difference between the latus rectum and 4DS.
COR. 3. Hence also if a body move in any conic section, and is forced
out of its orbit by any impulse, you may discover the orbit in which it will
afterwards pursue its Bourse. For bv compounding the proper motion oi
SEC. IV.] OF NATURAL PHILOSOPHY. 125
the body with that motion, which the impulse alone would generate, you
will have the motion with which the body will go off from a given place
of impulse in the direction of a right line given in position.
COR. 4. And if that body is continually disturbed by the action of some
foreign force, we may nearly know its course, by collecting the changes
which that force introduces in some points, and estimating the continual
changes it will undergo in the intermediate places, from the analogy that
appears in the progress of the series.
SCHOLIUM.
If a body P, by means of a centripetal
force tending to any given point R, move
in the perimeter of any given conic sec
tion whose centre is C ; and the law of
the centripetal force is required : draw
CG parallel to the radius RP, and meet
ing the tangent PG of the orbit in G ;
and the force required (by Cor. 1, and
CG3
Schol. Prop. X., and Cor. 3, Prop. VII.) will be as -
SECTION IV.
Of the finding of elliptic, parabolic, and hyperbolic orbits, from ttu.
focus given.
LEMMA XV.
Iffrom the two foci S, II, of any ellipsis or hyberbola, we draw to any
third point V the right lines SV, HV, whereof one HV is equal to the
principal axis of the figure, thai is, to the axis in which the foci are
situated, the other, SV, is bisected in T by t/ie perpendicular TR let
fall upon it ; that perpendicular TR will somewhere touch the conic
section : and, vice versa, if it does touch it, HV will be equal to the
principal axis of the figure.
For, let the perpendicular TR cut the right line
HV, produced, if need be, in R ; and join SR. Be
cause TS, TV are equal, therefore the right lines SR,
VR, as well as the angles TRS, TRV, will be also
equal. Whence the point R will be in the conic section, and the perpen
dicular TR will touch the same ; and the contrary. Q.E.D.
126 THE MATHEMATICAL PBINCIP, -ES [BOOK 1
PROPOSITION XVIII. PROBLEM X.
From a focus and the principal axes given, to describe elliptic and hy
perbolic trajectories, which shall pass through given points, and touch
right lines given by position.
Let S be the common focus of the figures ; AB A 33
the length of the principal axis of any trajectory ; r p T~*
P a point through which the trajectory should \ /R
pass ; and TR a right line which it should touch. / \
About the centre P, with the interval AB SP, \ S
~~yf
if the orbit is an ellipsis, or AB {- SP, if the
y>
G ^
orbit is an hyperbola, describe the circle HG. On the tangent TR let fall
the perpendicular ST, and produce the same to V, so that TV may be
equal to ST; and about V as a centre with the interval AB describe the
circle FH. In this manner, whether two points P, p, are given, or two
tangents TR, tr, or a point P and a tangent TR, we are to describe two
circles. Let H be their common intersection, and from the foci S, H, with
the given axis describe the trajectory : I say, the thing is done. For (be
cause PH -f- SP in the ellipsis, and PH SP in the hyperbola, is equal
to the axis) the described trajectory will pass through the point P, and (by
the preceding Lemma) will touch the right line TR. And by the same
argument it will either pass through the two points P, p, or touch the two
right lines TR, tr. Q.E.F.
PROPOSITION XIX. PROBLEM XI.
About a given focus, to describe a parabolic trajectory, which shall pass
through given points, and touch right lines given by position.
Let S be the focus, P a point, and TR a tangent of
the trajectory to be described. About P as a centre,
with the interval PS, describe the circle FG. From
the focus let fall ST perpendicular on the tangent, and
produce the same to V, so as TV may be equal to ST.
After the same manner another circle fg is to be de
scribed, if another point p is given ;
or another point v
is to be found, if another tangent tr is given; then draw
the right line IF, which shall touch the two circles YG,fg, if two points
P, p are given ; or pass through the two points V, v, if two tangents TR,
tr, are given : or touch the circle FG, and pass through the point V, if the
point P and the tangent TR are given. On FI let fall the perpendicular
SI, and bisect the same in K ; and with the axis SK and principal vertex K
describe a parabola : I say the thing is done. For this parabola (because
SK is equal to IK, and SP to FP) will pass through the point P ; and
/KS
SEC. IV.] OF NATURAL PHILOSOPHY. 127
(by Cor. 3, Lem. XIV) because ST is equal to TV. and STR a light an
gle, it will touch the right line TR. Q.E.F.
PROPOSITION XX. PROBLEM XII.
About a given focus to describe any trajectory given in specie which shah
pass through given points, and touch right lines given by position.
CASE 1. About the focus S it is reuired
to describe a trajectory ABC, pass
ing through two points B, C. Because the
trajectory is given in specie, the ratio of the
principal axis to the distance of the foci GAS H
will be given. In that ratio take KB to BS, and LC to CS. About the
centres B, C, with the intervals BK, CL, describe two circles ; and on the
right line KL, that touches the same in K and L, let fall the perpendicu
lar SG ; which cut in A and a, so that GA may be to AS, and Ga to aS,
as KB to BS ; and with the axis A., and vertices A, a, describe a trajectory :
I say the thing is done. For let H be the other focus of the described
figure, and seeing GA is to AS as Ga to aS, then by division we shall
have Ga GA, or Aa to S AS, or SH in the same ratio; and therefore
in the ratio which the principal axis of the figure to be described has to
the distance of its foci
; and therefore the described figure is of the same
species with the figure which was to be described. And since KB to BS,
and LC to CS, are in the same ratio, this figure will pass through thtpoints
B, C, as is manifest from the conic sections.
CASE 2. About the focus S it is required to
describe a trajectory which shall somewhere
touch two right lines TR, tr. From the focus
on those tangents let fall the perpendiculars
ST, St, which produce to V, v, so that TV, tv
may be equal to TS, tS. Bisect Vv in O, and j
erect the indefinite perpendicular OH, and cut I.
the right line VS infinitely produced in K and V
k, so that VK be to KS, and VA* to A~S, as the principal axis of the tra
jectory to be described is to the distance of its foci. On the diameter
K/J describe a circle cutting OH in H ; and with the foci S, H, and
principal axis equal to VH, describe a trajectory : I say, the thing is done.
For bisecting Kk in X, and joining HX, HS, HV, Hv, because VK is to
KS as VA- to A*S
; and by composition, as VK -f- V/c to KS + kS ; and
by division, as VA* VK to kS KS, that is, as 2VX to 2KX, and
2KX to 2SX, and therefore as VX to HX and HX to SX, the triangles
VXH, HXS will be similar
; therefore VH will be to SH as VX to XH ;
and therefore as VK to KS. Wherefore VH, the principal axis of the
described trajectory, has the same ratio to SH, the distance of the foci, as
12S THE MATHEMATICAL PRINCIPLES [BOOK 1.
K S
the principal axis of the trajectory which was to be described has to the
distance of its foci
; and is therefore of the same species. Arid seeing VH,
vH are equal to the principal axis, and VS, vS are perpendicularly bisected
by the right lines TR, tr, it is evident (by Lem. XV) that those right
lines touch the described trajectory. Q,.E.F.
CASE. 3. About the focus S it is required to describe a trajectory, which
shall touch a right line TR in a given Point R. On the right line TR
Jet fall the perpendicular ST, which produce to V, so that TV may be
equal to ST ; join VR, and cut the right line VS indefinitely produced
in K and k, so. that VK may be to SK, and V& to SAr, as the principal
axis of the ellipsis to be described to the distance of its foci ; and on the
diameter KA: describing a circle, cut the H
right line VR produced in H ; then with
the foci S, H, and principal axis equal to R
VH, describe a trajectory : I say, the thing .---
is done. For VH is to SH as VK to SK, V" "1
and therefore as the principal axis of the trajectory which was to be de
scribed to the distance of its foci (as appears from what we have demon
strated in Case 2) ; and therefore the described trajectory is of the same
species with that which was to be described ; but that the right line TR,
by which the angle VRS is bisected, touches the trajectory in the point R,
is certain from the properties of the conic sections. Q.E.F.
CASE 4. About the focus S it is r
required to describe a trajectory
APB that shall touch a right line
TR, and pass through any given
point P without the tangent, and
shall be similar to the figure apb,
described with the principal axis ab,
and foci s, h. On the tangent TR
let fall the perpendicular ST, which / .. ,.---"
"
produce to V, so that TV may be
equal to ST ; and making the an
gles hsq, shq, equal to the angles VSP, SVP, about q as a centre, and
with an interval which shall be to ab as SP to VS, describe a circle cut
ting the figure apb in p : join sp, and draw
SH such that it may be to sh as SP is to sp,
and may make the angle PSH equal to the
angle psh, and the angle VSH equal to the
angle pyq. Then with the foci S, H, and B
principal axis AB, equal to the distance VH,
describe a conic section : I say, the thing is
done ; for if sv is drawn so that it shall be to
SEC. IV.] OF NATURAL PHILOSOPHY. 129
sp as sh is to sq, and shall make the angle vsp equal to the angle hsq, and
the angle vsh equal to the angle psq, the triangles svh, spq, will be similar,
and therefore vh will be to pq as sh is to sq ; that is (because of the simi
lar triangles VSP, hsq), as VS is to SP? or as ab to pq. Wherefore
vh and ab are equal. But, because of the similar triangles VSH, vsh, VH
is to SH as vh to sh ; that is, the axis of the conic section now described
is to the distance of its foci as the axis ab to the distance of the foci sh ;
and therefore the figure now described is similar to the figure aph. But,
because the triangle PSH is similar to the triangle psh, this figure passes
through the point P ; and because VH is equal to its axis, and VS is per
pendicularly bisected by the rght line TR, the said figure touches the
right line TR. Q.E.F.
LEMMA XVI.
From three given points to draw to afonrth point that is not given three
right lines whose differences shall be either given, or none at all.
CASE 1. Let the given points be A, B, C, and Z the fourth point which
we are to find ; because of the given difference of the lines AZ, BZ, the
locus of the point Z will be an hyperbola
whose foci are A and B, and whose princi
pal axis is the given difference. Let that
axis be MN. Taking PM to MA as MN
is to AB, erect PR perpendicular to AB,
and let fall ZR perpendicular to PR ; then
from the nature of the hyperbola, ZR will
be to AZ as MN is to AB. And by the
like argument, the locus of the point Z will
be another hyperbola, whose foci are A, C, and whose principal axis is the
difference between AZ and CZ ; and QS a perpendicular on AC may be
drawn, to which (QS) if from any point Z of this hyperbola a perpendicular
ZS is let fall (this ZS), shall be to AZ as the difference between AZ and
CZ is to AC. Wherefore the ratios of ZR and ZS to AZ are given, and
consequently the ratio of ZR to ZS one to the other ; and therefore if the
right lines RP, SQ, meet in T, and TZ and TA are drawn, the figure
TRZS will be given in specie, and the right line TZ, in which the point
Z is somewhere placed, will be given in position. There will be given
also the right line TA, and the angle ATZ ; and because the ratios of AZ
and TZ to ZS are given, their ratio to each other is given also ; and
thence will be given likewise the triangle ATZ, whose vertex is the point
Z. Q.E.I.
CASE 2. If two of the three lines, for example AZ and BZ, are equal,
draw the right line TZ so as to bisect the right line AB ; then find the
triangle ATZ as above. Q.E.I.
130 THE MATHEMATICAL PRINCIPLES [BOOK I.
CASE 3. If all the three are equal, the point Z will be placed in the
centre of a circle that passes through the points A, B, C. Q.E.I.
This problematic Lemma is likewise solved in Apollonius s Book oi
Tactions restored by Vieta.
PROPOSITION XXL PROBLEM XIII.
About a given focus to describe a trajectory that shall pass through
given points and touch right Hues given by position.
Let the focus S, the point P, and the tangent TR be given, and suppose
that the other focus H is to be found.
On the tangent let fall the perpendicular
ST, which produce to Y, so that TY may
be equal to ST, and YH will be equal
to the principal axis. Join SP, HP, and
SP will be the difference between HP and
the principal axis. After this manner,
if more tangents TR are given, or more
points P. we shall always determine as
many lines YH, or PH, drawn from the said points Y or P, to the focus
H, which either shall be equal to the axes, or differ from the axes by given
lengths SP ; and therefore which shall either be equal among themselves,
or shall have given differences ; from whence (by the preceding Lemma).
that other focus H is given. But having the foci and the length of the
axis (which is either YH, or, if the trajectory be an ellipsis, PH -f SP ;
or PH SP, if it be an hyperbola), the trajectory is given. Q.E.I.
SCHOLIUM.
When the trajectory is an hyperbola, I do not comprehend its conjugate
hyperbola under the name of tins trajectory. For a body going on with a
continued motion can never pass out of one hyperbola into its conjugate
hyperbola.
The case when three points are given
is more readily solved thus. Let B, C,
I), be the given points. Join BC, CD,
and produce them to E, F, so as EB may
be to EC as SB to SC ; and FC to FD
as SC to SD. On EF drawn and pro
duced let fall the perpendiculars SG,
BH, and in GS produced indefinitely E
take GA to AS, and Ga to aS, as HB
is to BS ; then A will be the vertex, and Aa the principal axis of the tra
jectory ; which, according as GA is greater than, equal to, or less than
SEC. V.] OF NATURAL PHILOSOPHY. 131
AS. will be either an ellipsis, a parabola, or an hyperbola ; the point a in
the first case falling on the same side of the line GP as the point A ;
in
the second, going oft* to an infinite distance ;
in the third, falling on the
other side of the line GP. For if on GF the perpendiculars CI, DK are
let fall, TC will be to HB as EC to EB ; that is, as SO to SB ; and by
permutation, 1C to SC as HB to SB, or as GA to SA. And, by the like
argument, we may prove that KD is to SD in the same ratio. Where
fore the points B, C, D lie in a conic section described about the focus S,
in such manner that all the right lines drawn from the focus S to the
several points of the section, and the perpendiculars let fall from the same
points on the right line GF, are in that given ratio.
That excellent geometer M. De la Hire has solved this Problem much
after the same way, in his Conies, Prop. XXV., Lib. VIII.
SECTION V.
How the orbits are to be found when neither focus is given.
LEMMA XVII.
Iffrom 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 ang 7
ei,
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 tie 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 I ^^ p ; T
the other sides ; as PQ and PR to the side AC, and s
|
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^ IQ I3
those parallel sides will be one of the diameters of the 1L
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 XXI1L, Book III.,
of Apollonius s Conies) 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
132 THE MATHEMATICAL PRINCIPLES [BOOK I
CASE 2. Let us next suppose that the oppo
site sides AC and BD of the trapezium are not
parallel. Draw Be/ parallel to AC, and meeting
as well the right line ST in /, as the conic section
in d. Join 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), Et or PQ is to Tt as DN to NB. And ^^ Q N
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
N i)M 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 PR to the rectangle PS X PT. Q.E.D.
CASE 3. Let us suppose, lastly, the four lines
?Q, PR, PS, PT, 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, Pt
parallel to AB ; and because the angles of the
triangles PQ</, PRr, PSs, PTt are given, the ratios
of IQ to Pq, PR to Pr, PS to P*, PT to Pt
will b? also given; and therefore the compound
ed ratios Pk X PR to P? X Pr, and PS X PT to Ps X Pt are
given. But from what we have demonstrated before, the ratio of Pq X Pi
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 s niL 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 pas
sing 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 c1
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 ^ B
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,
SEC. V.] OF NATURAL PHILOSOPHY 133
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 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 bn to bd as
PR to PT. And therefore the equiangular trapezia ~Dnbd, DRPT, are
similar, and consequently their diagonals D6, 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 com
mon 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
(onger cut, but only touch, the curve in those coinciding points.
SCHOLIUM.
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 hap
pens 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 *he four points. If the two opposite an
gles of the trapezium taken together are equal c
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
