# The Mathematical Principles of Natural Philosophy (1729)/Book 1/Section 5

*How the orbits are to be found when neither focus is given.*

Lemma XVII

*If from any point* P *of a given conic section, to the four produced sides* AB, CD, AC, DB *of any trapezium ABDC inscribde in that section, as many right lines* PQ, PR, PS, PT *are drawn in given angles, each line to each side; the rectangle* PQ x PR *of those on the opposite sides* AB, CD, will be to the rectangle PS x PT of those on the other two opposite sides *AC, BD, in a given ratio.*

Case 1. Let us suppose first that the lines drawn to one pair of opposite sides are parallel to either of the other sides; as *PQ* and *PR* (Pl. 8. *Fig.* 4) to the side *AC*, and *PS* and *PT* to the side *AB*. And farther, that one pair of the opposite sides, as *AC* and *BD*, are parallel betwixt themselves; then the right line which bis€cts those parallel sides will be one of the diameters of the conic section. and will likewise bisect *RQ*. Let *O* be the point in which *RQ* is bisected, and *PO* will be an ordinate to that diameter. Produce *PO* to *K* so that *OK* may be equal to *PO*, and *OK* will be an ordinate on the other side of that diameter. Since therefore the points *A*, *B*, *P*, and *K* are placed in the conic section, and *PK* cuts *AB* in a given angle, the rectangle *PQK* (by prop. 17. 19. 21. & 23. book 3. of *Apolloniu'*s conics) will be to the rectangle *AQB* in a given ratio. But *QK* and *PR* are equal, as being the differences of the equal lines *OK*, *OP*, and *OQ*, *OR*; whence the rectangles *PQK* and *PQ* x *PR* are equal; and therefore the rectangle *PQ* x *PR* is to the rectangle *AQB*, that is, to the rectangle *PS* x *PT* in a given ratio. *Q. E. D.*

Case 2. Let us next suppose that the opposite sides *AC* and *BD* (Pl. 8. *Fig.*. 5.) of the trapezium, are not parallel. Draw *Bd* parallel to *AC* and meeting as well the right line *ST* in *r*, as the conic section in *d*. loin *Cd* cutting *PQ* in *r*, and draw *DM* parallel to *PQ*, cutting *Cd* in *M* and *AB* in *N*. Then (because of the similar triangles *BTt*, *DBN*) *Bt* or *PQ* is to *Tt* as *DN* to *NB*. And so *Rr* is to *AQ* or *PS* as *DM* to *AN*. Wherefore. by multiplying the antecedents by the antecedents and the consequents by the consequents, as the rectangle *PQ* x *Rr* is to the rectangle *PS* x *Tt*. so will the rectangle *NDM* be to the rectangle *ANB*, and (by case 1.) so is the rectangle *PQ* x *Pr* to the rectangle *Ps* x *Pt*, and by division, so is the rectangle *PQ* x *PT* to the rectangle *PS* x *PT*. *Q. E. D.*

Case 3. Let us suppoe lastly the four lines *PQ*, *PR*, *PS*, *PT* (Pl. 8. *Fig.* 6.) not to be parallel to the sides *AC*, *AB*, but any way inclined to them. In their place draw *Pq*, *Pr* parallel to *AC*; and *Ps*, *Pr* parallel to *AB*; and because the angles of the triangles *PQq*, *PRr*, *PSs*, *PTt* are given, the ratio's of *PQ* to *Pq*, *PR* to to *Pr*, *PS* to *Ps*, *PT* to *Pt* will be also given; and therefore the compounded ratio's *PQ* X *PR* to *Pq* x *Pr*, and *PS* x *PT* to *Ps* x Pt *are given. But from what we have* demonstrated before, the ratio of *Pq* x *Pr* to *Ps* x *Pt* is given; and therefore also the ratio of *PQ* x *PR* to *PS* x *PT*. *Q. E. D.*

Lemma XVIII.

*The same things supposed, it the rectangle* PQ *x* PR *of the lines drawn to the two opposite sdes of the trapezium is to the rectangle* PS *x* PT *of those drawn to the other two sides, in a given ratio; the point* P*, from whence those lines are drawn, will be placed in a conic section described about the trapezium.* (PL 8. Fig. 7.)

Conceive a conic sction to he described passing through the points *A*, *B*, *C*, *D*, and any one of the infinite number of points *P*, as for example *p'; I say the* point *P* will be always placed in this section. If you deny the thing, join *AP* cutting this conic section somewhere else if possible than in *P*, as in *b*. Therefore if from those points *p* and *b*, in the given angles to the sides of the trapezium, we draw the right lines *pq*, *pr*, *ps*, *pt*, and *bk*, *bn*, *bf*, *bd*, we shall have as *bk* x *bn* to *bf* x *bd*, so (by Lem. I7.) *pq* x *pr* to *ps* x *pt*; and so (hy supposition) *PQ* x *PR* to *PS* x *PT*. And because of the similar trapezia *bkAf*, *PQAS*, as *bk* to *bf*; so *PQ* to *PS*. Wherefore by dividing the terms of the preceding proportion by the correspondent terms of this, we shall have *bm* to *bd* as *PR* to *PT*. And therefore the equiangular trapezia *Dubd*, *DRPT* are similar, and consequently their diagonal *Dk*, *DP* do coincide. Wherefore *b* falls in the intersection of the right lines *AP*, *DP*, and consequently coincides with the point *P*. And therefore the point P where-ever it is taken falls to be in the assigned conic section. *Q. E. D.*

Cor. Hence if three right lines *PQ*, *PR*, *PS*, are drawn from a common point *P* to as many other right lines given in position *AB*, *CD*, *AC*, each to each, in as many angles respectively given, and the rectangle *PQ* x *PR* under any two of the lines drawn be to the square of the third in a given ratio: The point *P*, from which the right lines are drawn, will be placed in a conic section that touches the lines *AB*, *CD* in *A* and *C*; and the contrary. For the position of the three right lines *AB*, *CD*, *AC* remain the same, let the line *BD* approach to and coincide with the line *AC*; then let the line *PT* come likewise to coincide with the line *PS*; and the rectangle *PS*; *PT* will become , and the right lines *AB*, *CD*, which before did cut the curve in the points *A* and *B*, *C*, and *D*, can no longer cut, but only touch, the curve in those co-inciding points.

Scholium.

In this lemma, the name of conic section is to be understood in a large sense, comprehending as well the rectilinear section thro' the vertex of the cone, as the circular one parallel to the base. For if the point *p* happens to be in a right line, by which the points *A* and *D* or *C* and *B* are joined, the conic section will be changed into two right lines, one of which is that right line upon which the point *p* falls, and the other is a right line that joins other two of the four points. If the two opposite angles of the trapezium taken together are equal to two right angles, and if the four lines *PQ*, *PR*, *PS*, *PT* are drawn to the sides thereof at right angles, or any other equal angles, and the rectangle *PQ* x *PR* under two of the lines drawn *PQ* and *PR*, is equal to the rectangle *PS* x *PT* under the other two *PS* and *PT* the conic section will become a circle. And the same thing will happen, if the four lines are drawn in any angles, and the rectangle *PQ* x *PR* under one pair of the lines drawn, is to the rectangle *PS* x *PT* under the other pair, as the rectangle under the sines of the angles *S*, *T* in which the two last lines *PS*, *PT* are drawn, to the rectangle under the sines of the angles *Q*, *R*, in which the two first *PQ*, *PR* are drawn. In all other cases the locus of the point *P* will be one of the three figures, which pass commonly by the name of the conic sections. But in room of the trapezium *ABCD*, we may substitute a quadrilateral figure whose two opposite sides cross one another like diagonals. And one or two of the four points *A*, *B*, *C*, *D* may be supposed to be removed to an infinite distance, by which means the sides of the figure which converge to those points, will become parallel: And in this case the conic section will pass through the other points. and will go the same way as the parallels *in infinitum*.

Lemma XIX.

*To find a point* P *(Pl. 8. Fig. 8.) from which if four right line:* PQ*,* PR*,* PS*,* PT *are drawn to as many other right lines* AB*,* CD*,* AC*,* BD *given by position, each to each, at given angles, the rectangle* PQ *x* PR*, under any two of the lines drawn, shall be 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 rectangles, 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 you would find the point *P*. Let this cut the opposite lines *BD*, *CD*, in *A* 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 this ratio from the given ratio of *PQ* x *PR* to *PS* x *PT* the ratio of *PR* to *PT* will be given; and adding the given ratio's 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 thro' the point *D*, becomes a tangent. In which case the ultimate ratio of the evanescent lines *IP* and *PH* will be found asabove. 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 tangent; 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*. (Pl. 9. *Fig.* 1.) draw *AE* touching the locus, and through 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 diameter to which *BG*, and *FG* are ordinates. Let this *AG* meet the locus in *H*, and *AH* will be its diameter or latus transversum, to which the latus rectum will be as *BG* to *AG* x *GH*. If *AG* no where meets the locus, the line *AH* being infinite the locus will be a parabola; and its latus rectum corresponding to the diameter AG will be . But if it does meet it any where, the locus will be an hyperbola, when the points *A* and *H* are placed on the 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 lime time 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 problem of the ancients concerning four lines, begun by Euclid, and carried on by *Appolonius* and this not an analytical calculus, but a geometrical composition, such as the ancients required.

Lemma XX.

*If the two opposite angular points* A *and* P (Pl. 9. Fig. 2.) *of any parallelogram* ASPQ *touch any conic section in the points* A *and* P; *and the sides* AQ, AS *of one of those angler, indefinitely produced, meet the same the same conic sonic section in* B *and* C; *and from the points of coucourse* B *and* C *to any fifth point* D *of the conic section, two right lines* BD, CD *are drawn meeting the two other sides* PS, PQ *of the parallelogram, indefinitely produced, in* T *and* R; *the parts* PR *and* PT, *cut off from the sides, will always he one to the other in a given ratio. And vice versa, if those parts cut of are one to the other in a given ratio, the locus of the point* D *will be a conic section, passing through the the four points* A, B, C, P.

Case 1. Join *BP*, *CP*, and foom the point *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. 17.) the rectangle *DE* x *DF* will be to the rectangle *DG* x *DH*, in a given ratio. But *PQ* in to *DE* (or *IQ*) as *PB* to *HB*, and consquently as *PT* to *DH*; and by permutation, *PQ* is to *PT*, as *DE* to *DH*. Likewise *PR* is to *DF* as *RG* to *DC*, and therefore as (*IG* or) *PS* to *DG*; and, by permutation, *PR* is to *PS* as *DF* to *DG*; and, by compounding those ratio's, the rectangle *PQ* x *PR* will be to tie 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 therefore the ratio of *PR* to *PT* is given. *Q. E. D.*

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 lem. 18.) will lie in a conic section passing thro' the points *A*, *B*, *C*, *P*, as its locus. *Q. E. D.*

Cor 1. Hence if we draw *BC* cutting *PQ* in *r*, and in *PT* take *Pt* to *Pr* in the same ratio which *PT* has to *PR*: Then *Pr* 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 *Br*. Co tt. z. And vice verla, if B r is atangent, and the lines B D, C D meet in any point D of a conic section; P R will be to P Tas Pr to Pr. And on the contrary, if.PR is to PTasPr to Pr, (hm BD, and CD will meet in some point D of a conic section.

Cor. 2. One conic section cannot cut another conic section in more than four points. For, if it is possible, let two conic sections pass thro' 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 *Pq* to *PT*: Whence *PR* and *Pq* are equal one to the other, against the suppossition.

Lemma XXI.

*if two moveable and indefinite right lines* BM*,* CM *drawn through given paints* B*,* C*, at poles, do by their point of course* M *descrihe a third right line* MN *given by position; and other two indefinite right lines* BD*,* CD *are drawn, making with the former two at those given points* B*,* C*, given angler,* MBD*,* MCD*: I say that those two right lines* BD*,* CD *will lines their point of concourse* D *descrihe a conic section passing through the points* B*,* C*. And 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 always equal to the given angle *ABC*, as well as the angle *DCM* always equal to the given angle *ACB*: the point *M* will lie in a right line given by position, at its locus. *Pl. 9. Fig. 3.*

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 immoveable 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*, and making the angle *BPT* equal to the given angle 'BNM and the angle *CPR* equal to the given angle *CNM*, Wherefore since (by supposition) the angles *MBD*, *NBP* are equal, as also the angles *MCD*, *NCP*; 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 triangless *NCM*, *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 lem. 20.) 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* (Pl. 9. *Fig.* 4.) lies in a conic section passing through the given points *B*, *C*, *A*; and the angle *DBM* is always equal to the given angle *ABC*, and the angle *DCM* always equal to the given angle *ACB*, and when the point *D* falls successively on any two immovable point *p*, *P*, of the conic section, the moveable point *M* falls successively on two immovable points *n*, *N*: through these points *n*, *M* draw the right line *nM*, this line *nN* 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 demonstrated before, the point *D* will be also placed in a conic section, passing through the same five points *B*, *C*, *A*, *p*, *P*. when the point *M* is perpetually placed in a right line. Wherefore the two conic sections will both pass through the same given points, against corol. 3. lem. 20. 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 that shall pass through five given given points*. Pl. 9 Fig. 5.

Let the five given points be *A*, *B*, *C*, *P*, *D*. From any one of them as *A*, to any other two as *B*, *C*, which may be called the poles, draw the right lines *AB*, *AO*, 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 lines *BDE*, *CRD*, meeting with the last drawn lines *TPS*, *PRQ* (the former with the former, and the latter with the latter) in *T* and *R*. Then drawing the right line *tr* parallel to *TR*, cutting off from the right lines *PT*, *PR*, any segments *Pt*, *Pr*, 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. 20.) that point *d* is placed in a conic section passing through the four points *A*, *B*, *C*, *P*; and the lines *Rr*, *Tt* 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.*Pl. 9. Fig. 6.

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 the other in both cases. 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* and always fall in that indefinite right line *MN*; and, the intersection which is now supposed to be *d*, of the legs *BA*, *CA*, or *BD*, *CD*, will describe the trajectory required *PADdB*. For, (by lem. 21.) the point *d* will be placed in a conic section passing through the points *B*, *C*; and when the point *m* comes to coincide with the points *L*, *M*, *N*, the point *d* will (by construction) come to coincide 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 drawn which shall be a tangent to the trajectory in any given point *B*. Let the point *d* come to coincide with the point *B*, and the right line *Bd* 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. 19.

Scholium.

The former of these constructions (*Fig. 5.) will become* something more simple by joining *BP*, and in that line, produced if need be, taking *Bp* to *BP* as *PR* is to *PT*; and through *p* drawing the indefinite right line be parallel to *SPT*; and in that line *pe* taking always *pe* equal to *Pr*; and drawing the right lines *Be*, *Cr* to meet in *d*. For since *Pr* to *Pt*, *PR* to *PT*, *pB* to *PB*, *pe* to *Pt*, are all in the 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 XV.

*To describe a trajectory that shall pass through four given points, and touch a right line given by position.* Pl. 10. Fig. 1.

Case 1. Suppose that *HB* is the given tangent, *B* the point of contact, and *C*, *D*, *P*, the three other given points. Join *BC*, and drawing *PS* parallel to *BH*, and *PQ* parallel to *BC*, compleat the parallelogram *BSPQ*. Draw *BD* cutting *SP* in *T* and *CD* cutting *PQ* in *R*. Lastly, drawing any line *tr* parallel to *TR*, cutting off from *PQ*, *PS*, the segments *Pr*, *Pt* proportional to *PR*, *PT* respectively; and drawing *Cr*, *Bt*, their point of concourse will (by lem. 20.) always fall on the trajectory to be described.

*The same otherwise.*Pl. 10. Fig. 2.

Let the angle *CBH* of a given magnitude revolve about the pole *B*, as also the rectilinear radius *DC* both ways produced. about the pole *C*. 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 therefore the constructions there set down will become the same with the constructions 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*, (Pl. 10. *Fig.* 3.) given, being situated without the tangent *HI* join each two by the lines *BD*, *CP*, meeting in *G*, and cutting the tangent in *H* and *I*. Cut the tangent in *A* in such manner that *HA* may be to *IA*, as the rectangle under mean proportional between *CG* and *GP*, and a mean proportional between *BH* and *HD*, is to a rectangle under a mean proportional between *GD* and *GB*, and a mean pr0portional between *PI* and *IC*; and *A* will be the point of contact. For if *HX*, a parallel 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 will become to in a ratio that is compounded out of the ratio of the rectangle *XHY* to the rectangle *BHD*, or of the rectangle *CGP* to the rectangle *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 firft 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.* Pl. 10. Fig. 4.

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 in indefinite 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 tangents 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 any where else in the tangents, and through any of the points *H*, *I*, *K*, *L*, as *I*, situated in either tangent *HI* a right line *IT* is drawn, parallel to the other tangent *KL*, and meeting the curve in *X* and *I* and in that right line there be taken *IZ* equal to a mean proportional between *IX* and *IT*; the rectangle *XIT* or , will (by the properties of the conic sections) be to , as the rectangle *CID* is to the rectangle *CLD*, that is (by the construction) as is to , and therefore *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 will (by the properties of the conic sections) be to as is to , and consequently *IZ* will be to *IA*, 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 constructions, where it does not, are also known; and therefore, for brevity's sake, I omit any farther demonstration of them.

Lemma XXII.

*To transform figures into other figures of the same kind.* Pl. 10. Fig. 5.

Suppose that any figure figure, *HGI* is to be transformed. Draw, at pleasure, two parallel lines *AO*, *BL*, cutting any third line *AB* given by pssition, in *A* and *B*, and from any point *G* of the figure, draw out any right line *GD*, parallel to *OA*, till it meet the right line *AB*. Then from any given point *O* in the line *OA*, draw to the point *D* the right line *OD*, meeting *BL* in *d*, and 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 we therefore conceive the point *G* to be carried along by a continual 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 given *OAB* is compleated) the new ordinate radius.

I say, then, that if the point *G* is placed in a right line given by position, 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 will be likewise placed in a conic section. And here I understand the circle as one of the conic sections. But farther, if the point *G* is placed in a line of the third analytical 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 two lines in which the points *G*, *g*, are placed, will be always of the same analytical order. For as *ad* is to *OA*, so are *ad* to *OD*, *dg* to *DG*, and *AB* to *AD*; and therefore *AD* is equal to and *DG* equal to . 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 *DG* is expressed, those in determined lines *AD* and *DG* rise no higher than to one dimension, by writing this equation in place of *AD*, and in place of *DG*, a new equation will be produced, in which the new abscissa *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 second 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 which 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 approach 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 lines with which those points are joined will 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 curvilinear 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 difficult 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, beause their point of concourse is by this means made to go off *in infinitum*, and parallel lines are such 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 then this ellipsiss may be easily changed into a circle. So also a right line and a conic section, in the construction of plane problems, may be trasformed into a right line and a circle.

Proposition XXV. Problem XVII.

*To desecribe a trajectory that shall pass through two given points, and touch three right lines given by position. (Pl. 10, Fig. 6.)*

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 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 *bl* a right line parallel thereto, passing through thos points *a*, *b*, through which the conic section ought to pass in this new figure; and compleating the parallelogram *hikl*, let the right lines *hi*, *ik*, *kl* be so cut in *c*, *d*, *e*, that be 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 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 to the rectangle *ahb*, and to , and to and to the rectangle *alb*, are all in the same ratio; and therefore *hc* to the square root of *ahb*, *ic* to *id*, *kc* 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 antecedents *hu* + *kl*, to the sum of all the consequents . Wherefore from that given ratio we have the points of contact *c*, *d*, *c*, 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. 14. Q. E. F. But according as the points *a*, *b*, fall between the points *h*, *l*, or without them, the points *c*, *d*, *e*, must be taken either between the points *b*, *i*, *k*, *l*, or without them. If one of the points *h*, *l*, falls between the points *h*, *l*, and the other without the points *h*, *l*, the problem is impossible.

Proposition XXVI. Problem XVIII.

*To describe a trajectory that shall pass through a given point, and touch four right lines given by position.* Pl. 11. Fig. 1

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. 22.) into a new figure, and the two pairs of tangents each of which are concurred in the first ordinate radius will now become parallel. Let *hi* and *kl*, *ik* and *hl*, be those points of parallels compleat, the parallelogram *hikm*. 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 section mull pass in this new figure. Let this point be transferred by the inverse operation of lem. 22. 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 prob. 17. *Q. E. F.*

Lemma XXIIII.

*If two right lines as* AC*,* BD *given by position, and terminating in given points* A, B, *are in a given ratio one to the other, and the right line* CD, *by which the indetermined points* CD *are joined, is cut in* K *in a given ratio; I say that the point* K *will be placed in a right line given by position.* Pl. 11. Fig. 2.

For let the right lines *AC*, *BD* meet in *E*, and in *BE* take *BG* to *AE*, as *BD* is to *AC*, and let *FD* be always equal to the given line *EG*; and by construction, *EC* will be to *GD*, that is, to *EF*, as *AC* to *BD*, and therefore in a given ratio; and therefore the triangle *EFC* will be given in kind. Let *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. 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 position) of that parallelogram. *Q. E. D.*

Cor. Because the figure *EFLC* is given in kind the three right lines *EF*, *EL* and *EC*, that is *GD*, *HK* and *EC* will have given ratio's to each other.

Lemma XXIV.

*If three right lines, two whereof are parallel, and given by position, touch any conic section; I say, that the semidiameter 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.* Pl. 11. Fig. 3.

Let *AF*, *GB* be the two parallels touching the conic section *ADB* in *A* and *B*; *EF* the third right line touching the conic section in *I*, and meeting the two former tangents in *F* and *G*, and let *CD* he the semi-diameter of the figure parallel to those tangents; I say, that *AE*, *CD*, *BG* are continually proportional.

For if the conjugate diameters *AB*, *DM* meet the tangent *FG* in *E* and *H*, and cut one the other in *C*, and the parallelogram *IKCL* be compleated; from the nature of the conic sections, *EC* will be to *CA* as *CA* to *CL*, and so by division, *EC* - *CA* to *CA* - *CL* or *EA* to *AL*; and by composition, *EA* to *EA* + *AL* or *EL*, as *EC* to *EC* + *CA* or *EB*; and therefore (because of

*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 æquo* *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 æquo* *pertubate*) will be to *BQ* as *AP* to *BG*, and by division, as *FP* to *GQ* and therefore as *FP* 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 parallelogram indefinitely produced touch any conic section, and are cut by a fifth tangent; I say, that taking thosse segments of any two conterminous side which is intercepted between the point of contact and the third side, is to the other segment.* Pl. 11. Fig. 4.

Le the four sides *ML*, *IK*, *KL*, *MI* of the parallelogram *MLIK* touch the conic section in *A, B, C, D*; and let the fifth tangent *FQ* cut those sides in *F, Q, H* and *E*, and taking the segments *ME, KQ* 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 *EL* 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×ME*, as also the rectangle *KH×MF* equal thereto, will be given. For, by reason of the similar triangles *KQH*, *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×ME* will be equal to the rectangle *Kq×Me*, and *KQ* will be to *Me*, as *Kq* to *ME*, and by division as *Qq* to *Ee.*

Cor. 3. Hence also if *Eq*, *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 *Qq* is to *Ee*, as *KQ* to *Me*; the same right line will pass through the middle of all the lines *Eq*, *eQ*, *MK* (by lem. 22.) and the middle point of the right line

*MK*is the centre of the section.

Proposition XXVII. Problem XIX.

*To describe a trajectory that may touch five right lines given by position.* Pl. 11 . Fig. 5.

Supposing *ABG*, *BCF*, *GCD*, *FDE*, *EA* to be the tangents given by position. Bisect in *M* and *N*, *AF*, *BE* the diagonals of the quadrilateral figure *ABFE* contained under any four of them; and (by cor. 3. lem. 25) the right line *MN* drawn through the points of bisection will 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 quadrilateral figure *BGDF* contained under any other four tangents, and the right line *PQ* drawn through the points of bisection will pass through the centre of the trajectory. And therefore the centre will be given in the concourse of the bisecting lines. Suppose it to be *O*. Parallel to any tangent *BC* 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 described. Let 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* meeting in *R*; and the right line *OR* drawn and produced, will cut the parallel tangents *CF*, *KL*, in the points of contact. This appears from cor. 3. lem. 24. And by the same method the other points of contact may be found, and then the trajectory may be described by prob. 14. *Q. E. F.*

Scholium.

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 side of the centre. And an asymptote is to be considered as a tangent, and its infinitely 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 the preceding problem will be changed into the constructions of those problems wherein the asymptote is given.

After the trajectory is described, we may find its axes and foci in this manner. In the construction and figure of lem. 21. (*Pl.* 12. *Fig.* 1.) let those legs *BP*, *CP*, of the moveable angles *PBN*, *PCN*, by the concourse of which the trajectory was described, be made parallel one to the other; and retaining that position, let them revolve about their poles in that figure. In the mean while let the other legs *CN*, *BN* of those angles, by their concourse *K* or *k*, describe the circle *BKGC*. Let *O* be the centre of this circle; and from this centre upon the ruler *MN*, wherein those legs *CN*, *BN* did concur while the trajectory was described, let fall the perpendicular *OH* meeting the circle in *K* and *L*. And when those other legs *CK*, *BK* meet in the point *K* that is nearest to the ruler, the first legs *CP*, *BP* will be parallel to the greater axis

and perpendicular on the lesser; and the contrary will happen if those legs meet in the remotest point *L*. Whence if the centre of the trajectory is given, the axes will be given; and those being given, the foci will be readily found.

But the squares of the axes are one to the other as *KH* to *LH*, and thence it is easy to describe a trajectory given in kind through four given points. For if two of the given points are made the poles *C,* B*, the third will give the moveable angles* PCK*,* *PBK*; but those being given, the circle *BGKC* may be described. Then, because the trajectory is given in kind, the ratio of *OH* to *OK*, and therefore *OH* it self 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 *CK*, *BK*, when the first legs *CP*, *BP*, meet in the fourth given point, will be the ruler *MN*, 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 lemma's by the help of which trajectories given in kind may be described through given points, and touching given lines. Of such a fort is this, that if a right line is drawn through any point given by position. that may cut a given conic section in two points, and the distance of the intersections is bisected, the point of bisection will touch another conic section of the same kind with the former, and having its axes parallel to the axes of the former. But I hasten to things of greater use.

Lemma XXVI.

*To place the three angles of a triangle both in kind and magnitude, in respect of as many right line given by position, provided they are not all parallel among themselves, in such manner that the several angles may touch the several lines.* (Pl. 12. Fig. 2.)

Three indefinite right lines *AB*, *AC*, *BC*, are given by position, and it is required so to place the triangle *DEF* that its angle D may touch the line *AB*, its angle *E* the line *AC*, and its angle *F* the line *BC*. Upon *DE*, *DF* and *EF*, describe three segments of circles *DRE*, *DGF*, *EMF*, capable of angles equal to the angles *BAC*, *ABC*, *AVB* respectively. But those segments are to described towards such sides of the lines *DE*, *DF*, *EF*, that the letters *DRED* may turn round about in the same order with the letters *BACB*; and the letters *DGFD* in the same order with the letter *ABCBA*; and the letters *EMFE* in the same order with the letters *ACBA*; then completing those segments into entire circles, let the two former circles cut one the other in *G*, and suppose *P* and *Q* to be their centres. Then joining *GP*, *PQ* take *Ga* to *AB*, as *GP* is to *PQ*; and about the centre *G*, with the interval *Ga* describe a circle that may cut the first circle *DGE* in *a*. Join *aD* cutting the second circle *DFG* in *b*, as well as *aE* cutting the third circle *EMF* in *c*. Compleat the figure *ABCdef* similar and equal to the figure *absDEF*. I say the thing is done.

For drawing *Fc* meeting *aD* 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 triangle *anc* equiangular to the triangle *ABC*. Wherefore the angle *anc* or *FnD* is equal to the angle *ABC*, and consequently to the angle *FbD*; and therefore the point *n* falls on the point *b*. Moreover the angle *GPQ* which is half the angle *GPD* at the centre is equal to the angle *GaD* at the circumference; and the angle *GQP*, which is half the angle *GQD* at the centre, is equal to the complement to two right angles of the angle *GbD* at the circumference, and therefore equal to the angle *Gab*. Upon which account the triangles *GPQ*, *Gab*, are similar, and *Ga* is to *ab* as *GP* to *PQ*; that is (by construction) as *Ga* to *AB*. Wherefore *ab* and *AB* are equal; and consequently the triangles *abc*, *ABC*, which we have now proved to be similar, are also equal. And therefore since the angles *D*, *E*, *F*, of the triangle *DEF* do respectively touch the sides *ab*, *ac*, *be* of the triangle *abc*, the figure *ABCdef* may be compleated similar and equal to the figure *abcDEF*, and by compleating it the problem will be solved. *Q. E. F.*

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 triangle *DEF*, by the access of its point *D* to the side *EF*, and by having the sides *DE*, *DF* placed *in directum* to be changed into a right line whose given part *DF* is to be interposed between the right lines *AB*, *AC* given position; then by applying the preceding construction to this case, the problem will be solved.

Proposition XXVIII. Problem XX.

*To describe a trajectory given both in kind and magnitude, given parts of which shall be interposed berween three right lines given by position.* (Pl. 12. Fig. 3.)

Suppose a trajectory is to be described that may be similar and equal to the curve line *DEF*, and may be cut by three rights lines *AB*, *AC*, *BC*, given by position, into parts *DE* and *EF*, similar and equal to the given parts of this curve line.

Draw the right lines *DE*, *EF*, *DF*; and place the angles *D*, *E*, *F*, of this triangle *DEF*, so as to touch those right lines given by position (by lem. 26). Then about the triangle describe the trajectory, similar and equal to curve *DEF*. *Q. E. F.*

Lemma XXVII.

*To describe a trapezium given in kind, the angles whereof may be so placed in respect offout right lines given by position, that are neither all parallel among themselves nor converge to one common point, that the several lines.*. Pl. 13. Fig. 1.

Let the four right lines *ABC*, *AD*, *BD*, *CE*, be given by position; the first cutting the second in *A*, the third in *B*, and the fourth in *C*; and suppose a trapezium *fghi* is to be described, that may be similarl to the trapezium *FGHI*; and whose angle *L* equal to the given angle *F*, may touch the right line *ABC*; and the other angles *g*, *h*, *i*, equal to the other given angles *G*, *H*, *I*, may touch the other lines *AD*, *BD*, *CE*, respectively. Join *FH*, and upon *FG*, *FH*, *FI* describe as many segments of circles *FSG*, *FTH*, *FVI*; the first of which *FSG* may be capable of an angle equal to the angle *BAD*; the second *FTH* capable of an angle equal to the angle *CBD*; and the third *FVI* of an angle equal to the angle *ACE*. But the segments are to be described towards those sides of the lines *FG*, *FH*, *FI*, that the circular order of the letters *FSGF* may be the same as of the letters *BADB*, and that the letters *FTHF* may turn about in the same order as the letters *CDBC*, and the letter *FIVIF* in the same order as the letters *ACEA*. Complete the segments into entire circles, and let *P* be the centre of the first circle *FSG*, *Q* the centre of the second *FTH*. Join and produce both ways the line *PQ*, and in it take *QR* in the same ratio to *PQ* as *BC* has to *AB*. But *QR* is to be taken towards that side of the point *Q*, that the order of the letters *P*, *Q*, *R* may be the same, as of the letters *A*, *B*, *C*; and about the centre *R* with interval *RF* describe a fourth circle *FNc* cutting third circle *FVI* in *c*. Join *Fc* cutting the first circle in *a* and the second in *b*. Draw *aG*, *bH* *cI*, and let the figure *abcFGHI* be made similar to the figure *abcFGHI*; and the trpezium *fghi* will be taht which was requited to be described.

For let the two first circles *FSG*, *FTH* cut one the other in *K*; join *PK*, *QK*, *EK*, *aK*, *bK*, *cK*, and and produce *QP*, to *L*. The angles *FaK*, *FbK*, *FcK* at the circumferences, are the halves of the angles *LPK*, *LQK*, *LRK*, the halves of those angles. Wherefore the figure *PQRK* is equiangular and similar to the figure *abcK*, and consequently *ab* is to *bc* as *PQ* to *QR*, that is, as *AB* to *BC*. But by construction, the angles *fAg*, *fBh*, *fVi* are equal to the angles *FaG*, *FbH*, *FcI*. And therefore the figure *abcFGHI*. Which done, a trapezium *fghi* will be constructed similar to the trapezium *FGHI*, and which by its angles, *f*, *g*, *h*, *i* will touch the right lines *ABC*, *AD*, *BD*, *CE*. *Q. E. F.*

Cor. Hence a right line may be drawn whose part intercepted in a given order, between four right lines given by position, shall have a given proportion among themselves. Let the angles *FGH*, *GHI*, be so far increased that the right lines *FG*, *GH*, *HI*, may lie in direction, and by constructing the problem in this case, a right line *fghi* will be drawn, whose parts *fg*, *gh*, *hi*, intercepted between the four right lines given by position, *AB* and *AD*, *AD* and *BD*, *BD* and *CE*, will be one to another as the lines *FG*, *GH*, *HI*, and will observe the same order among themselves. But the same thing may be more readily done in this manner.

Produce *AB* to *K* (Pl. 13. Fig. 2.) and *BD* to *L*, so as *BK* may be to *B*, as *HI* to *GH*; and *DL* to *BD* as *GI* to *FG*; and join *KL* meeting the right line *CE* in *i*. Produce *iL* to *M*, so as *LM* may be to *iL* as *GH* to *HI*; then draw *MQ* parallel to *LB* and meeting the right line *AD* in *g*, and join *gi* cutting *AB*, *BD* in *f*, *h*. I say the thing is done.

For let *Mg* cut the right line *AB* in *Q* and *AD* the right line *KL* in *S*, and draw, *AP* parallel to *BD*, and meeting *iL* in *P*, and *gM* to *Lb* (*gi* to *bi*, *Mi* to *Li*, *GI* to *HI*, *AK* to *BK*) and, *AP* to *BL* will be in the same ratio. Cut *DL* in *R*, so as *DL* to *RL* may be in that same ratio; and because *gS* to *gM*, *AS* to *AP*, and *DS* to *DL* are proportional; therefore (*ex* *æquo*) as *gS* to *Lb*, so will *AS* be to *BL*, and *DS* to *RL*; and mixtly *BL* - *RL* to *Lh* - *BL*, as *AS* - *DS* to *gS* - *AS*. That is, *BR* is to *Bh*, as *AD* is to *Ag*, and therefore as *BD* to *gQ*. And alternately *BR* is to *BD*, as *Bh* to *gQ*, or as *fh* to *fg*. But by construction the line *BL* was cut in *D* and *R*, in the same ratio as the line *FI* in *G* and *H*; and therefore *BR* it to *BD* as *FH* to *FG*. Wherefore *fh* is to *fg* as *FH* to *FG*. Since therefore, *gi* to *hi* likewise is as *Mi* to *Li*, that is, as *GI* to *HI*, it is manifest that the lines *FL*, *fi*, are similary cut in *G* and *H*, *g* and *h'.* Q. E. F.

In the construction of this corollary, after the line *LK* is drawn cutting *CE* in *i*, we may produce *iE* to *V*, so as *EV* may be to *Ei* as *FH* to *HI*, and then draw *Vf* parallel to *BD*. It will come to the same, if about the centre *i*, with an interval *IH*, we describe a circle cutting *BD* in *X*, and produce *iX* to *Y* so as *iY* may be equal to *IF*, and then draw *Yf* parallel to *BD*.

Sir *Christopher Wren*, and Dr. *Wallis* have long ago given other solutions of this problem.

Proposition XXIX. Problem XXI.

*To describe a trajectory given in kind, that may be cut by four right lines given by position, into parts given in order, kind and proportion.*

Suppose a trajectory is to be described that may be similar to the curve line *FGHI* (Pl. 13 Fig. 4.) and whose parts. similar and proportional to the parts *FG*, *GH*, *HI* of the other, may be intercepted between the right lines *AB* and *AD*, *AD* and *BD*, *BD* and *CE* given by position, *viz*. the first between the first pair of those lines, the second between the second, and the third between the third. Draw the right lines *FG*, *GH*, *HI*, *FI*; and (by lem. 27.) describe a trapezium *fghi* that may be similar to the trapezium *FGHI*, and whose angles *f*, *g*, *h*, *i*, may touch the right lines given by position, *AB*, *AD*, *BD*, *CE*, severally according to their order. And then about this trapezium describe a trajectory, that trajectory will be similar to the curve line *FGHI*.

Scholium.

This problem may be likewise constructed in the following manner. Joining *FG*, *GH*, *HI*, *FI*, (Pl. 13. Fig. 4.), produce *GF* to *V*, and join *FH*, *IG*, and make the angles *CAK*, *DAL* equal to the angles *FGH*, *VFH*. Let *AK*, *AL* meet the right line *BD* in *K* and *L*, and thence draw *KM*, *LM* of which let *KM* make the angle *AKM* equal to the angle *GHI*, and be it self to *AKM* as *HI* is to *GH*; and let *LN* make the angle *ALN* equal to the angle *FHI*, and be it self to *AL*, as *HI* to *FH* But *AK*, *KM*, *AL*, *LN* are to be drawn towards those sides of the lines *AD*, *AK*, *AL*, that the letters *CAKMC*, *ALKA*, *DALND* may be carried round in the same order as the letters *FGHIF*; and draw *MN* meeting the right line *CE* in *i*. Make the angle *iEP* equal to the angle *IGF* and let *PE* be to *Ei*, as *FG* to *GI*; and through *P* draw *PQf* that may with the right line *ADE* contain an angle *PQE* equal to the angle *FIG*, and may meet the right line *AB* in *f*, and join *fi*. But *PE* and *PQ* are to be drawn towards those sides of the lines *CE*, *PE*, that the circular order of the letters *PEiP* and *PEQP* may be the same, asof the letters *FGHIF*, and if upon the line *fi*, in the same order of letters, and similar to the trapezium *FGHI*, a trapezium *fghi* is constructed, and a trajectory given in kind is circumscribed about it, the problem will be solved.

So far concerning the finding of the orbits. It remains that we determine the motions of bodies in the orbits so found.