# Lemma 24

##### 12 minutes • 2361 words

## Table of contents

## Lemma 24

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

## Lemma 25

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

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

### Corollary 1.

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

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

drawn through the

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

## PROPOSITION 27 PROBLEM 19

To
describe
a
Supposing
GCD, FDE,
trajectory that
ABG
EA to
BCF,
;
be the
tangents given by position.
and N, AF, BE,
Bisect in
the diagonals of the quadri
M
lateral
tained
them
;
XXV)
figure
ABFE
con-
under any four of
and (by Cor. 3, Lem.
MN
the right line
draAvn through the points
(,f
may touch jive
right lines given by position.SEC.
OF NATURAL PHILOSOPHY.
V.]
147
bisection will pass through the centre of the trajectory.
P and Q, the diagonals (if I may so call them) Bl),
Again, bisect in
GF
teral figure
line
of the quadrila
contained under any other four tangents, and the right
through the points of bisection will pass through the cen
EC OF
PQ, drawn
tre of the trajectory
;
and therefore the centre will be given in the con
course of the bisecting lines.
Suppose it to be O.
that the centre
at
BG
draw
such
distance
gent
Parallel to any tan
be placed in the
KL
O may
KL
will touch the trajectory to be de
middle between the parallels; this
Let this cut any other two tangents GCD, FJ)E, in L and K.
scribed.
Through the points G and K, F and L, where the tangents not parallel,
CL, FK
K and the right line
meet the parallel tangents CF, KL, draw GK, FL meeting in
OR drawn and produced, will cut the parallel tan
in
the
This appears from Gor. 2, Lem.
points of contact.
gents GF, KL,
XXIV. And by the same method the other points of contact may be
found, and then the trajectory may be described by Prob. XIV. Q.E.F.
;
SCPIOLTUM.
Under the preceding Propositions are comprehended those Problems
wherein either the centres or asymptotes of the trajectories are given. For
when points and tangents and the centre are given, as many other points
and as many other tangents are given at an equal distance on the other
And an asymptote is to be considered as a tangent, ami
remote extremity (if we may say so) is a point of contact.
Conceive the point of contact of any tangent removed in infinitum, and
the tangent will degenerate into an asymptote, and the constructions of
side of the centre.
its
infinitely
the preceding Problems will be changed into the constructions of those
Problems wherein the asymptote is given.
After the trajectory
is
described,
we may
and foci in this manmr. In the
construction and figure of Lem. XXI, let those
legs BP, CP, of the moveable angles PEN,
PCN, by the concourse of which the trajec-
find its axes
,
^

tory was described, be made parallel one to
the other
and retaining that position, let
:
them revolve about their poles
figure.
In the
mean while
I
,
C, in that
let the other legs
of those angles, by their concourse
or k, describe the circle BKGC.
Let
be the centre of this circle;
and from this centre upon the ruler MN, wherein those legs CN,
did
GN, BN,
K
O
BN
OH
concur while the trajectory was described, let fall the perpendicular
and L. And when those other legs CK,
meet
meeting the circle in
K
in the point
K
that
is
BK
nearest to the ruler, the
parallel to the greater axis,
first
and perpendicular on the
legs
CP,
lesser
;
BP
will be
and the con-148
THE MATHEMATICAL PRINCIPLES
[Book
I
trary will hajpen if those legs meet in the remotest point L.
Whence ii
the centre of the trajectory is
and those be-
given, the axes will be given
ing given, the foci will be readily found.
;
But
the squares of the axes are one to the
KH
other as
to
LH, and
thence
describe a trajectory given in
f
mr
given points.
For
it
easy to
is
kind through
two of the given
if
points are made the poles C, 13, the third will
but
give the moveable angles PCK,
those being given, the circle
may be
PBK
BGKC
;
Then, because the trajectory is
OH to OK, and
described.
given in kind, the ratio of
and therefore
OH
itself, will be given. About
the centre O, with the interval OH, describe another circle, and the
right
line that touches this circle, and
passes through the concourse of the legs
meet in the fourth given point, will
CK, BK, when the first legs
;
CP BP
by means of which the trajectory may be described
Whence also on the other hand a trapezium given in kind (excepting a
few cases that are impossible) may be inscribed in a given conic section.
There are also other Lemmas, by the help of which trajectories given in
kind may be described through given points, and touching given lines.
be the ruler
Of
MN,
such a sort
is this,
given by position, that
that if a right line is drawn through any point
may cut a given conic section in two points, and
the distance of the intersections
to ich
bisected, the point of bisection will
is
ano her conic section of the same kind with the former,
axes parallel to the axes of the former.
its
But
I
arid
havin^
hasten to things of
greater use.
Lemma
To
place 1ht
tude,
in,
lit
rev
angles of a triangle, given both in kind and
Three
may
the triangle
the line AB,
angle
EF,
many rigid lines given by position, -provided th]
among themselves, in such manner tfia t j ic spiral
touch the several lines.
indefinite right lines
given by position, and
its
F
DEF
its
that
AB, AC, BC,
are
is required so to place
its angle 1) may touch
the line AC, and
it
angle
the line BC.
E
Upon DE, DF, and
describe three segments of circles
DRE,
DGF. EMF, capable of
Rubles BAG, ABC, ACB
scribed
t
magni
respect of as
are not all parallel
angles
XXVI.
angles equal to the
But those segments are to be de
respectively.
that the letters
wards such sides of the lines DE,
;
;
DF EF3 EC.
OF NATURAL PHILOSOPHY.
V.I
DRED
1411
letters
I1ACB
turn round about in the same order with the letters
may
:
DGFD in the same order with the letters ABCA and the
EMFE in the same order with the letters ACBA then, completing
the letters
;
;
th se segmerts into entire circles let the two former circles cut one the
to be their centres.
other in G, and suppose P and
Then joining GP,
Q
PQ, take Ga
AB
to
as
GP
is to
PQ
interval Ga, describe a circle that
Join aD cutting the second circle
third circle
EMF
in
cut the
DFG
in
Complete the figure
c.
a&cDEF
to the figure
and about the centre G, with the
;
may
I say,
:
the thing
is
b,
first
circle
as well as
ABCdef
DGE
aE
in a.
cutting the
similar and equal
done.
For drawing Fc meeting D in n,
and joining aG bG, QG, QD. PD, by
;
construction the angle EaD is equal to
the angle CAB, and the angle acF equal
to
the angle
ACB;
and therefore the
equiangular to the triangle
triangle aiic
Wherefore the angle anc or FnD
equal to the angle ABC, and conse-
ABC.
is
<
uently to the angle
n
and there
F/>D
;
on the point b,
Moreover the angle GPQ, which is half
fore the point
GPD
the angle
to the angle
falls
at the centre,
GaD
is
equal
at the circumference

GQD
and the angle GQP, which is half the angle
at the centre, is equal
to the complement to two right
angles of the angle GbD at the circum
ference, and therefore equal to the angle Gba.
Upon which account the
GPQ,
triangles
Gab, are similar, and Ga is to ab as
Ga to AB. Wherefore ab and
(by construction), as
consequently the triangles abc,
similar, are also equal.
GP to PQ. that is
AB are equal; and
ABC, which we have now
;
proved to be
And
therefore since the angles I), E, F, of the
do
triangle
respectively touch the sides ab, ar, be of the triangle
afjc the
figure AECdef may be completed similar and equal to the figure
afrcDEFj and by completing it the Problem will be solved. Q.E.F.
DEF
/
COR. Hence a right line may be drawn whose parts given in length may
be intercepted between three right lines given by position.
Suppose the
the
of
the
arid
access
its
to
side
triangle DEF, by
EF,
point
by having
D
the sides
DF placed
part DE is to
DE,
whose given
given by position; and
AB
its
i>t
directum
to
be changed into a right line
AB
be interposed between the right lines
AC
to
is
be
between
the
given part
interposed
;
DF
BC, given by position; then, by applying the preceding
right lines
construction to this case, the Problem will be solved.
;THE MATHEMATICAL PRINCIPLES
[BOOK
1.
PROPOSITION XXVIII. PROBLEM XX.
To
describe a trajectory giren both in kind and magnitude, given
parts
of which shall be interposed between three right lines given by position.
Suppose a trajectory is to be described that
may
-and
be similar and equal to the curve line DEF,
may be cut by three right lines AB, AC,
BC, given by
similar and
curve
position, into parts
equal to the given
DE
and EF,
parts of this
line.
the right lines DE, EF, DF= and
the
angles D, E, F, of this triangle DEF, so
place
as to touch those right lines given by position (by
Lem. XXVI). Then about the triangle describe
Draw
the trajectory, similar and equal to the curve
DEF.