Superphysics

# Lemma 24

## Lemma 24

If three right lines, two whereof are parallel, and given by position, touch any conic section ; I say, that the semi-diameter of the section which is parallel to those two is a mean proportional between the segments of those two that are intercepted between the points of contact and the. third tangent. GB Let AF, be the two parallels touch ADB in A and B ing the conic section the third right line touching the conic section in I, and meeting the two former ; EF CD be the tangents in F and G, and let semi-diameter of the figure parallel to those tangents I say. that AF, are continually proportional. For if the conjugate diameters ; CD, BG DM G Q 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 CA - FG meet the tangent in AB, E ; ; ; CL, orEAto AL; and by composition, EC + CA EB EA to EA + AL or EL, as EC to and therefore (because of the similitude of the triangles EAF, ELI, ECH, EBG) AF is to LI as CH to BG. Likewise, from tli? nature of the conic sections, LI (or CK) is to CD as CD to CH and or ; ; aquo pertnrhatfy AF is to CD as CD to BG. Q.E.D. Hence if two tangents FG, PQ, meet two parallel tangents AF, therefore (ex COR. BG 1. F in tnrbot, ) P and Q,, will be to BQ as FO OG. and G, therefore as to and cut one the other in O; AF (ex cequo per- to BG, and by division, as FP to GQ, and AP Whence also the two right lines PG, FQ, drawn through the and points Q, will meet in the right line ACB passing through the centre of the figure and the points of contact A, B. COR. 2. P and G, F

## Lemma 25

If four sides of a parallelogram indefinitely produced touch any conic and are cut by a fifth tangent ; I say, that, taking those seg ments of any two conterminous sides that terminate in opposite angles section,

of the parallelogram, either segment is to the side [BooK from which it 1. is cut off as that part of the other conterminous side which is intercepted between the point of contact and the third side is to Uie other segment, Let the four ML, sides IK, KL, MI, touch the ML JK of the parallelogram conic section in A, B, C, I) fifth tangent E Q, H, and ; and let the FQ cut those sides in F, and taking the segments : KQ ME, of the sides Ml, KJ, or the of the sides KL, segments KH, is to MI as to ML, 1 s/.y, that and KQ; MF ME KH to KL For, by Cor. BQ to HL is to as (BK or) to MF. ME is to El as (AM or) BK to ME to MI as BK to KQ. Q.E.D. Also AM to AF and by division, KH to KL as AM of the preceding 1 and, by composition, ; KH BK AM as Lemma, is ; MF.

### Corollary 1.

Hence a parallelogram if IKLM described about a given conic KH KQ section X ME, as also the rectangle given, the rectangle will be For, by reason of the similar triangles equal thereto, given. are those MFE, equal. rectangles is COR. MI

Q? 2. And if a sixth tangent eq is X ME KQH drawn meeting the tangents Kl. KQ X ME will be equal to the rectangle Me as Kq to ME, and by division ns and e, the rectangle will be to and X Me, to Ee. in q KQ eQ, are joined and bisected, and a right line of bisection, this right line will pass through points to Me, the is to Ee as For since the centre of the conic section. COR. is 3. Hence, also, if

drawn through the

KQ MK same right line will pass through the middle of all the lines Eq, eQ, is the (by Lem. XXIII), and the middle point of the right line MK centre of the section.

## PROPOSITION 27 PROBLEM 19

To describe a Supposing GCD, FDE, trajectory that ABG EA to BCF, ; be the tangents given by position. and N, AF, BE, Bisect in the diagonals of the quadri M lateral tained them ; XXV) figure ABFE con- under any four of and (by Cor. 3, Lem. MN the right line draAvn through the points (,f may touch jive right lines given by position.SEC. OF NATURAL PHILOSOPHY. V.] 147 bisection will pass through the centre of the trajectory. P and Q, the diagonals (if I may so call them) Bl), Again, bisect in GF teral figure line of the quadrila contained under any other four tangents, and the right through the points of bisection will pass through the cen EC OF PQ, drawn tre of the trajectory ; and therefore the centre will be given in the con course of the bisecting lines. Suppose it to be O. that the centre at BG draw such distance gent Parallel to any tan be placed in the KL O may KL will touch the trajectory to be de middle between the parallels; this Let this cut any other two tangents GCD, FJ)E, in L and K. scribed. Through the points G and K, F and L, where the tangents not parallel, CL, FK K and the right line meet the parallel tangents CF, KL, draw GK, FL meeting in OR drawn and produced, will cut the parallel tan in the This appears from Gor. 2, Lem. points of contact. gents GF, KL, 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. ; SCPIOLTUM. Under the preceding Propositions are comprehended those Problems wherein either the centres or asymptotes of the trajectories are given. For when points and tangents and the centre are given, as many other points and as many other tangents are given at an equal distance on the other And an asymptote is to be considered as a tangent, ami remote extremity (if we may say so) is a point of contact. Conceive the point of contact of any tangent removed in infinitum, and the tangent will degenerate into an asymptote, and the constructions of side of the centre. its infinitely the preceding Problems will be changed into the constructions of those Problems wherein the asymptote is given. After the trajectory is described, we may and foci in this manmr. In the construction and figure of Lem. XXI, let those legs BP, CP, of the moveable angles PEN, PCN, by the concourse of which the trajec- find its axes , ^
tory was described, be made parallel one to the other and retaining that position, let : them revolve about their poles figure. In the mean while I , C, in that let the other legs of those angles, by their concourse or k, describe the circle BKGC. Let be the centre of this circle; and from this centre upon the ruler MN, wherein those legs CN, did GN, BN, K O BN OH concur while the trajectory was described, let fall the perpendicular and L. And when those other legs CK, meet meeting the circle in K in the point K that is BK nearest to the ruler, the parallel to the greater axis, first and perpendicular on the legs CP, lesser ; BP will be and the con-148 THE MATHEMATICAL PRINCIPLES [Book I trary will hajpen if those legs meet in the remotest point L. Whence ii the centre of the trajectory is and those be- given, the axes will be given ing given, the foci will be readily found. ; But the squares of the axes are one to the KH other as to LH, and thence describe a trajectory given in f mr given points. For it easy to is kind through two of the given if points are made the poles C, 13, the third will but give the moveable angles PCK, those being given, the circle may be PBK BGKC ; Then, because the trajectory is OH to OK, and described. given in kind, the ratio of and therefore OH itself, will be given. About the centre O, with the interval OH, describe another circle, and the right line that touches this circle, and passes through the concourse of the legs meet in the fourth given point, will CK, BK, when the first legs ; CP BP by means of which the trajectory may be described Whence also on the other hand a trapezium given in kind (excepting a few cases that are impossible) may be inscribed in a given conic section. There are also other Lemmas, by the help of which trajectories given in kind may be described through given points, and touching given lines. be the ruler Of MN, such a sort is this, given by position, that that if a right line is drawn through any point may cut a given conic section in two points, and the distance of the intersections to ich bisected, the point of bisection will is ano her conic section of the same kind with the former, axes parallel to the axes of the former. its But I arid havin^ hasten to things of greater use. Lemma To place 1ht tude, in, lit rev angles of a triangle, given both in kind and Three may the triangle the line AB, angle EF, many rigid lines given by position, -provided th] among themselves, in such manner tfia t j ic spiral touch the several lines. indefinite right lines given by position, and its F DEF its that AB, AC, BC, are is required so to place its angle 1) may touch the line AC, and it angle the line BC. E Upon DE, DF, and describe three segments of circles DRE, DGF. EMF, capable of Rubles BAG, ABC, ACB scribed t magni respect of as are not all parallel angles XXVI. angles equal to the But those segments are to be de respectively. that the letters wards such sides of the lines DE, ; ; DF EF3 EC. OF NATURAL PHILOSOPHY. V.I DRED 1411 letters I1ACB turn round about in the same order with the letters may : DGFD in the same order with the letters ABCA and the EMFE in the same order with the letters ACBA then, completing the letters ; ; th se segmerts into entire circles let the two former circles cut one the to be their centres. other in G, and suppose P and Then joining GP, Q PQ, take Ga AB to as GP is to PQ interval Ga, describe a circle that Join aD cutting the second circle third circle EMF in cut the DFG in Complete the figure c. a&cDEF to the figure and about the centre G, with the ; may I say, : the thing is b, first circle as well as ABCdef DGE aE in a. cutting the similar and equal done. For drawing Fc meeting D in n, and joining aG bG, QG, QD. PD, by ; construction the angle EaD is equal to the angle CAB, and the angle acF equal to the angle ACB; and therefore the equiangular to the triangle triangle aiic Wherefore the angle anc or FnD equal to the angle ABC, and conse- ABC. is &lt; uently to the angle n and there F/&gt;D ; on the point b, Moreover the angle GPQ, which is half fore the point GPD the angle to the angle falls at the centre, GaD is equal at the circumference
GQD and the angle GQP, which is half the angle at the centre, is equal to the complement to two right angles of the angle GbD at the circum ference, and therefore equal to the angle Gba. Upon which account the GPQ, triangles Gab, are similar, and Ga is to ab as Ga to AB. Wherefore ab and (by construction), as consequently the triangles abc, similar, are also equal. GP to PQ. that is AB are equal; and ABC, which we have now ; proved to be And therefore since the angles I), E, F, of the do triangle respectively touch the sides ab, ar, be of the triangle afjc the figure AECdef may be completed similar and equal to the figure afrcDEFj and by completing it the Problem will be solved. Q.E.F. DEF / COR. Hence a right line may be drawn whose parts given in length may be intercepted between three right lines given by position. Suppose the the of the arid access its to side triangle DEF, by EF, point by having D the sides DF placed part DE is to DE, whose given given by position; and AB its i&gt;t directum to be changed into a right line AB be interposed between the right lines AC to is be between the given part interposed ; DF BC, given by position; then, by applying the preceding right lines construction to this case, the Problem will be solved. ;THE MATHEMATICAL PRINCIPLES [BOOK 1. PROPOSITION XXVIII. PROBLEM XX. To describe a trajectory giren both in kind and magnitude, given parts of which shall be interposed between three right lines given by position. Suppose a trajectory is to be described that may -and be similar and equal to the curve line DEF, may be cut by three right lines AB, AC, BC, given by similar and curve position, into parts equal to the given DE and EF, parts of this line. the right lines DE, EF, DF= and the angles D, E, F, of this triangle DEF, so place as to touch those right lines given by position (by Lem. XXVI). Then about the triangle describe Draw the trajectory, similar and equal to the curve DEF.