The Mathematical Principles of Natural Philosophy

by Isaac Newton

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

*If from any point* P *of a given conic section, to the
four produced sides* AB, CD, AC, DB*, of any trapezium*
ABDC *inscribed 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 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 bisects 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. XVII., XIX., XXI. and XXIII., Book III., of Apollonius'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 A B, 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 of the trapezium are not parallel. Draw B*d*
parallel to AC, and meeting as well the right line ST in *t*,
as the conic section in *d*. Join C*d* cutting PQ in *r*,
and draw DM parallel to PQ, cutting C*d* in M, and AB in N.
Then (because of the similar triangles BT*t*, DBN), B*t*
or PQ is to T*t* as DN to NB. And so R*r* 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 R*r* is to the rectangle PS x T*t*, so will the
rectangle NDM be to the rectangle ANB; and (by Case 1) so is the
rectangle PQ x P*r* to the rectangle PS x P*t*; and by
division, so is the rectangle PQ x PR to the rectangle PS x PT.
Q.E.D.

Case 3. Let us suppose, lastly, the four
lines PQ, PR, PS, PT, not to be parallel to the sides AC, AB, but any
way inclined to them. In their place draw P*q*, P*r*,
parallel to AC; and P*s*, P*t* parallel to AB; and
because the angles of the triangles PQ*q*, PR*r*, PS*s*,
PT*t* are given, the ratios of PQ to P*q*, PR to P*r*,
PS to P*s*, PT to P*t* will be also given; and therefore
the compounded ratios PQ x PR to P*q* x P*r*, and PS x
PT to P*s* x P*t* are given. But from what we have
demonstrated before, the ratio of P*q* x P*r* to P*s*
x P*t* is given; and therefore also the ratio of PQ x PR to PS
x PT. Q.E.D.

*The same things supposed, if the rectangle* PQ x PR *of
the lines drawn to the two opposite sides 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.*

Conceive a conic section to be 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.
XVII) *pq x pr* to *ps x pt*; and so (by supposition)
PQ x PR to PS x PT. And because of the similar trapezia *bk*A*f*,
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 *bn* to *bd* as PR to PT.
And therefore the equiangular trapezia D*nbd*, DRPT, are
similar, and consequently their diagonals D*b*, 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, wherever 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 PS 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 remaining 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 x PT will become PS², 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 coinciding points.

In this Lemma, the name of conic section is to be understood in a
large sense, comprehending as well the rectilinear section through 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 the 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
first two 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*.

*To find a point* P *from which if four right lines*
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 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 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 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 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, 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 BG² to AG x GH. If AG nowhere 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
BG^{2}

AG. But if it does meet it 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 BG² 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 Apollonius; and this not an analytical calculus, but a geometrical composition, such as the ancients required.

*If the two opposite angular points* A *and* P *of
any parallelogram* ASPQ *touch any conic section in the
points* A *and* P*; and the sides* AQ, AS *of
one of those angles, indefinitely produced, meet the same conic
section in* B *and* C*; and from the points of
concourse* 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 be 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 *will be a conic section passing through the
four points* A, B, C, P*.*

Case 1. Join BP, CP, and from 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. XVII) the rectangle DE x DF will be to the rectangle DG x DH in a given ratio. But PQ is to DE (or IQ) as PB to HB, and consequently 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 compounding 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 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. XVIII) will lie in a conic section passing through 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 P*t* to P*r* in the same ratio which PT
has to PR; then B*t* 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*t*.

Cor. 2. And, vice versa, if B*t* is a
tangent, and the lines BD, CD meet in any point D of a conic section,
PR will be to PT as P*r* to P*t*. And, on the contrary,
if PR is to PT as P*r* to P*t*, 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 C*d* cut the right line PQ in *q*. Therefore PR is
to PT as P*q* to PT: whence PR and P*q* are equal one to
the other, against the supposition.

*If two moveable and indefinite right lines* BM, CM *drawn
through given points* B, C*, as poles, do by their point of
concourse* M *describe 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 angles,* MBD, MCD*: I say, that those two right lines*
BD, CD *will by their point of concourse* D *describe 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,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 immovable 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
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 angle 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 successively on two immovable points *n*,
N. Through these points *n*, N, draw the right line *n*N:
this line *n*N 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*, when the point M is
perpetually 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.

*To describe a trajectory that shall pass through five given points.*

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, 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 lines BDT, 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 P*t*, P*r*, proportional to
PT, PR; and if through their extremities, *t, r*, and the
poles B, C, the right lines B*t*, C*r* 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 R*r*, T*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.

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. 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 *d*,
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 *m*,
of the legs BA, CA, or BD, CD, will describe the trajectory required,
PAD*d*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
*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 B*d* 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.

The former of these constructions will become something more simple
by joining BP, and in that line, produced, if need be, taking B*p*
to BP as PR is to PT; and through *p* draw the indefinite
right line *pe* parallel to SPT, and in that line *pe*
taking always *pe* equal to P*r*, and draw the right
lines B*e*, C*r* to meet in *d*.
For since P*r* to P*t*, PR to PT, *p*B to PB, *pe*
to Pt, are all in the same ratio, *pe* and P*r* 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.

*
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.
Join BC, and draw PS parallel 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*r*, P*t* proportional to PR, PT
respectively; and draw C*r*, B*t* their point of
concourse *d* will (by Lem. XX) always fall on the trajectory
to be described.

Let the angle CBH of a given magnitude revolve about the pole B; as also the rectilinear radius BC, 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, given, being situated with out 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 a 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 proportional 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 HA² will become to AI² 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 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.

*
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 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 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 IY 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 proportional between IX and IY, the rectangle XIY or IZ², will (by the properties of the conic sections) be to LP² as the rectangle CID is to the rectangle CLD, that is (by the construction), as SI is to SL², 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 IZ² will (by the properties of the conic sections) be to IA² as GP² is to GA², 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.

*To transform figures into other figures of the same kind.*

Suppose that any figure HGI is to be transformed. Draw, at pleasure,
two parallel lines AO, BL, cutting any third line AB, given by
position, 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 O*d* 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
distraction'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 O*a* (by which the parallelogram OAB*a* is
completed) 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 *g* 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 O*d*
to OD, *dg* to DG, and AB to AD; and therefore AD is equal to
OA x AB

ad, and DG equal to
OA x dg

ad. 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 dimension, by
writing this equation OA x AB

ad in place of AD, and
OA x dg

ad 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
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
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 because their point of
concourse 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 construction of plane problems, may be transformed into a right line and a circle

*
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 *hl*
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 parallelogram *hikl*, let the right lines *hi,
ik, kl* be so cut in *c, d, e*, that *hc* 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, *hc*² to the rectangle *ahb*, and *ic²*
to *id²*, and *ke²* to *kd²*, and *el²*
to the rectangle *alb*, are all in the same ratio; and
therefore *hc* 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 antecedents *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, l,* or without them, the points *c, d, e,*
must be taken either between the points, *h,
i, k, l,* or without them. If one of the points *a, b*,
falls between the points *h, i*, and the other without the
points *h, l*, the Problem is impossible.

*
To describe a trajectory that shall pass through a given point,
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 radius,
will now become parallel. Let *hi* and *kl*, *ik*
and *hl*, be those pairs of parallels completing the
parallelogram *hikl*. 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 O*q*
being equal to O*p*, *q* will be the other point
through which the conic section 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. XVII.

*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* C, D *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.*

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 ratios to each other.

*
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.
*

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 be the semi-diameter of the figure parallel to those tangents; I say, that AF, 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 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 − 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 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 aequo
perturbatè*) 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 aequo perturbatè*) 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.

*
If four sides of a parallelogram indefinitely produced touch
any conic section, and are cut by a fifth tangent; I say, that,
taking those segments of any two conterminous sides that terminate
in opposite angles of the parallelogram, either segment is to the
side from which it 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 the other segment.
*

Let the four sides ML, IK, KL, MI, of the parallelogram MLIK touch the F 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 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 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 x ME will be equal to the rectangle K*q* x M*e*,
and KQ will be to M*e* as K*q* to ME, and by division as
Q*q* to E*e*.

Cor. 3. Hence, also, if E*q*, *e*Q,
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*q* is to E*e* as KQ to M*e*,
the same right line will pass through the middle of all the lines E*q*,
*e*Q, MK (by Lem. XXIII), and the middle point of the right
line MK is the centre of the section.

*To describe a trajectory that may 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 quadrilateral figure ABFE contained under any four of them; and (by Cor. 3, Lem. XXV) 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 con course 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, GL, FK meet the parallel tangents OF, KL, draw OK, 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. 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.

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 Problems 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. XXI, 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 B, C,
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 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 CK, BK, when the first legs CK, 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 Lemmas, by the help of which trajectories given in kind may be described through given points, and touching given lines. Of such a sort 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.

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

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, ACB respectively. But
those segments are to be 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; the letters DGFD in the
same order with the letters ABCA; 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 G*a*
to AB as GP is to PQ; and about the centre G, with the interval G*a*,
describe a circle that may cut the first circle DGE in *a*.
Join *a*D cutting the second circle DFG in *b*, as
well as *a*E cutting the third circle EMF in *c*.
Complete the figure ABC*def* similar and equal to the figure *abc*DEF:
I say, the thing is done.

For drawing F*c* meeting *a*D in *n*, and
joining *a*G, *b*G, QG, QD, PD, by construction the
angle E*a*D is equal to the angle CAB, and the angle *ac*F
equal to the angle ACB; and therefore the triangle *anc*
equiangular to the triangle ABC. Wherefore the angle *anc* or
F*n*D is equal to the angle ABC, and consequently to the angle
F*b*D; 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 G*a*D 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 G*b*D at the
circumference, and therefore equal to the angle G*ba*. Upon
which account the triangles GPQ, G*ab*, are similar, and G*a*
is to *ab* as GP to PQ; that is (by construction), as G*a*
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, bc* of
the triangle *abc*, the figure ABC*def* may be
completed similar and equal to the figure *abc*DEF, and by
completing 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 DE is
to be interposed between the right lines AB, AC given by position; and
its given part DF is to be interposed between the right lines AB, BC,
given by position; then, by applying the preceding construction to
this case; the Problem will be solved.

*
To describe a trajectory given 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 be similar and equal to the curve line DEF, and may be cut by three right 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, EP, DF: and place the angles D, E, F, of this triangle DEF, so as to touch those right lines given by position (by Lem. XXVI). Then about the triangle describe the trajectory, similar and equal to the curve DEF. Q.E.F.

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

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
similar to the trapezium FGHI, and whose angle *f*, 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 CBDC
and the letters FVIF in the game 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
the interval RF describe a fourth circle FN*c* cutting the
third circle FVI in *c*. Join F*c* cutting the first
circle in *a*, and the second in *b*. Draw *a*G,
*b*H, *c*I, and let the figure ABC*fghi* be made
similar to the figure *abc*FGHI; and the trapezium *fghi*
will be that which was required to be described.

For let the two first circles FSG, FTH cut one the other in K; join
PK, QK, RK, *a*K, *b*K, *c*K, and produce QP
to L. The angles F*a*K, F*b*K, F*c*K at the
circumferences are the halves of the angles FPK, FQK, FRK, at the
centres, and therefore equal to LPK, LQK, LRK, the halves of those
angles. Wherefore the figure PQRK is equiangular and similar to the
figure *abc*K, and consequently *ab* is to *bc*
as PQ to QR, that is, as AB to BC. But by construction, the angles *f*A*g*,
*f*B*h*, *f*C*i*, are equal to the angles
F*a*G, F*b*H, F*c*I. And therefore the figure ABC*fghi*
may be completed similar to the figure *abc*FGHI. 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
parts 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 directum*; 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 them
selves. But the same thing may be more readily done in this manner.

Produce AB to K and BD to L, so as BK may be to AB as HI to GH; and
DL to BD as GI to FG; and join KL meeting the right line CE in *i*.
Produce *i*L to M, so as LM may be to *i*L 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 M*g* cut the right line AB in Q, and AD the right line
KL in S, and draw AP parallel to BD, and
meeting *i*L in P, and *g*M to L*h* (*gi*
to *hi*, M*i* to L*i*, 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 *g*S to *g*M, AS to
AP, and DS to DL are proportional; therefore (*ex aequo*) as *g*S
to L*h*, so will AS be to BL, and DS to RL; and mixtly, BL − RL
to L*h* − BL, as AS − DS to *g*S − AS. That is, BR is
to B*h* as AD is to A*g*, and therefore as BD to *g*Q.
And alternately BR is to BD as B*h* to *g*Q, 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 is to BD as
FH to FG. Wherefore *fh* is to *fg* as FH to FG.
Since, therefore, *gi* to *hi* likewise is as M*i*
to L*i*, that is, as GI to HI, it is manifest that the lines
FI, *fi*, are similarly 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 *i*E to V, so as EV
may be to E*i* as FH to HI, and then draw V*f* 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 *i*X
to Y so as *i*Y may be equal to IF, and then draw Y*f*
parallel to BD.

Sir Christopher Wren and Dr. Wallis have long ago given other solutions of this Problem.

*
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, 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. XXVII) 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.

This problem may be likewise constructed in the following manner.
Joining FG, GH, HI, FI, 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, LN, of which let KM make the angle AKM equal to the
angle GHI, and be itself to AK as HI is to GH; and let LN make the
angle ALN equal to the angle FHI, and be itself 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 *i*EP
equal to the angle IGF, and let PE be to E*i* as FG to GI; and
through P draw PQ*f* 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 PE*i*P and PEQP may be the same as of 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.