math 〉〈 math 〉 by the same abscissa eb; which was to be found and demonstrated.

CONSECTARYS.

I. HEnce yov have first of all the reason of the name of the Ellipse, which Apollonius gave to this section; viz. because the square of the semiordinate IK is defective of, or less than the rectangle of the Latus Rectum and the abscis∣sa.

II. Since therefore the Latus Rectum here also as well as in the parabola and hyperbola, is found by making as b to c so od to 〈 math 〉〈 math 〉 (i. e. as BM parallel to the section is to the inter∣cept. diam. AM so the Latus Primarium EP to a fourth EL) now if any one had rather express this Latus Rectum after A∣pollonius's way, he will easily see that the quantity above found being multiplyed both Numerator and Denominator by b, that there will come out an equivalent one 〈 math 〉〈 math 〉, which instead of the former proportion will give this other,

as

• bb− to cd− so ob to a fourth;
• BM−▭AMC—Latus Transvers. to a fourth;

which is the same with that we have also found in the hyper∣bola, and which also Apollonius has Prop. 13. Lib. 1.

III. This Latus Rectum may also be had geometrically, if you find, 1. in the hyperbola a third proportional FH to the abscissa EI (Fig. 121.) and semiordinate IK (= EF.) 2. But EL a fourth proportional to DI (the difference of the Latus Transversum and the abscissa) and the found FH, or IS equal to it, and the Latus Transversum DE, is the Latus Rectum sought.

SCHOLIƲM.

FRom this third Consect. we may reciprocally, having the Latus Rectum and Transversum given, apply as many semiordinates to the ax as you please, and so draw the ellipsis thro' as many points given as you please, viz. if, taking any ab∣scissa EI, you make as DE to EL so DI to a fourth IS; then between this IS and the abscissa EI find a mean proportional IK, and that will be the semiordinate sought: And this Prax∣is also and the third Consect. may be abundantly proved by making use of a literal Calculus. For e. g. here a fourth proportional to ob, 〈 math 〉〈 math 〉 and ob−eb will be 〈 math 〉〈 math 〉 and a mean proportional between this fourth and eb will be 〈 math 〉〈 math 〉, &c.

Proposition IV.

IN a Parabola(α) 1.4 the squares of the ordinates are to one another as the ab∣scissa's.

Demonstration.

For if EF (Fig. 122.) be called ib, as above EI was cal∣led eb, since the Latus Rectum is 〈 math 〉〈 math 〉 the square of FG wil be = oicc. Therefore □ IK will be to □ FG as oecc—oicc

• e to i or
• eb to ib.

Q. E. D.

CONSECTARY.

HEence having drawn LO parallel to the ax or diameter EF, if it be cut by the transverse line EG in M and by the curve of the parabola in K; then will OL, ML, and KL be continual proportionals. For EF is to EN as FG to NM or IK, by reason of the similitude of the ▵ ▵ EFG and ENM. But the squares FG and IK are in duplicate proportion of EF to EN by Prop. 35. Lib. 1. and are also in the same propor∣tion as the abscissa's EF and EI by the pres. Therefore EF to EI is also in duplicate proportion of EF to EN i. e.

• EF is to EN as EN to EI Q. E. D.
• OL is to ML as ML to KL Q. E. D.

Proposition V.

IN the hyperbola and Ellipsis(α) 1.5 the squares of the Ordinates are as the rectangles contained under the lines which are intercepted between them, and the Vertex's of the Latus Transversum's.

Demonstration.

For, if EF (Fig. 118. and 120.) be called ib, as EI was above called eb, then will according to Prop. 2. and the 3d. dedu∣ction.

GF=oicd+eicd in the Hyperb.

oicd−eicd in the Ellips.

and the ▭ DFE=oibb+iibb in the Hyperb.

oibb−iibb in the Ellips.

Therefore the □ KI is to the square GF as oecd±eecd to oicd ± iicd i. e. as oe±ee to oi±ii.

and ▭ DIE is to the ▭ DFE, as oebb±eebb to oibb±iibb i. e. in like manner as

oe±ee to oi±ii. Q. E. D.

CONSECTARY I.

IN the Ellipsis this may be more commodiously expressed apart thus; the squares of the ordinates (KI and GF) are as the rectangles contained under the segments of the Diame∣ter (viz. DIE and DFE) in which sense this property is also common to the circle, as in which the squares of the ordinates are always equal to the rectangles of the segments.

CONSECTARY II.

THerefore, if the Latus Rectum be conceived to be apply∣ed in the hyperbola, so that N shall be the Focus; (see Fig. 123.) then will LN=〈 math 〉〈 math 〉, and its square be 〈 math 〉〈 math 〉. But as the □ KI to the square LN, so is the ▭ DIE to the ▭ DNE i. e. oecd+eecd to 〈 math 〉〈 math 〉 so is oebb+eebb to 〈 math 〉〈 math 〉. But now the ▭ of the whole DE and the part added EN into the part added EN, i. e. ▭ DNE (=〈 math 〉〈 math 〉 together with the square of half CE (=〈 math 〉〈 math 〉) is = □ compounded of half and the part added CN= 〈 math 〉〈 math 〉 by Prop. 9. lib. 1. Wherefore CN the distance of the Focus from the centre is = 〈 math 〉〈 math 〉. But 〈 math 〉〈 math 〉 is the fourth part of the ▭ of the Latus Transversum ob and the Latus Rectum 〈 math 〉〈 math 〉 (or the

fourth part of the figure, as Apollonius calls it) and 〈 math 〉〈 math 〉 is the □ of 〈 math 〉〈 math 〉 i. e. of half the Latus Transversum. Wherefore we have found the following Rule of determining the Focus in an hyperbola: If a fourth part of the figure (or the rectangle of the Latus Rectum into the Transversum) be added to the square of half the Latus Transversum, and from the sum you extract the square root; that will be the distance of the Focus from the center CN: And hence substracting half the Latus Transversum CE, you will have distance of the Focus from the Vertex EN.

CONSECTARY III.

IN like manner in the Ellipsis having drawn the ordinate LM (Fig. 124.) that the Focus may be in N, the □ LN would be 〈 math 〉〈 math 〉 as above, and by a like inference □ DNE=〈 math 〉〈 math 〉. But now □ DNE together with the square of the dif∣ference CN is equal to the □ of half CE by Propos. 8. lib. 1. and consequently the □ CN is = □ CE−□DNE, that is, 〈 math 〉〈 math 〉. Wherefore CN the distance of the Focus from the centre is = 〈 math 〉〈 math 〉. Wherefore we have found the following Rule to determine the Focus in the El∣lipse. If the fourth part of the figure (or the rectangle of the Latus Rectum into the Latus Transversum) be substracted from the square of half the Latus Transversum, and from the remain∣der you substract the square root; that will be the distance of the Focus from the Centre CN: And taking hence half the La∣tus Transversum CE, you'l have the distance of the Focus from the Vertex EN.

SCHOLIƲM I.

BOth the Rules are easie in the practice, for since 〈 math 〉〈 math 〉 is nothing but the square of CE, and 〈 math 〉〈 math 〉 nothing but the re∣ctangle of ¼ DE into LM; if between LM and ¼ DE or MO (Fig. 125.) you find a mean proportional MN, (and so whose □ is equal to that □) and in the hyperbola join to it at right angles MC=CE, the hypothenusa CN will be the distance sought of the Focus from the centre: And the same may be had in the Ellipsis, if (n. 2.) having described a semi-circle upon CM=CE you draw or apply the mean found MN, and draw CN.

SCHOLIƲM II.

HEnce also we have(α) 1.6 a new genesis of the Ellipse in Plano about the diameters given, from the speculations of Monsieur de Witt; viz. If about the rectilinear angle DCB (Fig. 126. n. 1. and 2.) consider'd as immove∣able, the rule NLK (which all of it will equal the greater semidiameter CB, and with the prominent part LK the lesser CD) be so moved that N going from C to D, and L from B to C may perpetually glide along the sides of the angle, the extreme point in the K in the mean while describing the curve BKE (and in a like application the other quadrants) and that this curve thus described will be the ellipsis of the ancients is hence manifest, because it has the second property of the ellipse just now described. For, 1. if the angle DCB or NCB be supposed to be a right one (as in Fig. 126. num. 1.) and the rule KN in the same station, it marked out the point K, and having apply'd the semiordinate KI, and drawn the perpendicular LM, from the square KL and the square CE (as being equal) substract mentally the equal squares LM and CI, and there will remain by virtue of the Pythagorick Theor. on the one hand □ KM and on the other by

Prop. 8. lib. 1. □ DIE equal among themselves. But now the square of KI is to the square of KM (i e. to the □ DIE) as the square of KN to the square of KL (i. e. as the square of CB to the square of CE) by reason of the similitude of the ▵ ▵ KLM and KNI; and since the same may be demonstra∣ted after the same manner of any other semiordinate Ki viz. that its □ Ki is to the □ DiE as the square CB to the square CE. It also follows, that the □ KI is to the □ DIE as □ Ki to the □ DiE, and alternatively, the square KI will be to the square Ki as the □ DIE to the □ DiE; which is the se∣cond property of the ellipse. 2. If the angle NCE be not a right one (as in Fig. 126. n. 2. and the like cases) having drawn NO and KP parallel to the rule nlB in the first station, [in which station the angle NCE, to which the flexible ruler is to be made, is determined, viz. by letting fall the perpendicular Bl from the extremity of one diameter upon the other, and moreover by adding or substracting the difference of the semi-diameters ln] having also drawn the Ordinate KIM, and PI parallel to CN; which being done the ▵ ▵ CBl and IKF, and also CBn and IKP will be similar. Where∣fore having joined NP, from the parallelism of the lines IP and NC and the similitude of the aforesoid ▵ ▵, as also of NCO and nCl, it will be easie to conclude that NCIP is a parallelogram. Wherefore since KN is = CE and □ KN = □ CE, having substracted the squares of the equal lines NP and CI, there will remain on the one hand □ KP on the other the □ DIE equal among themselves as above. Therefore the square of KI will be to the square of KP (i. e. to the □ DIE) as the square of BC to the square of Bn (i. e. to the square of CE) as in the former case: And since here also the same may be demonstrated after the same manner of any o∣ther semiordinate Ki; we may infer as above, that the □ □ KI and Ki are to one another as the rectangles DIE and DiE, &c.

But after what way the same ellipses may be described by these right lined angles without any of these rulers thro' infi∣nite points given, will be be manifest from the same figures to any attentive Person. For having once determined the angle NCE or nCD (num. 2. e. g.) if NL or nl be applyed where you please by help of a pair of compasses, and continued to

K, so that LK or lk shall be equal to lB, you will have eve∣ry where the point K, &c.

CONSECTARY IV.

SInce in the hyperbola (Fig. 123.) the □CN−□CE=□DNE, and in the ellipsis (Fig. 124.) □CE−□CN=□DNE, by vertue of Prop. 9 and 8. lib. 1 if for CN you put on both sides for brevity's sake m, then will the □ DNE in the hyperbola be rightly expressed in these terms 〈 math 〉〈 math 〉, and in the ellipsis in these 〈 math 〉〈 math 〉.

Proposition VI.

IN the parabola(α) 1.7 the Latus Rectum is to the sum of two semiordinates (e. g. IK+FG i. e. HO in Fig. 122.) as their difference (OG) to the difference of the abscissa's (IF or (KO.)

Demonstration.

For if the greater abscissa EF be made = ib, and the less EI=eb, the semiordinates answering to them FG and IK will be √ooic and √oecc as is deduc'd from Prop. 1. Where∣fore if you set in the same series

1	2	3
The Latus R.	−sum of the semiord.	−their diff.
〈 math 〉〈 math 〉	−√oicc+√oecc	−√oicc−√oecc
	4
	−diff. of the absciss.
	−ib−eb

and multiply the extremes and means, you'l have on both sides the same product oicc−oecc, which will prove by vertue of Prop. 19. lib. 1. the proportionality of the said quantities. Q. E. D.

SCHOLIƲM.

THis is that property of the parabola, whereon the Clavis Geometrica Catholica of Mr. Thomas Baker is founded, which as unknown to the ancients, nor yet taken notice of by Des Cartes, he thinks was the reason why that incomparable Wit could not hit upon those universal rules for solving all Equations howsoever affected. Concerning which we shall speak further in its place. We will only further here note, that Baker was not the first Inventor of this property, but had it, as he himself ingeniously confesses, out of a Manuscript communicated to him by Tho. Strode of Maperton, Esquire.

Proposition VII.

IN the hyperbola and ellipsis(α) 1.8 the Latus Rectum is to the Latus Transversum, 〈◊〉〈◊〉 the square of any semiordinate (e. g. IK in Fig. 118▪ and 120.) to the rectangle (DIE) con∣tained under the lines intercepted between it and the Vertex's of the Latus Transversum.

Demonstration.

For the Latus Rectum is on both sides 〈 math 〉〈 math 〉, the Latus Trans∣versum ob, &c. Wherefore if you make in the same series as the Latus R. to the Lat. Transv. so the □ IK to ▭ DIE

• 〈 math 〉〈 math 〉 in hyperb. 〈 math 〉〈 math 〉
• in ellips. 〈 math 〉〈 math 〉

The rectangles of the extremes and means will both be ooebcd±oeebcd, and so will prove the proportionality of the said quantities, by Prop. 19. lib. 1. Q. E. D.

CONSECTARY I.

HEnce having given in the ellipsis (see Fig. 124) the La∣tus Rectum and the transverse ax, you may easily obtain the second ax or diameter, if you make as the Lat. Transv. to the Lat. Rect. so the □ DCE to □AC 〈 math 〉〈 math 〉.

CONSECTARY II.

THerefore the □ of the whole AB will be = oocd=□ of the Latus Rectum into the Lat. Transv. (which Apllo∣nius calls the Figure) so that the second Ax (and any second Diameter) will be a mean proportional between the Latus Re∣ctum and the Latus Transversum. Hence in the hyperbola also the second or conjugate diameter may be called a mean proportional between the Latus Rectum and Transversum, i. e. √oocd or a line which is equal in power to the Figure, as A∣pollonius speaks.

SCHOLIƲM I.

HEnce may be derived another and more simple way of delineating organically the ellipsis in Plano about the given axes AB, DE (Fig 127.) which Schooten has given us▪ viz. by the help of two equal rulers CG and GK move∣able about the points G and G: If, viz. the portions CF and HK are equal to half the lesser ax AC, but taken with both the augments (viz. CF+FG+GH) may = ½ the greater ax CD or CI; and the point K moving along the produced line DE the point H may describe the curve EHAD. That this will be an ellipsis will be evident by vertue of this seventh Prop. from a property that agrees to this curve in all its points H. For having drawn circles about each diameter, and the lines IHN, FO perpendicular to CE; and having made the Latus Rectum EL, which is a third proportional to DE and AB by the second Consect. of this Prop. &c. by reason of the similitude of the triangles CFO, CIN, FO will be to FC as

IN to IC, and alternatively FO to IN as FC to IC i. e. as AC to CE or AB to DE. Therefore also the square of FO (or HN) will be to the square of IN, as the square of AB to the square of DE, by Prop. 22. lib. 1. i. e. as EL the Latus Rectum to ED the Latus Transversum, by Prop. 35. lib. 1. But the □ IN is = DNE from the proporty of the circle. Therefore □ FO (or of the semiordinate HN) is to the ▭ DNE as EL the Latus Rectum to ED the Latus Transv. therefore by ver∣tue of the pres. Prop. the point H is in the Ellipsis, and so any other, &c. Q. E. D.

CONSECTARY III.

NOW if in the ellipsis the □ of AC the second Ax (= 〈 math 〉〈 math 〉 by Consect. 1.) and □ CN the distance of the Fo∣cus from the centre (= 〈 math 〉〈 math 〉 by Censect 3. Prop. 5. the figure whereof you may see n. 124.) be joined in one sum; the □ AN will be = 〈 math 〉〈 math 〉, and so the line AN=〈 math 〉〈 math 〉 i. e. to half the Latus Transversum: So that hence having the axes given you may find the Foci, if from A at the interval CD you cut the transverse ax in N and N.

CONSECTARY IV.

NOW if, on the contrary, in an hyperbola (Fig. 123.) the □ AC or EF=〈 math 〉〈 math 〉 be substracted from the □ CF or CN=〈 math 〉〈 math 〉 by vertue of Consect. 2. Prop. 5. there will remain 〈 math 〉〈 math 〉 and its root 〈 math 〉〈 math 〉, i. e. half the Latus Trans∣versum CD: So that here also, the axes being given, you may find the Focus's, if from the vertex E you make EF a per∣pendicular to the ax = to the second Ax AC, and at the in∣terval CF from the centre C you cut the Latus Transversum continued in N and N.

SCHOLIƲM II.

BUT now, that the right lines KN and KN drawn from any other point (e. g. K) to the Foci, when taken toge∣ther in the ellipsis, but when substracted the one from the o∣ther in the hyperbola, are equal to the Latus Transversum DE, we will a little after demonstrate more universally, and also shew an easie and plain Praxis of delineating the ellipsis and hyperbola in Plano, having the axes and consequently the Fo∣ci given.

CONSECTARY V.

SInce we have before demonstrated Consect. 2. and 3. Prop. 5. that the □ DNE in the hyperbola and also in the el∣lipsis is = 〈 math 〉〈 math 〉; and here in Consect. 1. the □ of the second semi-diameter AC is also = 〈 math 〉〈 math 〉; it is evident that this □ AC is equal to the □ DNE.

CONSECTARY VI.

IT is hence moreover evident, if the square of half the trans∣verse diameter GE=〈 math 〉〈 math 〉 be compared with the square of half the second diameter AC or EF=〈 math 〉〈 math 〉, multiplying both sides by 4 and dividing by o; they will be to one another as obb to ocd i. e. further dividing both sides by b, as ob to 〈 math 〉〈 math 〉 the Latus Transversum to the Latus Rectum.

CONSECTARY VII.

BUT since also the □ DIE is to the □ IK as the Latus Transversum to the Latus Rectum, by vertue of the pre∣sent 7. Prop. the square of CE the transverse semidiam. will be

to the square of AC the second semidiam. (or by the 5th. Con∣sect. of this, to the □ DNE) as the □ DIE to the square of IK.

CONSECTARY VIII.

YOU may also now have the □ IK (which otherwise in the hyperb. is eocd+eecd, in the ellipse oecd−eecd, by vertue of Prop. 2. and 3.) in other terms, if you make as □ CE to the □ DNE so the □ DIE to a fourth; i. e.

as 〈 math 〉〈 math 〉 to 〈 math 〉〈 math 〉 by vertue of Consect. 4. Prop. 5.

(so oebb+eebb in the hyperb.

as 〈 math 〉〈 math 〉 to 〈 math 〉〈 math 〉 so oebb−eebb in the ellipsis.

For hence by the Golden Rule the square IK may be infer'd as a fourth proportional.

In the hyperbola 〈 math 〉〈 math 〉;

In the ellipsis 〈 math 〉〈 math 〉:

The use of which quantities will presently appear.

Proposition VIII.

THE

• Aggregate in the ellipse
• Difference in the hyperb.

of the right lines(α) 1.9 KN and Kn (Fig. 128.) drawn from the same point K to both the Focus's is equal to the transverse ax DE.

An Ocular Demonstration.

WHich consists wholly in this to find the lines KN and Kn by help of the right-angled triangles IKN and IKn (sc. the hypothenuses having the sides given) and afterwards see if the sum of both in the ellipse, and difference in the hyperbola be = ob, i. e. to the transverse ax DE.

I. In the Ellipsis.

Putting for CN (which above Prop. 5. Cons. 3. was found to be 〈 math 〉〈 math 〉) I say putting for it m, you'l have

IN=CI+CN=½ob−eb+m

In=Cn−CI=m−½ob+eb

Therefore □IN= 〈 math 〉〈 math 〉 □In= 〈 math 〉〈 math 〉

Add to each □ IK which was found in Prop. 7. Consect. 8. in the ellipsis = 〈 math 〉〈 math 〉 and you'l have □KN= 〈 math 〉〈 math 〉 and by extracting the roots (which is easie) you'l have

〈 math 〉〈 math 〉 and

〈 math 〉〈 math 〉;

Sum ob. Q. E. D.

II. In the Hyperbola.

Putting again m for CN (which above Cons. 2. Prop. 5. was found to be 〈 math 〉〈 math 〉) and you'l have

IN =CI+CN=½ob+eb+m

In=CI−Cn=½ob+eb−m. Therefore □IN= 〈 math 〉〈 math 〉 and □In= 〈 math 〉〈 math 〉

Add to both the □ IK which was found in Prop. 7. Consect. 8. in the hyperbola 〈 math 〉〈 math 〉 and you'l have □KN= 〈 math 〉〈 math 〉

□Kn= 〈 math 〉〈 math 〉 and extracting the roots out of these (which is easie) you'l have KN= 〈 math 〉〈 math 〉.

Kn= 〈 math 〉〈 math 〉 (which is a false or impossible root, for it would be CE−CN and moreover — another quantity.

Or Kn= 〈 math 〉〈 math 〉; which is a true and possible root.

The difference therefore of the true roots is = ob. Q. E. D.

SCHOLIƲM I.

WE first of all tried to make a literal Demonstration by using the quantity of the square IK as you have it expressed Prop. 2. and 3. and the quantity IN as it was com∣pounded of CI=ob+eb+CN= 〈 math 〉〈 math 〉, &c. but we found it very tedious in making only the squares of IN and In. Then for the surd quantity CN we substituted another, viz. m, and we produced the squares of IN and In as above, but we added the square of IK in its first value; and thus we obtain'd the squares KN and Kn, but in such terms, that the exact roots could not be extracted, but must be exhibited as surd quantities, and consequently we must make use of the rules belonging to them to find their sum or difference, which we laid down Cons. 3. Prop. 7. and Consect. Prop. 10. Lib. 1. which tho' it would succeed, yet would be full of trouble and tediousness. Therefore at length when we came to use those other terms which express the square IK, the business succeed∣ed as easie as we could wish, and that in a plain and easie way and no less pleasant, which I doubt not but will also be the opinion of the Reader, who shall compare this with other demonstrations of the same thing, which only lead indirectly to this truth, or with them, which de Witte has given us in E∣lem▪ Curvar. lin. p. m. 293. and 302. and which he thinks

easie and short enough in respect of others both of the ancients and moderns, and which we have reduced into this yet more distinct form, and accommodated to our schemes.

Preparation for the Hyperbola.

Make as

• CD to CN
• EE−CN

so CI to CM so that the □ of

• DCM
• MCE

will be = □

• NCI
• nCI.

Because therefore it will be by Consect. 7. Prop. 7. as □ CD to □ DNE, so the □ DIE to the □ IK.

And also by composition, as □ CD to

• □CD+□DNE
• i. e. □ CN per 9. lib I.

so DIE to DIE + □ IK.

Therefore by a Syllepsis, as □ CD to the □ CN so

• □ CD + DIE
• i. e. □ CI

to □ CN + DIE + □ IK.

But also by the Hypothesis. as the □ CD to the □ CN so □ CI to □ CM. Therefore □ CM is = □ CN + DIE + □ IK.

Demonstration.

Since therefore it is certain that the difference between DM and EM is the transverse ax DE; if it be demonstrated that DM is = KN and EM = Kn, the business will be done, be∣cause the difference between KN and Kn is also the transverse ax DE.

Resolve the □ KN.

It is certain that NIq + IKq=KNq.

Substitute for NIq, by the 7. Lib. 1. CIq+CNq+2NCI.

Preparation for the Ellipsis.

Make as CD to CN so CI to CM.

So that the □ DCM is = □ NCI.

Because therefore by Consect. 7. Prop. 7. as □ CD to □ DNE so □ DIE to the □ IK;

Then also by dividing, as □ CD to

• □ CD−DNE
• i. e. □ CN, by 8. l. 1.

so DIE to DIE−□IK.

Therefore by a Dialepsis, as □ CD to □ CN, so

• □CD−□DIE
• i. e. CI □ by 8. cit.

to □ CN−DIE+□IK.

But also by the Hypothesis, as □ CD to □ CN, so □ CI to □ CM: Therefore □ CM is = □ CN−DIE + □ IK.

Demonstration.

Since therefore it is certain that the sum of DM and EM is the transverse ax DE; if it be demonstrated that DM is = KN and EM = Kn, the business will be done, because the sum of KN and Kn is also equal to the transverse ax DE.

Resolve the □ KN.

It is certain that NIq+••Kq=KNq.

Substitute for NIq, by the 7. lib. 1. CIq+CNq+2NC••

Then will CIq+CNq+2NC••+IKq=KNq.

Substitute for C••q, by the 9. lib. 1. CDq+DIE; then will CDq+DIE+CNq+2NC••+Kq=KNq.

Resolve also □ DM.

It is certain that CMq+CDq+

• 2DCM
• 2NCI

= DM, by the 7. lib. 1.

Substitute for CMq its value by the Preparation, and you'l have CNq−D••E+••Kq+CDq+2NCI=DMq:

Which were before = KNq.

Therefore KN=DM; which is one.

In like manner resolve □ Kn.

It is certain that nlq+••Kq=Knq.

Substitute for nIq, by Consect. 1. Prop. 10. Lib. 1. C••q+CNq−2nCI, and you'l have

CIq+CNq−2nCI+IKq=Knq.

Substitute for CIq, by the 9. lib. 1. CDq+D••E, and you'l have 〈 math 〉〈 math 〉=Knq.

Resolve also the □ EM.

It is certain that 2CDq+2CMq−DMq=EMq per 13. lib. 1.

Then will CIq+CNq+2NCI+IKq=KNq.

Substitute for CIq by the 8. lib. 1. CDq−DIE; then will 〈 math 〉〈 math 〉.

Resolve also the □ DM.

It is certain that CMq

• 2DCM
• 2NCI

= DMq per 7. lib. 1.

Substitute for CMq its value from the preparation, and you'l have CNq−DIE+IKq+CDq+2NCI=DMq:

Which before were = KNq.

Therefore KN = DM; which is one.

In like manner resolve the □ Kn.

It is certain that nIq+IKq=Knq.

Substitute for nIq by Consect. 1. Prop. 10. lib. 1.

CIq+CNq−2NCI, and you'l have CIq+CNq−2NCI+IKq=Knq.

Substitute for CIq per 8. lib. 1. CDq−DIE, and you'l have CDq−DIE + CNq−2NCI+IKq=Knq.

Resolve also □ EM.

It is certain that 2CDq+2CMq−DMq=EMq per 13.1.

Substitute the value of DMq first found above, and you'l have CDq+CMq−2nCI=EMq.

Substiute for CMq the value as in the preparation, and you'l have CDq+CNq+DIE−2nCI+IKq=EMq: Which were before = Knq.

Therefore Kn=EM; which is the other.

SHCOLIƲM II.

HEnce you have the common mechanical ways of describ∣ing the ellipsis and hyperbola about their given axes; and the ellipsis, if the Foci N, N, (Fig. 129. n. 1.) are gi∣ven, or found according to Consect. 3. Prop. 7. and having therein stuck or fixed two pins, put over them a thread NFn tyed both ends together precisely of the length you design the greater ax DE to be of, and having put your pencil or pen in that ▵-string draw it round, always keeping it equally extended or tight. Now because the parts or portions of the thread re∣main always equal to the whole ax DE, what we proposed is evident by the present Prop. which may also be very elegantly described by a certain sort of Compasses, a description where∣of Swenterus gives us in his Delic. Physico-Math. Part. 2. Prop. 20. which may be also done by a sort of organical Mechanism, by the help of two rulers moveable in the Foci GN and Hn (n 2.) and equal to the transverse ax DE, and fastned a∣bove by a transverse ruler GH equal to the distance of the Fo∣ci, as may appear from the Figure. For if the style F be moved round within the fissures of the cross rulers Hn and GN the curve thereby described will be an ellipsis from the pro∣perty we have just now demonstrated of it, which it hath in every point F. For the triangles HGN and NHn, which have one common side HN, and the others equal by construction, are equal one to another, and consequently the angles FHN and FNH equal, so also the legs HF and FN, and so likewise FN and Fn together are equal to Hn = DE; which is the very property of the ellipse we are now treating of. But Van Schoo∣ten, who taught us this delineation, hints, that, if thro' the middle point •• of the line HN you draw the line IFL, it will touch the ellipsis in the point F; for since the angles IFH and IFN are equal, by what we have just now said, the vertical angle LFn of the one IFH, will be necessarily equal to the o∣ther IFN: But this equality of the angles, made by the line KL drawn thro' F, with both those drawn from the centres, is here a sign of contact, as is in the circle the equality of the angles with a line drawn from its one centre. So that after this way you may draw a tangent thro' any given point F of

[illustration]

Pag. 192.

129

130

131

132

133

134

the ellipsis without this organical apparatus of Rulers; viz. if, having drawn from both the Focus's thro' the given point F the right lines nH, NG equal to the Latus Transversum DE, you bisect HN in I and draw IFL: Or if the line that connects the extremes GH be produced to K, and you draw thence KFL, viz. in that case where GH and Nn are not parallel; other∣wise a line drawn thro' the point F parallel to them would be the tangent sought.

As to the hyperbola, there is a mechanick method of draw∣ing that also, not unlike the others, from a like property in that, communicated by the same Van Schooten, viz. If ha∣ving found the Focus's N and n (Fig. 129. n. 3.) you tye a thread NFO in the Focus N and at the end of the ruler nO of the length of the transverse ax DE; then putting in a pen or the moveable leg of a pair of compasses (nor would it be dif∣ficult to accommodate the practice we before made use of to this also) draw or move it within the thread NFO from O to E, so that the part of the thread NO may always keep close to the ruler as if it were glued to it. For if we call the length of the thread X, and the transverse ax ob as above, the ruler nO will be, by the Hypoth. = X + ob. Make now the part of the thread OF = ½ X, the remainder or other part will be NF = ½ X and nF = ½ X + ob, and the difference between FN and Fn, = ob. Make OF = ¾ X, then will FN be ½ X and Fn ¼ X + ob, the difference still remaining ob and so ad infinitum. In short, since the difference of the whole thread and of the whole ruler is ob, and in drawing them, the same OF is taken from both, there will always be the same difference of the re∣mainders. Hence also assuming at pleasure the points N and n you may describe hyperbola's so, the thread NFO be short∣er than the ruler nFO: For if it were equal there would be described a right line perpendicular to Nn, thro' the middle point C.

There yet remains one method of describing hyperbola's and ellipses in Plano, by finding the several points without the help or Apparatus of any threads or instruments, viz. in the ellipsis, having given or assumed the transverse axis DE and the Foci N and n (Fig. 130. n. 1.) if from N at any arbi∣trary distance, but not greater than half the transverse ax NF, you make an arch, and keeping the same opening of the com∣passes

you cut off, from the transverse ax, EG, and then, taking the remaining interval GD, from n you make another arch•• cutting the former in F, and so you will have one point of the ellipse, and after the same way you may have innume∣rable others, f, f, f, &c.

In like manner to delineate the hyperbola, having given o•• assumed the transverse ax DE and the Focus's N and n (n. 2.) if from N at any arbitrary distance NF you strike an arch, and keeping the same aperture of the compasses from the dia∣meter continued, you cut off EG, and then at the interval GD from n make another arch cutting the former in F, you will have one point of the hyperbola, and after the same way innumerable others, f, f, &c.

Proposition IX.

IF the secondary ax, or conjugate diameter of the hyperbola AB (Fig. 131.) be applied parallel to the vertex E, so that it may touch the hyperbola, and OE, EP are equal like BC and AC, and from the centre C you draw thro' O and P right lines running on ad insinitum, and lastly QR parallel to the Tangent OP; you'l have the following

CONSECTARYS.

I. THE parts QG and HR intercepted between the curve and those right lines CQ, CR will be equal; for by reason of the similitude of the ▵ ▵ CEP and CFR as also CEO, CFQ as CE is to EO (and EP) so will CF be to FQ and FR, and consequently these will be equal; and so taking away the semiordinates FG and FH which are also equal, the remainders GQ and HR will be also equal, and consequently the □ □ QGR, GRH, &c. all equal among themselves: Which we had already deduced before in Consect. 2. and 3. Def. 7.

II. The rectangle QGR will be = □ EO or EP = 〈 math 〉〈 math 〉 i. e. (as Apollonius speaks) to the fourth part of the figure: For by

reason of the similitude of the ▵ ▵ CEO, CFQ, CE will be to EO as CF to FQ: i. e. as the □ CE to the □ EO i. e. as (by Cons 2. 7.) the Lat. Transv. to the Lat. Rect.

• so the □ CF
• to th □ FQ

i. e. as (by the 7. Prop.) as the □ DFE to the □ FG

But now if from the □ CF you take the □ DFE, there will remain the □ CE, by Prop. 9. lib. 1. and if from the □ FQ you take the □ FG there will remain the □ QGR, by Prop. 8. lib. 1. wherefore that remaining □ CE to this remain∣ing □ QGR will be, as was the whole square CF to the whole square FQ, by Prop. 26. lib. 1. i. e. as was the □ CE to the □ EO; consequently the □ QGR and the square EO (to which the same square CE bears the same proportion) will be equal among themselves.

III. Since this is also after the same manner certain of any other rectangle ggr or grh, &c. it follows that all such rect∣angles are equal among themselves.

IV. Wherefore it is most evident, since the lines FR, fr, &c. and so GR and gr grow so much the longer, by how much the more remote they are from the vertex E; that on the contrary the lines QG and qg must necessarily so much the more decrease and grow shorter, and consequently the right line CQ approach so much nearer and nearer to the curve EG.

V. But that they can never meet or coincide altho' produ∣ced ad infinitum will thus appear; if it were possible there could be any concourse or meeting, so that the point G and Q or g and q could any where coincide, it would follow from Consect. 2. that as the □ DFE to the square FG so the square CF to the square FQ i e▪ to the same square FG; and so that the □ DFE would be = □ CF; which is absurd by Prop. 9 lib. 1. so that now it is evident that the lines COQ and CPR drawn according to Consect. 1. are really Asymptotes (i. e. they will never(α) 1.10 coincide (viz. with the curve of the hyperbola) as Apollonius has named them.

VI. Having drawn the right lines from G and g parallel to both the asymptotes, viz. GS and gs and likewise GT and gt, the rectangles TGS and tgs will be(α) 1.11 e∣qual among themselves. For by reason of the similitude of the ▵ ▵ TQG and tqg, first, TG will be to QG as tg to qg; and, by reason of the equality of the □ □ QGR and qgr, secondly, QG will be to gr reci∣procally as qg to GR, by Prop 19▪ lib. 1. and by reason of the similitude of the ▵ ▵ SGR and sgr, thirdly, as gr to gs so GR to GS, wherefore (since in two series

1. 2. 3. as TG to QG to gr to gs so tg to qg to GR to GS) you'l have ex aequo or by proportion of equality as TG to gs so tg to GS, by Prop. 24 lib. 1. Therefore, by Prop. 17. of the same, the □ of TG into GS=□ of tg into gs. Q. E. D.

SCHOLIƲM.

HEnce, lastly, we have a new genesis of the hyperbola in Plano about its given diameters from the speculations of(β) 1.12De Witt; if, viz. having drawn the lines AB and EF cross one another at pleasure (Fig. 132) to the angle BCF you conform the move∣able angle BCD (acd being to be delineated in the opposite hyperbola equal to the contiguous ACD) one of whose legs is conceived to be indefinitely extended, but the o∣ther CD of any arbitrary length; and to the end of it D ap∣ply the slit of a moveable ruler GD about the point G at any arbitrary interval GD (but yet parallel to the leg CB in this first station) and so carrying together along with it the moveable angle BCD about the line ECF, but so that the leg CD may always remain fast to it, and the other CB be inter∣sected in its progress by the ruler GDH, e. g. in b or β This point of intersection, thus continually moved on, will describe the curve bGβ, which we thus prove to be an hyperbola: Be∣cause the ruler GDH turning about the pole G, and carried from D e. g. to d or δ cuts the leg of the moveable angle CB

brought to the situation cb or γβ, and in the mean while remain∣ing always parallel to it self; and having drawn from the points of intersection b and β and G the lines GI, bK and βη parallel to the ruler CF, because e. g. in the second station, having ta∣ken the common quantity cD from the equal ones CD and cd, the remainders Cc and Dd are equal, and by reason of the si∣militude of the ▵ ▵ dcb and dDG,

• as Dd
• i. e. Cc
• or bK

to DG,

• so dc
• i. e. DC
• or GI

to cb; the rectangle of Kb into bc will be = □ of DG into GI, by Prop. 18. lib. 1. and in like manner, when in the third station having added the common line Dγ to the equal ones CD and γδ, the whole lines Dδ and Cγ are equal, and, by reason of the similitude of the ▵ ▵ βγδ and GDδ

• as Dδ
• i. e. Cγ
• or βη

is to DG so

• is γδ
• i. e. DC
• or GI

to γβ; the □ of ηβ into βγ=□ of DG into GI, by the same 18. Prop. Wherefore the three points b, G, β, (and so all the o∣thers that may be determined the same way) are in the hyper∣bola, whose asymptotes are CB and CF and its centre C, &c. by the present Prop. Consect. 6. Q. E. D.

You may also determine innumerable points of this curve separately without the motion we have now prescrib'd, viz. as the point a in the opposite hyperbola, if thro' any assumed point c in the asymptote CE you draw a parallel to the other asymptote CA, and having made cd equal to CD, from G thro' d draw Gda, and so in others.

CHAP. II. Of Parabolical, Hyperbolical and Elliptical Spaces.

Proposition X.

THE(α) 1.13 Parabolick Space (i. e. in Fig. 133. that comprehended under the right line GH and the parabola GEH) is to a cir∣cumscribing Parallelogram GK, as 4 to 6 (or 2 to 3) but to an inscribed ▵ GEH as 4 to 3.

Demonstration.

Suppose FH divided first into two then into four equal parts, and draw parallel to the ax EF the lines ef, ef, &c. dividing also EF into four parts, the first fg will be 3, the second 2, the third 1, by Prop. 34. lib. 1. but as ef is to ge so is ge to he, by Consect. 1. Prop. 4. Therefore he in the diameter EF is = o, in the first ef it is = ¼ (for as ef, 4, to ge, 1, so ge, 1, to he, ¼) in the second ef a portion of he is = 〈 math 〉〈 math 〉, in the third to 〈 math 〉〈 math 〉, and so the portions eh in the trilinear figure EhHK make a series in a duplicate arithmetical progression, viz. 1, 4, 9, 16: After the same manner, if the parts Ff, &c. are bisected, you'l find the portions eh in the external trilinear figure to make this series of numbers ⅛, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, and so onwards. Wherefore since the portions eh or the indivisibles of the trili∣near space circumscribed about the parabola are always in a duplicate arithmetical progression: the sum of them all will be to the sum of as many indivisibles of the parallelogram FK, ••qual to the line KH, i. e. the trilinear space it self to this pa∣ra••lelogram as 1 to 3, by Consect. 10. Prop. 21. lib. 1. Wherefore the semi-parabola ••EhH will be as 2, and the ▵ FEH as 1 ½; therefore the whole parabola as 4, and the whole ▵ GEH as 3, and the whole parallelogram GK as 6. Q E. D.

CONSECTARY I.

IT is evident(α) 1.14 that in the first division, the second line fh (i. e. that drawn from the middle of the base FH) is three such parts whereof FE is 4; for eh is 〈 math 〉〈 math 〉 i. e. 1, therefore fh is 3.

CONSECTARY II.

IT is also evident, that this demonstration will hold of any parabolick segment.

Proposition XI.

THE Elliptical Space(α) 1.15 comprehended by the Ellipsis DAEB (Fig. 127.) is to a circle described on the transverse ax DE, as the Axis Rectus or conjugate diameter AB to the transverse ax DE.

Demonstration.

THis is in the first place evident from the genesis of the ellipse we deduced in Schol. 1. Prop. 7. for in that de∣duction we shewed that FO, i. e. HN was to NI as AB to DE: Which since it is true of all the other indivisibles or or∣dinates HN and IN ad infinitum; it is manifest that the planes themselves constituted of these indivisibles will have the same reason among themselves, as the Axis Rectus AB to the trans∣verse DE. Q. E. D.

CONSECTARY I.

THerefore the quadrature of the ellipse will be evident, if that of the circle be demonstrated.

CONSECTARY II.

SInce a circle described on the least diameter AB will be to one described on the greater diameter DE, as AB to a third proportional by Prop. 35. lib. 1. it follows by vertue of the present Prop. that the ellipse is a mean proportional between the greater and lesser circle, i e. as the ellipse is to the great∣er circle so is the lesser circle to it, viz. as AB to DE.

CONSECTARY III.

HEnce you may have a double method of determining the area of an ellipse. 1. If having found the area of the greater circle, you should infer, as the greater diameter of the ellipsis to the less, so the area of the circle found to the area of the ellipse sought. 2. If having also found the area of the les∣ser circle, you find a mean proportional between that and the area of the greater.

SCHOLIƲM.

WE may also shew the last part of the second Consect. thus, 1. If having described the circle EadbE (Fig. 134.) about the least axis of the ellipse we conceive a regular hexagon to be inscribed, and an ellipse coinciding with one end E of its transverse ax, and with the other or opposite one D to be so elevated, that with the point d it may perpendicu∣larly hang over the circle, and further from all the angles of the figure inscribed in the circle you erect▪ the perpendiculars gG, bB, &c. it is certain that the sides ED and Ed of the triangle DEd will be cut by the parallel planes FGgf, &c. into proportional parts, and that those by reason of the simi∣litude of the ▵ ▵ FDG and fdg, and so also the other rectan∣gles will be among themselves as the intercepted parts of the lines ID and id, CI and ci, and in infinitum, (viz. of how many sides soever the inscribed figure consists:) Wherefore also all the parts of the ellipse taken together will be to all the parts of the circle taken together, i. e. the whole ellipse to the whole

[illustration]

Pag. 201.

135

136

137

138

139

140

141

circle as all the parts of the diameter ED or ab, i. e. as DE it self to AB. Q. E D.

CONSECTARY IV.

IT is also evident that both these demonstrations of the pre∣sent Prop. will be also the same in any segments of the el∣lipsis or circle.

Proposition XII.

ANY Hyperbolical space GEHG (Fig. 135) is to any Hyperbolick figure of equal heighth gEhg [whose Latus Rectum and Transversum are equal (as in the circle) and also equal to the Latus Transversum of the former DE, as the Axis Rectus (or conjugate) AB is to the Latus Transversum DE (as in the ellipsis.)

Demonstration.

By the Hypoth. and Prop. 7. and its second Consect. the □ Fg is = □ DFE. Wherefore this □ DFE i e. the □ Fg is to the □ FG as the Latus Transversam to the Latus Rectum of the hyperbola GEHG, by the same seventh Prop. i. e. (by Consect. 2 of the same) as the square of the Latus Transvers. DE to the square of the conjugate AB: Therefore the roots of these squares will be also proportional, viz. Fg to FG as DE to AB; and consequently (since the same is true of any other ordinates ad infinitum) the whole hyperbola gEhg will be to the whole one GEHG as DE to AB. Q. E. D.

CONSECTARY I.

THerefore having found the quadrature of such an hyper∣bola, whose Latus Rectum and Transversum are equal, you may have also the quadrature of any other hyperbola.

CONSECTARY II.

IT is evident that the same demonstration will hold in any other hyperbola's.

Proposition XIII.

ANY Parabolick segments upon the same base, and hyperbo∣lical and elliptical ones described about the same conjugate (one whereof shall be a right one, the other a scalene) and con∣stitued between the same parallels, are equal.

Demonstration.

I. It is evident of Parabola's; for both the right one GEHG, and the scalene one GEHG (Fig. 136. n. 1.) (for the demon∣stration of Prop. 10. will hold in both) is to a ▵ inscribed in them as 4 to 3. But the triangles GEH and GEH are equal, by Consect. 5. Def. 12. or Prop. 28. lib. 1. Therefore the Parabola's also.

Or thus, in the right parabola GEHG every thing is the same as in 1. and 4. Prop. of this Book, viz. EI=eb, EF=ib, the square IK=oecc, the □ FG=oicc. And because therefore in the scalene Parabola also the square FG remains =oicc, make FE=n, and find both the abscissa EI, and the □ answering to it IK.

1. For the abscissa; as FE to EI so FE to EI, per 〈 math 〉〈 math 〉 Consect. 4 Prop. 34 lib. 1.

2. For the □ IK; as FE to EI so □ FG to □ IK, 〈 math 〉〈 math 〉. per Prop. 4. of this.

Therefore the □ IK=□IK and IK=IK, and this in any case ad infinitum: Therefore the one parabola is = to the other. Q. E. D.

II. The business is much after the same way evident of el∣lipses and hyperbolas. For making all things in the ellipsis and right hyperbola (n. 2. and 3. Fig. 136.) as in Prop. 2, 3, 5, 7. viz. the □ IK oecd−eecd in the ellipsis, oecd+eecd in the hyperbola, the □ AB oocd by Consect. 2. Prop. 7. EI=eb, DE=ob, &c. if in oblique ones for the Latus Transversum DE you put n, and seek the Latus Rectum and abscissa EI, you may by means of these also have the square IK, by Prop. 2. and 3.

1. For the Latus Rectum. As n to √oocd so √ oocd to 〈 math 〉〈 math 〉 by Cons. 2.7▪

2. For EI the abscissa. As ob to eb so n to 〈 math 〉〈 math 〉=EI.

3. For the side RS □ deficient or exceeding, from Prop. 2. and 3. As n to 〈 math 〉〈 math 〉 so 〈 math 〉〈 math 〉 to 〈 math 〉〈 math 〉=RS.

Now the abscissa multiplyed by the Lat. Rect.	The abscissa multiplyed by RS.
〈 math 〉〈 math 〉 by 〈 math 〉〈 math 〉 gives □ oecd.	〈 math 〉〈 math 〉 by 〈 math 〉〈 math 〉 gives □ eecd.
The sum of these □ □ oecd + eecd in the hyperb. = □ IK by Prop. 2. evidently = □ IK	The difference of these □ □ oecd−eecd gives in the ellipsis □ IK by Prop. 3. evidently = □ IK.

Wherefore the lines IK and IK, and the whole KL and KL will be equal; and since the same thing is evident after the same way of all other lines of this kind ad infinitum, the elliptical and hyperbolical segments will be so also. Q. E. D.

CHAP. III. Of Conoids and Spheroids.

Proposition XIV.

A Parablick Conoid(α) 1.16 is subduple of a Cylinder, and in sesquialteran reason (or as 1 ½) of a cone of the same base and altitude.

Demonstration.

Because in the parabola the □ AD (Fig. 137.) is to the □ SH, as BD to BH, i. e. as 3 to 1, and so to the □ TI as BD to BI, i. e. as 3 to 2, by Prop. 4. of this; it is evident that these squares of SH and TI and AD and consequently of the whole lines also Sh, Ti, AC, and the circles answering to them will be in arithmetical progression, 1, 2, 3; and more∣over if there are new Bisections in infinitum, as the abscissa's so also the squares and circles of the ordinates, by vertue of the aforesaid fourth Prop. will always be in arithmetical Progres∣sion 1, 2, 3, 4, 5, 6, &c. It is evident that an infinite series of circles in the conoid, consider'd as its indivisibles, will be to a series of as many circles equal to the greatest AC, i. e. the co∣noid to the cylinder AF as 1 to 2, or as 1 ½ to 3, by Consect. 9. Prop. 21. or Consect. 4. Prop. 16. lib. 1. but to the same cylinder AF the inscribed cone ABC is as 1 to 3, by Prop. 38. lib. 1. therefore the cylinder, conoid and cone are as 3, 1 〈 math 〉〈 math 〉 and 1. Q.E.D.

Proposition XV.

THE half of(α) 1.17 any Spheroid, or any o∣ther segment of it is in subsesquialteran pro∣portion to the cylinder, and double of the cone having the same base and altitude.

Demonstration.

Having divided the altitude BD (Fig. 138.) e. g. into three equal parts, because in the ellipse as well as in the circle the square of AD is to the square of SH as the □ GDB to the □ GHB, i. e. as 9 to 5, and so to the square TI as 9 to 8, by Consect. 1. Prop. 5. of this; and in like manner if you make new bisections, the squares (and consequently the circles) of the ordinates go on or decrease by a progression of odd num∣bers, as 36, 35, 32, 27, 20, 11, and so ad infinitum, the bisections being continued on; as we have shewn in the sphere and circumscribed cylinder Prop. 39. lib. 1. and it will neces∣sarily follow here also (by vertue of Consect. 12. Prop. 21.) that the whole cylinder will be to the inscribed segment of the spheroid, as 3 to 2; and since the same cylinder is to the cone ABC as 3 to 1, also the segment of the spheroid will be to the cone as 2 to 1. Q. E. D.

Proposition XVI.

AN hyperbolical Conoid(α) 1.18 is to a cone of the same base and altitude, as the aggre∣gate of the ax of the hyperbola that forms it and half the Latus Transversum, to the aggregate of the said axis and Latus Transversum.

Demonstration, containing also the Invention of this Proportion.

Make (in Fig. 139.) CE=a, EF=b, OE=c; then will CF=a+b. Since therefore as CE to OE so CF to FQ 〈 math 〉〈 math 〉 the □ EO will = cc and □ FQ= 〈 math 〉〈 math 〉.

But as these squares so also are the circles of the lines EO and FQ to one another, by Prop. 32. lib. 1. and so the cone COP

will be as 〈 math 〉〈 math 〉, and the cone CQR as 〈 math 〉〈 math 〉 (viz. by multiplying the third part of the altitude CF by the base FQ:) Having therefore substracted the cone COP from the cone CQR, there will remain the truncated cone QOPR 〈 math 〉〈 math 〉, and from this solid truncated cone having further substracted the hollow truncated cone, which the space EHRP produced in the genesis of the conoid (and which ac∣cording to Consect. 2. Definit. 9. is as bcc) there will remain the hyperbolical conoid 〈 math 〉〈 math 〉 i. e. (by substituting now the values of the ax or abscissa EF, and of half the Lat. Transv. EC, and of the conjugate diam. OP, &c. found in the de∣monstrations of the preceding Chapter, viz. 〈 math 〉〈 math 〉 for a, eb for b, and √oocd for c or oocd for cc) the hyperbolical conoid will come out 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉. But the cone GEH (multiplying the third part of EF into the □ GH, i. e. ⅓eb into 4oecd+4eecd) is as 〈 math 〉〈 math 〉. Therefore the conoid is to the cone as 6eebocd+4e{powerof3}bcd to 4eebocd+4e{powerof3}bcd, i. e. (dividing on both sides by 4eecd) as ½ob+eb to ob+eb. Which was to be found and demon∣strated.

SCHOLIƲM.

IF any one had rather proceed herein by indivisibles, as in the precedent Prop. having divided the ax EF (Fig. 140.) again into three equal parts, and assuming the values of the lines determined in the hyperbola, viz. eb for the abscissa EF, ob for the transverse ax, 〈 math 〉〈 math 〉 for the Latus Rectum, oecd+eecd for the square of the semiordinate FG, &c. the lowest and greatest circle of the diameter HG will be as oecd+eecd, and, if you make

as the Latus Transv. to the Latus Rectum, so the □ DfE ob−〈 math 〉〈 math 〉 made of ob+⅔eb into ⅔ eb (i. e. ⅔oebb+〈 math 〉〈 math 〉eebb) to a fourth; there will come out ⅔oecd+〈 math 〉〈 math 〉eecd for the second circle of the diam. hg; and by the same inference (as ob to 〈 math 〉〈 math 〉 so ob+⅓eb into ⅓ eb to a fourth) for the third circle of the diameter HG ⅓oecd+〈 math 〉〈 math 〉eecd; so that these indivisibles [for which here and in the precedent also the partial circumscribed cylinders may be assumed] proceed in a double series of numbers, the first in a simple arithmetical progression 3, 2, 1, the latter in a du∣plicate Arithmetical progression of squares 9, 4, 1; and the same if you make further new bisections, will necessarily hap∣pen ad Infinitum, (the former numbers e. g. in the first bise∣ction will be 〈 math 〉〈 math 〉 ⅚ 〈 math 〉〈 math 〉 oecd the latter 〈 math 〉〈 math 〉 eecd, &c) it is manifest from the consectaries of Prop. 21. lib. 1. that the whole cylinder HK will in like manner be expressed by a double series of parts answering, in numbers to the indivisibles of the conoid made by any bisection, but in magnitude to the greatest of them all, and in the sum of its first series of parts will be to the sum of the first in the conoid, both being infinite, as 2 to 1 or 3 to 1 ½ oecd, by Consect. 9. of the said Prop. 21. and the sum of its latter to the sum of the former in the conoid will be as 3 to 1 eecd and so the whole cylinder to the whole conoid as 3 oecd+3 eecd to 1 ½ oecd + eecd i. e. (dividing by ecd) as 3o+3e to 1 ½o+e i. e. mul∣tiplying both sides by b) as 3ob+3eb to 1 ½ob+eb; and consequently the cone (which is ⅓ of the cylinder) to the co∣noid as ob+eb to 1 ½ob+eb. Q.E.D.

CONSECTARY.

HEnce also appears the proportion of the hyperbolick co∣noid to a cylinder of the same base and altitude, which we did not express in the Prop. viz. as the aggregate of the ax and half the Latus Transversum to triple the aggregate of the said ax and Latus Transversum.

CHAP. IV. Of Spiral Lines and Spaces.

Proposition XVII.

THE(α) 1.19 first spiral space is subtriple of the first circle, i. e. as 1 to 3.

Demonstration.

Having divided the circumference of the circle into (Fig. 141. n. 1.) three equal parts by lines drawn from the initial point, beginning from the first line BA, the line BC will be as 1, BD as 2, BA as 3, by Consect. 1. Def. 12. of this book, and consequently the sectors circumscribed about the spiral will be CBc as 1, DBd as 4, ABa as 9, by Prop. 32. lib. 1. and in like manner, if you make new bisections, the lines drawn from the point B to the spiral, will be 1, 2, 3, 4, 5, 6; but the circumscrib'd sectors, 1, 4, 9, 16, 25, 36; and so the circumscrib'd partial sectors ad infinitum will proceed in an order of squares, there being always as many sectors in the circle equal to the greatest of them. Therefore all the sectors that can be circumscrib'd ad infinitum about the spiral space, i. e. the spiral space it self (in which at last they end) to so many equal to the greatest, i. e. to the circle, is as 1 to 3, by Consect. 10. Prop. 21. lib. 1. Q. E. D.

CONSECTARY I.

SInce the first circle is to the second as 1 to 4 (i. e. as 3 to 12) by Def. 12. of this, and Prop. 31. lib. 1. and the first spiral space to the first circle as 1 to 3 by the present Prop. the same spiral space will be to the second circle as 1 to 12; and to the third by a like inference as 1 to 27, to the fourth as 1 to 48, &c.

CONSECTARY II.

THE first spiral line is equal to half the circumference of the first circle. For the lines or radii of the sectors, and consequently their peripheries or arches proceed in a sim∣ple arithmetical reason, as 1, 2, 3, 4, 5, 6, &c. while in the mean time the whole periphery of the circle contains so many arches equal to the greatest. Therefore the whole pe∣riphery of the circle is to an infinite series of circumscrib'd ar∣ches, i. e. to the spiral line it self, as 2 to 1, by Consect. 9. Prop. 21. lib. 1.

Proposition XVIII.

THE whole spiral(α) 1.20 space comprehended under the second right line EA and the second spiral EGIA (see Fig. 141. n. 2.) is to the second circle as 7. to 12.

Demonstration.

For having divided the circumference of the circle first into three equal parts, there will be drawn to the second spiral four right lines BE, BG, BI and BA being as 3, 4, 5, 6, and but only three sectors circumscrib'd, viz GBg, IBi and ABa, which proceed according to the squares of the three latter lines, viz. 16, 25, 36, so that the sum is 77, while the sum of three equal to the greatest is 108, and so the one to the other (dividing both sides by 9) as 12 to 8 〈 math 〉〈 math 〉▪ Having moreover bisected the arches and parts of the line BE, so that that shall be 6, the second BF will be 7, and so the other five 8, 9, 10, 11, 12; and the sectors answering to them (excepting the first) 49, 64, 81, 100, 121, 144, so that their sum shall be 559, while the sum of six equal to the greatest, i. e. the whole circle is 864, and so one to the other (dividing both by 72) as 12 to 7 〈 math 〉〈 math 〉. In the other bisection of the arches and the parts of the line BE, so that the one shall be 12, the second 13, &c. to the thirteenth BA which will be 24, the sum of twelve sectors will be found to be 4250; and the sum of

as many equal to the greatest 6912, and so the one to the other (dividing both sides by 576) as 12 to 7 〈 math 〉〈 math 〉 s. 〈 math 〉〈 math 〉. Therefore the proportion will be

• I. In the first case 12 to 7+1+½+〈 math 〉〈 math 〉 viz. 〈 math 〉〈 math 〉.
• II. In the second case 12 to 7+½+¼+〈 math 〉〈 math 〉 viz. 〈 math 〉〈 math 〉.
• III. In the third case 12 to 7+¼+⅛+〈 math 〉〈 math 〉, &c.

The first and second fractions thus decreasing by ½ the latter by ¼. Wherefore the proportion of the second circle to the second spiral space will be as

12 to 7+1+½+〈 math 〉〈 math 〉

−½−¼−〈 math 〉〈 math 〉

−¼ &c.−⅛ &c.−〈 math 〉〈 math 〉 &c.

By vertue of Consect. 3. and 8.=0=0 Prop. 21. lib. 1. i. e. as 12 to 7. Q. E. D.

CONSECTARY I.

BEcause the second circle is to the first spiral space as 12 to 1, by Consect. 1. of the preceding Prop. and to the se∣cond spiral space as 12 to 7, by the present. it will be to the second space without the first (viz. BCDEAIGE) as 12 to 6 i. e. as 2 to 1.

CONSECTARY II.

THerefore the second space separately to the first is as 6 to 1.

CONSECTARY III.

SInce in the trisection of both these circles, first and second, there arise six lines, and as many sectors, viz. three lines BC, BD, BE, i. e. 1, 2, 3, to which there answer three ar∣ches in the same progression within the second circle, and also as many equal to its greatest; therefore the sum of all the un∣equal arches will be 21, but the sum of the equal ones of both circles (each of which in the first are equivalent to 3, in the second to 6) will be 27. Wherefore the sum of both the Pe∣ripheries to the sum of all the circumscrib'd arches will be as

27 to 21, i. e. (dividing both sides by 9) as 3 to 2 ⅓. More∣over bisecting the arches of the circles and the parts of the line BA, there will arise six circumscribed unequal arches within the first circle, which are as 1, 2, 3, 4, 5, 6, and as many within the second 7, 8, 9, 10, 11, 12; the sum of all which is 78, while the sum of as many equal ones on both sides is 108. Wherefore the one will be to the other, i. e. the sum of both the peripheries to twelve circumscribed arches ta∣ken together, is now as 108 to 78, i. e. (dividing both sides by 36) as 3 to 2 ⅙. And making yet another bisection, the proportion will be found to be as 3 to 2 〈 math 〉〈 math 〉, &c. and hence at length may be evidently inferr'd; that the sum of both the peripheries will be to the sum of all the arches circumscribible ad infinitum, i. e. to the whole helix as

3 to 2+⅓

−⅙

−〈 math 〉〈 math 〉 &c.=0. that is, as 3 to 2. Q. E. D.

CONSECTARY IV.

THerefore, since the periphery of the second circle is dou∣ble of the first, that alone will be equal to the whole spiral.

CONSECTARY V.

THerefore, if the periphery of the second circle be 2, the periphery of the first will be 1, and the first spiral line ½ by Consect. 2. of the anteced. Prop. wherefore the second spiral alone will be 1 ½, and so the periphery of the second circle alone will be to the second spiral alone as 2 to 1 ½ i. e. as 4 to 3; and to the first alone as 4 to 1.

SCHOLIƲM I.

BUT as Consect. 4. may be also deduced after another way, viz. by comparing only the arches of the second circle with the correspondent circumscripts, but considering them as taken twice (because that circle is twice turned round while the whole helix or spiral is described) and finding in the first

trisection the proportion of double the second periphery to all the circumscripts as 12 to 7; and in the succeeding bisection as 12 to 6 ½; in the second bisection as 12 to 6 ¼, &c. and at length by inferring, that the second periphery is double of all the arches circumscribible about the whole helix ad infinitum, that is to the helix it self.

as 12 to 6+1

−½

−¼ &c.=0. i. e. as 12 to 6;

and consequently the simple second periphery will be to the whole helix as 6 to 6: Thus the 5. Consect. may be separate∣ly had after the same manner, if instead of the first trisection, you only bisect; (vid. Fig. 141. n. 3.) for so in the first bi∣section the arches circumscribed about the second spiral line would be separately two semi-circles Dd, 3 and Aa, 4, (for as the line BC is one, BE, 2, BD, 3, BA, 4; so the arch described by the radius BD is 3 and described by the radius BA=4,) and their sum 7; while the sum of two equal to the greatest is 8. In the second bisection (when BE is 4) BF and its arch is made 5, the arch BD 6, the arch BG 7, the arch BA 8, the sum 26; while the sum of so many quadrants equal to the greatest is 32. Thus in the third bisection the sum of eight Octants circumscrib'd about the second helix will be found to be 100, the sum of so many = to the greatest 128, &c. Wherefore the periphery of the second circle in the first case will be to the arches circumscrib'd about the se∣cond helix as 4 to 3+½; in the second as 4 to 3+¼; in the third as 4 to 3+⅛, &c. and so to all the arches circumscri∣bible in infinitum, i. e. to the second helix it self as

4 to 3+½

−¼

−⅛ &c. = 0. i. e. as 4 to 3. Q. E. D.

By the same method you may easily find the proportion of the third circle to the third spiral space, and of that periphery either to the whole spiral, or separately to the third, as will be evident to any one who trys.

I. For the third spiral space.

(Fig. 142)

BC 1	BF 4	BI 7	49
BD 2	BG 5	BK 8	64
BE 3	BH 6	BA 9	81

are the three first sectors circumscrib'd about the parts of the third helix. The sum of these three sectors is 194; and the sum of so many equal to the greatest 243. Therefore the first proportion of the one sum to the other will be as 243 to 194, i. e. (dividing both sides by 9) as 27 to 21 〈 math 〉〈 math 〉.

In the first bisection there will be seven lines:

BH 12,	BL 13	169	Sectors circumscribed about the parts of the third helix.
	BI 14	196
	BM 15	225
	BK 16	256
	BN 17	289
	BA 18	324
	Sum	1459

; while in the mean time the sum of as many equal to the greatest is 944, and so the second pro∣portion as 1944 to 1459 i. e. (dividing both sides by 72) as 27 to 20 〈 math 〉〈 math 〉.

In the second Bisection there will be thirteen lines, viz. BH 24, the rest 25, 26, &c. but the sum of the sectors, i. e. of the square numbers answering to the twelve latter will be found to be 11306; while in the mean time the sum of as ma∣ny equal to the greatest will be 15552, so that you will have the third proportion of this sum to the other, viz. as 15552 to 11306, i. e. (dividing both sides by 576) as 27 to 19 〈 math 〉〈 math 〉.

Therefore the I. proportion will be as 27 to 19+2+〈 math 〉〈 math 〉 i. e.

to 19+2+½+〈 math 〉〈 math 〉

II. — as 27 to 19+1+〈 math 〉〈 math 〉 i. e.

to 19+1+¼+〈 math 〉〈 math 〉

III. — as 27 to 19+〈 math 〉〈 math 〉—i. e.

to 19+½+⅛+〈 math 〉〈 math 〉.

Therefore the proportion of the third circle to the third spi∣ral space will be

as 27 to 19+2+½+〈 math 〉〈 math 〉

−1−¼−〈 math 〉〈 math 〉 i. e. as 27 to 19.

−½ &c. −⅛ &c. −〈 math 〉〈 math 〉 &c.

= 0. = 0. = 0. Q. E. D.

II. For the third spiral line.

If instead of the first trisection (as less commodious for the end proposed) you make use here also, as before, of bisecti∣on in the same figure, there will come out six lines from the point B to the helix, viz. Bm, 1, BE, 2, Bn, 3, BH, 4, Bo, 5, BA, 6; to which there answer as many semicircular arches in the same progression, and to the greatest of the two as many equal to 2, 4, 6; so that the sum of the unequal ones is 21, and of the equal ones 24, and so the proportion of three peripheries together to all the circumscripts together will be as 24 to 21 (and dividing both by 6) as 4 to 3 ½. In the se∣cond bisection the twelve unequal lines and arches make the sum 78, and as many equal to the greatest of the four will give the sum 96; so that the second proportion will be 96 to 78, i. e. (dividing both sides by 24) 4 to 3 ½. In the third bisection the proportion will come out as 384 to 300, i. e. (dividing both sides by 96) as 4 to 3 ½, &c. Therefore the proportion of the three circles together to the whole Helix will be as 4 to

3+½

−¼

−⅛ &c.=0. i. e. as 4 to 3 or 12 to 9.

Q. E. D.

CONSECTARY VI.

NOW, if the periphery of the first circle be made 2, the second will bee 4, the th••rd 6, and consequently the sum 12; it will be manifest that the third periphery separately will be to the whole helix as 6 to 9, i. e. as 2 to 3.

CONSECTARY VII.

AND because the second periphery (which is 4) is equal to the first and second helix together, by the above Con∣sect. 4. the remaining third spiral will be 5, and so the pro∣portion of the third periphery to it as 6 to 5.

CONSECTARY VIII.

WHerefore the proportions of each of the peripheries to their correspondent spirals will be in a progression of ordinal numbers, viz. so that the latter of every two will denote the periphery of a circle, and the former an inscribed spiral; and consequently the spiral lines will be in an arith∣metical progression of odd numbers, and the peripheries of the circles in a progression of even ones.

• 1—The first Spiral,
• 2—The first Periphery,
• 3—The second Spiral,
• 4—The second Periphery,
• 5—The third Spiral,
• 6 &c.—The third Periphery, &c.

SHCOLIƲM II.

THE seventh Consectary may also be easily deduced sepa∣rately this way: In the first bisection the line BA and its periphery is 6, the line Bo and its periphery 5, the sum of the circumscribed Peripheries 11; the sum of as many equal to the greatest 12. Therefore the periphery of the third circle will be to the two circumscripts as 12 to 11, i. e. as 6 to 5 ½. In the second bisection the four circumscribed quadrants will be 12, 11, 10, 9, their sum 42; and the sum of four equal to the greatest, i. e. the periphery of the third circle 48. Therefore the proportion is now as 48 to 42, i. e. (dividing both sides by 8) as 6 to 5 ¼. Thus you will have the third proportion as 192 to 164, i. e. (dividing both sides by 32) as 6 to 5 ⅛. Wherefore the proportion of the third periphery to the third helix or spiral is

as 6 to 5+½

−¼

−⅛ &c.

=0. i. e. as 6 to 5. Q. E. D.

CONSECTARY IX.

AS Consect. 8. supplies us with a rule to determine the pro∣portion of every spiral of every order to the periphery of the correspondent circle, viz. if the number of the order be doubled for the periphery of the circle, and the next antece∣dent odd number be taken for the spiral line; so what we have hitherto demonstrated supplys also another rule, to define the proportion of the spiral space in any order to its circle. For since the circles are in a progression of Squares 1, 4, 9, 16, &c. but the first circle is to the first space as 3 to 1 (i. e. a, 1 to ⅓) by Prop. 17. and the second to the second as 12 to 7 (i. e. as 4 to 2 ⅓) by Prop. 18. the third to the third as 27 to 19 (i. e. as 9 to 6 ⅓) by Schol. 1. of this. And contem∣plating both these series one by another,

• Of the circles, 1, 4, 9.
• Of the spaces, ⅓, 2 ⅓, 6 ⅓.

We see the numbers of the spaces are produced, if from the spuare numbers of the circles you substract their roots, and add to the remainder ⅓. Wherefore, if, e. g. we were to de∣termine the proportion of the fourth circle to the fourth spiral space; the square of 4 viz. 16 would give the circle; hence substracting the root 4, there wiill remain 12, and adding ⅓ you would have the fourth spiral space 12 ⅓; and in like man∣ner the spiral space 20 ⅓ would answer to the circle 25, &c. And that this is certain is hence evident, that if we mul∣tiply these numbers 16 and 12 ⅓, also 25 and 20 ⅓ by 3, that we may have those proportions in whole numbers, 48 and 37, 75 and 61, these are those very numbers Archimedes had hinted at in the Coroll. of Prop. 25.

[illustration]

Pag. 217.

142

143

144

145

146

CONSECTARY X.

NAY what we have now said, is that very Coroll. com∣prehending also that 25th. Proposition, viz. that a spi∣ral space of any order is to its correspondent circle, as the re∣ctangle of the semidiameters of this and the preceding circle together with a third part of the square of the difference be∣tween both semi-diameters to the square of the greatest semi∣diameter. For, if e. g. the proportion of the third spiral space to the third circle be required, since the semidiameter of this third circle is as 3, and the semidiameter of the second precedent one is 2, and so the difference 1; the rectangle of 2 into 3 i. e. 6, together with ⅓ of the square of the diffe∣rence will define the third spiral space 6 ⅓; since the third cir∣cle may be defined by the square of the semidiameter of the greater, viz. by 9, and so in the rest; as the numbers we have found shew, or further that may be found according to given Rules which may be here seen in the following Table.

Orders.	I	II	III	IV	V	VI	VII	VIII	XI	X
Circles.	1	4	9	16	25	36	49	64	81	100
The whole spaces, the preced. ones being included.	⅓	2 ⅓	6 ⅓	12 ⅓	20 ⅓	30 ⅓	42 ⅓	56 ⅓	72 ⅓	90 ⅓
Separate spaces the preced. ones be∣ing excluded.	⅓	2	4	6	8	10	12	14	16	18

CONSECTARY XI.

OUT of which table it is obvious to sight, that the se∣cond space excluding the first is sextuple of the first, as we have already deduced in Consect. 2. Prop. 18. and the third separate space double of the second, and the fourth triple of the same second, and the fifth quadruple, and so onwards.

SCHOLIƲM III.

AND this shall suffice for spirals, which comprehends not only the chief Theorems of Archimedes of spiral spaces, but also the chief of spiral lines (whereof Archimedes has left nothing.) If any should have a mind to carry on our me∣thod further, he may easily demonstrate after the same way what remains in Archimedes, and what Dr. Wallis in his A∣rithmetick of Infinites from Prop. 5. to the 38, and what o∣thers have done on this Argument.

CHAP. V. Of the Conchoid, Cissoid, Cycloid, Quadratrix, &c.

Proposition XIX.

THE first conchoid of Nicomedes Bbb (Fig. 110) on both sides of the perpendicular cDb approaches nearer always to the directrix or horizontal line AE, and yet will never coincide with it, altho' it be conceived to be produced on both sides ad infinitum.

Demonstration.

For since only Db is perpendicular to AE, and all the rest ab are so much the more inclined to it by how much the more remote they are from the middle one Db, and all in the mean while are equal both to it and to one another, by Def. 13. it is evident that the points b and B will come so much the near∣er to AE, by how much the farther they recede from the mid∣dle line Db. And yet because the lines BAC and bac are all right ones, whose points A, a, are in a right line AE it is e∣qually as impossible that the point b or B, which is always in the conchoid should ever touch this right line, as it is impos∣sible that the point C should be in it, by vertue of the afore∣cited Def. Q. E. D.

Proposition XX.

YET no other right line can be drawn between the dire∣ctrix AE and the conchoid, but what will cut it if pro∣duced.

Demonstration.

For if such a right line be made parallel to AE, as GH, and you make, as DI to IC so Db to a fourth, which will be greater than IC, as Db is greater than DI, and consequently, if making that an interval you draw the circular arch from C, it will necessarily cut the line GH e. g. in G. Drawing there∣fore CaG. you'l have as DI to IC so aG to GC, i. e. to that fourth proportional before found, by vertue of Prop. 34. lib. 1. but as DI to IC, so was also Db to the same fourth by Con∣struc. Therefore aG and Db, which have both the same pro∣portion to the same quantity, are equal; and consequently the point G is in the conchoid by virtue of Def. 13. and conse∣quently the right line GH being produced will cut that pro∣duced also, on both sides, by the same reason. Much more will it cut it on either side if it be not parallel to the directrix AE, which is very obvious. Therefore no right line can be drawn between the conchoid, &c. Q. E. D.

CONSECTARY.

HEnce, besides orher Problems, that may be very easily solved, which requires, having any rectilinear angle gi∣ven ABC (Fig 143.) and a point without it, from that point to draw a right line DEF, so that part of it EF, which is intercepted between the legs of the angle, shall be equal to a given line Z. For if you draw the perpendicular DGH from the given point D through the nearest leg of the angle BC, and make GH equal to the given line Z, and from the center C at the interval GH describe the conchoid IHK, which will be necessarily cut by the other leg of the angle by vertue of the present Prop. e. g. in F, the line DF being drawn will

give the intercepted part = GH by the nature of the conchoid, and consequently = to the given line Z.

SCHOLIƲM.

BY means of this Consectary Nicomedes solves that noble Problem of finding two mean proportionals, after this way, which we will here shew from Eutocius, but drawn into a compendium, and somewhat changed as to the order. Let two given lines AB and BC (Fig. 144.) between which you are to find two mean proportionals, be joined together at right angles, and divide both into two parts in D and E, and ha∣ving compleated the rectangle ABCL, from L thro' D draw LG to BC prolonged; so that after this way GB may become = AL or BC: Having let fall a perpendicular from E cut off from C at the interval CF=AD the part EF, and having drawn FG make CH parallel to it; and lastly thro' the legs of the angle KCH draw the right line FHK, so that the part HK shall be equal to the line CF, by the preceding Consect. and also draw the right line KM from K thro' L to the con∣tinued line BA: All which being done, CK and AM will be two mean proportionals between AB and BC; which after our way we thus demonstrate: By reason of the similitude of the ▵ ▵ MAL and LCK

MA is to LC or AB as AL or BC to CK

b−eb−c−ec and moreover,

as MA to AD so GC to CK i. e. FH to HK

b−½eb−2c−ec by reason of GF and CH be∣ing parallel, by Consect. 4. Prop. 34. lib. 1. therefore since HK is = AD=½eb, FH will be = A=b, and consequent∣ly MD=FK, viz. both b+½eb, and the square of both =bb+eb+¼eebb=□EF+EK by vertue of the Pythag. Theor. Now if to these equal quantities you add the equal □□ DX and EC=¼cc, their sum, viz. □MD+□DX i. e. □ MX will be bb+ebb+¼eebb+¼cc, equal to the sum of these, viz. □EF+□EC i. e. □ CF (by the Pythag. Theor. or EX by Construct.) +□KX; whence these two things now follow: 1. That the lines MX and KX are equal. 2. If from those equal sums you take away the common quan∣tities

¼eebb+¼cc, the remainders will be equal, viz. bb+ebb =ecc+eecc; and (since the part taken away, viz. bb is ma∣nifestly to the other part taken away, viz. ecc as the remain∣der ebb to the remainder eecc, and the whole with the parts taken away and the remainders are in the same proportion by Prop. 26. lib. 1.) separately also bb will = ecc and ebb=eecc. But from the latter equation it follows that

as eb to ec so ec to b by vertue of the 19. Prop. lib. 1. AB to CK so CK to MA

and by the same reason it follows from the former Equation as ec to b so b to c

CK to MA so MA to BC i. e. CK and MA are two mean Proportionals between AB and BC. Q. E. D.

From which deduction you have also manifest the foun∣dation of that mechanical way, which Hiero Alexandrinus makes use of in Eutocius, lib. 2. of the Sphere and Cylinder, and which Swenterus has put into his practical Geometry lib. 1. Tract. 1. Prop. 23. when, viz having joined in the form of a rectangle the given right lines AB and BC (Fig. 145.) and continued them at the other ends, he so long moves the ruler in L, having a moveable center, backwards and forwards, 'till XK and XM by help of a pair of compasses are found equal. To which, another way of Philo's is not unlike, and flows from the same fountain, wherein, having made on AC a semicircle, the moveable ruler in L is so long moved back∣warks and forwards, until LM and NK are found equal: Which seems to Eutocius to be more accommodated to practice, and easier to be perform'd by help of a ruler divided into small equal particles.

Proposition XXI.

IF from any point of the other diameter in the generating cir∣cle e. g. from G (Fig. 111. n. 1.) you draw a perpendicu∣lar GE thro' the cissoid of Diocles. the lines CG, GE, GD, and GH will be continual proportionals.

Demonstration.

For since GE and IF, as right sines, and also GD and IC as versed sines of equal arches by the Hypoth. are equal; you'l have as ID to IF (i. e. CG to GE) so IF to IC (i. e. GE to GD) per n. 3. Schol. 2. Prop. 34. lib. 1. But GD is to GH as ID to IF (i. e. as GE to GD) by the forecited Prop. 34. lib. 1. Therefore CG to GE, GE to GD, and GD to GH, are all in the same continual proportion. Q. E. D.

CONSECTARY.

HEnce it was easie for Diocles to find two mean proportio∣nals x and y between two given right lines V and Z; (Fig. 146) for he made (having first described his curve DHB) as V to Z so CL to LK, and having drawn CKH to the curve, and thro' H the perpendicular GE, he had between CG and GH two mean proportionals GE and GD by vertue of the pre∣sent Prop. when in the mean while CG the first would be to GH the last, as CL to LK, i. e. as the first V to the last Z given by vertue of the Constr. Therefore nothing remain'd but to make, 1. as CG to GE so V to X; and lastly, as GE to GD so x to y.

SCHOLIƲM.

IT may not be amiss to mention here another way of find∣ing two mean proportionals between any two given lines by the help of two Parabola's, which Menechmus formerly made use of, viz. by joining at right angles the given lines AB and BC (Fig. 147.) and prolonging them as occasion shall require thro' E and D; and then describing a Parabola about BE as its axis, so made that BC shall be its Latus Rectum, and in like manner describing another Parabola about BD as its ax, that shall have AB for its Latus Rectum, and that shall cut the for∣mer in F: Which being done, the semiordinate FE (or BD which is equal to it) being drawn to the point of Intersection F, will be the two mean proportionals sought. For by ver∣tue of the fourth Consectary of Prop. 1. of this Book, DF or

BE is a mean proportional between AB and BD, and in like manner EF or BD is a mean proportional between BE and BC, and consequently as AB to BE so BE to BD, and as BE to BD so BD to BC; Q. E. D.

To this way of Menechmus that of Des Cartes is not unlike, which he gives us p. m. 91. except only that instead of two Parabola's, he makes use only of one and a circle in room of the other: In imitation of whom Renatus Franciscus Slusius has since shewn infinite methods of doing the same thing by help of a circle, and either infinite Ellipse's or Hyperbola's, in his ingenious Treatise which he thence names his Mesolabium.

Proposition XXII.

ANY semiordinate of the Cycloid as BF (Fig. 148.) or bf is equal to its corresponding Sine in the generating circle as BD, bd, together with the arch of that sine AD or Ad.

Demonstration.

For the motion of the point A describing the semi-cycloid AFE, by Def. 11. is compounded of the motion of the orb (or wheel) B along the semi-circle ADC, and of the motion of the centre along the right line BC equal to CE, and conse∣quently to the semi circle it self, or motion of the orb, There∣fore as the point A moving to E by the motion of the orb (or wheel) moved or was carried from the diameter AC thro' the whole semi circle ADC 'till it came to AC again, and by the motion of the Centre passes thro' the whole space BG or CE, which is equal to the semicircular arch; thus the same A, when come to F, will have describ'd the quadrant AD, by moving from the diameter AC the quantity of the sine BD, and moreover by the motion of its centre (which is equal to the motion of the Orb) moves from AC the space of DF: And so the semiordinate BF will be equal to the arch AD and to its sine BD taken together; and in like manner the semiordinate bf will be equal to the arch Ad and its sine bd, &c. Q E. D.

CONSECTARY I.

HEnce may be easily assign'd by help of the cycloid a right line equal to the semi-periphery or any given arch AD or Ad; viz. CE double, or taken twice for the whole circum∣ference, and single for the semi-periphery or half circumfe∣rence, DF for the quadrant Ad, df for the arch AD, &c.

CONSECTARY II.

WHerefore the quadrature of the circle may be geome∣trically obtain'd according to Consect. 2. of Def. 15. lib. 1.

CONSECTARY III.

IF you take Be, be, double of the sines BD, bd, &c. so that all the indivisibles bd taken together, may be to all the in∣divisibles be taken together as BD to Be, a curve described thro' the points e will be an ellipsis by Prop. 11. and the curviline∣ar space ADCeA will be equal to the semicircle ACDA.

CONSECTARY IV.

AND since DF (i. e. De+eF) is equal to the quadrant DA, by vertue of the present Prop. BD+FG will be also equal to the quadrant (because the whole BG or CE is = to a semicircle) and consequently eF and FG will be equal; in like manner since df both above and below is equal to the arch dA, below bd+fg will be = to the remaining Arch dC: and above bd+ef (i. e. df) will be equal to the equal arch dA. Therefore ef above and fg below are equal, and (since the same may be shewn of all the indivisibles of the same sort throughout) the trilinear figure FGE will be equal to the tri∣linear eFA.

Proposition XXIII.

THE cycloidal space is triple of the generating circle i. e. the semi-cycloidal space AECA is triple of the semi-circle ADCA.

Demonstration.

Since the parallelogram BCEG is equal to the whole circle by Consect. 2. of Def. 15. lib. 1. i. e. to the semi-ellipse AeCA, by the present construction the Trapezium CeGE will be equal to the quadrant of the ellipse or the semi-circle. But the trili∣near space FEG is = to the trilinear space FAe, by the fourth Consect. of the preced. therefore also the trilinear space AeCEA is equal to the semi-circle. Therefore the whole cycloidal space is equal to the three semi-circles. Q. E. D.

Or thus.

Since the whole parallelogram AE is equal to two circles and the semi ellipsis AeCA to one; the remaining space AeCEC to one circle, and its half AeGC to a semi-circle. But the tri∣linear space AeF is equal to the trilinear space FGE by Consect. 4. of the preced. Therefore the one being substituted in the other's place the trilinear space AFEC will be equal to the se∣mi-circle: Therefore the remainder of the Parallelogram, i. e. the cycloidal space AFECA will be equal to three semi-circles. Q. E. D.

SCHOLIƲM.

TO these short Demonstrations, which we confess we owe for the most part to Hon. Faber, we will subjoin another somewhat more prolix, but yet not unpleasant, which we find in Carolus Renaldinus, lib. 1. de Resol. & Compos. Math. p. 299. But here we will give it the Reader more plain, and free from all Scruples, and likewise much easier. It is perform'd in these inferences, 1. That the right lined Parallelogram AbaB (Fig. 149 n. 1) is equal to the curvilinear space AbdaBDA.

2. That as that is divided into two equal parts by its right li∣ned diagonal Aa, so likewise is this by the semi-cycloid Aaa, so that the right lined triangle AaB is equal to the curvilinear space AaaBDA. 3. Therefore the one as well as the other is equal to the generating circle; and consequently, 4 If to this curvilinear space there be added the semi-circle ADBA the semi-cycloidal space AaaBA will be equal to three semi-cir∣cles. The first is evident, while if you take from the right lined parallelogram the semi-circle ADBA on the one side, and on the other add the semi-circle abda, there will arise the curvi∣linear Parallelogram. The third is evident from Consect. 2. Def. 15. lib. 1. Because the line Ba is equal to the semi-peri∣phery, which multiplyed by the diameter BC gives the area of the circle. The fourth is self evident; and so there remains only the second to be demonstrated, viz. That the curvilinear paral∣lelogram is divided into two equal parts by the cycloid, i. e. that the external trilinear Figure AadbA is eqaal to the inter∣nal one AaaBDA; which may be thus shewn: Having divi∣ded the base Ba into three equal parts, and drawn thro' them three semi-circles, and moreover the transverse right lines Dd and Ee thro' the intersections of the sem-icircles and the cy∣cloid; it is certain from the genesis of the cycloid, by vertue of the Cons. of Def. 11. that as the right line a1 is a third part of the whole aB, so the arch 1a is a third part of the ge∣nerating periphery, and by the same reason the arch 2a two thirds, and so the remaining arch a11 also ⅓; insomuch that the first arch 1a, and the last a11, and consequently their right sines af, ag, and likewise their versed ones f1, g11, are equal, and so the curvilined partial Parallelograms, both above and below, all upon equal bases, and of the same height (viz. the two linear ones ae21 and da 11. 1.) are equal among themselves, and so likewise the two pricked or pointed ones Da 2B and aeb1. Wherefore if now the base Ba (n. 2) be con∣ceived to be divided into six equal parts, and having drawn semi-circles, and transverse lines thro' their intersections with the cycloid, the arches and 〈 math 〉〈 math 〉 will be 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉 of a semi-circle,

[illustration]

Pag. 227.

147

148

149

150

and so the versed sines of each, i. e. the altitudes of the corre∣sponding parallelograms will be equal, and consequently the parallelograms of the internal and external trilinear space that are alike noted or signed will be equal to each other. Now this inscription of curvilinear parallelograms always respective∣ly equal both in number and magnitude, since it may be con∣tinued in both the trilinear figures ad infinitum; it will evi∣dently follow, that the trilinear figures themselves, whose in∣finite inscripts are always equal, will be likewise equal to one another.

Proposition XXIV.

THE base of the quadratrix AE (Fig. 150) and the se∣midiameter of the generating quadrant AD and the qua∣drant it self BD are in continual Proportion.

Demonstration.

For the quadrant DB is to the radius DA as the arch IB to the perpendicular He by Consect. 1. Def. 16 and Ib is to He as Ab to Ae by Prop. 34. lib. 1. But the arch IB (if it be con∣ceiv'd to be less and less ad infinitum) will at length coincide with Ib, as ending in the same moment in the point B, where∣in He will end in the point E, and so Ae will end in AE and Ab in AB. Therefore at length DB will be to DA as IB (i. e. Ib) to He, i. e, as Ab to Ae, i. e. as AB (or DA) to AE. Q. E. D.

SCHOLIƲM I.

CLavius about the end of the sixth Book of Euclid, and o∣thers, demonstrate this indirectly by a deduction ad Ab∣surdum, or concluding the opposite much after this manner: If DA or AB is not to AE as DB to DA, suppose it to be so to the greater Af or the less Ae. In the first case therefore, because AB is to Af as DB to DA per Hypoth. i. e. as Kf to Af the quadrant Kf and the radius AB or DA will be equal. But as BD is to IB so is Kf to Hf by reason of the similitude of the arches; and as BD to IB so also DA (or =Kf) to the sine He, by

Consect. 1. Def 16. Therefore the sine He and the arch Hf (to which the same Kf bears the same proportion) will be e∣qual; which is absurd. In the latter case, because AB would be to Ae as DB to DA by the hypoth. i. e. as Le to Ae, the quadrant Le and the radius AB or DA would be again equal. But as BD is to IB so is Le to Me by reason of the similitude of the arches; and as BD to IB so also is DA (ie=Le) to He by Consect. 1. Def. 16. Therefore the tangent He and the arch Me (to which the same Le bears the same proporti∣on) will be equal, which is again absurd. Wherefore BD is to DA, not as DA to a greater Af or a less Ae; therefore as DA to AE. Q. E. D.

CONSECTARY I.

WHerefore it is evident from what we have deduced, if by means of the base of the quadratrix AE you draw a quadrant, the side of the quadratrix DA will be equal to it, and consequently double of the semi periphery, and quadruple of the whole periphery.

CONSECTARY II.

IT is evident also that you may obtain a right line equal to the quadrant DB of any given circle, if, having described a quadratrix, you make as AE to AD so AD to a third equal to the quadrant DB: Which third proportional taken four times will be equal to the whole periphery.

CONSECTARY III.

YOU may also obtain a right line equal to any less arch, if you make, as DA to He so a third proportional found (i. e. the quadrant DB) to a fourth, by vertue of Cons. 1. Def. 16.

CONSECTARY IV.

THE quadrature of the circle therefore, by vertue of Cons. 2. Def. 15. lib. 1. as likewise the trisection of an angle, by vertue of Consect. 2. Def. 16. lib 2. may be Geometrical∣ly obtain'd, if the Quadratrix might be number'd among Geo∣metrical Curves.

SHCOLIƲM II.

CLavius was also of this Opinion in the book afore menti∣oned, who thought that if the quadratrix be excluded out of the number of geometrical curves, by the same reason you may also exclude the ellipse, parabola, and hyperbola, since they as well as this are commonly described thro' innu∣merable points. But by that great Man's leave, we may de∣ny this consequence, by the same reason as Des Cartes has de∣ny'd the converse of it in his Geom. p 18. and 19. by vertue of which he suspects the ancients took the conick sections, &c. for mechanick or non-geometrick lines, because they did the spiral, quadratrix, &c. for such. But this is the difference between the description of the quadratrix and the conick secti∣ons thro' points, that all and every of the points of the conick sections, relating to any given point of the axis, may be geo∣metrically determin'd; but all the points of the quadra∣trix promiscuously related to any point of the generating qua∣drant, cannot be geometrically determin'd, but only those which respect some certain point, from which the quadrant may be divided into two arches of known proportion. For if, e g. in the quadrant BD the point X be given at pleasure, it will be impossible by Clavius's Rule to define a point of the quadra∣trix answering to it, because the proportion of the arches DX and BX is unknown, and consequently neither can a propor∣tional section of the right line AD be made: Not to mention that the last point E (which is the primary and most necessary one to the quadrature) even by Clavius's own confession cannot be geometrically defined. We may pass the like judgment on Archimedes's spiral and such like curves, which are conceiv'd to be described by two motions independent on one another; as

will be manifest to any one who compares the genesis of the spiral with that of the quadratrix and what we have hitherto said. Whence neither will Monantholius's trisection of a gi∣ven angle (which he essays) by means of a spiral be enough geometrical; which in his Book de Puncto, Cap. 7. p. 24. he attempts to perform thus: To the centre of a described spi∣ral and its first helical or spiral line BA (Fig. 151.) he applies the angle ABC equal to the given one abc; then having drawn circles thro' F and A where the legs of the angle cut the spi∣ral, he divides the intermediate space DA into three equal parts in 1, and K: And then thro' these points he draws circles cut∣ing the helix in L and M; and lastly having drawn BLN, BMO, he easily demonstrates from the genesis of the spiral that the arches AO, ON, NC are equal. And so after the same manner not only any angle or arch, but the whole peri∣phery may be geometrically divided into as many parts as you please; only supposing that this spiral line may be numbred among geometrical ones; as we have heretofore hinted that the cycloid, conchoid, cissoid, and logarithmical curve, &c. might be; and we have above sixteen Years ago declared our opinion for it in our German Edition of Archimedes; and now are therein confirm d by those celebrated Mathematicians Leib∣nitz, Craige, &c. who number lines of this kind, altho' they cannot be expressed by our common equations, among geome∣trical ones, notwithstanding the contrary opinion of Des Car∣tes, &c. because they admit of equations of an indefinite or transcendent degree, and are capable of a Calculus as well as others, tho' it be of a nature and kind different from that commonly used. See the Acta Erud. Lips. ann. 84. p. 234. and ann. 86. p. 292. and 294.

CHAP. VI. The Conclusion, or Epilogue of the whole Work.

NOW we may at length understand what Honoratus Fa∣bri delivers concerning the distribution of figurate mag∣nitudes into certain Classes, in his Synopsis Geom. p. 57. and the following.

1. The first Classe contains elementary figures, or equal in∣divisibles, such as, 1. All Parallelograms, as the Square, Ob∣long, Rhombus and Rhomboid, the elements whereof are e∣qual right lines, as in Def. 12. lib. 1. 2. Convex or con∣cave Surfaces, the elements whereof are curve lines moved thro' right lines by a parallel motion; among which are chiefly reckon'd cylindrical surfaces, whereof see Def. 16. lib. 1. about the end. 3. Parallelepipeds, and among them the cube, whose indivisibles are squares, or other Parallelograms. 4. Prisms made by the motion of a Triangle, Trapezium, or any Polygonous Body, along a right line, all the indivisibles whereof are consequently similar and equal to the generating plane.

2. The second Classe contains Figures whose Elements de∣crease in a simple arithmetical Progression; such are, 1. Tri∣angles, as is evident from Prop. 37. lib. 1. 2. The circle, and its Sectors, as resolvible into concentrick Peripheries ac∣cording to Cons. 1. and 3. of the aforecited Prop. 3. The Cylinder as resolvible into concentrick cylindrick Surfaces, as its indivisibles. 4. The Surface of a Cone, whose elements are circular Peripheries, and also of the Pyramid whose indivi∣sibles are similar angular Peripheries every where increasing in arithmetical Progression. 5. The Parabolick Conoid, whose indivi∣sibles are Circles according to the proportion of the abscissa's in arithmetical Progression, by vertue of Prop. 14. lib. 2. &c.

3. The third Classe contains elementary Figures increasing in duplicate arithmetical Progression; such are, 1. The Py∣ramid and Cone, the first whereof may be resolv'd into an∣gular Planes, the second into circular ones increasing accord∣ing to a series of square numbers; as is evident from Prop. 38.

lib. 1. and its Consectary. 2. The trilinear parabolick Space, as defin'd Prop. 10. lib. 2. by the letters Eh HK. 3. The Sphere, as far as it may be resolved into spherical concentrick Surfaces, every one whereof may be consider'd as a base, tak∣ing the semidiameter for the altitude. 4. The Cone, as re∣solvible into parallel conical Surfaces describ'd by the parallel indivisibles of the Triangle. 5. The remainder of a Cy∣linder after an Hemisphere of the same base and altitude is tak∣en out, according to Schol. 1. of Prop. 39. lib. 1.

4. The fourth Classe would comprehend all magnitudes re∣solvible into elements or indivisibles increasing in triplicate, quadruplicate, &c. Arithmetical Progression; such we have not treated of, but may be found among Planes terminated by Curves of superior Genders; see Fabri's Synopsis, p. m. 67.

5. The fifth Classe is of those Magnitudes, whose indivisi∣bles decrease, proceeding from a square number by odd num∣bers, as, 36, 35, 32, 27, 20, 11, &c. such are, first, an Hemisphere, as is evident from Prop. 39. lib. 1. 2 An Hemi∣spheroid, as in Prop. 15 lib 2. 3. A Semi-parabola, as may be gather'd from the demonstration of Prop. ••0 lib. 2. For since the indivisibles of the circumscrib'd trilinear figure eb are found in a duplicate arithmetical Progression, ¼, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, the indivisibles of the semi-parabola will necessarily be 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, &c.

6. We may make a sixth Classe of those Magnitudes whose indivisibles decrease in a like Progression, not of the numbers themselves descending by odd steps from a given square, but of their roots, which are for the most part surd ones; such as is first, the Semi-circle, as is evident from Prop. 43. lib. 1. and by vertue of Prop. 5. lib. 2. and also the semi-ellipse, &c.

7. The seventh Classe comprehends those Magnitudes, whose Elements are in a Progression of a double series of num∣bers, as in the Parabolick Conoid, as may be seen in the Scho∣lium of Prop. 16. lib. 2.

But, to omit the other Classes of Magnitudes of a superiour Gender, the consider••tion whereof these Elements either have not touch'd on, or only by the by; (which any one who pleases may see in Faber s Synopsis, especially those which he compre∣hends under the sixth and seventh Classes, p. 70 and the fol∣lowing) about those we have here particularly noted, there re∣main

only two things to be taken notice of. 1. That since in the first Classe we place Parallelograms and Cylinders, in the second Triangles, in the third Pyramids and Cones, in the fifth Hemispheres, in the sixth semi-circles, &c. We may with Hon. Faber call the first Classe, that of Cylindrical or Parallelogrammatick Figures; the second, the Classe of Tri∣angular Figures; the third, of Pyramidals; the fifth, of He∣mispherical Figures; the sixth of semicircular ones, &c. 2. That having ranged or reduced after this manner homoge∣neous Figures, or those of like condition, to a few Classes, their dimension, and consequently almost the whole business of measuring may be very compendiously reduc'd to a few Rules; whereof we will here give the Reader a short Speci∣men, in the following

CONSECTARYS.

I. THE dimension of Parallelogrammatick Figures, i. e. of those of the first Classe, may be had, by multiply∣ing the whole base by the whole altitude: See lib. 1. Def. 12. Cons. 7. Def. 18. Cons. 6. Def. 16. Cons 3. and 4.

II. The dimension of Triangular Figures, i. e. of those of the second Classe, may be had by the multiplication of the whole Base by half the Altitude, or of half the Base by the whole altitude; [see lib. 1. Def. 12. Consect. 8. Def. 15. Con∣sect. 2. Def. 18. Consect. 4. lib. 2. Prop. 14.] and their Pro∣portion is to their respective circumscribing Parallelograms, as 1 to 2; [see besides the Prop. already cited, lib. 1. Prop. 37. and its first Consect.]

III. The dimension of Pyramidals, i. e. of magnitudes of the third Classis, may be obtain'd by the multiplication of the Base by the third part of the of Altitude; [see lib. 1. Def. 17. Consect. 3. and 4. and Def. 20. Consect. 1. &c.] and their Proportion to the corresponding Figures of the Classe of the same Base and Altitude is as 1 to 3. [see besides the Prop. al∣ready cited Prop. 38. lib. 1. and its Cons. Prop. 39. and Schol. 1. lib. 2. Prop. 10. &c.

IV. The Proportion of Hemispherical Magnitudes, i. e. of the fifth Classis to corresponding ones of the first Classe of the same Base and Altitude is as 2 to 3; [see lib. 1. Prop. 39. lib. 2. Prop. 10. and 15.] and so their dimension may be had by multiplying theit Base by ⅔ of their Altitude.

V. The Proportion of semi-circular Magnitudes, i. e. of those of the sixth Classe to so many corresponding ones of the first Classe of the same Base and Altitude cannot be expressed by whole numbers or by a small fraction [see lib. 1. Prop. 43. and lib. 2. Prop. 11.] and consequently their exact numeral dimension cannot be had.

Notes

• (α) 1.1

  I. Pro∣perty of the Pa∣rab. Apoll. Prop. 11. Lib. 1.
• (α) 1.2

  I. Pro∣perty of the Hy∣perb. Apoll. Prop. 12. Lib. 1.
• (α) 1.3

  I. Pro∣pert. of the El∣lipse of Apoll. Prop. 13. Lib. 1.
• (α) 1.4

  2. Property of the ••arab. 20. Prop. Apoll. Lib. 1. Con.
• (α) 1.5

  Proper∣ty of the Hy∣perbola and Ellips. Apol. 21. Lib. I.
• (α) 1.6

  De Witt Elem. Curv. Lib. 1. Cap. 3. Prop. 13.
• (α) 1.7

  3. Pro∣perty of the Pa∣rab.
• (α) 1.8

  The 3. Property of the Hyperb. and El∣lips. Apollon. lib. I. Prop. 21. pars prior.
• (α) 1.9

  Apollon. Lib. 3. Prop. 51. and 52.
• (α) 1.10

  Apoll. Prop. 1. lib. 2.
• (α) 1.11

  Apollon. Prop. 12. lib. 2.
• (β) 1.12

  De Witt Elem. Curv. lib. I. cap. 2. prop. 3.
• (α) 1.13

  Archim. de Quadratur. Parab. Prop. 17. and 24.
• (α) 1.14

  Archim. Prop. 19. with the Coroll.
• (α) 1.15

  Archim. lib. de Conoid. &c. Prop. 5.
• (α) 1.16

  Archimed. Prop 23. and 24. (al. 26. and 27.)
• (α) 1.17

  Archim. 29. and 30. (al. 32. and 33.)
• (α) 1.18

  Archi••. Prop. 27. and 28. (al. 30. and 31.)
• (α) 1.19

  Archim. Prop. 24. de Spi∣ral.
• (α) 1.20

  Archim. Prop. 25.