Principia Mathematica (Motte Translation, 1848) — part 10 of 45

LEMMA XIX. Tb find a point P from ivhich if four right lines PQ,, PR, PS, PT are drawn to as many other right lines AB, CD, AC, BD, giveyi by posi- tion, each to each, at given angles, the rectangle PQ, X PR, under any tivo of the lines draiun, shall he to the rectangle PS X PT, under the other two, in a given ratio. Suppose the lines AB, CD, to which the two right lines PQ, PR, containing one of. the rect- angles, are drawn to meet two other lines, given by position, in the points A, B, C, D. From one of those, as A, draw any right line AH, in which sl you would find the point P. Let this cut the opposite lines BD, CD, in H and I ; and, because all the angles of the figure are given, the ratio of PQ to PA, and PA to PS, and therefore of PQ to PS, will be also given. Subducting tins ratio from the given ratio of PQ X PR to PS X PT, the ratio of PR to PT will be given; and ad- ding the given ratios of PI to PR, and PT to PH, the ratio of PI to PH. and therefore the point P will be given. Q.E.I.

Cor. 1. Hence also a tangent may be drawn to any point D of the locus of all the points P. For the chord PD, where the points P and D meet, that is, where AH is drawn through the point D, becomes a tangent. In which case the ultimate ratio of the evanescent lines IP and PH will be found as above. Therefore draw CF parallel to AD, meeting BD in F, and cut it in E in the same ultimate ratio, then DE will be the tan- gent ; because CF and the evanescent IH are parallel, and similarly cut in E and P.

Cor. 2. Hence also the locus of all the points P may be determined. Through any of the points A, B, C, D, as A, draw AE touching the locus, and throuo-h any other point B parallel to the tangent, draw BF meeting the locus in F ; and find the point F by this Lemma. Bisect BF in G, and, drawing the indefinite line AG, this will be the position of the dia- meter to which BG and FG are ordinates. Let this AG meet the locus

Sec. v.]

OF NATURAL PHILOSOPHY.

135

in H^ and AH will be its diameter or latus trans- versum. to which the latus rectum will be as BG^ to AG X GH. If AG nowhere meets the locuS; the line AH being infinite, the locus will be a par- abola ] and its latus rectum corresponding to the

diameter AG will be— r-^-. But if it does meet it.; AG ;

anywhere, the locus will be an hyperbola, when the points A and H are placed on the same side the point G ; and an ellipsis, if the point G falls between the points A and H ; unless, perhaps, the angle AGB is a right angle, and at the same time BG^ equal to the rectangle AGH, in which case the locus will be a circle.

And so we have given in this Corollary a solution of that famous Prob- lem of the ancients concerning four lines, begun by Euclid, and carried on by Apollonius ; and this not an analytical calculus, but a geometrical com- position, such as the ancients required.

LEMMA XX. Tf the ttvo opposite angular points A and P of any parallelogram ASPQ. toiich any conic sectio7i in the points A and P ; and the sides AQ., AS of one of those angles, indefinitely produced, meet the sams conic section in B a7id C ; and from> the points of coficoirrse B and C to any fifth point D of the conic section, two right lines BD, CD are drawn 'meet- ing the two other sides PS, PQ, of the parallelogram, indefinitely pro- duced in T and R ; the parts PR and PT, cut off from, the sides, ivill always he one to the other in a given ratio. And vice versa, if those parts cut off are one to the other in a given ratio, the locus of the point D loill he a conic section passiiig through the four poi?its A, B, G, P. Case 1. Join BP, CP, and from the point q D draw the two right lines DG, DE, of which the first DG shall be parallel to AB, and meet PB, PQ, CA in H, I, G ; and the other DE shall be parallel to AC, and meet PC^ PS, AB, in F, K, E ; and (by Lem. XYH) the rectangle DE X DF will be to the rect- angle DG X DH in a given ratio. But PCI is to DE (or Id) as PB to HB, and con- secjuently as PT to DH ; and by permutation PQ. is to PT as DE to DH. Likewise PR- is to DF as RC to DC, and therefore as (IG or) PS to DG ; and by permutation PR is to PS as DF to DG ; and, by com- pounding those ratios, the rectangle PQ. X PR will be to the rectangle PS X PT as the rectangle DE X DF is to the rectangle DG X DH, and consequently in a given ratio. But PQ. and PS are given, and there- fore the ratio of PR to PT is given. Q.E.D.

136

THE AtATHEMATICAL PRINCIPLES

[Book I.

Case 2. But if PR and PT are supposed to be in a given ratio one to the other, then by going back again, by a like reasoning, it will follow that the rectangle DE X DF is to the rectangle DG X DH in a given ratio ; and so the point D (by Lein. XYIII) will lie in a conic section pass- ing through the points A, B, C, P, as its locus. Q..E.U.

Cor. 1. Hence if we draw BC cutting PQ, in r and in PT take P^ to Pr in the same ratio which PT has to PR; then B^^ will touch the conic section in the point B. For suppose the point D to coalesce with the point B, so that the chord BD vanishing, BT shall become a tangent, and CD and BT will coincide with CB and B^.

Cor. 2. And, vice versa, if B^ is a tangent, and the lines BD, CD meet in any point D of a conic section, PR will be to PT as Pr to Vt. And, on the contrary, if PR is to PT as Pr to P^, then BD and CD will meet in some point D of a conic section.

Cor. 3. One conic section cannot cut another conic section in more than four points. For, if it is possible, let two conic sections pass through the five points A, B, C, P, O ; and let the right line BD cut them in the points D, d, and the right line Cd cut the right line PQ. in q. Therefore PR is to PT as P^ to PT : whence PR and P^' are equal one to the other, against the supposition.

LEMMA XXL

If two moveable, and indefinite right lines BM, CM draxon through given 'points B, C, as poles J do by their point of concourse M describe a third right line MN give7i by position ; and other tivo indefinite right lines BD,CD are draw?i, making with the fortner two at those given points B, C, given angles, MBD, MCD : / say, that those two right lines BD, CD loill by their point of concourse D describe a conic section passing through the points B, C. Knd, vice versa, if the right lines BD, CD do by their point of concourse D describe a conic section passing through the given points B, C, A, and the angle DBM is alivays equal to the given angle ABC, as loell as the angle DCM always equal to the given angle AGE, the point M will lie in a right line given by position, as its locus. For in the right line MN let a point

N be given, and when the moveable point

M falls on the immoveable point N, let

the moveable point D fall on an immo- vable point P. Join CN, BN, CP, BP,

and from the point P draw the right lines

PT, PR meeting BD, CD in T and R, c

and making the angle BPT equal to the

given angle BNM, and the angle CPR

Sec. y.]

OF NATURAL PHILOSOPHY.

137

equal to the given angle CNM. Wherefore since (by supposition) the an- gles MBD, NBP are equal, as also the angles MOD, NOP, take away the angles NBD and NCD that are common, and there will remain the angles NBM and PBT, NCM and PCR equal; and therefore the triangles NBM, PBT are similar, as also the triangles NCM, PCR. Wherefore PT is to NM as PB to NB ; and PR to NM as PC to NC. But the points, B, C, N, P are immovable : wherefore PT and PR have a given ratio to NM, and consequently a given ratio between themselves; and therefore, (by Lemma XX) the point D wherein the moveable right lines BT and CR perpetually concur, will be placed in a conic section passing through the points B, C, P. Q.E.D.

And, vice versa, if the moveable point D lies in a conic section passing through the given points B, C, A ; and the angle DBM is always equal to the given an- gle ABC, and the angle DCM always equal to the given angle ACB, and when the point D falls successively on any two immovable points p, P, of the conic section, the moveable point M falls suc- cessively on two immovable points n, N. Through these points n, N, draw the right line wN : this line ?iN will be the perpetual locus of that moveable point M. For, if possible, let the point M be placed in any curve line. Therefore the point D will be placed in a conic section passing through the five points B, C, A, p, P, when the point M is perpetually placed in a curve line. But from what was de- monstrated before, the point D will be also placed in a conic section pass- ing through the same five points B, C, A, p, P, when the point M is per- petually placed in a right line. Wherefore the two conic sections will both pass through the same five points, against Corol. 3, Lem. XX. It is therefore absurd to suppose that the point M is placed in a curve line. Q.E.D.

PROPOSITION XXII. PROBLEM XIV.

To describe a trajectory thai shall pass through five given points. Let the five given points be A, B, C, P, D. c

From any one of them, as A, to any other g\ ^^- — --c — ,-.v .„iT

two as B, C, which may be called the poles,

draw the right lines AB, AC, and parallel to

those the lines TPS, PRQ through the fourth

point P. Then from the two poles B, C,

draw through the fifth point D two indefinite a Q

lines BDT, CRD, meeting with the last drawn lines TPS, PRQ (the

138

THE MATHEMATICAL PRINCIPLES

[Book I.

former witli the former, and the latter with the latter) in T and R. Then dra.wing the right line tr parallel to TR, cutting oiF from the right lines PT, PR; any segments Vt, Vr, proportional to PT, PR ; and if through their extremities^ t^ r, and the poles B, C, the right lines Bt, Cr are drawn, meeting in d, that point d will be placed in the trajectory required. For (by Lem. XX) that point d is placed in. a conic section passing through the four points A, B, C, P ; and the lines Rr, T^ vanishing, the point d comes to coincide with the point D. Wherefore the conic' section passes through the five points A, B, C, P, D. Q.E.D.

The same otherwise.

Of the given points join any three, as A, B, C ; and about two of them B, C, as poles, making the angles ABC, ACB of a given magnitude to revolve, apply the legs BA, CA, first to the point D, then to the point P, and mark the points M, N, in which the other legs BL, CL intersect each other in both cases, c* Draw the indefinite right line MN, and let those moveable angles revolve about their poles B, C, in such manner that the intersection, which is now supposed to be m, of the legs BL, CL, or BM, CM, may always fall in that indefinite right line MN ; and the intersection, which is now supposed to be c?, of the legs BA, CA, or BD, CD, will describe the trajectory required, PAD^B. For (by Lem. XXI) the point d will be placed in a conic section passing through the points B, C ; and when the point ni comes to coincide with the points L, M, N, the point d will (by construction) come to coin- cide with the points A, D, P. Wherefore a conic section will be described that shall pass through the five points A, B, C, P, D. Q.E.F.

Cor. 1. Hence a right line may be readily draAvn which shall be a tan- gent to the trajectory in any given point B. Let the point d come to co- incide with the point B, and the right line B<i will become the tangent required.

Cor. 2. Hence also may be found the centres, diameters, and latera recta of the trajectories, as in Cor. 2, Lem. XIX.

SCHOLIUM.

The former of these constructions will be- c come something more simple by joining W?, g and in that line, produced, if need be, taking Bj9 to BP as PR is to PT ; and through p draw the indefinite right ^inejt?e parallel to S PT, and in that line pe taking always pe equal to Pr ; and draw the right lines Be, Gr

-iT

Sec. y.]

OF NATURAL PHILOSOPHY.

1S9

to meet in d. For since Pr to Vt, PR to PT, pB to PB, jje to Vt, are all in tlie same ratiO; pe and Pr will be always equal. After this manner the points of the trajectory are most readily found, unless you would rather describe the curve mechanically, as in the second construction.

PROPOSITION XXIII. PROBLEM XY.

To describe a trajectory that shall pass through four given points ^ and

touch a right line given by position.

Case 1. Suppose that HB is the given tangent, B the point of contact, and C, D, P, the three other given points. .lo'n BC, and draw PS paral- lel to BH, and PQ parallel to BC ; complete the parallelogram BSPQ.

Draw BD

cutting

SP in T, and CD

cutting PQ, in R. Lastly, draw any

line tr parallel to TR, cutting off

from^PQ, PS, the segments P?^, P^ proportional to PR, PT respectively ;

and draw Cr, Bt their point of concourse d will (by Lem. XX) always fall

on the trajectory to be described.

The same otherwise..

Let the angle CBH of a given magnitude re- volve about the pole B, as also the rectilinear ra- dius DC, both ways produced, about the pole 0. Mark the points M, N, on which the leg BC of the angle cuts that radius when BH, the other leg thereof, meets the same radius in the points P and D. Then drawing the indefinite line MN, let that radius CP or CD and the leg BC of the angle perpetually meet in this line; and the point of concourse of the other leg BH with the radius will delineate the trajectory required.

For if in the constructions of the preceding Problem the point A comes to a coincidence with the point B, the lines CA and CB will coincide, and the line AB, in its last situation, will become the tangent BH ; and there- fore the constructions there set down will become the same with the con- structions here described. Wherefore the concourse of the leg BH with the radius will describe a conic section passing through the points C, D, P, and touching the line BH in the point B. Q..E.F.

Case 2. Suppose the four points B, C, D, P, given, being situated with- out the tangent HI. Join each two by the lines BD, CP meeting in G,

and cuttinp; the tano^ent in H and I.

Cut the tangent in A in such manner

140

THE MATHEMATICAL PRINCIPLES

[Book L

that HA may be to lA as the rectangle un- der a mean proportional between CG and GP; and a mean proportional between BH and HD is to a rectangle under a mean pro- portional between GD and GB, and a mean proportional betweeen PI and IC, and A Avill be the point of contact. For if HX, a par- allel to the right line PI, cuts the trajectory in any points X and Y, the point A (by the properties of the conic sections) will come to be so placed, that HA^ will become to AP in a ratio that is compounded out of the ratio of the rec- tangle XHY to the rectangle BHD; or of the rectangle CGP to the rec- tangle DGB ; and the ratio of the rectangle BHD to the rectangle PIC. But after the point of contact A is found, the trajectory will be described as in the first Case. Q.E.F. But the point A may be taken either between or without the points H and I, upon which account a twofold trajectory may be described.

PROPOSITION XXIV. PROBLEM XVI. To describe a trajectory that shall pass through three given points, and

touch two right lines given by position.

Suppose HI, KL to be the given tangents, and B, C, D, the given points. Through any two of those points, as B, D, draw the indefi- nite right line BD meeting the tangents in the points H, K. Then likewise through any other two of these points, as C, D, draw the indefinite right line CD meeting the tan- gents in the points I, L. Cut the lines drawn in R and S, so that HR may be to KR as the mean proportional between BH and HD is to the mean proportional between BK and KD ; and IS to LS as the mean proportional between CI and ID is to the mean proportional between CL and LD. But you may cut, at pleasure, either within or between the points K and H, I and L, or without them ; then draw RS cutting the tangents in A and P, and A and P will be the points of contact. For if A and P are supposed to be the points of contact, situated anywhere else in the tangents, and through any of the points H, I, K, L, as I, situated in either tangent HI, a right line lY is drawn parallel to the other tangent KL, and meeting the curve in X and Y, and in that right line there be taken IZ equal to a mean pro- portional between IX and lY, the rectangle XIY or IZ^, will (by the pro- perties of the conic sections) be to LP^ as the rectangle CID is to the rect- angle CLD, that is (by the construction), as SI is to SL^, and therefore

s----'

Sec. v.] of natural philosophy. 141

IZ is to LP as SI to SL. Wherefore the points S, P, Z, are in one right line. Moreover, since the tangents meet in G, the rectangle XIY or IZ^ will (by the properties of the conic sections) be to lA^ as GP^ is to GA^, and consequently IZ will be to lA as GP to GA. Wherefore the points P, Z, A, lie in one right line, and therefore the points S, P, and A are in one right line. And the same argument will prove that the points R, P, and A are in one right line. Wherefore the points of contact A and P lie in the right line RS. But after these points are found, the trajectory may be described, as in the first Case of the preceding Problem. Q.E.F.

In this Proposition, and Case 2 of the foregoing, the constructions are the same, whether the right line XY cut the trajectory in X and Y, or not ; neither do they depend upon that section. But the constructions being demonstrated where that right line does cut the trajectory, the con- structions where it does not are also known ; and therefore^ for brevity's sake, I omit any farther demonstration of them.

LEMMA XXIL To transform, figures into other figures of the same kind.

Suppose that any figure HGI is to be ^

transformed. Draw, at pleasure, two par- /

allel lines AO, BL, cutting any third line ^ I

AB, given by position, in A and B, and from /\ .Aj^ i^--yS^ any point G of the figure, draw out any / \ \ / / N^^ right line GD, parallel to OA, till it meet / \ \ / - / \

the right line AB. Then from any given / '^'V/ / '

point O in the line OA, draw to the point / K^ / I

D the rio'ht line OD, meetino; BL in d : and / . '- — ^^^; 1

from the point of concourse raise the right

line dg containing any given angle with the right line BL, and having such ratio to Od as DG has to OD ; and g will be the point in the new figure hgi, corresponding to the point G. And in like manner the several points of the first figure will give as many correspondent points of the new figure. If AYC therefore conceive the point G to be carried along by a con- tinual motion through all the points of the first figure, the point g will be likewise carried along by a continual motion through all the points of the new figure, and describe the same. For distinction's sake, let us call DG the first ordinate, dg the new ordinate, AD the first abscissa, ad the new abscissa ; O the pole, OD the abscinding radius, OA the first ordinate radius, and Oa (by which the parallelogram OABa is completed) the new ordinate radius.

I say, then, that if the point G is placed in a right line given by posi- tion, the point g will be also placed in a right line given by position. If the point G is placed in a conic section, the point g will be likewise placed

142 THE MATHEMATICAL PIIINCIPLES [BoOK 1.

in a conic section. And here I nnderstand the circle as one of the conic sections. But farther, if the point G is placed in a line of the third ana- lytical order, the point g will also be placed in a line of the third order, and so on in curve lines of higher orders. The U\o lines in which the points G, g, are placed, will be alwa.ys of the same analytical order. For as ad is to OA, so are Od to OD, dg to DG, and AB to AD ; and there-

OA X AB OA X ds'

fore AD is equal to — , , and DG equal to — j~^~' Now if the

point G is placed in a right line, and therefore, in any equation by which the relation between the abscissa AD and the ordinate GD is expressed, those indetermined lines AD and DG rise no higher than to one dimen-

• T T .V r QA X ^B . . _^ , OA X dg

sion, by writing this equation -^ m place of AD, and -> — -

in place of DG, a new equation will be produced, in which the new ab- scissa ad and new ordinate dg rise only to one dimension ; and which therefore must denote a right line. But if AD and DG (or' either of them) had risen to two dimensions in the first equation, ad and dg would likewise have risen to two dimensions in the se.cond equation. And so on in three or more dimensions. The indetermined lines, ad, dg in the second equation, and AD, DG, in the first, will always rise to the same number of dimensions ; and therefore the lines in Avhich the points G, g, are placed are of the same analytical order.

I say farther, that if any right line touches the curve line in the first figure, the same right line transferred the same way with the curve into the new figure will touch that curve line in the new figure, and vice versa. For if any two points of the curve in the first figure are supposed to ap- proach one the other till they come to coincide, the same points transferred will approach one the other till they come to coincide in the new figure ; and therefore the right lines with which those points are joined will be- come together tangents of the curves in both figures. I might have given demonstrations of these assertions in a more geometrical form ; but I study to be brief.

Wherefore if one rectilinear figure is to be transformed into another, we need only transfer the intersections of the right lines of which the first figure consists, and through the transferred intersections to draw right lines in the new figure. But if a curvilinea^r figure is to be transformed, we must transfer the points, the tangents, and other right lines, by means of which the curve line is defined. This Lemma is of use in the solution of the more difiicult Problems : for thereby we may transform the proposed figures, if they are intricate, into others that are more simple. Thus any right lines converging to a point are transformed into parallels, by taking for the first ordinate radius any right line that passes through the point of concourse of the converging lineS; and that because their point of con-

Sec. Y.] of natural philosophy. 143

course is by this means made to go off in infinitum ; and parallel lines are such as tend to a point infinitely remote. And after the problem is solved in the new figure; if by the inverse operations we transform the new into the first figure, we shall have the solution required.

This Lemma is also of use in the solution of solid problems. For as often as two conic sections occur, by the intersection of which a problem may be solved, any one of them may be transformed, if it is an hyperbola or a parabola, into an ellipsis, and then this ellipsis may be easily changed into a circle. So also a right line and a conic section, in the construc- tion of plane problems, may be transformed into a right line and a circle.

PROPOSITION XXY. PROBLEM XVIL To describe a trajectory/ that shall .pass through two given points, and

touch three right lines given by position.

Through the concourse of any two of the tangents one with the other, and the concourse of the third tangent with the right line which passes through the two given points, draw an indefinite right line ; and, taking this line for the first ordinate radius, transform the figure by the preceding Lemma into a new figure. In this figure those two tangents will become parallel to each other, and the third tangent will be parallel to the right line that passes through the two given points. Suppose hi, kl to be those two parallel tangents, ik the third tangent, and hi a right line parallel thereto, passing through those points a, b, .\ ,

through which the conic section ought to pass \ ^

in this new figure ; and completing the paral- lelogram hikl, let the right lines hi, ik, kl be so cut in c, d, e, that he may be to the square root of the rectangle ahb, ic, to id, and ke to kd, as the sum of the right lines hi and kl is

to the sum of the three lines, the first whereof ~^~~a ; - \f

is the right line ik, and the other two are the square roots of the rectangles ahb and alb ; and c, d, e, will be the points of contact. For by the properties of the conic sections, hc^ to the rectan- gle ahb, and ic^ to id^, and ke^ to kd^, and el^ to the rectangle alb, are all in the same ratio ; and therefore he to the square root of ahb, ic to id, ke to kd, and el to the square root of alb, are in the subduplicate of that ratio ; and by composition, in the given ratio of the sum of all the ante- cedents hi + kl, to the sum of all the consequents ^ahb + ik -- ^/alb. Wherefore from that given ratio we have the points of contact c, d, e, in the new figure. By the inverted operations of the last Lemma, let those points be transferred into the first figure, and the trajectory will be there described by Prob. XIV. Q.E.F. But according as the points a, b, fall between the points h, I, or without them, the points c, d, e, must be taken

144 THE MATHEMATICAL PRINCIPLES BoOK I.]

/

dther between the points, h, i, k, I, or withont them. If one of the points a, b, falls between the points h, i, and the other without the points h, /, the Problem is impossible.

PROPOSITION XXVL PROBLEM XVIII.

To describe a trajectory that shall pass through a given pointy and touch

four right lines given by position.

From the common intersections, of any two of the tangents to the common intersection of the other two, draw an indefinite right line ; and taking this line for the first ordinate radius, transform the figure (by Lem. XXII) into a new figure, and the two pairs of tangents, each of / "^^

which before concurred in the first ordinate ra-

dius, will now become parallel. I^et hi and kl, hi l\

ik and hi, be those pairs of parallels completing the parallelagram kikl. And let p be the point in this new figure corresponding to the given point in the first figure. Through O the centre of the figure draw pq : and Oq being equal to Op, q will be the other point through which the conic sec- tion must pass in this new figure. Let this point be transferred, by the inverse operation of Lem. XXII into the first figure, and there we shall have the two points through which the trajectory is to be described. But through those points that trajectory may be described by Prop. XVIL

LEMMA XXIII.

If tiDO right lines, as AC, BD given by positio7i, a?id terminating in given points A, B, are in a give?! ratio one to the other, and the right line CD, by which the indetermined points C, D are joijied is cut in K in a given ratio ; I say, that the point K will be placed in a right line given by position.

For let the right lines AC, BD meet in c/

E, and in BE take BG to AE as BD is to Xi\

AC, and let FD be always equal to the given y/^ I \

line EG- ; and, by construction, EC will be y/^ ,-''''f'V"'' \

to GD, that is, to EF, as AC to BD, and _a X^'"" ''"'"'/ ' \ therefore in a given ratio ; and therefore the J>-''' ,,■-'''' I \

triangle EFC will be given in kind. Let E n g b i' n

CF be cut in L so as CL may be to CF in the ratio of CK to CD ; and because that is a given ratio, the triangle EFL will be given in kind, and therefore the point L will be placed in the right line EL given by position. Join LK, and the triangles CLK, CFD will be similar : and because FD is a given line, and LK is to FD in a given ratio, LK will be also given.

Sec. v.] of natural philosophy. 145

To this let EH be taken equal, and ELKH will be always a parallelogram. And therefore the point K is always placed in the side HK (given by po- sition) of that parallelogram. Q.E.D.

Cor. Because the figure EFLG is given in kind, the three right lines EF, EL, and EC, that is, GD, HK, and EC, will have given ratios to each other.

LEMMA XXIY.

If three right lines, two whereof are parallel, and given by position, touch any conic section ; I say, that the sefrd-diameter of the section which is parallel to those tioo is a mean proportional betiveen the segments of those two that are intercepted betiveen the points of contact and t/ie third tangent.

Let AFj GB be the two parallels touch- ^E

ing the conic section ADB in A and B ; /

EF the third riffht line touchins; the conic v — ■ -^ /^i-

section in I, and meeting the two former \ j/^ I \

tangents in F and G, and let CD be the \ /f j \

semi-diameter of the figure parallel to /^T^ ifc /^

those tangents ; I say, that AF, CD, BG / ° A / /

are continually proportional. / ^ / ^.X

For if the conjugate diameters AB, DM ^ Q b

meet the tangent FG in E and H, and cut one the other in C, and the parallelogram IKCL be completed ; from the nature of the conic sections, EC will be to CA as CA to CL ; and so by division, EC — CA to C A — CL, or EA to AL ; and by composition, EA to E A -f AL or EL, as EC to EC + CA or EB ; and therefore (because of the similitude of the triangles EAF, ELI, ECH, EBG) AF is to LI as CH to BG. Likewise, from the nature of the conic sections, LI (or CK) is to CD as CD to CH ; and therefore {ex cequo perturbate) AF is to CD as CD to BG. Q.E.D.

Cor. 1. Hence if two tangents FG, PQ, meet two parallel tangents AF, BG in F and G, P and Q., and cut one the other in O ; AF {ex cequo per- tij,rbatp) will be to BQ, as AP to BG, and by division, as FP to GQ,, and therefore as FO to OG.

Cor. 2. Whence also the two right lines PG, FQ, drawn through the points P and G, F and Q., will meet in the right line ACB passing through the centre of the figure and the points of contact A, B.

LEMMA XXV.

If four sides of a parallelogrwm indefinitely produced touch any conic section, and are cut by a fifth tangent ; I say, that, taking those seg- ments of any tioo conterminous sides that terminate iii opposite angles^

LO

146

THE MATHEMATICAL PRINCIPLES

[Book I.

of the parallelogram^ either segment is to the side from ivhich it is cut off -as that part of the other conterminous side lohich is intercepted hetiveen the point of contact and the third side is to the other segment» Let the four sides ML, IK, KL, MI, ^ ^

of the parallelogram MLIK touch the ~ '

conic section in A, B, O, D ; and let the

fifth tangent FQ. cut those sides in F,

Q,, H, and E ; and taking the segments

ME, KQi of the sides MI, KI, or the

segments KH, MF of the sides KL,

ML, I say, that ME is to MI as BK to

KQ,; and KH to KL as AM to MF.

For, by Cor. 1 of the preceding Lemma, ME is to EI as (AM or) BK to

BQ; and, by composition, ME is to MI as BK to KQ. Q.E.D. Also

KH is to HL as (BK or) AM to AF ; and by division, KH to KL as AM

to MF. Q.E.D,

Cor. 1. Hence if a parallelogram IKLM described about a given conic

section is given, the rectangle KQ, X ME, as also the rectangle KH X MP

equal thereto, will be given. For, by reason of the similar triangles KQ.H,

MFE, those rectangles are equal.

Cor. 2. And if a sixth tangent eq is drawn meeting the tangents KI,

MI in q and e, the rectangle KQ, X ME will be equal to the rectangle

K§' X Me, and KQ will be to Me as K^' to ME, and by division as

Q^' to Ee.

Cor. 3, Hence, also, if E^', eQ, are joined and bisected, and a right line

is drawn through the points of bisection, this right line will pass through

the centre of the conic section. For since Q^' is to Ee as KQ to Me, the

same right line will pass through the middle of all the lines 'Etq, eQ, MK

(by Lem. XXIII), and the middle point of the right line MK is the

centre of the section.

PROPOSITION XXVIL PROBLEM XIX.

To describe a trajectory that Qnay touch five right lines given by position.^

Supposing ABG, BCF, GCD, FDE, EA to be the tangents given by position. Bisect in M and N, AF, BE, the diagonals of the quadri- lateral figure ABFE con- tained under any four of them ; and (by Cor. 3, Lem. XXT) the right line MN drawn through the points of

Sec. Y.] of natural philosophy. 14;^

bisection Tvill pass through the centre of the trajectory. Again^ bisect in P and Q, the diagonals (if I may so call them) BD, GF of the quadrila- teral figure BGDF contained under any other four tangents, and the right line PQ, drawn through the j^oints of bisection will pass through the cen- tre of the trajectory ; and therefore the centre will be given in the con- course of the bisecting lines. Suppose it to be O. Parallel to any tan- gent BO draw KL at such distance that the centre O may be placed in the middle between the parallels ; this KL will touch the trajectory to be de scribed, l^et this cut any other two tangents GCD, FDE, in L and K. Through the points C and K, F and L, where the tangents not parallel, CL, FK meet the parallel tangents CF, KL, draw CK, FL mxceting in R ; and the right line OR drawn and produced, will cut the parallel tan- gents OF, KL, in the points of contact. This appears from Cor. 2, Lem. XXIV. And by the same method the other points of contact may be found, and then the trajectory may be described by Prob. XIV, Q.E.F.

SCHOLIUM.