Lemma 20 Icon


If the two opposite touch any conic angular points A and P of any parallelogram ASPQ and the sides AQ, AS section in the points A and P

of one of those angles, indefinitely produced, meet the same conic section and from the points of concourse, B and C to any fifth conic section, two right lines BD, are drawn meet- the point of tlie two other sides PS, the ing of parallelogram, indefinitely pro and PT, cut off from the sides, will duced in and the parts in B and C ; CD D PQ T R PR ; And vice versa, if those ratio. a cut one to the other in are parts off given ratio, the locus of the point D will be a conic section passing through the four points A, B, C, F CASE 1. Join BP, CP, and from the point always D be one to the other in draw the two right the first DG CA in H, DE shall be parallel to PS, AB, in F, K, E the rectangle DG, DE, be parallel to shall meet PB, PQ, lines a given ; I, G ; of which AB, and and the other AC, and meet PC, and (by Lem. XVII) DE X DF will be to the rect DG X DH in a given ratio. But PQ is to DE (or IQ) as PB to HB, and con sequently 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 and by permutation PR is to PS as DF to DG and, by com to DG 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. angle

and consequently in fore the ratio of PR

But PQ and PS

given. given ratio. to PT is are given, and thereTHE MATHEMATICAL PRINCIPLES 136 CASE But 2. if PR PT and are supposed to be in a given ratio one to the other, then by going back again, by a like reasoning, it will follow is to the in a given that the rectangle X X rectangle DE rati) ; and ing through the points COR. Pr 1. in the section in (by Lem. XVIII) B, C, P, as A., its locus. BT shall become a tangent, and CD CB BT 2. And, vice D in some point 3. four points. if D One versa, if Bt PR is to PT CD meet will And, meet a tangent, and the lines BD, will be to as Pr to Pt. is PR of a conic section, any point on the contrary, COR. Q.E.I). ; BD in will lie in a conic section pass Hence if we draw BC cutting PQ in r and in PT take Pt to same ratio which PT has to PR then Et will touch the conic For suppose the point D to coalesce with the point the point B. vanishing, B, so that the chord and Bt. will coincide with and COR. DH DG DF D so the point as Pr to Pt, PT then BD and CD of a conic section. conic section cannot cut another conic section in For, if it is possible, let more than two conic sections pass through the h ve points A, B, C, P, O and let the right line BD cut them in the Therefore points D, d, and the right line Cd cut the right line PQ, in q. PR is to PT as Pq to PT whence PR and Pq are equal one to the other, ; : against the supposition.


CM drawn through given If two moveable and indefinite right lines BM, concourse their describe a third do as point of by poles, points B, C, and other two line ; indefinite right lines given by position right two at those given points the with are former drawn, making BD,CD those that two right lines BD, I say, B, C, given angles, MBD, CD will by their point of concourse D describe a conic section passing through the points B, C. And, vice versa, if the right lints BD, CD do by their point of concourse D describe a conic section passing M MN MCD : DBM DCM the angle is always through the given points B, C, A, and as the angle as well to the ABC, always giren angle equal will lie in a right line the point the to ACB, angle given equal M given by position, as its locus. MN N M let a point For in the right line be given, and when the moveable point falls on the immoveable point N. let D the moveable point Join vable point P. and from the point fall on an immo ON, BN, CP, BP, P draw the right lines PT, PR meeting BD, CD in T and R, and making the angle BPT c jual to the given angle BNM, and the angle CPR

CNM. Wherefore since (by supposition) the an are equal, as also the angles MOD, NCP, take away the and that are common, and there will remain the angles equal to the given angle gles MBD, NBP NBD NOD NBM and PBT, NCM angles and PCR equal; PBT NM and therefore the triangles NBM, PCR. Wherefore PT is to are similar, as also the triangles NCM, to and to as as to NM PC NC. But the points, B, C, PR PB NB N, P are immovable= wheiefore PT and PR have a given ratio to NM, and consequently a given ratio between themselves; and therefore, (by ; BT D CR wherein the moveable right lines and Lemma XX) the point perpetually concur, will be placed in a conic section passing through the Q.E.D. points B. C, P. if the moveable point a conic section passing through the given points B, C, A and the angle And, vice versa, D lies in ; DBM gle is always equal to the given an ABC, and equal to the the angle given angle DCM always ACB, and when D falls successively on any two immovable points p, P, of the conic falls suc section, the moveable point the point M cessively on two immovable points /?, N. Through these points ??, N, draw the right line nN this line nN will be the perpetual locus of that moveable point M. For, if possible, let the : M D Therefore the point be placed in any curve line. will be placed point in a conic section passing through the five points B, C, A, p, P, when the But from what was de is perpetually placed in a curve line. point M D will be also placed in a conic section pass is per ing through the same five points B, C, A, p, P, when the point Wherefore the two conic sections will both petually placed in a right line. monstrated before, the point M It is pass through the same five points, against Corol. 3, Lem. XX. is placed in a curve line. therefore absurd to suppose that the point

PROPOSITION 22 Problem 14

Describe a trajectory that shall pass through Jive given points.

Let the five given points be A, B, C, P, D. c

From any one of them, as A, to any other s v two as B, C, which may be called the poles, draw the right lines AB, AC, and parallel to TPS, PRO, through the fourth Then from the two poles B, C, those the lines point P. draw through lines the fifth point BDT, CRD, D two indefinite meeting with the last drawn lines TPS, PRQ

former with the former, and the latter with the drawing the right line tr parallel to TR, t, r, T and R. Then from the right lines PT, PR and if through latter) in cutting off Pt, Pr, proportional to and the poles B, C, the right lines PT, PR, any segments their extremities, lit, Cr are drawn, meeting in d, that point d will be placed in the trajectory required. For that point d is placed in a conic section passing through (by Lena. XX) and the lines R/ , TV vanishing, the point d the four points A, B, C, P 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 C ; the given points join any three, as A, B, and about two of them 15, C, as poles, ACB making the angles ABC, of a given apply the legs BA, magnitude CA, first to the point D, then to the point P, and mark the points M, N, in which the other to legs CL BL, revolve, intersect each other in both cases. Draw the indefinite right line those moveable angles revolve MN, and about C let their poles B, C, in such manner that the intersection, which is now supposed to be ???, of the legs BL, CL or CM, may always fall in that indefinite BM ; right line MN legs BA ^A, For (by ; 7 and the intersection, which or BD ; CD, is now supposed to be d, of the trajectory required, PADc/B. will be placed in a conic section passing- will describe the Lem. XXI) the point d m comes to coincide with through the points B, C and when the point the points L, M, N, the point d will (by construction) come to coin Wherefore a conic section will be described cide with the points A, D, P. ; that shall pass through the five points A, B. C, P, D. Q,.E.F. COR. 1. Hence a right line may be readily drawn 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, arid the right line Bt/ Avill become the tangent required. COR. 2. Hence also may be found of the trajectories, as in Cor. 2, the centres, diameters, and latera recta

Lem. XIX. SCHOLIUM. The former of these constructions will come something more simple by joining and in that BP line, as PR produced, if need be, PT and be-

aking rough p draw the indefinite right inc j0e parallel to S PT, and in that line pe taking always pe and draw the right lines Be, Cr equal to Pi Bp to , is to ; t

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

PROPOSITION 23 Problem 15

describe a trajectory that shall

pass through four given points, and touch a right line given by position. CASE Suppose that 1. given tangent, and C, BH, and to the complete is the the point of contact, the three other given Jo n BC. and draw points. lel P, 1., HB B PQ IS BD SP in T, and Lastly, BC ; BSPQ. parallelogram Draw cutting cutting PQ, in R. paral to parallel CD draw any parallel to TR, cutting off from PQ, PS, the segments Pr, Pt proportional to and draw Cr, Bt their point of concourse d will (by on the trajectory to be described. line tr The same PR, PT respectively Lem. XX) always fall ; otherwise. CBH of a given magnitude re volve about the pole B ; as also the rectilinear ra- d us 1C, both ways produced, about the pole C. 1 et tl e angle : Mark the points M, N, on which the leg BC of the angle cuts that radius when the other ; leg thereof, meets the same radius in the points P and D. Then drawing the indefinite line MN, BH let that radius CP or CD and the leg BC of the angle perpetually meet in this Ikie; and the with the point of concourse of the other leg BH radius will delineate the trajectory required. For if in the constructions of the preceding Problem the point comes to a coincidence with the point B, the lines and CB will coincide, and A CA the line AB, in its last situation, will fore the constructions there set down become the tangent BH and there become the same with the con ; will BH with Wherefore the concourse of the leg the radius will describe a conic section passing through the points C, D, in the point B.

P, and touching the line structions here described. BH CASE Suppose the four points B, C, D, P, given, being situated with- ont the tangent HI. Join each two by the lines BD, CP meeting in G, 2. and cutting the tangent in H and I. Cut the tangent in A in such mannr:

HA may be to IA as the rectangle un 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 1C, and A will For if HX, a par be the point of contact. that der a the right line PI, cuts the trajectory and Y, the point A (by the allel to in any points IT X X HA 2 will properties of the conic sections) will come to be so placed, that in a ratio that is compounded out of the ratio of the rec become to AP XHY CGP the rectangle BHD, or of the rectangle to the rec ratio of the rectangle to the rectangle PIC. tangle DGB; But after the point of contac.t is found, the trajectory will be described as tangle to BHD and the A But the point in the first Case. and or without the points may be described. H A may be taken either between upon which account a twofold trajectory I,