A 544533
The Gift of
WILLIAM H. BUTTS, Ph.D.
A.B. 1878 A.M. 1879
Teacher of Mathematics
1898 to 1922
Assistant Dean, College of Engineering
1908 to 1922
Professor Emeritus
1922
RADCLIFFE
OBSERVATORY
OXFORD.
!
•
A. Robertion.
Observatory, Oxford.
Emerson.
williane
+ Cyclomathesis
Q A
V.5
ARTES
1817
SCIENTIA
VERITAS
LIBRARY
OF THE
UNIVERSITY OF MICHIGAN
FLUNIOUS UNUM
TUEBOR
51-QUÆRIS PENINSULAM-AMŒNAM
CIRCUMSPICE
Emerson, Willi ane
r
Cyclomathesis
THE
ARITHMETIC
OF
O F
INFINITES,
AND THE
DIFFERENTIAL METHOD;
ILLUSTRATED, BY
EXAMPLE S.
Hæ tibi erunt Artes.
VIRG.
.9
RADCLIFFE
OBSERVATOR
LONDON:
Printed for J. NOURSE, in the Strand,
Bookfeller in Ordinary to his MAJESTY.
M DCC LXVII.
OXFORD
Den hib
Milr
[iii]
Profion Williin. H. Bitio
10.14-1985-
THE
PREFACE.
THE method which the antients generally made
ule of for finding the areas of figures, or the fo-
lidities of bodies, was by the help of other infcribed
and circumfcribed figures; after fuch a manner, that
the infcribed figures being continually divided into very
Small parts, the fum of the infcribed parts, approach-
ed fo near the true figure, as to fall short of it by a
quantity less than any given quantity. And in like
manner the circumfcribed figure was divided into fo
many small parts, that their fum exceeded the given
figure, by a quantity alſo less than any given one. So
they always had the value of the figure propofed,
between the two values of the infcribed and circum-
fcribed figures; and by this method found out its con-
tent. But this way being for the most part ſo very
laborious, that it put the fucceeding geometers upon
finding out fhorter methods; and thus came up the
Method of Indivifibles, or the Arithmetic of In-
finites. This method folves thefe forts of Problems
with admirable brevity, and extreme facility; and
that by a help of a few Propofitions, which reduces
the whole bufinefs to the fumming up the powers of
an arithmetical progreffion, confifting of an infinite
number of terms. As we ſhall have frequent occafion
for this afterwards; I have here delivered that me-
thod
A 2
4
PREFACE.
THE
thod at large, with many examples to fhew the ufe of
it. The whole is contained in Sect. I. of the follow-
ing Book.
I have likewife given the principles of the diffe-
rential method, and the Interpolation of Series, in
Sect. II. an art extremely useful in almost all parts
of the Mathematics, efpecially in Aftronomy. For it
is frequently required to find the pofition of a planet
or comet at fome given time when an obfervation
cannot be made; but by help of other obfervations, we
can find it to that time, by the interpolation of feries.
This method I have also made very clear and plain,
and illuftrated the fame with variety of examples.
W. Emerfon.
[ 1 I
]
SECT. I.
The Arithmetic of Infinites.
T fumming up
HE Arithmetic of Infinites is the art of
fumming up the powers of a ſeries of
quantities in arithmetical progreffion;
whofe number of terms is infinite, and their com-
mon difference infinitely ſmall.
It is alfo called the Method of Indivifibles, be-
caufe magnitudes are here fuppofed to be refolved
into their indivifible parts, or fo far as there is any
occafion to confider them as fuch.
PROP. I.
If a ſeries of equal quantities be added together,
the fum of all the terms is equal to any one of them
multiplied into the number of terms.
For a+a+a... to ʼn terms is
n nxa.
PRO P. II.
If there be a ſeries of quantities in arithmetical
progreffion, and continued ad infinitum; as o, a, 2a,
3a, 4a... to na; the fum of all the terms is ina,
or to half so many times the greatest.
For by (Prop. 7. Arith. Progreffion) the fum
=o+na
naxn+I
xn+1, or
but fince n is infi-
2
2
Na
n
nite, n+1 =n. therefore the fum ==
xna =
X
2
laft term.
B
Cor.
ล
ARITHMETIC
Cor. The greatest term multiplied by the number of
terms, is equal to twice the fum of the feries.
PROP.
III.
If there be a ſeries of Squares, whofe roots are in
arithmetical progreffion beginning at o, and continued
2
I
2
2
ad infinitum, as o, a², 2a, 3a, 4a
за
the fum of all the terms is nxna",
times the greatest.
=
3
For let so+a²+2a²+za
2
I
2
to na ;
or / fo many
3
2
to na", L =
laft or greateſt term. Then by the Method of In-
crements, s=n+1.az², putting n+1.az.
But (by Cor. Ex. 10. Prop. XIII. Increments), the
xn+1
&c. Or writing z for
n+1.x 2
integral of x =
x, and 2 for n, the int. of x2 =
3
Z
3Z Z
2
&c. But
fincen is infinite, zis infinite,and z³ is infinitely greater
2
than z²; whence int. of z², that is s=
1.
and naa; therefore s =
א
W
3º
Z3
32
But n=1,
3
n+1.a
зв
But as n is infinite, n+1=n;
2
n+1.a
X
n+1.a
3a
2
and s=a na
х
= n × na = {xnL.
na
за
Cor. The greatest term multiplied by the number of
terms, is equal to thrice the fum of the feries.
PROP. IV.
In a series of cubes, whofe roots are in arithme-
tical progreffion, beginning at o, and infinitely conti-
nued; as o, a³, za³, za³,
3
за
3
to na³ ¿ the fun
of
Sect. I. OF INFINITES.
1
+
3
1
of all the terms is = nxna, or ſo many times the
greatest term.
3
3
3
3
For put so+a³ +za³ +зa³... +ña³, L=
greateſt term, then by the method of increments
s=n+1.a = ׳, putting n+1.a=z. But (by Cor.
Ex. 10. Prop. XIII. Increments). The integral of
3
xn+1
&c. or putting z for x,
xn
2012 =
n+I.x
2
24
23
for n; the int. of z³-
42
22 |
x, and
3
&c. But n = 1,
and z=na=a. Alſo fince n is infinite, z is infi-
nite, and 24 is infinitely greater than z³; and n+1=n.
па
Z4
4
4
Whence int. z3 or s
n+1.a =
na
42
40
4a
49-
x na³ = nxL.
Cor. The greatest term multiplied by the number of
terms, is equal to four times the fum of the feries.
P.RO P. V.
If there be a ſeries of mth powers, whofe roots are
in arithmetical progreffion, beginning at o, and conti-
nued to infinity, as o, a", za", 3a", 4a".
n
M+I
the fum of all the terms is =
Xfo many times the greatest term.
× na
to nam,
I
m =
m+ I
m
Let s=0+a+za™ +za”..: +na”, L = laſt
term; and by the method of increments, s=n+1.a
=zm. And (by the fame Cor. Ex. 10. Prop. XIII.
Increments) s= int. x =
zm+1
m
Z &c. But
M+IZ
B 2
2
N=1,
ARITHMETIC
n1, and znaa; and fince n is infinite, n+1.a
na, and zm vanishes, being infinitely lefs than
m+i
Z
zm+I
m+1
; therefore s
n+1.a
m+x.a
m+1.a
m+1
na
na
X-m
n
I
na.
X
m
m+1.a
m+1.a
m+I
na
M+I
XnL.
Cor. The greateſt term multiplied by the number of
terms, is equal to m+1 times the fum of the feries.
PRO P. VI.
To refolve a Problem by the Arithmetic of Infinites.
The chief uſe of this method is to find the con-
tent of geometrical quantities, as of fuperficies and
folids, by reducing them to their indivifible parts,
or firſt elements; and then finding the fum of all
thefe, as directed by the following
RULE.".
Suppofe a line to be compofed of an infinite
number of points; a furface of an infinite number
of lines, drawn parallel and equidiftant from one
another; and a folid of an infinite number of
planes, parallel and equidiftant; then obferve what
fort of a feries it is that makes up that quantity.
Find the fum of that feries by fome of the forego
ing propofitions, which will be the content required.
SCHOLIU M.
Here lines are fuppofed to be made up of an in-
finite number of equidiftant points; plain figures
of an infinite number of parallel lines, like the
threads in a piece of cloth. And folids of an in-
finite
Sect. I.
5
OF INFINITES.
finite number of planes, like the leaves in a book. Fig
Yet theſe points, lines and furfaces, are not really
fuch, but are called ſo by reaſon of the fimilitude.
For theſe points are in ftrictnefs, infinitely fhort
lines; the lines, infinitely narrow parallelograms,
and the planes that compofe the folids, infinitely thin
folids; fo that whenever we ſpeak of points, lines,
or planes, compofing any geometrical magnitude,
it muſt always be underſtood in the ſenſe here men-
tioned. Moreover, if all theſe parallel lines that
make up a furface be perpendicular to the axis of
a curve; then thefe parallel lines will cut the curve,
or any other line, at oblique angles, and the points
ſuppoſed to conſtitute that curve, or that line, will
be greater than the correfpondent points in the axis,
in fome certain proportion; which proportion is to
be inveſtigated by the property of the figure.
Ex. I.
To find the area of the Parallelogram ABCD.
Draw AP perp. to the baſe CD; divide AP into
an infinite number of equal parts, thro' all which
ſuppoſe ſo many parallels drawn to the baſe CD.
Then all theſe equal lines compofe the area of the
parallelogram, and their number is denoted by AP.
Then by Prop. I. the fum of all is one of them
multiplied by the number of all; that is
or the baſe multiplied into the height.
Ex. 2.
CDXAP,
To find the content of a cylinder ABCD.
Draw AP perp. to the baſe, which divide into an
infinite number of equal parts; thro' which draw
as many planes parallel to the bafe CD. Then all
thefe equal planes compofe the folidity of the cy-
linder; whoſe number is AP. Whence by Prop. I.
the fum of all is one CD x their number AP;
that
B 3
1.
2.
ARITHMETIC
Fig. that is, the cylinder
3.
4.
plied by the height.
CDXAP, or the baſe multi-
And the fame way the folidity of a parallelopi-
pedon or of a prifm is calculated.
Ex. 3.
To find the area of a triangle ABC.
Draw AP perp. to the baſe BC; divide AP into
an infinite number of equal parts; thro' which
fuppofe lines drawn parallel to BC. Then theſe
lines make up the triangle, and are a ſeries of terms
in arithmetical progreffion beginning at o, and end-
ing at BC; and the number of them is denoted by
AP. Then (by Prop. II.) the fum of all the terms
APXBC; that is, the area of the triangle
is
=
2
APXBC
2
Ex. 4
To find the folidity of a cone ABC.
Draw AP perp. to the bafe BC; divide AP into
an infinite number of equal parts; thro' which draw
planes parallel to the bafe BC. Then theſe planes
make up
up the cone; and their diameters are a feries
of quantities in arithmetical progreffion, beginning
at o, and ending at BC; therefore the planes or
circles themſelves are proportional to a feries of
fquares, whofe roots are in arithmetical progreffion.
Whence (by Prop. III.) the fum of all the ſeries is
=AP× baſe BC; that is, the cone is height
> baſe.
•
1
3
And the fame way it is proved, that a pyramid
is the height x by the baſe.
Ex.
Sect. I.
7
OF INFINITES.
Ex. 5.
To find the furface of the right cone ABC.
Draw the axis AP, and DF perp. to it, and put
AP=b, BP=b, AB=s, AD=x, c=3.1416; then
by fimilar triangles, AP (b): PB (b) : ; AD (*) :
DF =
bx
Ђ
and
;
2cbx
b
= periphery at F. Then the
fum of all the peripheries at F makes up the fur-
face of the cone. But the points in AB being
greater than thofe in AP, in the ratio of AB to
AP. Therefore the periphery at F muſt be in-
2scbx 25cb
creaſed to
or
bb
bb
Xx, and then the fum of
all theſe peripheries make up the furface. But the
fum of all the xs is 0, 1, 2, 3, &c. to AP; that
is (by Prop. II.) APXAP or AP². Therefore
25cb
the conic furface is
ference of BCXAB.
bb x bb = bcs circum-
½ =
Otherwife.
Put BPb, ABS, AFz, then by fimilar
bz
triangles AB (s): BP (b):: AF (x): FD = ;
S
zcbz
ther = periphery at F; and the fum of all
S
theſe peripheries make up the furface.
the furface. Now all the
z's conftitute a ſeries of terms in arithmetical pro-
greffion, whofe laft term, and alfo the number of
terms is s and therefore the fum (by Prop. II.) is
ss. Therefore the fum of all the peripheries, or
2cb
S
the furface of the cone is = מss = cbs as be-
fore.
Fig.
5.
B 4
Ex.
8
ARITHMETIC
Fig.
6.
Ex. 6..
To find the folidity of any fegment of a sphere
ADF.
Draw the diameter AB, and DEF perp. to it; and
put radius AC=r, AE =x, DE=y, c = 3.1416;
then by the nature of the circle yy=2rx-xx. Then
if AE be divided into an infinite number of equal
parts, and planes be drawn thro' them parallel to
DF; the fum of all theſe circular planes make up
the fegment. Now one of theſe circles is cyy or
2crx-cxx ; and all the x's conftitute a ſeries of terms
in arithmetic progreffion, whofe fum (by Prop. II.)
is xx. Alfo the fum of all the xx's (by Prop. III.)
is x³, where the greateſt term is xx, and number
of terms, x.
Therefore the fum of all the cir-
I
I
cles conftituting the fegment, is 2crxxx-cx*³,
or crxx
CX3
3
Therefore the fegment DAF =
crx-cxx xx: that is, the fegment DAF is equal
to a cylinder whoſe height is AE, and baſe crx-cxx,
or cyy+cxx = circle DF+ circle AE.
I
2
I
ठ
2
3
Cor. The hemifphere, as likewife the fphere is the
circumfcribing cylinder.
For when xr,the fegment becomes crr-crr
xr = +crrxr.
Ex. 7.
To find the furface of the fegment of a fphere DAF.
Let CA=r, AE=x,DE=y, c= 3.1416; then
the circumference at D2cy. And the fum of all
the peripheries make up the furface of the feg-
ment DAF. But the points in AD being greater
than thoſe in AE, in proportion of the tangent TD
to
Sect. I. OF INFINITES.
9
to TE, or of DC to DE; therefore y: r:: 1:
= magnitude of a point at D.
or 2cr is the true periphery at D.
all theſe peripheries
Whence 2cy X
ર
y
And the fum of
furface of the fegment; and
the number of them is denoted by AE. Therefore
(by Prop. I.) that fum = 2crxAE; that is, the
furface of the fegment DAF = 2crx.
Cor. The furface of the hemifphere is
2crr; and
of the whole fphere 4crr, or four great circles of the
Sphere.
****
***
##
SECT.
10
THE
SECT. II.
The Differential Method.
'HE Differential Method, is the art of working
with the differences of quantities. By this me-
thod any term of a feries may be found from the
feveral orders of differences being given; or vice
verfa, any difference may be found from having
the terms of the feries given: it likewife fhews how
to find the fum of fuch a feries. And it gives
rules to find by interpolation, any intermediate
term, which is not expreffed in the feries, by hav-
ing its place or pofition given.
V
When any ſeries of quantities is propoſed, take
the firſt term from the ſecond, the fecond from the
third, the third from the fourth, &c. then all
theſe remainders make a new ſeries, called the first
order of differences. In this new ſeries take the firſt
term from the fecond, the fecond from the third,
the third from the fourth, &c. as before; and
theſe remainders make another feries, called the
Jecond order of differences. In like manner, in this
feries, take the firſt term from the fecond, the fe-
cond from the third, &c. And thefe will make a
feries, called the third order of differences. And
after this manner you may proceed as far as you
will. Thus in the following propofition, A, b, c, d, e,
&c. is the feries; B, B2, B3, B4, &c. the firft order
of differences; C, C2, C3, &c. the fecond order
of differences; D, D², &c. the third order; E,
&c. the fourth order, and fo on. But the firſt terms
of theſe ſeveral orders of differences, as B, C, D,
E, &c. are thoſe that are principally made uſe of
in calculations by this method.
PROP.
Sect. II. THE DIFFERENTIAL METHOD. II
PROP. I.
If there be any feries A, b, c, d, e, &c. and if
there be taken, the first differences B, B², B³, &c.
the fecond differences C, C2, C3, &c. the third dif
ferences D, D2, D3, &c. and fo on.
Then if T ftand for the first term of the nth dif
ferences; I fay, ±T=A-nb+nx
X
3
d + nx
NI
N I N2
-2 N3
2
+T, when n is even,
X
3
N-I
2
c-nx
2
X e-&c. that is,
4
and -T when n is odd.
The feveral Orders of
differences being taken
as before directed, will
ftand thus. Then
A
B
b
C
B
D
Cz E
B³
3
D2
d
C3
B+ &c.
e
C
d
"
e, &c.
feries
A
Ift diff.
2d diff.
3d diff.
4th diff.
b
b-A c-b d-c C e-d, &c.
"
2
c—2b+A, d—2c+b, e—2d+c, &c.
d—3c+3b—A, e-3d+3c-b, &c.
e-4d+6c46+A, &c.
That is, Bb-A, C—c—2b+A, D=d−3c+36
—A, E—e—4d+6c-4b+A, &c. or-B = A—b,
+C=A=2b+c,-DA-3b+3c-d, +E=A
—4b+6c—4d+e, &c. where putting T fuccef-
fively equal to B, C, D, E, &c. and n = 1, 2, 3,
4, &c. the Prop. will be evident.
Cor. Hence A≈A, the firſt term.
B=-A+b, the firft difference.
C= A-2b+c, the 2d difference.
DA+363c+d, the 3d difference.
E=
A-4b+6c4d+e, the 4th difference.
FA+56-10c+10d-5e+f, the 5th difference.
&c.
PROP.
12 THE DIFFERENTIAL
PROP. II.
If A, b, c, d, e, &c. be any feries; and there
be taken B, C, D, E, &c. the first of the feveral
orders of differences.
Then I fay the nth term of the feries, will be A+
C+
n 2 N3
X
2- I
N-- I N-2
n- I
B +
X
X
I
I
2.
I
2
3
N- I
D+
X
X
X
E+ &c.
I
2
3.
4
N2 N3 n4.
For from the equations in the laft Prop. viz.
B = b—A, C = c−2b+A, &c. we have by tranf-
pofing, c
b=A+B=A+26+C=A+2A+2B+C
(expunging b); that is,
c=A+2B+C, d=A-3b+3c+D=A-3A-3B
+3A+6B+3C+D (expunging b and c); that is,
d = A +3B+3C+D. Alſo e = —A+4b—6c+4¢
+E (expunging b, c, d) -A+4A+4B-6A
-12B−6C+4A+12B+12C+4D+E; that is,
e=A+4B+6C+4D+E, &c.
Then puting A, b, c, d, &c. for the nth term;
and n fucceffively 1,2, 3, 4, &c. the feries will
be evident.
Cor. 1. If d', d", d'", &c. be the first of the first,
Second, third order, &c. of differences; then
The nth term of the feries, A, b, c, d, &c. will
N-I N2
N→→ I
n-
n-
2
n- I
Be=A+
ďt
X ・d" +
X
I
I
2
I
2
n
X d" +
2-3 d'"""
.n- I
N2
N3 24
n
31111
X
X
X
•4d" +
I
2
3
4
3
&c.
For Bd', C=d", D=d", &c. And the coef-
ficients are the unciæ of the n—1th power.
Cor. 2. Hence alfo it follows, that any term of a
given feries may be accurately determined; if the dif-
ferences of any order happen at last to be equal.
Cor.
Sect. II.
13
METHOD.
Cor. 3. Hence A≈A, the first term.
b=A+B, the 2d term.
<=A+2B+C. the 3d term.
d=A+3B+3C+D, the 4th term.
e=A+4B+6C+4D+E, the 5th term.
ƒ=A+5B+i0C+10D+5E+F, the 6th term.
g=A+6B+15C+20D+15E+6F+G, the 7th term.
&c.
PROP.
III.
If a, b, c, d, e, &c. be any feries; and d', d",
d"", &c. the first of the feveral orders of differences;
Then
The Sum of n terms of the feries, is =na+nx
NI
NI N-2
·ď + n x
X
d" + nx
2
2
X
n = 3 d!!! +nx
3
NI N2 N-3 1-4
NI N2
2
X
3
X
X
X
d"
4
2
3
4
5
+ &c.
For in the ſeries of quantities,
o, a, a+b, a+b+c,a+b+c+d, &c.
ift diff. are a
b
C
d
&c.
?
2d diff.
d'. d'a
>
d3
&c.
3d diff.
4th diff.
d"2
&c.
&c.
,
th
Therefore (by Cor. 1. Prop. II.) the n+1 term
of the feries, o, a, c+b, a+b+c, a+b+c+d, &c.
or the term of the feries, a, a+b, a+b+c
n
th
NI
2
g+b+c+d, &c. is = o+na+nx d' + n x
NI N
2
X d" + &c. But the nth term of the
3
feries a, a+b, a+b+c, &c. is the fum of n
terms of the feries, a, b, c, d, &c. and therefore
N I
equal to na+nx
d'+nx
2
N
2
7Z2
X d" + &c.
3
PROP.
14 THE DIFFERENTIAL
PROP. IV.
If there be feveral fets of quantities in arithmetic
progreffion, whofe number is m, and the common dif-
ferences p, q, r, s, &c. and if a new feries be form-
ed by multiplying the refpective terms of all thefe fe-
ries into one another; then I fay the mth order of dif-
ferences will be equal to one another, and to pqrs, &c.
Xinto 1×2×3×4, &c. to m.
1. Suppose two feries;
as a, a+p, a+2p, a+3p, a+4p, &c.
and b, b+q, b+2q, b+3q, b+49, &c.
then by multiplying we have ab, a+pxb+q,a+2p
xb+29, &c. that is,
ab
feries
ab+pb+aq+pq
ab+2pb+2aq+4Pq
ab+3pb+3aq+9Pq
&c.
1 Diff.
2 Diff.
pb+aq+pq
pb+aq+3pq
2pq
pb+aq+5pq
2pq
therefore the 2d differences are equal, and equal to
PqX1X2, where m-2.
2. Suppose three feries.
as a, a+p, a+2p, a+3p, a+4p, &c.
b, b+q, b+2q, b+39, b+49, &c.
c, c+r, c+2r, c+3r, c+4r, &c.
The terms of theſe being multiplied, produce a fe-
ries whofe terms are,
abc
a+pXb+qXc+r
a+2pxb+2q Xc+2r
a +3 p× b + 3q × c + 3r
a+4pxb+49 × c+4r
&c.
thefe
Sect. II.
15
METHOD.
theſe being multiplied at length, and fubtracted
from one another, will give the following feries,
which is the firſt order of differences,
abr+bcp+aqc+ pqc+ aqr+ pbr+pqr
abr+bcp+aqc+3pqc+3aqr+3pbr+7pqr
abr+bcp+aqc+5pqc+5agr+5pbr+19pqr
abr+bcp+aqc+7pqc+7agr+7pbr+37 p q r
&c. And the differences of thefe, or the fe
cond differences being taken, we have
2pqc+2aqr+2pbr+6pqr
2pqc+2aqr+2pbr+12pqr
2pqc+2agr+2pbr+18pqr
6pqr
6pgr
and the third difference will be
here m 3, and the mth difference is pqrx1X2X3.
3. Suppofe four feries.
pqr×1×2×3.
As a, a+p, a+2p, a+3p, a+4p, &c.
b, b+q, b+aq, b+39, b+49, &c.
c, c+r, c+2r, c+3r, c+4r, &c.
d, d+s, d+25, d+3s, d+45, &c.
and when multiplied, produce
abcd
+a+2p × b+29 × c+2r x d + 2 s
+a+3p× b+39 × c+3r ×d+35
+a+4p × b+49 × c+4r ×d+4s
+a+5p × b+5q × c+5r × d+55.
&c.
Let ABCD be any term;
and the next, A+pXB+qxC+rx D+s; that is,
AB+Aq+pB+pq×CD+Cs+rD+rs=ABCD+
ABX Cs+rD+rs+CDx Aq+pB+pq + Aq+pB
+pq xCs+rD+rs. From this, fubtract ABCD, and
the
16
DIFFERENTIAL
THE
the remainder is the firft differenceABxCs+rD+rs
+ CD× Aq+pB+pq +Aq+pB+pq×Cs+rD+rs.
To abridge the work, keep in only fuch terms as in-
volve the moſt of the quantities ABCD, for the reſt
will vaniſh before you come to the laſt differences.
By this means the firft difference becomes only ABX
Cs+rD+CD× Aq+Bp; that is, the firſt difference
ABCs+ABDr+CDAq+CDBp; then the
(1)
next term is evidently A+pxB+qx C + rxs +
A+pxB+qxD+s×r+A+p×C+r xD+s xq
+B+qxC+r×D+s×p=AB+Aq+pB×C+r
×s+ AB+Aq+pB ×D+sxr+AC+Ar+pC ×
D+s×q+BC+Br+qCxD+sxp= (2) ABCs
+ABrs+ Aq+pB ×C+r × s + ABDr+ABsr+
Aq+pBxD+s x r+ACDq+ACsq+ Ar+pCx
D+s× q+BCDp+BCsp+ Br+qCxD+s × p;
from this fubtracting the equation (1), and the fe-
cond difference is ABrs + Aq+pBxC+r xs+
ABsr+ Aq+pBxD+sx r+ACsqx Ar+pC x
D+s xq+BCsp + Br+qCxD+s xp=ABrs+
ACqs+BCps, +ABrs+ĀDqr+BDpr, +ACsq+
ADrq+CDpq, +BCsp+BDrp+CDpq, &c. ´by
abridging (3) 2ABrs+2ACqs+2BCps+2ADqr
+2BDpr+2CDpq, this is the 2d difference; and the
next term is 2rs× A+p×B+q +2qsx A+pxC+r
+2ps × B+qxC+r+2qr× A+p×D+s+2prx
B+q×D+s+2pqx C+r xD+s =(4)2rsxAB+
2rs× Aq+pB +29sAC +29sx Ar+pC +2psBC+
2psx Br+qC+2qrAD+2qrx As+pD +2prBD+
2 prx Bs+qD +2pqCD+2pqx Cs+rD, &c. by a
bridging; from this laft equation taking equation
(3), the remainder is the 3d difference 2rsx Aq+pB
+2qsx Ar+pC +2ps×Br+qC +2qrx As+pD +
2 prx Bs +qD +2pqX Cs+rD = (5) 6qrs A+6prsB
+6pqs C
Sect. II.
17
METH O D.
+6pqsC+6pqrD. Then the next term is (6) 6qrsx
A+p+6prs× B+q+6pqs×C+r+6pqr× D+s;
from this equation, take equation (5), and there
remains 6pqrs+6pqrs + 6pqrs + 6pqrs=24pqrs the
fourth and laſt difference, fo the fourth difference
is = pqrs×1×2×3×4; and fo on for more progreſ-
fions.
-712
m
n2
-m
m
Cor. 1. In the ſeries a +a+p +a+2p +
a+3P, &c. the last and equal differences will be p
× into 1×2×3×4, &c. to m terms.
For then p, q, r, s, &c. are all equal.
Cor. 2. In any adfected equation of the mth power,
putting fucceffively any numbers of a feries in arithme-
tical progreffion, for the unknown quantity; the mth
differences will be equal.
الله
For the lower powers all vaniſh before you come
to the laſt difference; and therefore that difference
depends entirely on the higheſt power in the equa-
tion, which is the mth power.
PROP. V.
Let A, b, c, d, e, &c. be any feries; B, C, D,
E, &c. the firſt of the ſeveral orders of differences;
and let their poſition be fuppofed to be fuch, that
they may be at a units diftance from one another;
and let x be the diſtance of any termy (to be inter-
polated;) then I fay,
X-I
2
y=A+xB+xx: C+xx
-X
2
X-- I X--2
3
X
D+x
x-3 E + &c.
X--- I
2
X
X
X
2
3
4
or y = A+B+PC+
X 2
3
QD +
RE
3
4
+ SF+&c. where P, Q, R, &c. are the co-
X-4
5
efficients of B, C,D, &c.
C
For
18
THE DIFFERENTIAL
For (Prop. II.) the nth term of the feries is
NI N2
= A +” —¹ B + X C, &c. and the
th
I
I
n
2
*+1" term = A + −B+nx
+I
n- I
C, &c. which
2
is the term at the nth diſtance from A; then put
x=n, and the term at x diſtance from A, will be
X-I
2
P=A+B+xx ·C + ≈ X
X-I X2
2
X
3
D,
&c. And as this holds exactly, when P is any of
the terms A, b, c, d, e, &c. Therefore it will be
nearly true, when P is any intermediate term.
Cor. Hence if feveral equidistant ordinates in any
curve be given; the quantity of any other ordinate,
whofe pofition is given, may be found by this Prop.
that is, fuch an ordinate may be interpolated.
PRO P. VI. Prob.
If there be a ſeries of equidistant terms; a, b, c,
d, e, &c. whose first differences are small; to find
any term wanting in that feries, having any number
of terms given.
RULE.
Take the equation which ftands againſt the
number of given terms, in this table; and by re-
ducing the equation, that term will be found.
nla -b=0.
2a-2b+c=0.
3a3b+3c―do.
4a4b+6c-4d+e=0.
54-5b+10c-10d+5e-f=0.
6a-66 +15c-20d+15e-6f+g=0.
7a-76+21c-35d+35e—21f+78—b=0.
8 686 +28c56d+70e−56f+28g—8b+i=@.
N-I
N- I
na-nb-nx
2
2
X d+&c.=0.
1-2
3
For
Sect. II.
19
METHO D.
For (by Prop. I.) the quantities ſtanding againſt Fig.
the numbers 1, 2, 3, 4, &c. are the firft of the
firft, fecond, third, fourth, &c. differences. Now
as the first differences are ſmall, the other differences
will decreaſe, and the laſt of them may be taken
for nothing. And the more terms there are given,
the nearer the equation approximates; likewiſe
the quantity fought will be determined more ex-
actly, as it is nearer the middle of the equation ;
the error being almoft reciprocally as the coefficient.
For Ex. Suppoſe b, c, d, e, f, g, are given to
find a; againſt 6 you have a—6b+15c~20d+15e
-6f+go, which reduced gives a = 6b-15c+20d
-15e+6ƒ—g. If the third c had been fought, and
20d—15e+6f—g—a+6b
the reft given; then c=
PROP.
VII.
15
If a, b, c, d, e, &c. be any feries; and fuppofe 8.
the feveral terms placed at unequal diſtances from one
another. To find any intermediate term, whofe place
or poſition among the rest, is given.
Suppoſe the ſeries a, b, c, d, e, &c. placed as
ordinates in a curve; and let MN, MP, MQ,
MR, MS, &c. be the correfpondent bafes, for
which put n, P, Q, R, S, &c. refpectively; alfo
put NP = p, PQ=q, QR = r, RSs, &c. and let
x repreſent any bafe, and y its correfpondent term
in the feries.
Then affume y=A+Bxxn + C xx~n ×
x-P+Dxxnxx-P xx-Q + Exx¬n ×
x-P x *~Qxx-R, &c. continued to as many
terms as there are ordinates, or terms of the ſeries.
Confequently x will afcend to a power, one lefs than
the number of terms.
C 2
Then
.20
DIFFERENTIAL
THE
Fig,
8.
Then put a, b, c, d, e, &c, fucceffively for y.
and n, P, Q, R, S, &c. for x.
And theſe will produce as many equations as
there are unknown quantities, A, B, C, D, E,
&c. which therefore will be determined as follows.
a = A
b=A+BxPn
c = A + BxQ¬n +CxQ-nxQ-P
d = A+BxR―n+CxR-nxR-P+ D x
R—Ź × R—P × R—Q
e=A+BxS—n +C× S—n× S—P +Dx S—n
XS-PXS-Q+ExS n x S-P
(
x S-QXS-R,
&c. Then inftead of P-n, Q-n, Q-P, &c.
put their values p, p+q, q, &c. and you'll have
the following equations.
@ = A
a
b = A + B p
c = A+Bxp+q + C x p + q x q
4
d = A+Bxp+q+r +Cxp+q+r xq+r+Dx
p + q + r xq + r x r
e = A + Bxp+q+r+s+ C × p+q+r+s X
q+r+s+Dxp+q+r+sxq+r+s
xr+s+Exp+q+r+sxq+r+s
xr+sxs.
Then take each equation from the following one,
and you'll have,
b-a = Bp
ε → b = Bq+Cxp + q xq
i
d-c = Br+Cxp+2q+rxr+Dxp+q+r x
q + r x r
!
Sect. II.
21
METHOD.
e-d
Bs+Cxp+29+2r+s xs+Dxp+q+rx Fig.
q+2r+sxs+q+r+sxr+sxs+E
xp+q+r+sxq+r+sxr+s xs.
And dividing by the refpecting coefficients of B,
8.
and then
b-a
B.
P
c-b
=B+Cxp+q=B1.
9
dc
=B+Cxp+2q+r+Dxp+q+r x q + r
=B2.
1
e-d
===B+Cxp+2q+2r+s + D × p+q+r
S
Xq+2r+s+q+r+sxr+s+Ex
p+q+r+sxq+r+sxr+s=B3,
&c.
And again fubftracting each equation from the
next,
B1-B =Cxp+q.
B2-B1=Cxq+r +Dxp+q+rxq+r.
B3—B2=Cxr+s+ Dx p+2q+2r+s xr+s
&c.
+Exp+q+r+sxq+r+sxr+s.
And dividing by the coefficients of C,
BI-B
p + q
B2-BI
q + r
B3-B2
r+s
&c.
= C.
·=C+Dxp+q+r=Ci.
= C + D xp+2q+2r+s+ Ex
p+q+r+sxq+r+s=C2.
C 3
And
22
THE DIFFERENTIAL
Fig. And taking the differences of thefe equations,
8. CI-CDxp+q+r.
C2—C1 = Dxq+r+s+Exp+q+r+sxq+r+s.
&c.
And dividing by the coefficient of D,
CI-C
p + q + r
C2-CI
g+r+s
=D,
=D+Exp+q+r+s=D1,
&c. And taking the differences as before,
Di-D=Exp + q + r+s,
ánd
DI-D
p + q +r+s
= E, &c.
And thus you may proceed as far as you will;
whence to find the intermediate term, whoſe baſe
is x, we have this
RULE.
Let a, b, c, d, e, &c. be the feries, M any
fixed point, and
for MN, NP, PQ, QR, RS, &c.
put
N
2 B
,
r
, S., &c. and
let Ba
BI-B
C
>
-b
BI =
p + q
B2-BI
CI-C
q
C1:
>
d-c
qtr. Di
B2 =
r
જ
Cz
B3-B2,
q+r+s
D=
p+g+r E=
q
DI-D
C2-CI p+q+r+s
- &c.
e-d
r+s &c.
B3
&c.
&c.
Then the term anfwering to the bafe x, is
y=A+Bxx n +Сxx nxxn+p+Dxxnx
x−n+pxx-n+p+q+Ex x− n x x − n + p x
Xn+p+qxx-n+p+q+r, + &c.
where
Śect. II.
METH O D..
23
where M may be taken at pleaſure, and may be Fig.
either without or among the ordinates. And here 8.
x is fuppofed to lie towards the right hand; but if
it be taken to the left from M, it must be made
negative. And if any of the terms of the feries
a, b, c, &c. be negative, the ordinate reprefenting
it, muft lie below the bafe MS.
Cor. Suppofing B, C, D, E, &c. found as be- 9.
fore, and let M fall upon N at the first ordinate,
then will no, and let p, q, r, s, &c. be the feve-
ral distances of the ordinates from each other. Then
putting x for any baſe reckoned from M; then will the
correfponding term to be interpolated be,
y=A+Bx+Cxxx—p+Dxxx−pxx~p—q+Exx
x-pxx p~qxx p q r + &c. continued to
as many terms, as there are terms in the given feries.
SCHOLIU M.
They that would fee other methods of interpo-
lating the terms in a given feries, may confult the
Philofophical Tranfactions, No. 362; or Sterling's
Summiation of feries, engliſhed by Mr. Holliday.
¡
PRO P. VIII.
To find a curve line of the parabolic kind, which shall
pass through any given number of points.
RULE.
Let the points be A, B, C, D, E, &c. and 19.
from them to any right line MT given in pofition,
let fall as many perpendiculars, AM, BP, CQ,
DR, ES, &c. Then to find other points through
which the curve muſt paſs; affume any point X,
and put x for the bafe MX, and y for the corref
ponding ordinate XY. Then if the diſtances of
the ordinates AM, BP, CQ, &c. be all equal; find
C
the
24
THE DIFFERENTIAL
Fig. the value of y by Prop. IV. But if their diſtances
10. be unequal, find the value of y by Prop. VII.
JI.
which gives the point Y, through which the curve
is to pafs; and thus you may find as many points
as you will; through all which draw the curve re-
quired.
Cor. Hence, the interpolation of the terms of a
feries comes to the fame thing, as finding the ordi-
nates in a parabolic curve of fome degree or other.
As fuppofe the equation to the parabolic curve be
y=A+Bx+Cxx+Dx³ +Ex¹, &c. where the pow-
ers of x must afcend to as many dimenfions as there are
ordinates wanting 1. Therefore if fuck a parabola
be defcribed as above directed, whofe feveral ordinates
repreſent the particular terms of a given feries; then
to find any other term of that feries, is no more than
finding the correfpondent ordinate of the curve.
PROP. IX.
To find the area of any given curve.
RULE.
Thro' any number of points in it, defcribe a
parabolic curve (by Prop. VIII.); and find its
area, which will be nearly equal to the area of the
curve propoſed.
Thus let ADF be the curve, and fuppofe any
number of equidiftant ordinates to be erected, as
AM, BP, CQ, DR, &c. and thro' A, B, C, D,
&c. fuppofe a parabolic curve to pafs, and let any
baſe MX = x, and its ordinate XYy; and the
equation of the parabolic curve is y = A+Bx+
Cx²+Dx³ &c. continued to as many terms as
there are ordinates. Then to find the area of this
parabola; fuppofe x divided into an infinite num
ber of equal parts, according to the Arithmetic of
Infinites. Then (by Prop. I. Sect. I.) the fum of
all
Sect. II.
25
METHOD.
Bxx
all the Bx is =
all the A's is = Ax.
And (Prop. II.) the fum of Fig.
And (Prop. III.) the ſum of
11.
2
CX3
all the Cxx =
And (Prop. IV.) the fum of
3
Dx4
all the Dx'
and ſo on. Therefore (by
2
4
Prop. VII.) the fum of all the y's, or of all the
A+Bx+Cx²+Dx3, &c. that is, the area of the
curve is Ax+
Bx² Cx3 Dx4
2
3
+
+
3
, &c.
4
A+÷Bx+÷Cx²+‡Dx³ &c. × *, or put L for
* the length of the bafe; then the area =
A+÷Bx+÷Cx²++Dx³, &c. × L.
Now let a, b, c, d, e, &c. be the ordinates of
the given curve placed at equal diftances; and let
MP, PQ, QR, &c. each = 1. Then from the
nature of the parabolic curve, (y=A+Bx+Cx²,
&c.) putting x = 0, 1, 2, 3, 4, &c. and you'll
have the following equations.
a=A
b=A+ B+ C+ D+ E, &c.
c=A+2B+ 4C+ 3D+ 16E, &c.
d=A+3B+ 9C+27D+ 81E, &c.
e=A+4B+16C+64D+256E, &c.
&c.
Theſe are continued to as many terms as there
are equations or ordinates. Then if the quantities
A, B, C, D, &c. be expunged, by help of theſe
equations, out of the area (A+÷Bx+÷Cx², &c.
×L); you'll have the area expreffed by the ordi-
nates, a, b, c, d, &c.
I. For one ordinate.
Here a A, xo, and AL area. Therefore
AL area.
2. For
26
THE DIFFERENTIAL
2. For two ordinates.
Here x = 1,a=A, b =A+B, therefore B = bα,
ba
and the area = A+÷BxxL = a+ XL; that
2
a+b
is, the area =
L.
2
3. For three ordinates.
Here 2, and a = A
b=A+ B+ C
c=A+2B+4C
ba = B+ C
c-b=B+3C
c—2b+a = 2C
1/2 c − b + 2a = C.
B — b—a—-C — — 232a+26—2—2c.
=
Then the area A+Bx2++CX4XL becomes
a
~ 3a+2b — 2 c
2
+ } a − b + c { x L area.
4
a+c+46
or
{ a + 3 b + { cx L =
L the area.
6
4. For four ordinates.
Here 3, area = A+B+3C+22D.
and a - A
z-a
dc
4
b=A+ B+ C+ D
C=A+2B+4C+ 8D
d=A+3B+9C+27D
B+ C+ D
B+3C+ 7D
B+5C+19D
Sect. II.
27
METHO D.
c-2b+a = 2C+6D
d-2c+ b = 2C+12D
d—3c+3b—a = 6D
c—2b+a―d+3c—3b+a = 2C
or 2a-56+4c— d= 2C
and a-2b+2c2d = C
I
- a + b c + ÷d = D
2
b—a—C—D
I I
C—D =B = —÷÷÷a+3b—— 2c+3d.
whence a
that is
4.
2/2
- = a + 2 b — 2 c + d)
+3a2b+6c3dXL area.
27
9
33 b
3 a + 2/2 2 6 — 23/3/27 c + & d
a+d+3b+3c,
-L
L= area.
8.
+
5. For five ordinates.
Here x4, and the area=A+2B+C+16D
D=
256 E.
S
and a A
3
E
b = A+ B+ C+ D+
c=A+2B+ 4C+ 8D+ 16E
₫ = A +3B+ 9C+27D+ 81E
e =A+4B+16C+64D+256E
d
then ba
B+ C+ D+ E
c-2b+a=
d3c+3ba =
e-4d+6c-4b+a =
2C+ 6D+ 14E
6D+ 36E
24E
whence E=ed+ /c1b + 2 = a.
d—3c+36—a—36E — — že + Zd—2c+36 —zza.
6
•
I
C— — 26+ a — 3D —7E — 1e−3d+ 12c−1b + 15a.
C=
2
4.
3
B = b—a—C—DE ——÷E + ‡d — 3c +46-232152α.
Theſe
28 THE
DIFFERENTIAL
Theſe values of A, B, C, &c. fubftituted in the
value of the area will give the area =e+d+
736 + 156 + 3a: X L; that is, the area
2
7
7×a+e+32xb+d+12c
90
6. If fix ordinates be given.
3
Here x = 5, and the area=A+B+C+
125D+125E+3125F; and proceeding as before,
4
6
we ſhall find the area =
50Xc+d
19×a+f+75xb+e+
288
L.
288
For a A
b=A+ B+ C+ D+
E+
F
c=A+2B+ 4C+ 8D+ 16E+
32F
d=A+3B+ 9C+ 27D+ 81E+ 243F
e=A+4B+16C+ 64D+256E+1024F
ƒ=A+5B+25C+125D+625E+3125F.
ولا
4,
19
Then for
288
19
75
5_b₂
50 50
75
e
and
288 288c, 288 d 288
288f; put their reſpective values, from the equa-
tions above, and we ſhall have
19 A
288
75 A+ 75B+ 75 C+
+288
288 288
5 A+ 10 B+
+288 288
50 A+B+
+288 288
75 A+
+
288
300B +
288
75 D+
75 E+
75
F.
288
288
288
200
800
1600
C+400 D+
E+
F
288
288
288
288
450 C+
C+ 135° D+
288
4050 E+ 12150F.
288
288
288
C+
4800+ 19200E+
76800F.
288
288
288
1
475 C+
288
1200
288
+12A+ 25-B+
288 288 288
2375D + 11875E+ 59375F.
288
288
The
Sect. II.
29
METHOD.
288
288
The fum of all theſe is the area, viz.
A+ 720 B+ 2400C + 9000 D + 36000E+150000F
288
288
288
288
288
or A + 2B + 83C + 314D + 125E + 520§F,
the area, as it ought to be. And thus you
proceed to find the area, when more ordinates are
given.
Cor. 1. If we put A for the fum of the firſt and
laft ordinates, B the fum of the fecond and last but
one, С the fum of the third and last but two, &c.
and where the number of the ordinates is odd, let the
laft capital denote the middle ordinate. Alfo put L
for the length of the baſe, or the distance of the firſt
and laft ordinates; you'll have the following table of
areas against the respective number of ordinates.
nates
Ordi-Areas
IAL
2
2
AxL
3A+4B
A+
6
-XL
A+3B
XL
8
57A+32B+12C
XL
90
619A+75B+50C
XL
288
41A+216B+27C+272D
840
-XL
8751A+3577B+1323C+2989D×I
17280
989A+5888B-928C+10496D-4540E,
28350
XL
Cor.
30 THE DIFFERENTIAL
:
Cor. 2. After the fame manner, if a, b, c, d,
&c. reprefent any equidistant fections of a ſolid; A
the sum of the firſt and laſt, В the fum of the fecond
and last but one, &c. the foregoing table will give the
folidity, against the proper number of fections, very
nearly.
Cor. 3. To reduce the curve to another with fewer
ordinates; take the fum of the first and last for a new
ordinate, the fum of the fecond and laſt but one, for
another new ordinate next it, &c. and lastly, double
the middle ordinate if odd, for the last of them. You
will have a new curve with fewer ordinates; whofe
area is eaſier found, and is nearly equal to the area of
the former.
Moreover, when the ordinates of this new curve
are bad; if there be taken the sum of the first and
fecond ordinates, the fum of the third and fourth, the
fum of the fifth and fixth, and fo on. Or if you take
the fum of the three first, and the fum of the three
next, and ſo on; or if there be taken the ſum of each
four, or the fum of each five ordinates, &c. for the
ordinates of another curve, it's area will be equal ta
the area of the given one. And in each curve the baſe
must be diminiſhed in proportion to the number of or-
dinates that were added together: and thus a proper
number of ordinates, in any curve being given; its
quadrature will be reduced to the quadrature of ano-
ther curve, with fewer ordinates.
Cor. 4. Or when nine ordinates give not the area
exact enough, erect more; and divide the area into
feveral parts, and find all the parts feparately. And
it will be most expeditiously done, if you take no more
than three or four ordinates at once.
PROP.
Sect. II.
31
METHOD.
PROP.
X.
To refolve Problems by the differential Method.
RULE.
This is done by making uſe of the foregoing Pro-
pofitions, according as the nature of the queftion
requires.
When the firſt of any order of differences is
wanted, it is to be found by Prop. I.
When any term of a feries is required, it will be
had by Prop. II.
If you want the fum of any number of terms,
it is found by Prop. III.
To interpolate any term of a ſeries, then Prop.
V, VI, VII, VIII, and their Corollaries, muſt be
made ufe of, as occafion requires.
And to find the areas of curves; proceed by
Prop. IX.
When the feveral orders of differences happen
to be very great, it will be convenient to take the
Logarithms of the quantities concerned, whoſe
differences will be fmaller; and at laft when the
Logarithm is obtained, the number or quantity
belonging to it, and which was required, will eafily
be had.
In working by the differences of quantities, the
labour may in many caſes be abridged, by putting
a number of cyphers, for fo many terms at the
beginning of the feries; by which means you'll get
feveral of the differences equal to o, and fo the
anfwer will be obtained in fewer terms.
Ex.
32
THE DIFFERENTIAL.
Ex. 1.
To find the first of the 8th order of differences in
the geometrical progreſſion 1, 3, 9, 27, 81, &c.
Here (by Prop. I.) n = 8, A, b, c, &c. = 1,
3, 9, &c. whence
cô
T = 1—86+28c-56d, &c. — 1—8×3+28×9
-56×27+70×81-56×243 +28 × 729-8x
2187 +6561=32896-32640 = 256; that is,
256 is the firſt of the eight order of differences.
Ex. 2.
What is the first of the 5th order of differences, of
this feries, 1, 1, 4, 4, 16, &c.
Put A, b, c,
then (Pr. I.)
I
I
89
I
&c. n = 5.
f
d, &c. = 1, 4, 42 82
T=1- 5b+10c10d + 5e — ƒ
=1—2 1/2 + 2 1/2 — 14 +1% −32=313—3 3 5 = 36 i
that is, T-38.
T=
I
6
5
Ex. 3.
What is the first of the fourth order of differences
of the ſeries of cubes, 1, 8, 27, 64, 125, &c.
Let A, b, c, d, &c. = 1, 8, 27, 64, &c.
4. Then (Prop. I.) T = A—4b+6c—4d+e
=1−32+162—256+125 = 288—288 = o, ſo
that To, the fourth order of differences.
Ex. 4.
What is the 20th term of the ſeries 1, 8, 27, 64,
125, &c.
Here put 1, 7, 12, 6, for A,
B, C, D; n = 20; then (Prop.
A
I
B C
II.) the 20th term
will be
A
8 7
D
12
19.18
+19B + ·C+
19.18.17 D
19
6
2
2.3
27
18
Q.
=1+133 + 2052 +5814 =
8000, the 20th term of the fe-
ries.
64
37 24
6
61
125
Ex.
Sect. II.
33
METH O D.
Ex. 5.
What is the 20th term of the triangular numbers,
1, 3, 6, 10, 15, 21, &c.
Here A, B, C, D, are e-
qual to 1, 2, 1, 0; and
n = 20; therefore (by Prop.
II.) the 20th term = A+19B
+19×9C1 + 38 + 175
=210.
A
I
3
6
B C D
2
I
3
4
ΤΟ
5
15
To find the fum
progreffion 1, 2, 3,
Ex. 6.
of n terms of the arithmetical
4, 5, &c.
I,
Put a, d', d"= 1, 1, 0, then (by
Prop. III.) the fum ofn terms of the
A
d'
I
d"
I
NI
feries is n+n.
= n + \nnin =
2
3
nn + n
n+I
I
2.
2
2
Ex. 7.
To find the fum of n terms of the fquare numbers,
I, 4, 9, 16, 25, 36, &c.
—
Here a, d', d", d"" 1, 3,
2, 0; whence (by Prop. III.)
the fum of n terms is
a
d'
n+3R
I
d"
N-I
n x n - IXn-2
3
4
2
X
+
2
3
5
Q
9
2
2n+ 3nn 3n
nn3n+2
+
n
16
7
2
2
3
923
2nn-6n+4
·? +
25 9
n
6
6
D
2nn
34
DIFFERENTIAL
THE
2nn + 3n+1
6
n terms.
n+I 2n+I
n = n x
X
the fum of
2
2
3
Ex. 8.
To find the fum of n terms of the cube numbers, I,
8, 27, 64, 125, &c.
Here fubftituting the
values of a, d', d', d",
by Prop. III. we have
the fum of n terms =
a
ď
I
d"
7
d""
8
12
N I
n+nx
×7+12N
19
6
2
27
18
NI n
2
X
X
+6nx
64
37
6
24
2
3
61
N I N24
N3
125
X
X
2
3
4
7nn-7n
N46N³+IIN²-бn
=nt
+2n³ — 6n²+4n+-
2
4
n¹ + 2n³ + n²
the fum of n terms.
2
4
Ex. 9.
To find the fum of n terms of the biquadrate num
bers, 1, 16, 81, 256, &c.
Here fubftituting
a
ď
for a, d', d", d"",
O
d"
d'"'
&c. their values o,
I
I
1, 14, 36, &c. and
15
the ſum of n terms
16
of the feries 1, 16,
8 1
81, &c. being the
175
14 36 240
50 60
65 110 84
d'
24
fame as the fum of
256
194
n terms of the
625
369
feries 0, 1, 16, &c.
therefore (by Prop. III.) fubftituting n+1 for n,
the
Sect. II.
35
METH O D.
the fum will be
N4 — 2223 — N² + 2n
I
24
12 4 = = = n² +
24 25+
·ns →
5
nn + n
2
×36+
I
+
N3N
6 X14+
n5—5n4+5n³+5n²—6n
I
120
X
n + + 1 = n³n, the fum of n terms
3
30
required, of the feries 1, 16, 81, &c.
Ex. 10.
To find the fum of n terms of the feries 1, 6, 20
50, 105, 196, 336, 540, 825, &c.
This is the fame thing
as finding the fum of n+3
terms of the ſeries o, ọ,
0, 1, 6, 20, 50, &c.
Therefore writing o for a,
d', d"; and 1, 2, for d",
d'"", and n+3 for n; we
have (by Prop, III.) the fum
n + 3x n + 2x n + 1 x n
2.3.4.
n+3xn+2xn+1xnxn−Į
a
ď
d"
2"
dille
I
2
3
I
I
4
2
5
6
5
9
14
20
+
2.3.4.5
2n5+15n4+40m³ +45m² +18n
120
X 2 =
the fum ofn terms
of the feries 1, 6, 20, 50, &c.
Ex. II.
Given the logarithms of 101, 102, 104, and 105;
to find the log. of 103.
Here are 4 quantities given; therefore by Prop.
VI. againſt 4 you have a-4b+6c—4d+e=0;
where c is fought, and all the reft are given; there-
4xb+d-a+e
C
fore we have c =
6
D 2
36
DIFFERENTIAL
THE
a = 2.0043214
b= 2.0086002
d = 2.0170333
e 2.0211893
b+d= 4.0256335
4
4xb+d=16.1025340
ate= 4.0255107
6) 12.0770233
2.0128372 the log. 103.
Ex. 12.
There are given the cube roots of 45, 46, 47, 48,
and 49; to find the cube root of 50.
Here five quantities are given; whence (by
Prop. VI.); againſt 5 you have a-5b+10c10d
+5e-fo, where f is required; therefore f=
a−5b+10c—10d+5e.
N. Log.
453.556893 = a.
a = 3.556893
463.583048 = b.
10c
IOC
36.088260
473.608826 c.
5e
18.296530
48/3.634241 = d.
57.941683
49/3.659306 = e.
17.915240
f
36.342410
54.257650
ƒ= 3.684033cube root of 59.
Ex
Sect. II.
37
METHOD.
Ex. 13.
Given the natural tangents of 88°, 54′, 55′, 56',
$7', 58', 59', and 89°; to find that of 88°, 58', 18".
Tangents
A=52.080673 801436
b=52.882109 82647
B
C
D
25042
8
E
1193 F
C
c = 53.708587
26235
76
G
d = 54.561300
852713
1269
27504
83
7
I
880217
e = 55.441517
28856
1352
8
91
f=56.350590 909073
1443
30299
g
8 = 57.289962
939372
Here taking 88° 54′ from 88° 58′ 18″, the re-
mainder is 4' 18" 4.3; therefore put x = 4.3,
and (by Prop. V.) y = A+4.3xB +
4.3X3·3C +
2
4.3X3.3X2.3
6
4.3X3.3X2.3X1.3
D.+
+
24
4.3X3.3X2.3X1.3X.3
F+
120
4.3X3.3X2.3X1.3X.3X.-.7 G.
720
and collecting the terms, 52.080673
3.446175
177673
6489
134
Tan. 88 58′ 18″ 55.711144
Ex. 14.
Given the log. fines of 1° 0′, 1° 1′, 1° 2′, 1° 3′;
to find the fine of 1° 1′ 40″.
Sines
B
A = 8.2418553
C
b
8.2490332
71779
D
1168
C
c = 8.2560943
70611
38.
d = 8.2630424
69481
1130
D 3
Then
38
THE DIFFERENTIAL
Then fubtracting 1° o' from 1° 1' 40", there re-
2
mains 1′ 40″ ====*; then (Prop. V.)the fine to
3
x;
5
be interpolated is y = A+B+C-D; that is,
3
9
+ 8.2418553
+119631
6.49
2
+ 8.2538184
651
S. 1° 1' 40″ = 8.2537533
Ex. 15.
I
509 SI
I
539 54
Given the feries
I
I
30, 31, 32, 33, 34; to find
53.
that which stands in the middle between
I
52
and
To fave the trouble of reduction, take the lo-
garithms of the numbers.
Log. A
B.
I
50°
1.6989700
C
86002
D
I
57
1.7075702
I
52
1.7160033
84331
1671
E
66
I
82726
1605
8.
58
53
1.7242759
I
54°
1.7323938
81179
1547
Then put x = 2 the diſtance of the number to
2
I
50
be interpolated, from ; then by Prop. V. y =
A+B+C+DE.
I
the log. of
52.5
1.6989700
215005
+ 3133
2 I
I
1.7204727
In numbers
+3133
1.7201594 being
the number fought
Ex.
Sect. II.
39
METH O D.
Ex. 16.
Fig.
Given three distances of the fun from the tropic, 8.
and the times of obfervation. To find the time of the
Solstice.
viz. Dec. 20.
21
39
6
24
· 74.
Comparing this with Prop. VII. we have a
b=6, c=74; and p=1, q=3; then B
39,
ba
Ρ
~~~--~33, BI
BI-
cb
68
BI-B
=
= 223; C =
9
3
p+q
553/3/
I
=
4
I I
1311; then (Cor. Pr. VII.) y = 39—
2
33*+1311*X*—1, or y = 39-4611x+1311xx
o, becauſe y is nothing at the time of the fol-
ftice. Therefore xx-3.3713≈ ——2.8024, and
*—1.6856 ±√.1383.3724, and x 1.3132,
or 2.0580.
x=
Ex. 17.
Given a comet's distance from the fun on the follow-
ing days at 12 at night, to find its diftance Dec. 20.
Decem. 12
21.
24
26..
dift. 301 = a
620=b
·
715 = c
772 = d
Here 8, the time after Dec. 12; and by
Pr. VII. p = 9, 9 = 3, r = 2; and B – 319
q =
9
35.4444; BI = = 31.6666, B2=57-28.5;
95
3
=-3148, CI =-
2
3.7778
C =
3.1666
12
5
.3185
−6333 D =
14
= 0227; then
D 4
y =
40
THE DIFFERENTIAL
y = 301+35.4444X8-3148×8x-1-.0227X-I
X-4 = 301+283.5555+2.5184-.0908; that is,
y=586.9831 the comet's diſtance Dec. 20, at 12
at night.
Ex. 18.
Given five places of a comet; to find it Nov. 16d.—17h.
d.
h.
Nov. 3 17, place Leo 29
51
5
10
15
16
Virgo 3 23
15
32
18 21
Libra 18
52
20 17
28 10
Proceeding by Prop. VII. we ſhall have
Nov. 3 17
5 15
30 16
18 21
20 17
Leo 29 51 a
Virgo 3 23=b
15 32 = 0
=
I 22
221.9166 p.
5
8
1
I
I
Ι
I = 5.0417 = 9.
58.2083 = r.
20 1.8333 = s.
=
3.5333b-a.
3 32
12
9
12.15
Cab.
Libra 18 52d
33 20
33.3333 d—c.
—
9 18 = 9.3
=ed.
28 10=e
B1.8436,C .0814,D .0028,E-.0003
B12.4099,C1.1246,DI=.0016
B24.0609,C2.1008
B35.0729
from Nov. 16 17
take Nov. 3 17
remains 13
o therefore 13, whence
Cor. 1. Pr. VII.) ya+1.8436×13+.0814X13X
11.0834+.0028X13X11.0834×6.0417—.0003×13
X
Sect. II.
4T
METHO D.
X11.0834×6.0417×—2.1666 = a+38°.6982 =
Leo 29° 51'+38° 41′ 53″ Libra 8° 32′ 53″, its
place November 16d. 17h.
Ex. 19.
2
6
In the given ſeries 1,7×1,1×3×1,—×÷×1
XI,
8 6
5
--x--x+x2x1, &c. to find the term in the
X X
7 5 3 I
middle between the ſecond and third.
A
B
1
C D
I
E &c.
2
3
I
उ
23
-
3332/55
उड
3 / 3/3
23
35
8
1 6
35
then y
2
IS
IOS
5
By Prop. V.
Alfo
x = 1½/.
3
P = 3, Q 89
R:
3
M2
I 289
I
T=-
7
S=
3
2569
V = 1024, &C.
TO24
A+B+C―D+,?E—₂}F+1024G
3
28
or y=1+11/
3/18 - 80
I
3
256
I
&c.
7
896 768
II264 &C.
that is y = 2.357.
Ex. 20.
Fig.
Given the lengths of three equidistant ordinates 10.
(5, 7, and 8); and the length of the baſe MQ = 10;
to find the area of the curve AMQC.
By Cor 1. Prop. IX. AAM+CQ =5+8
= 13. BBP7, then for three ordinates, the
A+4B 13+28
area=
6
L=
41
6 X 10 = = x 10 = 683.
Ex. 21.
6
Given the length of four equidistant ordinates (5, II.
7, 8, 10), and baſe MR = 10, to find the area of
the curve AMRD.
Here AAM+DR = 15, B=BP+CQ =
15. Then for four ordinates, the area (Cor. 1.
A+ 3B
IX.) is =
8
75, as required.
L=
15+45
60
8 X10=8X10=
Other-
420
THE DIFFERENTIAL
Fig.
II.
12.
Otherwife,
Put AM = 5+10=15, and BP =7+8 =
15; then put A = AM + BP = 30; and the area
A
for thefe two ordinates is
2
Ex. 22.
30
XL = ×5=75.
2
Given the lengths of five equidiftant ordinates, (10,
11, 14, 16, 16), and the length of the baſe MS, 20;
to find the area.
Here AAM+ES = 26, BBP+DR =
27, C = 14, L = 20. Then (by Cor. 1. Pr. IX.)
7A+32B+12C,
the area is
90
-L=
182+864+168
X
90
1214
20
×20=2693.
90
Or thus,
Reduce it to 3 ordinates, putting AM
10+16
26, BP 11+16=27, CQ = 2×14= 28.
Then AAM+CQ= 54, B27; then the
162
area=
A+4B
6
X 10 =
IO
54+108
6
X 10 = 6 x 10
X
= 270.
Or thus,
In the laſt curve, the three ordinates are 26, 27,
28; make AM 26+27= 53, BP = 27+28
=55, L= 5; and put A = AM+BP = 108;
A
=
108
then the area = L= X 5 = 270.
2
2
Ex. 23.
Given the lengths of fix equidiftant ordinates, (10,
12, 11, 7, 0, 9,); bafe 20; to find the area.
Here A 10-9 1, B 12+0 = 12, C =
= =
11+7=18, L = 20; then (Cor. 1. Pr. IX.),
area
Sect. II.
43
METHOD.
area =
19+900+900
XL =
288
X
19A+75B+50C
288
1819
20 = ×20 = 1262; where note, as the or-
288
dinate VF is negative; the area SVF below the
bafe, is alfo negative.
Or thus,
Divide the area into two parts, one with four
ordinates, the other with three. For the four or-
dinates 10, 12, 11, 7; we have A = 17, B = 23;
17+69
and the area =
A+3B
8
86
X 12 =
8 X 12 =
8
X12
129.
For the three ordinates 7, 0, -9; we have
A=2, Bo; whence the area =
=-=-=x8=-2.
A+3B
8
L=
Then the whole area 129-2 127.
Ex. 24.
Let 18 ordinates be given in a curve, 0, 2, 3, 4,
6, 5, 3, I, I, -3, -4, -5, -5, -3, -I,
0, 2, 5; and the bafe 17; to find the area.
The ſhorteſt way is to take three or four ordi-
nates at once. Thus for the four ordinates o, 2,
3, 4; we have A = 4, B=2+3 = 5; and the
A+3B 4+15
area =
8
L
57
8 x3= 8 = 73
Then for the area with the four ordinates 4, 6, 5.
3; here A4+3=7, B=6+5=11; and
the area =
7+33
8 X3 = 15.
Again, with the four ordinates 3, 1, −1, −3;
here A = 0, B = o, and the area ➡o.
Then
44
THE DIFFERENTIAL METHOD.
Then with the four ordinates -3, -4, -5,
-5; here A—8, B = −9; and the area =
-8-27
X3=
8
35
8
×3=-13/3/
Then with the four ordinates -5, -3, -1,
o; here A5, B=-4; and the area -
—5—12
17
8 X3=- 8 x3=-63.
=
Laſtly with the three ordinates o, 2,5; here A5,
B2; and the area =
24
5+8
13
6 X2= = 43.
3
8
623,
249
Then the fum of all theſe areas is 7+15—13%
~~~ 6 3 3 + 4 3 = 6232 ; that is the whole area
eſteeming theſe parts negative, that are below the
baſe.
Ex. 25.
Given 8 ordinates of a hyperbola between the af
fymptotes, 10, 11, 12 IN 1 10, 10, 17; to find
I
ΙΟ
T39
I O
I O
ΙΟ
the area.
Here
a
1.00000
b
Τ
I
I
I O
C
I 2
0.90gog
0.83333
d
I O
Τ
I O
e
O
TS
I O
I
I O
IT
0.76923
0.71428
0.66666
0.625
=0.58823
then A =a+b= 1.58823, B = b+g= 1.53409.
C=c+f=1.5, D=d+e=1.48351; then
(Cor. 1. Pr. IX.) the area =
7 51 A+ 3577 B+1323 C29+89DL, where L = 7;
17280
that is, the area =
1192.76+5487.44+1984.5+4434.21
17280
5.3062, as required.
× 7=
THE
Fig.1.
2.
B
A
P
C
D
B
P
6.
5.
D
E
F
F
C
B
P
B
8.
9
M
N P
Q
3.
B
E
9.
B
7.
T
d
4.
la
p.
p
9
71
S
R
S
M
Р
X
R
S
10. B
C
E
DYE
F
A
B
TIN TE
M
P
R
S
M
P
RX
11.
V
12.
Hill
Ar. Infinites, and Dif Method.
4
the End.
Cher
श्र
OF
THE
ELEMENTS
OF THE
CONIC SECTIONS;
DEMONSTRATED
In THREE BOOK S.
Book I. Of the ELLIPSIS.
Book II. Of the HYPERBOLA.
Book III. Of the PARABOLA.
Et fi omnia à veteribus inventa effent: hoc tamen femper
novum erit, ufus et inventorum ab aliis fcientia et difpofitio.
SENEC.
THE
PREFACE.
IT
T is not my defign in this Treatife, to lay down all the
properties of the Conic Sections, but ſhall confine myſelf to
thofe that are most useful. For to fay all that has been writ-
ten, or may be written concerning thefe curves, would far
exceed my intended brevity, and be attended with no advan-
tage. Therefore ſetting afide all trifling and uſeleſs propofi-
tions, I have endeavoured to explain the chief and most confi-
derable properties of theſe curves, and I think I have omit-
ed but few things that are either useful or curious.
in
1
The most useful and remarkable curves, next the circle, are
the Conic Sections. Theſe curves make a confiderable branch
geometry. They are of great use in feveral parts of the
mathematics, particularly in Dialling, for delineating the
figns; likewife in the Projection of the fphere, where feve-
ral of the circles are projected into Conic Sections; and the
like in Perſpective; alfo in Optics, to reflect or refract the
rays to a geometrical focus; all Aftronomy is built thereon ;
the Conftruction of algebraic equations depends much upon
them; they are very useful in natural Philofophy for invef-
tigating the phoenomena of nature. A man cannot be a pro-
ficient in the abftrufer parts of the mathematics, without the
knowledge of the Conic Sections, which leads the way to the
higher geometry. And it is abfolutely neceſſary to know the
fundamental properties, for all fuch perfons, as would make any
advances in natural philofophy, and for understanding the
amazing diſcoveries of the last age, in the mechaniſm of the
univerfe.
As to this Treatife, I have comprehended all I had to fay
on this fubject, in three Books, for the three Conic Sections
Separately: which is better, at least for beginners, than
treating them altogether, in a more abstract and general man-
ner. The feveral propofitions are delivered in fuch order, as
they
ii
THE PREFACE
they seemed most naturally to follow from one another; and,
for the feveral Sections, are laid down nearly in the fame or-
der, and almoft in the fame words, that the relation they have
to one another, may the better appear.
In the Hyperbola, many of the propofitions comprehend a
great many particular cafes, which would require a vast num-
ber of cuts to defcribe them all, as they relate to the oppofite
Sections, or to the conjugate hyperbola's. Therefore I have
treated them in as general a way as I could. And any parti-
cular cafe may be easily inferred from the general propofition,
mutatis mutandis, attending duly to the nature of that pro-
pofition. And to affift the reader's imagination, in the moſt re-
markable propofitions, I have denoted fuch particular caſes by
fmall letters of the Alphabet. And in many inftances I have
done the fame thing with the other Sections, where it appear-
ed neceffary.
I have chofen to demonſtrate theſe their properties from their
fimple defcription upon a plane, without making use of the
cone, as being easier to understand. For when the demon-
ftrations are taken from the Sections of a cone, they are more
difficult to comprehend, by reafon there are so many interfec-
tions of planes with planes, and planes with folids, that the
reader is confused with them. Whereas thefe properties are
eafily drawn from their mechanical deſcription upon a plane,
and the demonftrations are more fimple and natural.
At the end of each Book I have given a collection of the
moſt uſeful Problems. And last of all I have shewn what
fort of curves thefe are, which are made by cutting a cone by
a plane in fuch and fuch pofitions; and have demonftrated
that these are the very fame curves, we have here been treat-
ing on. So I hope the reader will be furniſhed here with
thing that is neceſſary for him, in this branch of geometry.
every
W. Emerfon.
BOOK
[ 3 ]
CONIC SECTIONS.
BOOK I.
Of the ELLIPSIS.
I
DEFINITION S.
DEFIN. I.
F two pins be fixed at the points F, S; and a
thread PSFP, put about them and knotted at P;
then if the thread be drawn tight, and the point P
and the thread be moved about the fixed centers F,
S; the point P will defcribe the curve PDpBEAP,
called an Ellipfis.
4
DE F. II.
The points or centers F, S, are called the Foc.
DE F. III.
The line AB, drawn thro' the foci to the curve,
is called the tranfverfe Axis.
DE F. IV.
The point C in the middle of the axis AB, is
the Center.
The line DE,
pendicular to the
jugate Axis.
DE F. V.
(drawn thro' the center C) per-
tranfverfe AB, is called the con-
Fig.
1.
E
DEF.
4
THE ELLIPS I S.
Fig.
DE F. VI.
Any line TO, drawn thro' the center C to the
2. curve, is called a Diameter. And the extrémity T
(or O) its vertex.
DE F. VII.
If TO be a diameter, then the diameter GK,
drawn parallel to the tangent at its vertex T, is
called its conjugate. And the two diameters TO,
GK, are faid to be conjugates to one another.
DE F. VIII.
The line LR (drawn thro' the focus F, perpen-
dicular to the tranfverfe axis AB,) is called the pa-
rameter or latus rectum.
DE F. IX.
A line drawn from any point of the curve (as HI)
perpendicular to the tranfverfe axis, is called an
Ordinate to the tranfverfe. And in general, any
line drawn from the curve to any diameter TO, pa-
rallel to its conjugate GK, (as HN,) is an ordinate
to that principal diameter TO. If it go quite thro'
the figure, as Hb, it is called a double Ordinate.
DE F. X.
A right line meeting the ellipfis in one point M,
but does not cut it, is called a Tangent to it in that
point, as TM.
DE F. XI.
The part of the diameter between the vertex and
the ordinate, is called the Axfciffa: TN, AI. And
the Vertex is the extremity of any diameter.
PROP.
B. I.
5
THE ELLIPSI S.
Fig.
I.
PROP.
The fum of the lines FP, SP, drawn from the foci, I.
to any point of the curve, is equal to the tranfverfe
axis AB.
For by conftruction, PF+PS = AF+AS =
AF+AF+FS ≈ 2AF+FS. And the fame PF+
PS 2BS+FS; therefore 2AF+FS 2BS+
FS; and 2AF2BS, or AF BS. Whence
PF+PS 2AF+FS AF+BS+FSAB.
Cor. The two foci are equally distant from the ver-
texes, and alfo from the center : AF BS, and
FC SC.
=
For it is proved that AF BS; and fince AC
= CB (def. 4), therefore AC—AF — CB—BS, or
FC
SC.
PROP.
II.
A line drawn from the end of the conjugate axis, to
the focus, is equal to half the tranfverfe; DF
CA.
Draw DS to the other focus. Then the two
right angled triangles CDF and CDS are fimilar
and equal. For SC CF, the angles at C are
right, and CD common; therefore SD DF; and
fince the fum SD+DF the tranfverfe (Prop. I.),
one of them DF half the tranfverfe CA.
Cor. The distance of the foci is a mean proportional
between the fum and difference of the tranfverfe and
conjugate axes, SF² = BA+DE × BA—DE.
2
For CA DF² = DC²+CF2; and CF2 =
CA²—CD² — CA+CD × CA—CD; and 4CF²
or SF² = 2ÇA+2CD × 2CA-2CD.
E 2
PROP.
3.
6
THE ELLIPSI S.
Fig.
3.
3.
PROP. III.
The rectangle of the focal distances from either ver-
tex, is equal to the fquare of the femiconjugate: AFX
FB DC².
For DC² = DF²-CF² = (Prop. II.) CA²-
CF² = (Geom. I. 12.) CA+CF x CA-CF =
BC+CF × CA-CF = BFXFA.
PROP.
IV.
As the tranfverfe axis to the conjugate, fo the con-
jugate to the latus rectum of the tranfverfe: AB: DE
:: DE: LR.
=
2
For SL+LF BA=2CA (Prop. I.); and
SL = 2CA-LF, and by fquaring (Geom. I. 11.),
SL² = 4CA²-4CAXLF+LF".
4CA×LF+LF. And in the
right angled triangle SLF, SL - SF²+LF²;
whence 4AC2-4ČA×LF+LF2 = SF2+LF2,
and 4AC²-4CA×LF — SF² = 4CF², and 4AC²
=4CA×LF+4CF² = 4CA×LF+4DF2—4DC².
and 4AC²+4DC 4CAXLF+4DF2; but CA?
= DF² (Pr. II.); therefore 4DC² = 4CA×LF =
2CAX2LF; that is, DE BAXLR.
Cor. 1. As the femitranfverfe is to the femi-conju
gate, fo the femiconjugate to half the latus rectum ;
CA: DC:: DC: LF.
Cor. 2. As the femitranfverfe, to the distance of
the focus from the center; fo is the fame distance, to
the difference between the femitranfverfe and half the
latus rectum: FC² CAXCA—LF.
2
2
For CF DF-DC (Pr. II.) CA-
CD² = CA?-ÇA×LF (Pr. IV. Cor. 1.)
2
Cor:
B. I.
7
THE ELLIPSIS.
Cor. 3. The rectangle BFA = half the tranfverfe Fig.
X half the latus rectum CAXFL. By Cor. 1. 3.
and Prop. III.
=
SCHOLIU M.
Since the tranfverfe axis is to the conjugate, fo
the conjugate to the latus rectum, of the tranfverſe
axis. Therefore in any other diameters, the third
proportional, to the diameter and its conjugate, is
called the latus rectum of that diameter.
PRO P. V.
From any point M in the curve, drawing the lines 4.
MF, MS, to the two foci; and the ordinate MP per-
pendicular to the tranfverfe axis BA; it will be,
As the femitranfverfe, CA:
to the distance of the focus from the center, CF ::
So the distance of the ordinate from the center, CP:
to half the difference of the lines MF, MS, or
=
2
MS-MF
2
2
For make SD = CA, then SM CA+DM,
and FM = 2CA-SM — CA-DM. In the right
angled triangle SMP, SM² or CA²+2CA×DM+
DM' (Geom. I. 10.) = SP²+PM² = CF+CP
+PM² = CF²+2CFxCP+CP²+PM², and in
the right angled triangle FMP, FM or CA--
2CA×DM+DM² = FP²+PM² = CF-CP
+PM² (Geom. I. 11.) CF-2CFXCP+CP²+
PM²; then fubtracting the latter equation from the
former, SM-FM24CAXDM = 4CFX CP,
and CFXCP CAXDM. But fince SM CA
+DM, and FM CA-DM; therefore SM-
FM 2DM; therefore CF x CP CA ×
=
SM-FM
2
=
=
X
2
E 3
Cor.
8
THE ELLIPSI S.
Fig. Cor. 1. If F, S be the foci, MP an ordinate;
4. then it is CA: CF:: CP: CA-MF or SM-CA.
For CFXCP = CA×DM, and DM = SM—
CA CA-FM.
Cor. 2. If F, S be the foci, MP an ordinate; then
the difference of the Squares of the lines SM, FM;
that is, SM²-FM² = 4CF×CP.
2
Cor. 3. If F, S be the foci, MP an ordinate;
then CAXSM-FM
For SM2 FM²
2CFxCP.
SM+FM x SM-FM
(Geom. I. 12.) — 2CA × SM—FM
2CA × SM-FM = 4CFX CP,
and CA X SM-FM = 2CF×CP.
SCHOLIU M.
If PM falls on the other fide of F, as pm, then
Cp-CF, and its fquare the fame as before,
and the reft of the demonftration the fame.
pF
PROP. VI.
If an ordinate MP be drawn to the tranfverfe
axis; it will be,
As the fquare of the tranfverfe, BA² :
to the fquare of the conjugate, NE² :
So the rectangle of the fegments of the tranf. BPA:
to the Square of the ordinate, PM2.
For make SD = CA, then DM is half the dif-
ference of SM and MF; therefore by Prop. V.
CA : CF : : CP : DM, and (Propor. 13.)
CA: CA+CF or BF:: CP: CP+DM, and (Pro.4.)
CA: CP:: BF: CP+DM, and (Propor. 13.)
CA: CA+CP or BP: BF: BF+CP+DM.
But BF BC+CF SD+CF; and BF+CP
+DM = SD+CF+CP+DM = SM+CS+CP
=SM+SP; whence
CA: BP :: BF: SM+SP.
Again,
B. I.
9
THE ELLIPSIS.
Again, fince CA: CF:: CP: DM; then
(Pro. 13.) CA; (CA-CF) AF:: CP: CP-DM; and
(Propor. 4.) CA : CP :: AF : CP—DM, and
(P. 13.)CA:(CA-CP) PA:: AF: AF-CP+DM.
But AFCA-CF — SD-SC; therefore AF
-CP+DM SD-SC-CP+DM SM-SP;
therefore
CA: PA:: AF: SM-SP, and we had
before, CA: BP :: BF:SM+SP; then multiply-
ing theſe proportions together (Propor. 18), we have
CA: BPXPA:: BFXFA: SM-SP (Geom. 12.)
But (Prop. III.) BF×FA = CN², and Geom. II.
21. Cor. 1.) SM²-SP² PM'; therefore CA² :
BPA :: CN² : PM², or alternately
CA²: CN²::BPA: PM²
2
2
2
or BA (4CA): NE² (4GN2): BPA: PM*.
Cor. 1. CA: CN2:
BPA: PM².
Cor. 2. As the tranfverfe BA :
to its latus rectum : :
So the rectangle BPA :
to Square of the ordinate PM*.
Fig.
4.
NE2
For (Prop. IV.)latus rectum =
whence fince
AB'
BA: EN2 :: BFA : PM2.
therefore BA :
NE 2
BA
or latus rectum:: BFA: PM
2
(Propor. 6.).
Cor. 3. The rectangles of the fegments of the tranf
verfe, are as the fquares of the ordinates.
2
For every rectangle is to the fquare of its ordi-
nate, in the given ratio of CA to CN', or of BA
to the latus rectum.
Cor. 4. As the Square of the femitranfverfe CA²:
Rectangle of the focal distances from vertex BFA::
So rectangle of the fegments BPA:
Square of the ordinate PM².
E 4
PROP.
ΙΟ
THE ELLIPSIS.
Fig.
5.
6.
T
PROP.
VII.
If the ordinate MG be drawn perpendicular to the
conjugate DE, it will be,
=
As the fquare of the femiconjugate, DC:
to fquare of the femitranfverfe, CA² : :
So the rectangle of the fegments of conj. DGE :
to fquare of the ordinate, GM2.
For by Cor. 1. Pr. VI. CA²: CD² :: BPA : PM².
But BPA AC+CP X AC-CP (Geom. I. 12.)
AC²-CP2 AC-GM2; therefore CA:
CD´ :: AC²—CP² : PM, and alternately CA² :
CA-CP²:: CD: PM'; and by divifion CA²:
€A²~CP²
PM²
CP2: CD2: CD2-PM or CD CG; and al-
ternately CA²: CD':: CP: CD2-CG or CD+CG
× CD-CG that is, EGXGD; or CA²:CD': :
GM: DGE.
Cor. The whole ordinate to the tranfverfe or con-
jugate axis, is biffected by that a is; and equal or-
dinates are equally diftant from the center.
For the rectangle of the fegments of the axis
being in a given ratio to the ſquare of the ordinate ;
and that rectangle being the fame, for the ordinates
on both fides; thefe ordinates muſt be equal. Like-
wife that rectangle remains the fame, at equal dif-
tances on each fide the center; and therefore the
ordinates are equal at equal diſtances.
PROP. VIII.
If MP be an ordinate, to the tranfverfe; F, S the
foci; and you make CA: CF:: CP: CI; then the
rectangle SMX MF CA-CI2.
For make SD = CA, then DM =
SM-FM
2
and by Prop. V. DM =
CFXCP
CA
= CI by con-
ftruction;
'B. I.
II
THE
ELLIPSI S.
ftruction; and SM×MFCA+DM × CA—DM Fig.
= CA-DM2 CA-CI2.
=
Cor. 1. If CA: CF :: CP: CI; then FM =
AI, and SM = BI.
For DM CI, and SM
=
ÈM
CA+DMCA
+CI BI, and therefore FM = AI.
Cor. 2. CA: CF:: CP: CA-FM or SM-CA.
PROP. IX.
6.
If F, S be the foci, C the center, MP an ordinate 6.
to the tranfverfe, CE the femiconjugate; then CM²
= CA²+CE’—SMF.
2
=
2
For in the triangle SMF, (Geom II. 28), SM
+MF² = 2CF2+2CM². But SM+MF = SM²
+2SM x MF+MF2; therefore SM+MF² =
SM+MF² 2SMX MF BA-2SMXMF,
whence BA2-2SM× MF2CF²+2CM². There
fore 2CM² BA22CF-2SMX MF = 4CA²
-2CF2-2SMX MF; and CM 2CA-CF2
2 =
-SMXMF CA²+CE-SMxMF.
Cor. 1. If CA: CF:: CP: CI; then CM = EL.
It follows from this and the laſt Prop.
Cor. 2. If CA: CF:: CP: CI; then rectangle
BPA rectangle SIF+PM*.
For BPA
=
2
CA-CP2 - CA2-CM²+PM*
CA²—CM²+PM²
SMF-CE+PM² = (VIII) CA²-CI²-CE
+PM² = CF³—CI²+PM² = SIF+PM²,
PROP. X.
If from the foci, two lines SM, FM be drawn to
any point in the curve, as M; they will make equal
angles with the curve in that point.
Take m infinitely near M in the curve, and draw
Sm, Fm; and from S and F defcribe the finall
arches
12
THE ELLIPSIS.
Fig. arches Mr, mt, thro' M and m. Then fince SM
7. +MF = Sm+mF, by Pr. I. that is, SM+Mt+
tF = Sr+rm+mF; and taking away the equal*
quantities SM, Sr, as alfo tF, mF; there remains
Mt = rm. Therefore in the infinitely fmall tri-
angles Mrm, and Mtm, right angled at r and t, we
have Mt mr, and the hypothenufe Mm common;
therefore the fides Mr and mt are equal, and the
oppofite angle Mmr and mMt; that is, MmS and
MF are equal; but being infinitely near M, the
angle MS GMS the external angle at M;
whence mMF = GMS.
Cor. If a line MO be drawn perpendicular to the
tangent MT in M; the angles FMO and SMO are
equal.
For they are the complements of the equal angles
FMT and SMG.
PROP.
XI.
8.
If MP be an ordinate to the tranfverfe; and MO
perpendicular to the tangent at M; and S, F the foci;
then CA : CF::CP:CÓ.
2
For (Geom. II. 25.) in the triangle SMF, where
MO biffects the angle SMF, it is SM: MF:: SO:
OF; and by compoſition (Propor. 13.) SM+MF:
SM-MF SO+OF: SO-OF; that is, 2CA:
SM-MF:: 2CF : 2CO :: CF : CO, and
SM-MF
CA² :
V.)
2
SM-MF
2
X CA :: CF : CO. But (Prop.
× CA=CF× CP; therefore CA² :
CFXCP:: CF: CO:: CF2: CFXCO. And al-
ternately CA²: CF² :: CF×CP: CFXCO :: CP :
CO.
Cor. 1. CA: CD² : : CP : PO.
För
B. I.
13
THE
ELLIPSIS.
For CA
2
CF :: CP: CO; and by divifion, Fig.
CA² : CA²-CF2 or CD²:: CP : CP-CO or PO. 8.
CP:CP-CO or
Cor. 2. 2CA: latus rectum :: CP: PO.
For CA: CD:: CP: PO, and 2CA:
or the latus rectum : CP: PO.
Cor. 3. CA: MF:: CF: OF.
2CD2
CA
For we had 2CA: SM-MF:: CF: CO, and
by divifion 2CA: 2CA-SM+MF:: CF: CF-
CO; that is, 2CA: SM+MF-SM+MF:: CF:
OF; that is, 2CA: 2MF, or CA: MF:: CF : FO.
PROP.
XII.
If S, F be the foci, and MO perpendicular to the 8:
tangent in M; then MO² SMX MF-SOXOF.
For fince MO biffects the angle SMF; there-
fore (Geom. II. 26.) MO +SOF SMF; and
MO² SMF-SOF.
2
Cor. 1. MO=
CD
CA
SMXMF; or CA:
CD²:: SMF : MO².
For (Geom. II. 25.) SM : MF:: SO: OF, and
by compofition, SM+MF or 2CA: MF:: SO+
OF or SF: OF.
or CA CF :: MF: OF, in like
:
manner CA CF :: SM: SO, and multiplying,
CA: CF:: SMF: SOF;
2
and CA: CA-CF2 or CD2: :SMF:SMF-SOF.
that is, CA²: CD² : : SMF : MO², by this Prop.
Cor. 2. SMF = MO²+BOA—CD².
For SMF MO²+SOF. But SOF SC²-
CO² = BC2-CD2-CO² = BQA-CD².
PROP.
14
THE ELLIPSIS.
Fig.
9.
10.
PRÓ P. XIII.
If thro' any point M, the line SG, equal to BÁ,
be drawn from the focus S, and FG from the other
focus F; and the line MD be drawn to the middle of
FG; then MD will touch the curve in M.
For draw FM, then MG AB-SM FM;
therefore MD is perpendicular to FG, whence FM
- GM.
To any other point m, in the line DM; draw
Sm, Gm, Fm; then in the triangles GmD, FmD
the angles at D being right, and the including
fides equal, the hyp. mGmF. In the triangle
SGm, the fum of the two fides Sm+mG is greater
than the baſe SM+MG, (Geom. II. 5.); that is,
Sm+mF is greater than SM+MF; and therefore
by conſtruction of the ellipfis, the point m is with-
out it; and fo are all points except the point M.
Cor. 1. The tangent at M, biſſects the external
angle FMG, made by two lines drawn from the
focus.
Cor. 2. The tangents at A and B are perpendicular
to the tranfverfe axis AB; likewife a tangent at the
end of the conjugate is perp. to the conjugate.
PROP. XIV.
If PM be an ordinate, AD a line drawn from
the ends of the tranfverfe and conjugate axes, and
PQ parallel to AD; then PM² CD²—CQ®
DQXQE.
For by the fimilar triangles ACD, PCQ; AC:
CD:: PC: CQ; and AC: CD:: PC: CQ2; and
CD²
(Propor. 13. laft cafe) AC: CD:: AC-PC2:
CD-CQ. But (Prop. VI.), AC: CD:: BPA
CD²
or
B. I.
15
THE ELLIPSI S.
or AC²-PC¹: PM²; therefore PM² = CD¹-CQ Fig.
or CD+CQ × CD—CQ.
PRO P. XV.
If MP be an ordinate to the tranfverfe axis, and II.
MH be drawn perpendicular to the curve at M,
cutting the tranfverfe and conjugate axes at O and
H. It will be as OP: OC :: CD²: CF¹. F being
the focus.
Z
:
For (by Prop. XI.) CA² : CF²::CP: CO. And
by divifion, CA-CF CF :: CP-CO: CO;
that is, CD: CF:: OP: CO.
CF²
Cor. 1. CD: CF:: MO: OH: : MP: CH::
PO: CO.
This follows from the fimilar triangles MOP
and HOC.
Cor. 2. If L be the latus rectum of AB; then,
L:AB-L::MO:OH::OP:OC :: MP : CH.
For CD² : CF² or CA²-CD² : : MO: OH, &c.
CD2
CA
(by Cor. 1.); therefore : CA
CD2
::MO:
CA
OH (Propor. 5. Cor. 1.); that is, (by Pr. IV. Cor. 1.)
L: CA-L:: MO: OH; or L: BA-L:: MO:
OH, &c.
Cor. 3. CA: CD::CH+PM: PM:: CP: OP.
For CD: CF2 :: MP: CH; and compounding,
CD² : CD²+CF² (CA²):: MP: MP+CH:: CP:
OP.
PROP. XVI.
If PM be an ordinate to the tranfverfe, MT a
tangent at M; then CP, CA, CT are continually
proportional; C being the center.
For draw MO perp. to MT, then (Prop. XI.
Cor. 1.) CA: CD2 :: CP: PO:: CP²: CPX PO,
and
12.
16
ELLIPSIS.
THE
Fig. and drawing PQ parallel to AD, then CA² : CD² : :
12. CP2: CQ2, by fimilar triangles; whence CP2: CPX
PO : : CP² : CQ; therefore CP×PO = CQ²; by
::
Ax. 7. Proportion. But in the right angled trian-
gle OMT, where MP is perp. to OT, OPX PT
PM² (Geom. II. 20. Cor. 2.) = CD²—CQ²
(Prop. XIV.) = CD-CPXPO, by what is juft
proved. Therefore OPXPT+OP×CP = CD²;
that is, OPXCT = CD².
13.
But OPXCT: CPXCT:: OP: CP:: (Pr. XI.
Cor. 1.) CD CA; that is, CD2: CPXCT::
CD2: CA2.
Therefore (Ax. 7. Proport.) CPxCT = CA²;
or CP: CA : CT÷÷÷÷.
Cor. 1. If C be the center, PM an ordinate, MT
a tangent, MO perp. to MT; then OP, CD, CT,
are continually proportional, or OP×CT = CD².
Cor. 2. If C be the center PM an ordinate; then
BP: PA :: BT: AT; and therefore BT is har-
monically divided in the points P, A.
For fince CT: CA: CA: CP; therefore
(Propor. XIII.) CT+CA: CT-CA:: CA+CP:
CA–CP;
that is, BT: AT:: BP: AP.
PRO P. XVII.
If MG be an ordinate to the conjugate, MV a
tangent at M, C the center; then CG, CD, CV,
are continually proportional.
By fimilar triangles CV: VG:: CT: GM or
CP (laft Prop.) CA: CP or GM² (Prop. VII.)
:: CD2: DGE or CDCG².
And by divifion CV: CV-VG or CG:: CD:
CG²; whence (Propor. 12.) CVXCG² = CGX
CD or CV×CG = CD2; therefore CG: CD:
CV are
Cor.
2.
M
Fig. 1.
D
P
H
L
A
B
B
B
E
4.
M
3.
D
B
S
F
I
S
F
B
m
R
5.
D
E
G
M
m
7.
E
M
8.
B
6.
E
Το
B
S
co PF
F
E
A
A
CIP F
11. D
M
9.
m
10.
D
M
M
B
PE
B
H
Conic Sections
E
E
PL.I pa 1.6.
UNIV
M
OF
B. I.
THE ELLIPSIS.
17
13.
Cor. 1. If C be the center, MG an ordinate to Fig.
the conjugate, MV a tangent, and MH perp. to the
tangent MV; then HG, CA, CV, are continual
proportionals.
For CGXCV = CD2; but CGXCV: HGXCV::
CG or PM: HG (by fimilar triangles)::OP: GM or
CP (Prop. XI. Cor. 1.) :: CD²: CA²; that is,
CD²: HGXCV:: CD²: CA; therefore (Ax. 7.
Propor.) HGXCV
2
CA¹.
Cor. 2. If C be the center, MG an ordinate to
the conjugate, MH perp. to the tangent MV; then
EG: GD:: EV: DV; and EV is harmonically di-
vided in the points G, D.
For fince CV: CD:: CD: CG, by compofition
and
divifion, CV+CD:CV-CD::CD+CG:CD—CG,
that is, EV: DV : : EG : DG.
PROP. XVIII.
If a femicircle AFD be defcribed on the tranfverfe 14.
or conjugate axis AD of an ellipfis AKD, and the
ordinate PM be drawn cutting the circle in I; the
tangents MT, IT, drawn from the points M, I,
will interfect the axis in the fame point T.
For (by Prop. XVI, and XVII.) the tangent MT
will interfect the axis in T, fo that CP, CA, CT,
be in geometrical progreffion. And (Geom. IV.
26.) If a tangent IT be drawn to touch the circle
at I, it will interfect the diameter in T, fo that
CP, CA, CT be in geometrical progreffion, the
fame as in the ellipfis.
PROR
81
THE ELLIPSI S.
Fig.
1.5.
16.
PROP. XIX.
If on the tranfverfe axis AB as a diameter a cir-
cle AKB be defcribed; and the ordinate PM be drawn
cutting the circle in N; then PN will always be to
PM, in the given ratio of the femitranfverfe CA, to
the femiconjugate CD.
For (Geom. IV. 17.) PN² = BPA; and (Prop.
VI.) CA²: CD²:: BPA or PN2: PM²; and there-
PN²
fore CA: CD:: PN: PM.
2
!
Cor. If a circle be defcribed on the conjugate axis
DE as a diameter, and an ordinate QR be drawn
cutting the circle in I; then QI will always be to QR,
in the given ratio of the femiconjugate CD to the femi-
tranfverfe CB.
=
There-
For (Geom. IV. 17.) QI DQE. And (Prop.
VIL) CD²: CB²:: DQE or QI² : QR².
fore DC: CB:: QI: QR.
PROP. XX.
:
If the line FD be drawn from the focus, perpendi
cular to any tangent MD; then the interfection D
will be in the circumference of the circle ADB, de-
fcribed on the diameter BA the tranfverfe axis.
For produce SM from the other focus, and FD
till they interfect in G. Draw CD, FM. Then
in the right angled triangles FMD, GMD. The
angle FMD (by Prop. XIII. Cor.) GMD, and
MD being common; therefore FD DG, and
FM MG; and fince FC = CS; therefore CD
is parallel to SG (Geom. II. 12. Cor. 1.). There-
fore the triangles FCD and FSG are fimilar; there-
SM+MG
=
=
fore, finçe FD=4FG, CD=4SG=
2
B. I.
19
THE ELLIPS I S.
SM+MF
2
= ¿BA = CA; therefore D is in the
circumference ADB.
And by the fame reaſoning, if S is the other fo-
cus, and SH be perp. to MH; the point H is in
the circumference of the ſame circle ADB.
•
Cor. 1. If a circle be defcribed on the tranfverfe
AB, and a tangent at M, cuts it in D, H; and the
lines FD, SH, be drawn from the foci, to the points
D, H; then FD, SH are perp. to the tangent HMD.
Cor. 2. If F, S be the foci, C the center, MD @
tangent at M; then if CD be drawn parallel to SM,
to cut the tangent in D; then CD = CA half the
tranfverfe.
Both thefe corollaries follow from the demon-
ftration of this Prop.
Cor. 3. If a tangent HD interfects the circle AHB
in D and H; the perpendiculars to it, DF, HS, will
pass thro' the foci F and S.
Cor. 4. If CI be drawn parallel to the tangent at
M; then IM
CA.
For then IM CD, and CD = CA, by Cor. 2.
=
PROP. XXI.
If S, F be the foci, HT a tangent at M; if
SH, FD be drawn perp. to the tangent MT; then
SHXFD BSXSA.
Deſcribe a circle ARH about the tranfverſe, and
produce HS to cut the circle at R; and draw CR.
Then fince <H is right, RHD (Geom. IV. 14.)
is a femicircle = BHA; therefore RB = DA, and
<RCB = < DCA. Therefore in the triangles
SCR, FCD, the fides SC, CR, are equal to FC,
CD, refpectively, and the included angles at C
equal; therefore SR DF. (But Geom. IV. 20.
Cor. 2.) HSXSR BSXSA; that is, HSX DF
= BSXSA =BF×FA.
=
Cor.
Fig.
16.
17.
20
THE ELLIPS I S.
Fig. Cor. The rectangle HSFD = Square of the ſe-
17. miconjugate. This is plain by Prop. III.
17.
18.
PRO P. XXII.
If HT be a tangent; FD, SH perpendiculars on
it, from the foci; then HM: MD:: HT: TD.
For defcribing the circle BHA, about the tranſ-
verſe BA, and drawing SM, FM; then the tri-
angles SMH, FMD are fimilar, for the angles at
H and D are right, and SMH = FMD (by Prop.
X.); therefore HM: MD:: HS: DF. Alſo the
triangles TSH and TFD are fimilar; therefore HS:
DF: TH: TD; whence HM: MD:: TH: TD.
Cor. Hence HM: MD:: TC+CF: TC-CF.
For HM MD:: TH: TD:: (by fimilar tri-
TH:TD::(by
angles) TS: TF:: TC+CS: TC-CF.
PROP. XXIII.
If HMD be a tangent at M, and SH, FD, per
pendiculars, on it from the foci S, F, CE the femi-
conjugate. Then the rectangle HMD SMF-ČE.
For deſcribing the circle BHDA, and drawing
the ordinate PM, and producing it to the circle at
N, and V; we have (Geom. IV. 20. Cor. 2.), HMD
= NMV = NP+PM × NP-PM — NP²—PM²
NP+PM×NP—PM
BPA-PM² (Geom. IV. 17.) BPA-CM²+
CP²; but BPA — CA²—CP², and BPA+CP²
CA². Alſo CM² = CA²+CE²—SMF (Prop.
IX.); therefore HMD CA2-CM CA²—
= = =
CA-CE+SMF SMF-CE.
2
2
PROP
B. I.
21
THE ELLIPSI S.
XXIV.
PROP.
Fig.
If S, F be the foci, HMD a tangent at M; FD, 18.
SH perpendiculars on the tangent HD, CE the ſemi-
conjugate. It will be SM : MF: : CE² : FD²; and
MF: MS :: CE: SH'.
For (by Prop. X.) the angle SMH = FMD;
therefore the triangles SMH, FMD are fimilar;
whence SM: MF::SH: FD::SHXFD : FD² : :
(Prop. XXI. Cor.) CE² : FD².
And FM: MS:: FD: SH:: FDxSH or CE2:
SH².
PROP. XXV.
If MP be an ordinate to the tranfverfe, HMT a 19.
tangent at M, C the center, AK, CG, BH perp. to
AB. Then AK: PM:: CG: BH.
For (Prop. XVI.) TC : CA :: CA: CP. And
by compounding, TC: TC+CA or TB:: CA:
CA+CP or BP. And (Propor. XIII.) TC: TB::
TC-CA: TB-BP; that is, TC: TB::TA:TP.
But all the triangles TAK, TPM, TCG, TBH
are fimilar; therefore TA: TP :: AK: PM, and
TC: TB:: CG: BH; whence AK: PM :: CG:
BH.
Cor. TA: TP::TC: TB,
PROP.
XXVI.
If PM be an ordinate, C the center, AK, CG,
BH perpendicular to the tranfverfe BA; HMT a
tangent at M; then AKXBH = PMX CG = CD¹,
the Square of the femiconjugate.
For (Prop. XXV.) AKXBH = PMXCG; and
(Prop. XVII.) PM: CD:: CD: CG; whence PM
XCG
CD3.
F 2
Cor.
19.
22
ELLIPSIS.
THE
Fig.
=
Cor. 1. AKXBH PMXCGBA x latus
19. rectum BSA, S being the focus.
20.
This follows from Prop. IV. Cor. 1. and Prop. III,
Cor. 2. Lines drawn from H and K to the focus,
make a right angle HSK.
For by Cor. 1. HB: BS :: SA: AK; therefore
(Geom. II. 16.) the triangles HBS, SAK are fi-
milar, and < BHS
ASK.
-
But HSK = HSA-ASK HSA-BHS =
HBA a right angle.
Cor. 3. Hence a circle defcribed on the diameter
HK will pass thro' the foci S, F.
PROP. XXVII.
If S, F be the foci, HMD a tangent at M, MO
perpendicular to it, and OG perp. to SM; then MG
is half the latus rectum.
For draw SH, FD perp. to HD; then (by Prop.
XXI. Cor.) SH×FD = CE², the fquare of the
femiconjugate. And by the fimilar triangles SMH,
FMD, MOG, MS: SH:: MO : MG,
and MF: FD:: MO: MG;
:
and multiplying, MF×MS:: SH×FD:: MO²:MG².
or (XXI. Cor.) MFXMS MO¹ :: CE² : MG².
But (XII. Cor. 1.) MF×MS: MO² ::CA²: CE².
therefore CA: CE :: CE² : MG²
:
2
or CA CE::CE: MG latus rectum,
by Prop. IV. Cor. 1.
Cor. 1. If MO be perp. to the tangent at M; L
=latus rectum of the tranfverfe axis; then AB-L:
L::reclangle SOF: MO.
2
For we had before SMF: MO²:: CA²: CE² or
CAX÷L::CA: 4L; and by divifion SMF-MO² :
MO² :: CA—÷L: L :: BA—L:L; but (Prop.
XII.) SMF-MO³ — SOF.
Cor.
Fig.12.
D
M
M
G
13.
B
C OP
A
T
B
P
T
H
E
Ε.
K
14.
K
M
F
K
M
M
15.
A 4
R
B
A
B
P
T
H
16.
M
E
D
B
S
18.
H
יד
A
B
G
H
M
C PF A
17.
M
K
19.
B
PF
Pl.II. pa.22.
UN
OF
MI
B. I.
23
THE ELLIPSI S.
Cor. 2. If MH be perp. to the curve at M, cut- Fig.
ting the two axes in O and H; and S, F be the foci. 11.
Then SOF MOH.
For SOF: MO² :: BA-L: L:: (Prop. XV.
Cor. 2.) HO MO:: MO×HO: MO²; therefore
SOF MOH.
Cor. 3. Therefore a circle may be deſcribed from
Some point of the conjugate CD; which will pass thro'
the foci, S, F, and alſo thro' M and H.
PROP. XXVIII.
If FH be an ordinate at the focus, TH a tangent 21.
at H, PM any other ordinate, continued to the tan-
gent at G; then this line PG = FM, the line drawn
from the focus to the top of the ordinate.
I
Z
1. For FH is the latus rectum; and (XXVI.
Cor. 1.) FHXCN
XCA; whence CN
I
latus rectum x CAFH
2
CA.
2. Draw AL, BZ, perp. to AB; then (Prop.
XVI.) CA: CF:: CT: ĈA. And by divifion
(CA-CF) AF: CF :: (CF-CA) TA: CA; and
compounding, AF: (AF+FC) ÁC:: AT: (AT
+AC) TC:: (fimilar triangles) AL: CN or AC;
therefore AF = AL.
3. Likewife (XVI. Cor. 2.) AF: FB:: TA:
TB: (fimilar triangles) AL or AF: BZ; there-
fore BZBF.
4. By Prop. XVI, CA² = CFXCT
CF+FT
CFXFT.
CF x
CF+CFT, and CA-CF2 or CD² =
5. By fimilar triangles, TF: FH:: TC: CN or
CA :: TCXCF or CA: CA×CF:: CA : CF;
that is, TF: FH :: CA: CF.
6. By the laſt art. CA: CF :: TF: FH:: (fimi-
lar triangles) TP : PG.
F 3
And
24
ELLIPSI S.
THE
Fig.
And (Prop. XVI. and VIII.CT: CA
21. Cor. 2.) CA: CF::
22.
CA}
SCP: FM-CA
(compounding) CT+CP or TP : FM; therefore
TP: PG :: TP: FM, whence PG FM.
Hence are drawn the following corollaries,
Cor. 1. If TZ be the focal tangent ; AL, CN,
BZ perpendiculars on AB; then CN
-
Cor. 2. AL = AF, and BZ = BF.
Cor. 3. CFXFT = CD².
Cor. 4. TF: FH:: CA: CF.
Cor. 5. TA: AF:: TP: FM.
CA.
For by fimilar triangles, TA: (AL) AF :: TP :
(PG) FM.
Cor. 6. CA: CF :: TP : FM.
For CA: CF:: TF: FH:: TP PG or FM.
PROP. XXIX.
If AB be the tranfverfe axis, C the center, F the
focus; and if CF: CA:: CA: CT; and TE be perp.
to TB. Then if from any point M, MF be drawn
to the focus, and ME parallel to TB; it will be CF :
CA::FM: ME.
For if the ordinate FH be drawn, and the tan-
gent TH, then (Prop. XVI.) it will be CF: CA::
CA: CT. And therefore (Prop. XXVIII. Cor. 6.)
CF: CA :: FM: FP, or CF : CA:: FM : EM.
Cor. MF is to ME always in a given ratio, where-
ever the point M is taken.
SCHOLIU M.
The line TE is by the writers on conic fections,
called the Directrix.
PROP.
B. I.
25
THE ELLIPSI S.
PRO P. XXX.
Fig.
If BA, DE, be the tranfverfe and conjugate axes; 23.
and if the difference between the femitranfverfe and
Semiconjugate, be fet from the point G in the conju-
gate, to I in the tranfverfe, and continued to M, till
IM = femiconjugate CD. Then M will be in the
curve of the ellipfis.
For draw MP perpen. to AB, and MR parallel
to it, then the triangles GRM and MPI are fimi-
lar; therefore GM: RM :: IM: IP; that is, CA:
CP IM: IP; and CA: CP2:: IM2: IP2. And
by diviſion CA: CA-CP2 or BPA :: IM or
CD2: IM²-IP2.
But IM²-IP
2
2
2
PM. Therefore CA: CD2 ::
CD²
BPA: PM²; which is the property of the ellipfis
by Prop. VI.
Cor. If from any point M in the ellipfis, MI =
CD (the femiconjugate) be drawn to the tranfverfe,
and continued to the conjugate at G; then MG is
equal to the femitranfverfe CA.
PRO P. XXXI.
If two circles BDA, KLV be defcribed about the 24.
tranfverfe and conjugate diameters; and CD be drawn,
and from D, DP perpendicular to BA; and from L,
LI perp. to DP; then the point I will be in the el-
lipfis.
For fince LI is parallel to CA, CD: CL:: DP:
IP (Geom. II. 12. Cor. 4.); that is, CA: CG::
DP: IP, which is the property of the ellipfis by
Prop. XIX.
F 4
PRO P.
26
THE ELLIPSI S.
Fig.
25.
26.
PROP. XXXII.
Any diameter MN is biffected in the center C.
From M draw the ordinate MP on the tranfverfe
AB; take CQ = CP, and erect QN perp. to AB
to cut MN in N. Then the triangles CPM and
CQN are fimilar, for all the angles are equal, alfo
CP; therefore CN
CQ
CM.
PROP. XXXIII.
If ML, VR be two conjugate diameters, and
from the ends thereof, there be let fall the perpendi-
culars MP, VQ ordinates, to the tranfverfe axis AB;
then CQ2 AP×PB.
For (Prop. XVI.) CP: CA:: CA: CT.
And CP: (CA-CP or) PA:: CA: (CT-CA
or) AT: (Propor. XIII.) CP+CA or BP: PA+
AT or PT; whence CPXPT APXPB-AC-
CP². And AC = CPXPT+CP2.
2 =
Alfo, fince CV is parallel to the tangent MT
(def. 7.), the triangles TPM and CQV are fimilar.
And AQB AC-CQ CPXPT+CP----
CQ, and (Prop. VI. Cor. 3.) APB : AQB:: PM² :
QV² (fimilar triangles): : TP²: CQ; that is, CPX
PT: CPXPT+CP-CQ²:: TP: CQ, and by
permutation CP×PT: TP² :: CP×PT+CP².
CQ₂: CQ; and by adding,CPXPT+TP:: TP²::
CP X PT + CP2: CQ; and by permutation,
CP²
CP+PT × PT: CP+PT x CP:: TP² : CQ²;
or TP CP:: TP² : CQ², whence TPXCQ²
CPXTP², or CP×TP = CQ², or AP×PB=CQ².
:
Cor. 1. AQXQB = CP².
ForCP — AC²_CPT-CA’—CQ²=CA+CQ
X CA-CQ = AQxQB.
Cor.
B. I.
27
THE
ELLIPSI S.
2
Cor. 2. CP²+CQ₂ = APB+AQB = CA².
For APB
Z
CQ AC-CP2,
and AQB = CP₁ = AC-CQ, and adding,
APB+AQB = CP²+CQ² = 2AC²—CP²—CQ²
and 2CP2+2CQ² = 2AC²
or CP² + CQ₁ = AC².
CP+CQ
Cor. 3. If ML, VR be two conjugate diameters,
MP, VQ ordinates to the tranfverfe axis; then
PM²+QV² = CD, the Square of the conjugate.
For CA: CD: BPA or CQ: PM²:: BQA
or CP² : QV²,
and CA: CD²::CQ+CP2: PM²+QV, and al-
ternately
2
CA²: CQ²+CP or CA :: CD: PM²+QV,
therefore (Propor. 2.) PM²+QV² = CD².
Cor. 4. CA: CD::CP: QV :: CQ : PM.
For CA: CD::: BPA or CQ2: PM².
and CA²:CD² :: BQA or CP² : QV²;
whence CA: CD :: CQ : PM :: CP: QV.
PROP. XXXIV.
Fig.
26.
The fum of the Squares of two conjugate diameters, 26.
is equal to the sum of the squares of the two axes:
ML²+VR² = AB+DE2.
2
2
For CM²
2
CP+PM², and CV CQ²+
CM²+CV²
QV²; whence CM'+CV
CP²+CQ₂+PM²+
QV²
V² = (Pr. XXXIII. Cor. 2.) CA²+PM²+QV²
= (ib. Cor. 3.) CA²+CD².
PROP.
XXXV.
If MN, KR be two conjugate diameters; then the 27.
rectangle of the distances of the foci, from the vertex,
of the diameter MN is equal to the fquare of its fe-
miconjugate CK: FMXMS CK².
-
For (Geom. II. 28.) FM+MS
2FC2; but FM+MS
2MC²+
2AC, which fquared is
FM²+
28
THE ELLIPS I S.
í
Fig. FM²+MS+2FMxMS = 4AC, from which fub-
27. tracting the firſt equation, 2FM×MS = 4AC²
2MC²-2FC², or FMX MS = 2AC-MC-FC².
But AC-FC² = CD², whence FMS AC²+
CD²-MC²; but (Prop. XXXIV.) AC²+CD²
CM²+CK², therefore FMS CK².
=
28.
Shorter thus,
By Prop. IX. CM² — CA²+CD²
fore SMF = CA² + CD²
SMF, there-
= CA² + CD² - CM²; but (Prop.
XXXIV.) ADz+CD² = CM²+CK², and AC²+
CD2-CM² CK SMF.
2
=
Cor. 1. If F, S, be the foci, then FMXSM =
MCX latus rectum of the diameter MC.
I
For KRMNX latus rectum of MN (Prop.
IV. Schol.), and dividing by 4, KC² = MCX;
latus rectum
SMF.
3
Cor. 2. If MO be perp. to the tangent at M ; then
CA: CD:: CK : MO.
2
For (Prop. XII. Cor. 1.) CA²: CD²:: SMF or
CK2: MO², and CA: CD:: CK: MO.
PRÒ P. XXXVI.
If F, S, be the foci; TH a tangent at any point,
M; FT, CI, SH perpendiculars on the tangent;
MC, CK femidiameters conjugate to one another;
then FM is to FT, or SM to SH, as CK to CD the
femiconjugate axis.
For fince CF = CS, FT+SH = 2CI. The
triangles FMT and SMH are fimilar; whence MF:
FT: MS: SH (Propor. XIII.) MF+MS:
FT+SH: 2AC :: 2CI :: AC : CI.
Again, AC CI:: MF: FT
:
and AC: CI:: MS: SH by fimilar triangles,
there-
B. I.
29**
THE ELLIPSI S.
therefore AC² : CI² : : MS× MF: SH×FT,
Fig.
but (Prop. XXXV.) MS×MF = CK², and (Prop. 28.
XXI. Cor.) SH×FT=CD², therefore AC² : CI² : :
CK2: CD2 :: MF: FT.
CK² CD²
and AC: CI:: CK: CD :: MF:FT:: MS: SH.
Cor. MF FT:: AC: CI:: CK: CD.
From the demonftration.
PRO P. XXXVII.
All circumfcribing parallelograms, whofe fides are
parallel to two conjugate diameters, are equal to the
rectangle of the two axes. GHNQ = RTXZ.
For let CI be perp. to GH which touches the
ellipfis in M; then (Prop. XXXVI. Cor.) BC:
CI:: CK: CD, therefore CIXCK BCXCD;
that is, MCKH = DCBT, and 4MCKH=4DCBT;
that is, GHNQ = RTXZ.
PROP. XXXVIII.
29.
If BA be the tranfverfe axis, MV any diameter; 30.
AE, MT, two tangents at A and M, interfecting
BA and VM in T and E; then the triangle CAE
= CMT.
Draw the ordinates MP, AD, to CA and CM;
then the triangles CPM, CAE are fimilar, and
likewife CDA, CMT, are fimilar; whence CD:
CM :: CA: CT :: (Prop. XVI.) CP: CA::
CM: CE; therefore (Geom. II. 12. Cor. 1, 4.)
the lines DP, MA, ET are all parallel to one ano-
ther; whence (Geom. II. 10), triangle AME =
triangle AMT; to which add the triangle CMA,
then triangle CAE triangle CMT.
Cor.
30
THE ELLIPS I S.
Fig.
Cor. 1. The lines DP, MA, ET are parallel to
30. one another.
31.
From the demonftration.
Cor. 2. The triangle AIT = triangle MIE.
This appears by taking CAIM from the equal
triangles CAE, and CMT.
Cor. 3. Triangle MPT = triangle ADE = tra-
pezoid MPAE= trapezoid MDAT.
For fince AME AMT, add the triangle AMP;
then PMT MPAE, or add AMD (AMP),
then ADE = (MPT =) ADMT.
Cor. 4. Triangle CDA = triangle CPM.
For CD: CM:: CP: CA; whence CDA =
CPM (Geom. II. 17. Cor. 1.)
PROP.
XXXIX.
If BA be the tranfverfe axis, MV a diameter;
MP, QL ordinates; AE, MT, tangents; QIF perp.
to AB; then the triangle QIR = trapezoid IAEF.
By fimilar triangles CA: AE:: CP: PM :: CI:
IF; and CA: AE:: (Propor. XIII.) CA+CP:
AE+PM:: CA+CI: AE+IF.
But rectangle BIA = AI × CA+CI, and rec-
tangle BPA = AP × CA+CP; therefore BIA :
BPA :: AI × CA+CI : AP × CA+CP :: AI ×
AE+IF.
AE+PM
:: trapezoid AEIF :
2
PAEM.
: APX
2
The triangles MPT, QIR are fimilar, whence
triangle MPT : QIR:: MP: QI:: BPA: BIA::
trapezoid PAEM: trapezoid AEIF; but MPT =
PAEM by Prop. XXXVIII. Cor. 3. therefore
QIR = AEIF.
The
&
B
Fig. 20.
H
E
G
1
G
Z
M
*
IM
H
со F A
22.
E
M
B
A F
PC
B
B
25.
27 M
A
R
P
K
F
S
B
N
23. D
R
E
F
C P
21.
24.
G
1
P
BK
C
P VA
26.
R
B
E
H
M
D
K
T
B
Η
F
с
S
29.
R
28.
T
D
G
B
E
2
X
PL.III. pa.30.
UNIL
OF
MIC
B. I.
31
THE
ELLIPSIS.
¡
The demonſtration is the fame thing for the point Fig.
9, uſing the ſmall letters inſtead of the great ones. 31.
Cor. The triangle QLF the trapezoid LMTR.
For (Prop. XXXVIII.) CAE = CMT, fubtract
CLR from both, and then LRAE LMTR, but
LRAELRIF+FIAELRIF÷RQI=LQF
= LMTR.
PROP. XL.
2
=
Let CK be the femiconjugate to MC, LQ an or- 31.
dinate thereto; then CM²: CK :: rectangle MLV :
LQ Square.
Draw the tranfverfe axis AB, and KO perp. to
it. Then (Cor. Prop. XXXIX.) FQL = LMTR,
and for the fame reaſon, OKC CMT. But
triangle CMT: CLR :: CM²: CL, and CMT :
CMT-CLR:: CM³: CM²-CL²; that is, triangle
CMT : trapezoid LMTR :: CM²: rectangle VLM.
But the triangles OKC, FQL are fimilar; there-
fore CK: LQ²: : triangle OKC : triangle FQL;
or CK² : LQ² : : triangle CMT: trapezoid LMTR::
CM² : rectangle VLM; or CK²: CM² : : LQ² : rec-
tangle VLM.
MV²
Cor. 1. As the tranfverfe MV: to its latus rectum ::
fo the rectangle MLV: to fquare of the ordinate LQ².
For MV: latus rectum :: MV2: MV × latus
rectum or 2CK (Sch. Prop. IV.): rectangle
MLV: LQ.
2.
:
Cor. 2. The rectangles of the fegments of any dia-
meter, are as the fquares of their ordinates.
Cor. 3. The diameter biffects all its double ordi-
nates; and the ordinates are equal at equal distances
from the center.
For
32
THE ELLIPSI S.
Fig. For the proportion being the fame for the ordi-
31. nates, at the fame point, on each fide; thefe ordi-
nates muſt be equal. And fince all the terms of
the proportion remain the fame at equal diftances
from the center; the ordinates will be equal at e-
qual diſtances.
32.
PRO P. XLI.
If two right lines GI, HP be drawn parallel to
33. two conjugate diameters DK, MV, to interfect in
R; then CM²: CK:: rectangle HRP: rectangle
GRI.
For (Prop. XL.) PN:: IO:: DNK: DOK;
and (fig. 3.) PN² : PN²—IO² :: DNK : DNK-
DOK. But PN-IO² = PN²-RN²
PRH.
And DNK-DOK = DN×NO+OK-DN+NO
xOK = DNxNONOXOK = DN—OK×NO
= GR×NO = GRI.
PN²
24
Alfo (fig. 33.) PN: IO-PN :: DNK: DOK
-DNK. But IO-PN = RN-PN2 PRH;
and DOK-DNK DOX ON+NK-DO+ON
—
× NK = DOXON-ON×NK
=DN-OK X ON
DO-NK X ON
GRXON GRXRI.
Therefore PN2: PRH:: DNK: GRI,
PN²
or PRH: GRI:: PN2: DNK :: CM2: CD2.
PN²
CM²
Cor. Hence if two other lines be drawn parallel
to HP, GI; the rectangles of their fegments will be
as the rectangles PRH and GRI parallel to them.
PROP
B. I.
33*
THE ELLIPSIS.
Fig.
PROP. XLII.
If a line as HG interfect any diameter AB in D, 34.
and a diameter FK be drawn parallel to HG; then 35.
the rectangle ADB : rectangle GDH: : as CB² : CF².
Draw the conjugate MV, and AI parallel to GH;
and the triangles CDE, CAI, are fimilar, whence
CAXCE
CI
=DC, and CI : AI : ;
CI: CA :: CE :
AIXCE
CE: DE =
CI
; alfo (by Prop. XL.) MC* :
CE2XCF
2
CF² :: CM²-CE²; EHCF" —
CM²
Hence (fig. 34.) rectangle ADB = AC-CD² =
2
AC2
CA2XCE
CI-CE2
CI2
= CA² × X
CI.Again CF:
CM²
CF2XCM². But the rectangle GDH
AI: CM-CI, whence CM XAI² =
CFXCM²-CF'xCI2, and CM XAI+CFXCI² =
CE2XCF2 AI2XCE²
EH-ED²
= CF2
CM²
= CF2
CI2
CE²xCF²×CI²+AI³×CE²×CM³
CM²XCI
= CF2
CE³×CF²×CM²
CE¹×CF²
CM2XCI2
= CF2
= CF2 x
C12
CI²-CE2
CI2
CA: CF2.
therefore rectangle ADB : GDH::
And in fig. 35, the demonſtration is the fame,
only ADB = CD-AC², or — ADB AC
CD2, and -GDH EH-ED; whence the fame
conclufion will follow.
PROP.
34
THE ELLIPSIS.
Fig.
36.
37.
38.
39.
PRO P. XLIII
If two lines GH, LE interfect one another in D,
and the femidiameters CF, CK be drawn parallel to
them; the rectangle GDH is to the rectangle LDE::
CF² : CK².
Thro' D draw the diameter AB; then by the laſt
Prop. rectangle GDH: rectangle 'ADB:: CF2:
CB². And rectangle ADB : rectangle LDE :: CB²:
CK. Therefore GDH: LDE:: CF2: CK².
Cor. 1. If any line AB cut two parallels GH, EF,
in D and L; the rectangle ADB : rectangle GDH::
rectangle ALB: rectangle ELF.
For EF, GH, being parallel to the fame dia-
meter; theſe rectangles will be as the fquares of
the femidiameters parallel to theſe lines.
Cor. 2. If two parallel lines EF, GH, cut other
two parallels MN AK; their feveral rectangles will
be in the fame ratio; ABK: GBH:: MDN: GDH::
ACK: ECF:: MLN: ELF.
For each pair will be feverally proportional to
the fquares of the femidiameters parallel to thefe lines.
PRO P. XLIV.
If DM be a tangent at M, and DN any line cut-
ting the ellipfis in O and N; and if the femidiameters
CF, CK be drawn parallel to DM, DN; then DM²:
CDXDN
CD x DN:: CF² : CK².
Although this may be inferred from Prop. XLII,
by fupppofing the points G, H, to coincide in M
(fig. 35.); yet I fhall give the demonſtration of it
here in particular, in refpect of the diameter DACB.
Thro'
B. I.
THE ELLIPS I S.
35
Thro' the center C draw DAB, and draw AI
rallel to DM.
CI²
By fimilar triangles CI: AI :: CM²: DM",
and (Prop. XL.) AI²: MIV:: CF² : CM²,
Therefore,
(Propor. XVII.) CI²: MIV :: CF² : DM²,
Again by ſimilar triangles,
pa-
CI CM²:: CA: CD, and
by divifion, CI² : CM²-—CI² : : CA² : CD²—CA².,
2
that is, CI: MIV: CA': ADB.
whence CA: ADB:: CF: DM.
But (Prop. XLII.) CA²: ADB:: CK²: ODN.
whence CF2: CK :: DM²: ODN.
24
Fig.
39.
Cor. 1. If DM, DN be two tangents, and CF, 49.
CK be drawn parallel thereto from the center C; then
DM: DN :: CF : CK.
For drawing DCB; DM: ADB:: CF: CB2,
and ADB DN² :: CB2: CK; whence DM²:
:
DN² :: CF²: CK², and DM: DN:: CF : CK.
Cor. 2. If two parallel tangents DM, TG, meet a 41.
third tangent DNT; then DM: DN:: TG: TN.
For if CK be drawn parallel to DT, and CF to
DM and TG; it will be DM: DN:: FC: CK,
and TG TN:: CF: CK; therefore DM: DN::
TG:TN.
Cor. 3. If PQ be parallel to the tangent DT, and 42.
RS to the two tangents DM, TG; then POQ:
ROS :: DN² : DM² : : TN²: TG².
For drawing CK, CF parallel to PQ, RS, by
Prop. XLIII. CK²: CF² : : POQ : ROS (and Cor.
1.) DN: DM (Cor. 2.) :: TN: TG.
1
Cor. 4. If M, N, G, be the points of contact of 43.
three tangents, MD, DT, TG; and MD, TG, OË
be parallel to the femidiameter CF; and DT, HR
parallel
G
36
THE
ELLIPSIS.
Fig. parallel to the femidiameter CK; then CF2: CK ::
IOE: NO² :: GT² : NT² :: GR²: LRP : : MH² :
`: ::
PHL.
43.
All this appears from what was before demon-
ftrated.
PROP.
XLV.
44.
45.
If DM, DN, touch the ellipfis in M and N, and
MN be drawn, and any line RL parallel to DN,
cutting DM, MN, in G and H; then GR, GH,
GL are continually proportional.
For (Prop. XLIV. Cor. 4.) RGL GM::
DN: DM²:: (fimilar triangles) GH: GM; there-
fore RGL GH², and RG: GH:: GH: GL.
Cor. If FP be parallel to DN, and touches the
ellipfis in P, and cuts DM in O, then OP
OF.
For then GL will fall on FP, R and L upon P,
G upon O, and H upon F; whence FOGH,
GR and GL OP; whence OP FO², and
OP — FO.
PROP. XLVI.
If MT, a tangent in M, cut the diameter BA in
T, and the ordinate MP be drawn; then CP, CA,
CT, are continually proportional.
For draw AK, BH tangents at A and B, to cut
MT in K and H. Then fince AK, PM, BH
are parallel, BP : AP::HM: MK :: (Prop. XLIV.
Cor. 2.) HB AK (fimilar triangles): : TB: TA.
By divifion BP-AP: BP :: TB-TA: TB; that
is, 2PC: BP:: BA: TB; and taking half of the
antecedents PC: PB:: BC: BT. And by divifion,
PC: PB-PC:: BC: BT-BC; that is, PC: BC::
BC: CT.
Cor.
B
D
Fig 30
D
M
E
32.
35.
P I
d
37.
G
1
2
B
P
LA
T
33.
M
D
H
C
M
H
E
L
I
B
B
F
h
38.
M
31.
L
R
P
M
F
K
34.
H
M
H
R
36.
G
e
M
E
G
39.
B
N
C
M
I
FK
h
H
F
B
K
F
B
N
Pl.N.pa.36.
UNIV
M
OF
B. I.
37
THE ELLIPS I S.
Cor. 1. If MT be a tangent at M, MP an or- Fig.
dinate; BP: PA:: TB: TA; and therefore BT is 45.
barmonically divided in A and P.
This appears from the demonſtration.
Cor. 2. If MT be a tangent at M, MP an ordi-
nate; then PC: PA:: PB: PT.
For by the demonftration, PC: PB:: BC: BT;
and by divifion PC: PB:: BC-PC: BT-PB : :
AP: PT,
Cor. 3. If MT be a tangent at M, and MP an
ordinate; then TA:TP:: TC : TB.
For fince CT: CA:: CA: CP, CT-CA: CT::
CA-CP : CA; that is, AT: CT:: AP: AC; and
(Propor. XIII.) AT:CT:: AT+AP:CT+AC,
or AT: CT::TP: BT.
Cor. 4. Two tangents drawn from equal ordinates
on each fide the curve, meet in the diameter.
PRO P.
XLVII.
If AB be a diameter, and AK, BH two parallel 46.
tangents, at A and B, and HT a tangent at M, cut-
ting the others in K and H; MP an ordinate, CD a
femidiameter parallel thereto; then AK, CD, BH are
in continual proportion.
Continue CD to G. And by fimilar triangles
AK: PM: TA: TP:: (Prop. XLVI. Cor. 3.)
TC: TB:: (fimilar triangles) CG: BH. Alfo fince
the tangent MG cuts CD in G; therefore (Prop.
XLVI.) PM: CD:: CD: CG. Therefore (Propor.
XVII.) AK: CD:: CD: BH.
Cor. 1. If AB be any diameter, CD its femiconju-
gate, PM an ordinate; AK, BH, HM, tangents at
A, B, M; then AKXBH PMXCG
ABX the latus rectum.
4
CD² =
G 2
For
38
ELLIPSIS.
THE
Fig. For fince AK: PM:: CG: BH; therefore AKx
46. BH PMX CG
47.
48.
=
ABX latus rectum.
CD = (Schol. Prop. IV.)
Cor. 2. If the two parallel tangents AK, BH, be
cut by the two tangents KH, GR, then AKXBH =
BGXAR.
For they are both equal to CD¹.
Cor. 3. AK: BG:: KO: OH.
For AK: BG:: AR: BH:: AR-AK: BH-
BG :: KR:GH (fimilar triangles): : KO: OH.
Cor. 4. A line drawn thro' RH wou'd interfect the
line drawn thro' KG, in the diameter AB produced.
Cor. 5. If the tangent GR was to touch the oppofite
fide BDA, as gr; it will still be AKXBH = Bg×Ár.
PROP. XLVIII.
If two tangents HK, HE, touching the ellipfis in
D and R; cut any diameter AB produced, in K and
E, fo that AE = BK; and if FG be a third tan-
gent; then the rectangle EFxGK = KHXRE, a
given quantity.
Compleat the parallelogram KHEL, by draw-
ing the lines EL, KL parallel to KH, HE, which
will touch the ellipfis in O, P. Draw the diame-
ter RP; then the figure RAC is fimilar and equal
to PBC, and REC to PKC, whence RE = PK.
Then in reſpect of the diameter RP it will be (by
Prop. XLVII. Cor. 3.) PK or RE: FR :: KG :
GH; and by compounding, RE: FE :: KG: KH;
whence KGXEF REXKH, a given quantity.
Cor. 1. VEXKZ = KH×RE a given quantity.
For by fimilar triangles VE: EF :: GK: KZ,
whence VEXKZ = EF×GK = KH×RE.
Cor.
B. I.
39
THE
ELLIPSI S.
▼
Cor. 2. And if the third tangent (FG) touch the Fig.
curve on the other fide (OP), and interfect the tangents 48.
HK, HE produced; the rectangle EF×GK will ſtill
be the fame.
For the figure on the other fide of KE, will be
the fame as VEKZ, and therefore that rectangle
is the fame as EVXKZ = KHXRE.
PROP. XLIX.
If two lines TF, TG, touch the ellipfis in H and 49.
K, and if HK be drawn, and biffected in O; then if
the line TO is drawn thro' O, it will biffect all the
lines IN within the ellipfis, which are parallel to HK.
That is, if HO = OK, then IP = PN.
Thro' the center C, draw FG parallel to HK;
then fince HO = OK, FCCG. And fince DE
paffes thro' C, it will be a diameter, and HO, OK
ordinates parallel thereto; and IP, PN, being pa-
rallel thereto, will alſo be ordinates; therefore (Prop.
XL. Cor. 3.) IP = PN.
Cor. 1. If LM be drawn parallel to HK; the
parts intercepted between the curve and tangent, on
each fide, are equal; LI MN.
For fince HO OK; therefore LP PM, and
IP being
PN, the remainder IL = NM. Like-
wife fince CDCE, therefore DF EG, &c.
Cor. 2. The tangent at A is biffected in that point.
Cor. 3. HO, OK, as alfo IP, PN are ordinates
to the diameter paffing thro' T, the interfection of the
tangents.
G 3
PROP.
40
THE ELLIPSIS.
Fig.
50.
51.
PROP. L.
If two lines TL, TM, touch the ellipfis in H and
K, and HK be drawn thro' the points of conta&;
and from T, any line TI be drawn, cutting HK in D,
and the curve in F and I; then TF: TI:: DF: DI.
Thro' F, I, draw AC, LM parallel to HK, then
(Prop. XLIX. Cor. 1.) LI NM, and AF =
BC, whence MI LN, and CF AB.
By fimilar triangles AF: LI:: TF: TI.
and CF: MI:: TF: TI.
and multiplying, AF×CF : LIXMI :: TF² : TI³,
that is, AFX AB: LIXLN:: TF: TI.
But (Prop. XLIV. Cor. 3.) AFX AB:LIXLN::
HA: HL: (fimilar triangles) DF: DI; therefore
TF2 TI²:: DF2: DI, and TF: TI:: DF: DI.
Cor. If it be TF : TI :: DF : DI; the line join-
ing the points of contact, (of the tangents TH, TK,)
will pass thro' the point D.
PROP.
LI.
If MT, AH be two tangents at M and A; and
52. if the femidiameters CM, CA be drawn, and pro-
duced to interfect the tangents in H and T; then the
triangle MNH = triangle ANT.
Draw PM parallel to AH; then by fimilar tri-
angles CM: CH:: CP: CA:: (Prop. XLVI.) CA:
CT; and alternately CM: CA:: CH: CT; whence
(Geom. II. 17. Cor. 1.) the triangle CAH tri-
angle CMT. Then fubtracting (or adding) the
quadrilateral figure CANM, the triangle MNH
ANT.
Cor. The triangle CAH triangle CMT.
From the demonftration.
SCHOL
D
42.
Fig. 40.
41.
D
N
N
R
N
T
T
P
G
QG
Ꮐ
C
C
C
M
M
M
K
K
B
D
0
N
T
H
43
G
C
F
M
K
P
SF
44.
R
E
F
H
G
M
D
K
46.
H
M
K
B
45.
A
B
C P
H
47:
B
{
H
**
48.
G.
R
•E
D
7'
}
f
D
B
I
T
PlV. pa. 40.
OF
B. I.
41
THE ELLIPSI S.
SCHOLI U M.
What has been demonftrated in Prop. XXXVIII,
and Cor. 1. in regard to the tranfverfe axis; is
here proved univerfally for any diameter whatever.
PROP. LII.
Fig.
52.
If ML, RV be two conjugate diameters, and MP, 53.
VQ ordinates to any other diameter AB, DE the
conjugate of AB; then CQ² = AP×PB,
This Prop. is demonftrated from Prop. XLVI,
and Prop. XL. Cor. 2. in the fame manner as
Prop. XXXIII. is from Prop. XIII. and Prop. VI.
Cor. 3. And the fame corollaries will follow, viz.
Cor. 1. AQxQB = CP².
Cor. 2. CP²+CQ
APB+AQB = CA².
Cor. 3. PM²+QV² = CD².
Cor. 4. CA: CD:: CP: QV:: CQ: PM.
Cor. 5. Hence alfo the triangle CPM - CQV.
For imagine PT = PC, then triangle PTM =
PCM, and (Cor. 4.) CP or PT : QV:: CQ: PM;
therefore. (Geom. XVII. Cor. 1.) CQV = TPM
= CPM.
PROP. LIII.
If ZH be a tangent at any point A; ML, VR 54.
two conjugate diameters cutting the tangent in Z and
H; DE the conjugate to AB; then the rectangle
ZAH CD2.
Draw the ordinates MP, VQ, to the diameter
AB; then by fimilar triangles CP: PM:: CA: AZ,
whence CPX AZ = PMX CA (Cor. 4. Prop. LII.)
= CQXCD; therefore CP: CQ:: CD: AZ.
G4
Again,
42
THE
ELLIPSIS.
Fig. Again, CQ: QV:: CA: AH, and CQXAH=
54. QV×CA = (ib.) CPxCD; therefore CP:: CQ:
AH: CD:: CD: AZ; whence AZXAH = CD².
55.
Cor. If SN be drawn parallel to CD; then SFX
FG+FN.CD².
For CA: AZ:: CF: FG (by fimilar triangles).
and CA: AH:: CF: FS,
whence CA: AZXAH or CD: CF2: FSXFG.
CD² CF²
and dividing, CA: CD²:: CA-CF or AFB:
CD-SFG :: (Prop. XL.) AFB FN'. Therefore
CD-SFGFN2, and FN2+SFG = CD².
PROP.
LIV.
24
If TP, TQ be two tangents at P, Q; and PQ
be drawn, and any line LD, cut PQ and the curve in
S, A, B; then ŠL² : SD²: : rectangle ALB : rec-
tangle BDA.
From the center C, draw CG, CM, CF parallel
to TP, TQ, LD; and draw LI parallel to QT.
Then the triangles SLI, SQD are fimilar, as alfo
PIL, PQT; whence SD: SL :: DQ : LI. But
the ratio of DQ to LI is compounded of the ratio
of DQ to PL, and (PL to LI or) PT to TQ;
therefore DQ: LI :: DQ×PT : PLXTQ; there-
fore SD: SL :: DQ×PT : PL×TQ. Take the
line E a mean proportional between AL and LB;
and H a mean proportional between BD and DA;
then ALB E', and ADB = H².
Then (Prop. XLIV.) GC: FC:: LP: ALB or
E. And FC: MC:: BDA or H: DQ'. There-
fore extracting the roots, GC: FC:: LP: E, and
FC: MC:: H: DQ; therefore GC: MC:: PLXH:
DQXE; but (Prop. XLIV. Cor. 1.) PT : QT ::
GC: MC :: PLXH: DQXE, and multiplying
means and extremes, PTXDQXE = QTXPL×H.
whence
B. I.
43
THE
ELLIPSI S.
whence H: E:: DQXPT: PLXTQ:: SD: SL; Fig.
and H or ADB: E2 or ALB:: SD2: SL2.
Cor. Hence in any circumfcribed trapezium LNDO,
the diagonals LD, NO, interfect in the fame point S,
where the lines PQ, KR, interfect, that join the op-
pofite points of contact.
For if LD cuts the tangents PL, QD; we have,
SL² : SD² : : ALB or EĚ : ADB or HH; and
SL: SD:: E: H, and SL+SD or DL: SL::
E+H: E.
2
Again, for the tangents KL and RD, it will be
SL²: SD² :: ALB : ADB; whence as before it
will be, LD:SL::E+H : E. Therefore PQ
interfects LD in the fame point S where KR inter-
fects it; or the diagonal LD paffes through the
common interſection S, of the lines PQ, KŘ. And
by the ſame reaſoning the diagonal NO paffes thro'
the fame interfection S.
PROP.
LV.
55.
If GM be any diameter, GT, MD, two tangents, 56.
at G and M, DT a tangent at N; then if NC be
drawn, and its femiconjugate CK; the rectangle DNT
= CK2.
For draw the ſemiconjugate CF parallel to MD;
then
(Prop. XLIV. Cor. 1.) CF: CK :: DM: DN
and CF: CK: TG: TN; then
multiplying, CF2: CK:: DMX TG: DNXTN.
CF²:CK²
But (XLVII.) CF² = DM×TG, therefore CK'
= DNXNT.
PROP.
44.
THE ELLIPSIS.
Fig.
PROP.
LVI.
57.
58.
If AF, AG be two tangents, and FG drawn thro'.
the points of contact. And thro' A, the interfection
of the tangents, AV be drawn parallel to GF. Then
if from any point in it as V, there be drawn VL thro'
O, the middle of FG, to cut the ellipfis in P and L.
Then will VP: VL:: OP: OL.
From L draw AQRL, and from the interfection
Q, draw QSP parallel to FG. Then (Prop. L.)
AQ: AL::RQ : RL, and AL×RQ = RLXAQ.
The ratio of PQ to SQ is compounded of the ra-
tio of PQ to OR and OR to SQ. But by fimilar tri-
angles, PQ: OR:: LQ: LR; alfo OR: SQ:: AR:
AQ; therefore PQ: SQ:: LQXAR :LRXAQ; and
by divifion, PS: SQ :: LQXAR-LRXAQ:LRX
AQ. But LQXAR-LR×AQ = LR+QR×AR
·LRXAQ = QR×AR+LR×QR = AL×QR
= LRXAQ, by what went before; therefore PS
=SQ; and confequently, as QS is an ordinate
(def. 9.), P is in the curve. Laftly, by reafon of
the parallels FG, PQ, VA; VP: VL::AQ:AL::
(Prop. L.) RQ: RL:: OP: OL.
PROP.
LVII.
If FA, GA be two tangents at F and G; and if
GF be drawn, and AV parallel to it. And from ang
point V in the line AV; let the tangents VH, VI, be
drawn; then the line drawn thro' the points of con-
tact, HI, will paſs thro' the point Ọ, the middle of FG..
Thro' O draw VL cutting the ellipfis in P and
L; then in reſpect of the ordinate FG, it is (Prop.
LVI.) VP: VL::OP: OL. And in refpect of
the
1
F H
B
Fig. 49.
51.
T
50.
1
ID
F
M
K
E
N
n
K
H
M
B
P
B
R
·53..
G.
D
E
P
B
M
Cy
7
B
T
B
P
A
D
R
A
F
B
I
N
55.
I
T
D
K
a
E
H
M
56.
52
54.
F
N
1
1
N
G
T
P1.VI. pa.44 ·
UNIL
OF
M\
B. I.
45
THE
ELLIPSIS.
the line HI, it is (Prop. L.) VP: VL:: OP: OL; Fig.
therefore O is common to both lines GF and HI;
therefore HI paffes thro' O.
Cor. 1. If FA, GA be any two tangents, and FG
be drawn thro' the points of contact; and likewiſe HV,
IV, two other tangents, and HI be drawn thro' their
points of contact; draw AV thro' the concourfe of
the tangents, A, V. Then O, the interfection of FG,
HI, will be the point thro' which all lines paſs, which
join the points of contact of any two tangents, drawn
from any point in the line ĂV.
Cor. 2. If GO be an ordinate to any diameter BE,
GA a tangent at G, meeting the diameter BE at A;
and if AV be drawn parallel to GO; alfo through
O, draw any line HI; then two tangents drawn from
H and I will meet fomewhere in the line AV, as at V.
Otherwiſe it could not be VP: VL :: OP : OL,
as by Prop. LVI.
Cor. 3. If IH be continued to D, and the tangents
DP, DL drawn; then LP paffing thro' the points of
contact, will pass thro' O.
PROP.
LVIII.
58.
If VH, VI be two tangents at H, I; and IHD 58.
be drawn thro' the points of contact; and any line VPOL
be drawn thro' V; and the tangents PD, LD be
drawn; they will meet in the line ÏD.
For thro' O (the interfection of VL and IH) draw
FG, fo that GO OF; and draw VAD parallel
to FG. Then (Prop. LVII. Cor. 2.) PD, LD, (tan-
gents at P, L) will meet ſomewhere in the line VA,
as at D. Then if DO, does not pafs thro' the
points of contact H, I; let a line drawn from D to
N paſs thro' the points of contact. Then (Prop. L.)
we fhall have VP: VL:: NP: NL. But by reafon
of
46
ELLIPSIS.
THE
Fig. of the tangents HV, IV, we have VP: VL::OP:
58. OL; therefore the former proportion cannot be
true, except N and O coincide. Therefore DO
paffes thro' the points of contact, H, I; that is,
the tangents PD, LD, meet in the line IH.
59.
60.
Cor. If FA, GA be two tangents, and GFB be
drawn thro' the points of contact; and from any point
B, the tangents BH, BI are drawn. Then IH drawn
thro' the points of contact, will paſs thro' A.
For a line drawn thro' A will pafs thro' the points
of contact H, I, and therefore no other can.
PROP. LIX.
If any line MN, be drawn thro' the focus F, and
the tangents MR, NR, drawn from M and N; then
the line RF, drawn from the interfection of the tan
gents to the focus, will be perpendicular to the line MN.
Make CF, CA, CT continual proportionals; at
T erect TL perp. to the tranfverfe axis BAT. Then
(by Prop. XVI.) a tangent drawn from the end of
the ordinate erected at F, will cut the axis at T.
And (by Prop. LVII. Cor. 2.), two tangents drawn
from M and N will meet fomewhere in the line TL
as at R. Produce SM to G, ſo that SG may be
equal to BA; draw GF which will be perp. to MR
at D, by Prop. XIII. Draw ML parallel to BA.
The triangles SFG, MOG are fimilar, whence
SG or BA : SF:: MG: MO; that is, CA: CF::
MF: MO. But (Prop. XXIX.) CA: CF:: ML:
MF; therefore ML: MF:: MF: MO, and MLX
MOMF². But the triangles MDO and MLR
are fimilar, being right angled at D and L; whence
MD: MO:: ML: MR, therefore MDXMR =
MLX MOMF; and MD: MF:: MF: MR;
therefore the triangles MDF, MFR. are fimilar,
whence the angle MFR = < MDF a right
angie.
=
Cor.
B. I.
47
THE ELLIPSIS.
Cor. If a circle be defcribed about the diameter MR, Fig.
it will pass thro' the focus F, and thro' the points L, G. 60.
For MFR and MLR are right angles, and the
triangle MGD being equal and fimilar to MFD, is
contained in the oppofite femicircle.
i
LEMM A.
If EM, EN be two lines meeting in E, and if it 61.
be ČA: CM: DB : DN, a given ratio, and AB, 62.
CD, MN be drawn. And if it be made CA: DB::
AE: BG:: EP GE. Then if OHK be drawn
thro' the middle of PE, EG; it will biſſect all the
lines AB, CD, MN, &c.
=
For take any of the lines, as MN; and make
NF GE, and draw MF. Make ES EG, and
draw PS, and OR, EI, HQ parallel thereto; alfo
draw VEL parallel to OH, then GN EF, RE =
EH, EI = HV; but by the fimilar triangles OEI,
OQH; it is OE: OQ:: EI or HV: HQ; but
fince RE EH, OE is EQ, or OE=OQ,
=
therefore HV = HQ, and HQ is biffected in
V. But fince EM is to GN or EF, and alſo EO to
ER, or EQ to EH, as AC is to BD; therefore
the triangles EHQ, EFM are fimilar, whence MF
is parallel to HQ. Therefore fince HQ is biffected
by EL, MF will be biffected in L ; therefore draw-
ing LK parallel to EN; it will biffect MN at K;
and meet OHK at K, becauſe LK EHEG
FN. And thus any other line as CD is proved
to be biffected by HK.
I
2-
Cor. 1, Hence if a line biffects two of the lines AB,
CD, MN, &c. it biffects all the reſt.
Cor 2. The line VEL is the locus of all the middle
points of the lines MF. And HOK is the locus of
the middle points of all the lines AB, CD, MN, &c.
SCHO-
48
THE EL LIP SÍS.
Fig.
61.
62.
63.
64°
SCHOLIUM.
Any line MN might be divided in a given ratio,
by the right line OHK; if inſtead of making EH
HG, and EO = OP, we make EH to HĞ, and
EO to OP in the ratio of MK to KN as defigned.
PROP. LX.
If a trapezium be circumfcribed about an ellipfis, as
ABFE; and if the diagonals AF, BE, be biffected;
the line drawn thro' the points of biffection, will pass
thro' the center C.
Thro' the points of contact of any two oppofite
fides, AE, BF, draw the line HI, and the diame-
ter GD parallel to it. Then (by Prop. XLVIII.
and Cor. 2.) GAXDB GEXDF, whence GA :
DF:: GE: DB. Therefore if AF, GD, EB be
drawn; then (by the Lemma Cor. 1.) the line that
paffes thro' the middle of AF, EB, paffes thro
the middle of GD. But as GD is parallel to HI,
the middle of GD is the center C.
Cor. And if any four lines touch the ellipfis, as AE,
EF, FB and BA; and the diagonals EB, AF be
drawn to the oppofite interfellions. The line drawn
thro' the middle points of thefe diagonals will pass thro'
the center.
For draw HI thro' the points of contact of two
oppofite fides EG, FB; and GD parallel to it
thro' the center. Then (Prop. XLVIII.) GE×DF
= GAXDB; and GE: DB:: GA: DF by divi-
fion EA BF; therefore if GD, EB, AF be
drawn, then (Cor. 1. Lem.) the line paffing thro'
the middle of AF, EB, paffes thro' C the center.
PROP.
1
58.
Fig. 57.
F
P
E
G
A
B
L
F
59.
61.
P
H
SRE
H Ꮐ
B
63.
H
F
A
I
G
H
E
B
S
T
G
60.
Ο
D
62.
R F
BHD N
Ꮐ
H
•E
64.
B
D
C
P
Pl.VII. pr. 48.
UNIL
OF
CH.
B. I. THE ELLIPSIS.
49
PRO P. LXI.
Fig.
If a circle be defcribed about the conjugate axis DE, 65.
and MG be any right line; and if MN, GQ be
drawn parallel to the tranfverfe axis AB, and QP pa-
rallel to GM; and if it be made, as AC: CD: :
MN: ON: PN: IN:: GL: QL. Then I fay
GM will be to LO in a given ratio, wherever MG
be placed, in a parallel fituation.
:
Produce MG, NQ to F; then fince NM: NO::
QG: QL, OL produced will fall in F. And be-
caufe PQ is parallel to MG, it will be NF: NQ::
NM NP (by conſtruction) NO: NI, or NF:
NO:: NQ: NI; therefore (Geom. II. 12. Cor. 1.)
FO (or LO) and QI are parallel, and OLQI is a
parallelogram; therefore QI LO; whence MG
will be to OL, as QP to QI.
Cor. If M (the end of MG) be in the curve of
the ellipfis; then O (the end of OL,) will be in the
periphery of the circle.
"
This appears from Prop. XIX. Cor.
PRO P. LXII.
If AB be any diameter, and if thro' any point of 66:
the curve as P, two lines APH, BPN, be drawn,
and alfo the ordinate DF, interfecting the lines in E
and G. Then DE, DF, DG will always be in con-
tinual proportion.
Draw the conjugate IL, and PO parallel to it,
and the tangents AN, BH. Then by fimilar tri-
angles,
AB: BH:: AO: OP
and BA : AN : : BO : OP
then
THE
:
ELLIPS I S.
50
Fig. then by multiplying, AB: ANXBH:: AOXOB:
66. OP² (Prop. XL) AB': IL"; therefore ANXBH
::
AB² IL²
= IL².
67.
2
Again (Prop. XL.), AB: ADB :: IL²: DF².
and by fiinilar triangles AB: AD :: BH : DE
and AB: BD :: AN: DG.
and multiplying, AB² : ADB :: AN×BH: DG×DE.
therefore IL: DF2:: AN>BH: DG×DE.
But IL' ANXBH, therefore DF = DGXDE,
whence DE: DF:: DF: DG.
Cor. If from the ends of any diameter AB, two
lines AH, BN be drawn, thro' any point P, of the
curve, to interfect the tangents AN, BH in N and
H. The rectangle of the tangents is equal to the
Square of the conjugate: ANXВH = IL².
From the demonſtration.
PROP. LXIII.
If C be the center, F the focus, and if CF: CA::
CA: CT, and the directrix TO be drawn perp. to
CA. And if any line MNO be drawn; and from
the focus F, MFH, FN and FO be drawn. Then
FO biflects the angle NFH.
Draw MP, NQ perp. to TO, and ND parallel
to MF. Then by fimilar triangles, MP: NQ ::
MO: NO:: MF: ND; and alternately MP: MF::
NQ: ND. But (Prop. XXIX.) CA: CF :: MP:
MF:: NQ: NF; therefore ND NF; therefore
< NFD = NDF DFH, by reaſon of the pa-
rallels ND, FH.
PROP.
B. I.
51
THE ELLIPSIS.
Fig.
PRO P. LXIV.
If AB, ED be the tranfverfe and conjugate axes, 68.
and if the diameters GCK, and HCL be drawn pa-
rallel to AD, AE; thefe diameters will be equal, and
conjugate to one another.
For fince EA
=
AD DB = BE, ADBE is
a parallelogram, and BE parallel to DA which is
parallel to KG; therefore the angle EBA GCA.
And fince BC CA, therefore EF FA; and
confequently GK is the diameter of EF, FA, and
HL (parallel to EA) its conjugate. Likewife the
<GCA == EBA = DBA = LCA, confequent-
ly CG CL, or GK = HL.
Cor. 1. There are no other conjugate diameters e-
qual to one another but GK, HL.
For GK grows greater in approaching to A, and
its conjugate LH grows lefs towards E; or GK
grows lefs towards E, and its conjugate LH grows
greater towards A, and G and L cannot both ap-
proach to A.
Cor. 2. The equal diameters GK, LH comprehend
the least angle of any two conjugates.
For whilſt CL approaches to A, CG approaches
to E, and the angle GCL continually increaſes, till
L come to A, and G to E, where it becomes a
right angle ECA.
PRO P.
LXV.
If AB be any diameter, BG a tangent at B, and 69.
equal to the latus retium, and if AG be drawn cut-
ting the ordinate PM (produced) in D. Then PM² =
BPX PD.
By Cor. 1. Prop. XL. AB: BG:: APB: PM².
But from the fimilar triangles APD, ABG; AB:
BG::
H
52
ELLIPSI S.
THE
Fig. BG: AP: PD :: AP× PB: PD× PB; therefore
69. APB: PM²:: APB: PDX PB; whence PM² =
PDXPB.
70.
Cor. The Square of the ordinate, PM² is leſs
than the rectangle BPXBG (the rectangle under the
parameter and abfciffa), by the rectangle PBXFG,
which is fimilar to the rectangle ABXBG. And by
reaſon of that defect, the curve is called an ELLIPSIS.
For if DF be parallel to AB, the rectangle DFG
is fimilar to ABG; becauſe from the fimilar tri-
angles ABG, DFG, it is AB: BG:: DF: FG.
SCHOLIU M.
The rectangle under AB the diameter, and BG
its parameter, is called the Figure of that diameter.
PROP.
LXVI.
1
If OT be a tangent at M; and a line MF be
drawn from the point of contact to the focus; and CO
be drawn from the center parallel to MF, to inter-
felt the tangent in O. Then CO = CB half the
tranfverfe axis.
Draw the ordinate MP perp. to the axis BA.
Then (Prop. VIII. Cor. 2.) CFXCP = CA²—CA
XFM, and CAXFM = CA²-CFXCP. But (Prop.
XVI.) CA² = CPXCT; therefore CAXFM =
CPXCT-CPXCF = TFXCP. But by fimilar
triangles, TC: CO :: TF : FM
TC: CO:: TF: FM:: TFXCP or
CAXFM: FMX CP; whence COXCA×FM =
TCXCPXFM, and CO×CA = CTXCP = CA²,
and CO - CA.
Cor. If MT be a tangent, F the focus, MP an
ordinate; then CAXFM = TF×CP.
PROP.
B. I. THE ELLIPS I S.
53
€24
PRO P. LXVII.
Then AB,
Fig.
If F be the focus, C the center, QZ any diameter, 71.
MN parallel to it drawn thro' the focus.
QZ, MN, are in continual proportion.
Draw the diameter CLD, thro' the middle of
MN; and MP parallel to it, an ordinate to QZ;
draw MO the tangent at M to cut CQ in O. Then
(Prop. LXVI.) CO = CA. And (Prop. XVI.)
CP, CQ, CO are in continual proportion; that is,
LM, CQ, CA, are in continual proportion. And
doubling the terms, MN, QZ, AB, are propor-
tionals.
Cor. CQ LMXCA.
PROP. LXVIII.
If any line GP be drawn thro' the focus F; the 72.
rectangle GFP = GP× latus rectum.
I
Draw HK parallel to GP thro' the center C,
and CD biffecting GP in L. Then (XLII.) CH2:
GFP :: CB²: AFB::(IV. Cor. 3.) CB² : CB× latus
rectum :: CB : L :: CBxGL : LXGL.' But
(LXVII. Cor.) CH GLXCB; therefore GFP
= GLX÷L =
latus rectum.
2
GPX
I
2
PROP. LXIX.
73.
If QT, PT be two tangents; and if QF, PF, 74.
be drawn from the points of contact to the focus. Then 75.
the line TF will biſſect the angle QFP.
Make CF CA :: CA: CG, and draw the di-
rectrix MGR perp. to BG. Produce TF to K,
and draw LR, KR tangents at L, K (Prop. LVIII.
H 2
and
54
ELLIPSIS.
THE
Fig. and Cor.). Let QD, PE, be perp. to MR. Then
75. (Prop. XXIX. Cor.) QF: PF :: QD: PE : : (fi-
milar triangles) QR : PR :: (Prop. L.) QN : PN.
Therefore (Geom. II. 25. Cor. 1.) FN biffects the
angle PFQ.
76.
PRO P. LXX.
If a trapezium ABDC be infcribed in an ellipfis;
and any point E be taken in the curve, from which
two lines EN, EH, are drawn parallel to any two
of the adjoining fides, as AB, AC; to interſect the
oppofite fides in N, Q; and H, R. Then taking the
rectangles on the oppofite fides; ERXEH will always
be to ENXEQ, in a given ratio.
Draw DFG, BPL parallel to AC; thro' L and
C, draw OCLS. Thro' the middle of AC, BL,
draw the diameter MK, which will biffect the pa-
rallels DF, TE; and alfo SG, OR terminated at
AB, CL. Then will SD = FG, OT = ER,
which would alſo be ſo if SG, RH were tangents.
By fimilar triangles, and the parallel lines; HO:
SD: OC: SC:: AR: AG; and alternately HO:
AR or EQ:: SD or FG: AG.
Again, BP or ER: PN :: DG : GB; and by
multiplying, HO×ER: EQ×PN :: FG×DG : AG
xGB :: (Prop. XLIII. Cor. 2.) ER×RT or ERX
EO: RAXRB or EQXEP. And by compound-
ing, FG×DG : AGÒGB :: HO×ER+EO×ER :
EQ×PN+EQ×EP :: EHXER: EQXEN. But
it is plain, as long as the points A, B, D, C, are fix-
ed, that FGD to AGB is a given ratio; and there-
fore that of REH to QEN.
Cor. 1. The fame things ſuppoſed; it will be, EH:
EN :: EO : EP : : HO : PN.
For EHXER: EQ×EN :: FG×GD: AG×GB::
HOXER: EQ/PN; therefore EHXER: HOXER::
EQXEN:
Fig: 65.
D
M
PI
A
L
G
B
A
B
H
68.
E
E
G
NA
66.
B
E
F I\P
D
69.
M
H
67.
M.
N
H
F
متر
M
C
P
A
A
p
B
m
E
70.
C P JA
O
71.
M
72.
H
P
73.
G
H
I
A
F
F
B
K
74.
IM
P
B
C
R
K
75.
1
D
IM
G
D
F
P
K
T
P
T
R
Pl.VIII.pa.5.¡ .
UNIL
OF
B. I. THE
55
ELLIPSIS.
EQ×EN : EQ×PN; and by equal divifion EH : Fig.
HO: EN PN; and by divifion EH: EO:
EN: EP.
:
Cor. 2. If the points A, B, C, E, be fixed, and D
any how changed; it will still be EH: EN; and HO:
PN, in the given ratio of EO to EP or RT to RB.
Cor. 3. If the points A, B, coincide; AG becomes
a tangent; and if C and D alfo coincide; DH_be-
comes a tangent, and then Q, N coincide.
76.
Cor. 4. If BC be drawn to cut EH in I. And 77.
if it be EH: EI; EN EV; then if BV be drawn,
it will touch the curve in B.
:
For (by Cor. 2.) If D be moveable, it will be
EH: EN EO: EP, wherever D is. Therefore
ſuppoſe D to move round, till it coincide with B,
then BND will become a tangent at B, and N will
fall upon V; that is, BV will be a tangent.
Cor. 5. If you make EH: EN :: EO: EP; and 78,
draw OC, BP, they will meet at V in the curve.
For fuppofing V a point in the curve, it will be
(Cor. 2.) EH: EN :: EO: EP. Therefore V can-
not be out of the curve.
Cor. 6. If there be five points B, D, V, C, E,
in the curve. Thro' any point E, draw EPN, EOH,
cutting BD, BV, DC, VC, in N, P, H, and O;
fo that EH: EN :: EO : EP. Then if CA, BA
be drawn parallel to EH, EN, they will meet at A
in the curve.
This is the reverſe of the laſt Cor. according to
which, if A were out of the curve, V would not
be in it.
Cor. 7. If A, B, coincide, then AB becomes a tan-
gent. And fo of any other two points.
H 3
PROP.
56
THE ELLIPSIS.
Fig.
79.
30.
PROP.
LXXI.
If a trapezium ABDC be infcribed in an ellip-
fis; and from any point E of the curve, four lines
EL, EK, EO, EM; be drawn to the four fides
of the trapezium, in any angles whatever, equal or
unequal; cutting them in L, K, O, M. And
likewife from any other point e, four more lines ol,
ek, eo, em, be drawn to the fides of the trapezium,
refpectively parallel to the former, cutting them in
1, k, o, m. Then taking the rectangles of the op-
pofite fides, the rectangle ELXEM: elxem:: EKX
EO: ekxeo.
Thro' E, e, draw EQN, eqn,
and ERH, erb, parallel to AC.
triangles.
parallel to AB;
Then by fimilar
ER: er: EL: el, and EH : eb :: EM: em,
and multiplying, ERXEH : erxeb :: ELXEM :
elxem. Again
eq
EQ: eg :: EK: ek, and EN : en:: EO: eo ;
and multiplying, EQXEN: eqxen: EKxEO:
ekxeo. But (Prop. LXX.) ERXEH: erxeb::EQ
XEN eqxen. Therefore EKXEO: ekxéo :: ERX
:
EH: erxeb:: ELXEM: elxem.
Cor. 1. If ABCD be an inſcribed trapezium, and
two oppofite fides AB, CD be produced to interfect in
X, and the diagonals AC, BD be drawn to interfect
in Y; and XeYF be drawn. Then EX: eX::EY:
eY.
For (in fig. 79.) fuppofe EL, EK, EO, EM,
and el, ek, eo, em, all to lie in one line Ee; and
that AB, CD interfect in X, and AC, BD, in Y; and
that Ee paffes thro' X, Y, (fig. 80.) which is a par-
ticular cafe of this Prop. then L, 1, M, m, coin-
cide in X; and K, k, 0, 0, meet in Y.
Then
the
B. I. T HE
THE
57
ELLIPSIS.
2
2
the proportion EL×EM: elxem :: EK×EO : ek×eo, Fig.
becomes EX: eX²:: EY eY", whence EX: 80.
eX:: EY: eY.
Cor. 2. And if two tangents be drawn from X, 81.
and the line joining the points of contact paſſes thro³
Y. It will be EX: eX:: EY: eY.
For let C approach to D, and A to B, till they
coincide; then the lines AC, BD coincide in one,
joining the points of contact, paffing through Y;
therefore it will ftill be EX: eX :: EY: eY.
PRO P. LXXII.
If BA be the tranfverfe axis; CD the femicon- 82.
jugate; S, F, the foci, PM an ordinate. Draw
DF, and make DO CP, and draw OI parallel to
DC; then SM
เ
BI.
For (Cor. 1. Prop. V.) CA: CF :: CP : SM-
CA; therefore multiplying means and extremes,
CFXCP CA×SM-CA²; and CAXSM = CA²
+CFXCP. But by reaſon of the parallels, DF :
CF:: DO: CI; that is, CA: CF CP: CI;
whence multiplying, CFXCP = CAXCI; there-
fore CAXSM CA²+CAXÇI, and by equal di-
vifion, SM CA+CI BC+CI BI.
::
Cor. 1. The fame conftruction remaining, FM = AI.
Cor. 2. If the circle CR be defcribed with the ra-
dius SC, to cut SM in N, and NL be drawn perp.
to BA. Then BC-SL, CD, and SM are continu-
ally proportional. But if PM falls between B and S,
it is BC+SL.
For by fimilar triangles SM SP:: SN or SC:
SL; whence SLXSM = SC×SP = CFXSP = CF×
SC+CP = CF²+CFXCP; but we had before CF
XCP CAXSM-CA; therefore SLXSM =
CF+
H 4
58
THE ELLIPSI S.
Fig. CF+CAXSM-CA", and CAXSM-SLXSM =
82. CA-CF CD; whence CA-SL: CD:: CD:
SM.
Cor. 3. Put z for the cofine of the angle ASM,
then SM =
CD:
CA-zxCS
And when ASM is great-
er than a right angle, then it is +zxCS.
For it is (by trigonometry) Rad. (1): SN or
SC:: ș.SNL (z) : SL = zxSC.
PROP.
LXXIII.
83.
As the tranfverfe axis, is to the conjugate: fo the
area of a circle whofe diameter is the tranfverfe, to
the area of the ellipfis.
Draw NMP parallel to the conjugate CD; then
(by Prop. XIX.) the ordinate of the circle and
ellipfis NP, MP; are every where to one another
as CE or CA to CD. Therefore if AB be divid-
ed into an infinite number of equal parts, and or-
dinates erected, according to the method of indi-
vifibles. Then it will be as CE: CD:: fo any or-
dinate PN to the correfpondent ordinate PM : ;
(Prop. X.) fum of all the ordinates PN; to the
fum of all the ordinates PM: that is, as the area
of the circle, to the area of the ellipfis.
:
Cor. 1. The area of a circle defcribed on the con-
jugate is to the ellipfis as the conjugate, to the
tranfverfe.
::
For the fame reafon as before. For all the cor-
reſponding ordinates are in that proportion, by Cor.
Prop. XIX.
Cor. 2. The correfpondent fegments of the circle
and ellipfis, are in the fame proportion as the wholes.
Cor.
H
of
Fig. 76.
T
E
R
L
S
D
K
M
H
C
V
78.
E
P
B
N
80.
M
H
P
G
E
hm
k
79.
C
E
B
E
81.
77
B
P1.IX.pa.58.
UNIL
A
OF
CH.
B. I.
59
THE ELLIPSI S.
1
Cor. 3. The ellipfis is a mean proportional, be- Fig.
tween two circles defcribed on the tranfverfe and con- 83.
jugate diameters.
::
For (Cor. 1.) the circle on the conjugate: el-
lipfis CD: CB:: (this Prop.) ellipfis circle on
the tranfverfe.
:
Cor. 4. The ellipfis is equal to a circle whofe dia-
meter is a mean between the two axes.
For circles being as the fquares of the diame-
ters; ABXDH will be the fquare of the diame-
ter of the mean circle, which is equal to the el-
lipfis. And hence,
Cor. 5. An ellipfis is to the rectangle of the two
axes: as any circle to the Square of its diameter.
Cor. 6. The areas of ellipfes are to one another as
the rectangles of their tranfverfe and conjugate axes.
PROP. LXXIV.
2
The folidity of a spheroid, generated by an ellipfis 84.
revolving about either axis, is the circumfcribing
cylinder.
3
Let C be the center; CA =t, the axis of re-
volution; CDc; AP = x, PM y, p =
3.1416. Suppoſe AC divided into an infinite num-
ber of equal parts, and planes be ſuppoſed to paſs
thro' them, then according to the method of Indi-
vifibles, all theſe circular planes make up the fo-
lid. Now (by Prop. VI. and VII.), tt: cc:: 2tx-xx:
yy
CC
x = π × 2tx-xx = PM², and
tt
pcc
tt
× 2tx-xx =
pxPM², which is one of the circles conftituting
pcc
the folid. Now the fum of all the × 2tx (Ar.
tt
Inf.
60
THE ELLIPSIS.
Fig.
pcc
2tx²
84.
Inf. Prop. II.) is
X
And the fum of all
tt
2
pсс
3tt
3. Therefore the foli-
I
pcc
the xx (Inf. Prop. III.) is
tt
рсс
dity of the fegment AMF is = xxtx-
CCXX
tt
3. XX
pxxy+6 And when AP becomes AC,
2
6tt •
1
the folidity =
pxCD2
CA2
!
2
× __AC³ = 13pxCD³×CA.
3
But the folidity of the cylinder, whoſe baſe is DE
and height CA, is = pxĊD¹×CA, whence the fphe-
roid is the cylinder.
2
3
Cor. 1. The folidity of the fegment AMF = a
cylinder whofe height is AP or x, and baje circle
CC
MF + X circle AP.
6tt
2
2
Cor. 2. The folidity of the spheroid DAE, is to
the ſphere whofe radius is AC; as DC to AC²; and
their correfpondent parts are in the fame ratio.
For the hemiſphere is 3pxAC³ (Arith. Infin. Ex.
6.); and the fegment, of the height AP, is
ptx―pxx × ×, which is to the ſpheroidal fegment,
3
as tt to cc.
Cor. 3. The folidity of the
is to a cylinder whofe bafe is
MF, and height CP.
middle zone MDEF
circle DE+ circle
2
3
Put v CP t-x, then x t-v, put this
for x in the ſegment, and it becomes
alm
pccxx
tt
-Xt- 3
pcc
2
I
tt
XtX
2 pcct
3
t +v=
34
pccv3
3
- pcc v +
3tt
fubtract
B. I.
6E
THE ELLIPSIS.
fubtract this from the whole DAE = pcct, and Fig.
3
pccv3
84.
there remains, pccv-
for the zone DMFE,
3tt
But pccv
pcc v³
2
1
3tt
pccv +
pccv
13
3
pccv3
2
I
CCVV
2
tt
3
pccv +3 pv x cc
the baſe
tt
3
I
DE × v + the baſe MF × v.
3
Cor. 4. A fpheroid whofe axis is the tranfverfe: is
to a spheroid whofe axis is the conjugate as the con-
jugate to the tranfverfe.
:
For they are to one another as cct to ttc.
PROP. LXXV. Prob
To defcribe an ellipfis by motion.
I Way.
Let CP, PD be two equal rulers moveable about 85.
P, E a fixed point in PD. Then if CP be moved
about the fixed point C, whilft the end D moves
along the right line AB. Then the point E will
defcribe an ellipfis AGEB.
=
It is plain by the conſtruction, that CG CP
-PE, and CBCP+PE, make CO CB, and
draw EK parallel to AB. Then fince CP = PD,
therefore PK PEPO. And fince the line PL
biffecting the angle KPE is perp. to KE and bif-
fects it; therefore OE which is parallel to it, will
be perp.
to KE or to AB. Therefore (Prop. XXXI.)
the point E is in the ellipfis.
Cor. The difference between the femitranfverfe and
Semiconjugate, or CB-CG is = 2 PE.
2 Way.
62
THE ELLIPSIS.
Fig.
23.
86.
2 Way.
Take the ruler MG, equal in length, to the
femitranfverfe CA; and make MI femiconju-
gate
CD; then if the ruler be fo moved that the
end G may flide along CE, whilft the point I
flides along CA. The point M by its motion will
defcribe the ellipfis AMD.
This conftruction is plain from Prop. XXX.
3 Way.
Take two rulers FE, SD, each equal to the
tranſverſe AB, and a third ruler DE equal to FS
the diſtance of the foci. And let DE be movea-
ble about the points D, E. Then cauſe the rulers
EF, DS, to move round, about the fixed points
F, S. Then the interfection P will defcribe an ellipfis
APB.
Draw the tangent GPH, this (Prop. XIII. and
Cor. 1.) biffects the angle at P, and alſo DF, ES.
Therefore the triangle FPG is fimilar and equal to
the triangle DPG; and SPH to EPH; therefore
PEPS, and FP+ PSFP+ PE = FE =
AB by conſtruction. Therefore P is in the curve
of an ellipfis by Prop. I.
Cor. If two lines FE, SD, equal to the tranfverfe
AB, be drawn thro' any point P in the ellipfis, and
the ends D, E, joined; then DE FS the distance
of the foci.
=
For the fide DP PF, and PE PS, and
=
<P vertical; therefore DE FS.
SCHOLI U M.
The ellipfis may alfo be defcribed by defini-
tion 1.
PROP
Fig. 82.
R
N
83.
E
N
M
M
B
A
S
LC IPF
A
H
DA
84.
M
F
P
B
P
D
C
E
86.
AE
87.
G
P
85.
H
P
G
E
A
K
F
S
P Q R T V
X
LID B
F
B
N
89.
F
E
F
88.
My
M
B
L
P
D
E
m
dn
m
m
m
m
I
B
F
f
F9
F
M
M
M
M
P
91.
M
90.
G
B
FFF
C
E
E
B
p
PLX. pa. 62.
UN
OF
Mi
CH.
B. I.
63
THE ELLIPSIS.
PROP. LXXVI.
Prob.
Fig.
To defcribe an ellipfis, by finding feveral points in
the curve.
I Way.
Let AB be the tranſverſe axis; F, S, the foci. 87.
Make PX AB, and divide PX into as many
equal parts as you pleafe, as PQ, QR, RT, TV,
&c. Then take any diſtance from P (greater than
AF), as PV; and fetting one foot in F, defcribe
the arch D. Then take the remainder XV, and
fet one foot in S, and croſs the former arch at D;
then D is one point in the curve. Do fo for all
the other points R, T, &c. finding as many points
D. Then a curve drawn thro' all the points D,
making no angles, as ADB, will be the ellipfis re-
quired.
This conſtruction is plain from Prop. I.
the circle AEFB.
2 Way.
Let AB be the tranfverſe or conjugate axis; di- 88.
vide AB into an infinite number of equal parts CL.,
LL, LL, &c. On the diameter AB, deſcribe
On C the middle of AB, make
CE perp. to AB. And thro' all the points L, draw
the lines FL, FL, FL, &c. parallel to CE. Then
in all the lines FL, make as the femitranfverfe
CB, to the femiconjugate CD; fo LF to LN; and
produce NL to n, fo that Ln LN. Then a re-
gular curve drawn thro' all the points N, n, as
ADNBndA will be the ellipfis required.
This conftruction is evident by Prop. XIX. and
Cor.
3 Way.
About the axis AB, defcribe the circle AEB, 88.
divide CB into any number of equal parts CL,
LL,
04
ELLIPSI S.
THE
Fig. LL, LL, &c. and draw LF, LF, LF, &c. paral-
88. lel to CE. Draw DB, and Lo, Lo, Lo, &c. paral-
lel thereto. Or rather from L neareft to C, draw
Lo parallel to DB, and divide CD into parts, e-
qual to Co, draw or, or, or, &c. parallel to AB,
and each equal to its correfpondent LF; then r, r,
r, &c. are in the curve of the ellipfis.
89.
For by the property of the circle ALB = LF²
=(by conftruction) or; therefore (Prop. XIV.) r
is in the curve of the ellipfis.
PROP. LXXVII.
Prob.
Given two conjugate diameters; to defcribe the el-
lipfis by points.
I Method.
Let AB, DE be the diameters; thro' A draw
FF parallel to DE. And AO perpendicular to it,
and equal to CD; and draw OĈ. About the cen-
ter O with radius AO defcribe the circle NAN.
From C draw any number of right lines CF, CF,
&c. to interfect FF; then from O to theſe inter-
fections, draw OF, OF, &c. cutting the circle in
N, N, &c. From N, N, &c. draw lines parallel
to OC, to cut CM, CM, &c. in M, M, &c.
Then M, M, M, &c. will be in the curve of the
ellipfis.
And if MC, MC, &c. be produced till Cm =
CM; then the points m, m, &c. will be in the el-
lipfis.
For draw MP, NQ parallel to FF, and join
PQ; then by fimilar triangles, CA: CP:: CF :
CM :: OF: ON :: OA:OQ; therefore PQ is
parallel to OC, and to MN; therefore NQ MP.
Then fince CA: CP :: OA or CD: OQ; there-
fore CA: CP² :: CD² or ON²: OQ²; and by
divifion CA: CA-CP2: . ON: ON-OO² or
NQ²;
B. I.
65
THE ELLIPSIS.
NQ₂
NQ²; that is, CA²: APB: CD2: PM". There- Fig.
fore (Prop. XL.) the point M is in the ellipfis. 89.
And fince Cm CM; therefore (Prop. XXXII.)
m is alfo in the curve.
Let AB, DE,
2 Method.
be the two diameters. With radius 90.
CB and center C, defcribe the circle BGO; let CO
be perp. to CB; divide CB into any number of
equal parts, at F, F, F, &c. and divide CD into
the fame number of equal parts at P, P, P, &c. at
F, F, &c. draw FG, FG, &c. perpend. to BC.
Thro' P, P, &c. draw PM, PM, &c. parallel to
AB, each equal to the correſponding line FG. Then
the points M will be in the ellipfis.
And if MP be produced beyond DC, fo as to
be equal to PM, theſe points will all be in the
ellipfis BD.
For ſuppoſe BD drawn, and FP parallel to it.
Then fince CF is the fame part of CB, as CP is
of CD, then CF: CP:: CB: CD; whence CB2:
CD² :: CF²: CP². And by divifion CB2: CD² : :
CB-CF2: CD-CP; but BC2-CF2 BFA
2
(nat. circle) FG² = (conſtruction) PM²; there-
fore CB2: CD2 :: PM2: CD2-CP2 or DPE or
CD2: CB2 :: DPE : PM. Whence (Prop. XL.)
M is in the curve of the ellipfis.
3 Method.
Let AB, DE be the two diameters. Draw the 91.
tangent AL, find a third proportional to AB, DE,
which fet from A to G, this is the latus rectum of
the diameter AB; draw GF parallel to AB. Thro'
A, draw any number of right lines AF, AF, &c.
Make AL, AL, &c. equal to the correfponding
GF, GF; and from B, draw BL, BL, &c. to in-
terfect AF, AF, &c. Then the interfections M,
M,
66
ELLIPSIS.
THE
Fig. M, &c. of the correſpondent lines AF, BL, will
91. be in the ellipfis.
92.
93.
93.
And a like conftruction may be made by means
of the tangent Bg, at the end B.
ing the fame for both.
For draw the ordinate PM.
triangles, GF GA :: AP: PM.
The reafon be-
Then by fimilar
And BA: AL
or GF:: BP: PM; and by multiplying, BA: GA::
APB: PM². Therefore (Prop. XL. Cor. 1.) the
point M is in the ellipfis.
PROP. LXXVIII.
Prob.
A right line DE being given in an ellipfis; to find
its diameter, and the center.
Draw HI parallel to DE; biffect DE in O, and
HI in G. Thro' O, G, draw AOGB, which is
the diameter. Biffect AB in C, and C is the center.
For the diameter biffects all the whole ordinates
(Prop. XL. Cor. 3.); and the diameter is biffect-
ed in the center (Prop. XXXII.)
PRO P. LXXIX. Prob.
Any diameter AB being given, in an ellipfis; to
find its conjugate.
Draw HI parallel to AB. Biffect AB, HI, in
C and G. Thro' C and G draw the conjugate DE.
PROP. LXXX.
Prob.
Any diameter AB being given; to draw an ordinate
to it from a given point in the curve as H.
Thro' the center C, draw HK, and KL parallel
to AB; then draw LH which will be the double
ordinate required.,
For
B. I.
67
THE
ELLIPSI S.
For fince HCCK, and AC, LK parallel, Fig.
therefore HO — OL.
Or thus,
Draw HI- parallel to AB. Draw DE through
the middle of AB, HI; and draw HL parallel to
it; then HO is the ordinate fought.
For DE is the conjugate, by the laft Prop. and
HO is parallel to it.
PRO P. LXXXI. Prob.
An Ellipfis being given to find the two axes.
Draw two parallel lines thro' the ellipfis as FK,
ML, biffect them, and thro' the middle points
draw IN, biffect IN in C, for the center. On the
center C deſcribe an arch of a circle, to cut the
ellipfis in two points G, and H. Draw GH and
biffect it; then thro' C and the middle point of GH,
draw BA for the tranfverfe axis; alfo through C
draw DE parallel to GH, for the conjugate.
For AB paffing thro' C muſt be a diameter; and
fince CG is = CH, CB paſſing thro' the middle of
GH, will be perp. to it, and alſo perp. to the dia-
meter DE (which is parallel to FG); therefore AB,
DE are the axes.
PRO P. LXXXII.
Prob.
93..
94.
An ellipfis being given; to find two diameters, 95.
making a given angle with one another, if poffible.
Draw any
diameter FG, upon FG defcribe the
fegment of a circle FNG, to contain the angle
given (or its fupplement, for it will be all one),
to cut the ellipfis in N; draw NG, NF and biffect
them, and thro' the points of biffection draw the
I
diame-
68
THE ELLIPSI S.
Fig. diameters AB, DE; theſe will be the diameter
95. required. If the circle does not cut the ellipfis,
the problem is impoffible.
For fince GN, FN are biffected, they will be
ordinates to CD, CB. And fince FG is alfo bif-
fected in C, therefore CD is parallel to FN, and
CB to GN; whence <BCD = FNG = the given
angle.
PROP.
LXXXIII. Prob.
From any given point M in an ellipfis, to draw
a tangent.
I Way.
96. From the given point M to the center draw
MC, and find its conjugate CB (by Prop. LXXIX.);
then from M draw MG parallel to CB, for the
tangent.
97.
96.
2 Way.
Draw the ordinate MP to the diameter CF; then
take CP: CF CG continual proportionals; and
draw MG for the tangent, by Prop. XLVI.
3 Way.
Draw two lines MF, MS, from the given point,
to the foci; make MD ME of any length, and
draw DE, then from the point M draw MT pa-
rallel to ED, for the tangent at M.
For fince EM MD, the <D=<E. And
fince MT is parallel to ED, the TMD MDE
= MED = EMR; therefore (Prop. X.) MT is a
tangent at M.
4 Way.
If KF be one of the axes, defcribe a circle about
KF as a diameter, and where it cuts the ordinate
PM (produced), as at m, draw the tangent mG to
the circle; then draw MG which will touch the
ellipfis at M, by Prop. XVIII.
5 Way:
B. I.
69
THE ELLIPSIS.
Fig.
5 Way.
Let F, S, be the foci, produce FM to G, fo 97.
MS, draw GS and biffect it in O. Then
that MG
MO drawn, will touch it in M, by Prop. XIII.
PRO P.
LXXXIV. Prob.
From a point given without the ellipfis, to draw a
tangent to it.
1 Way.
Let G be the point given, thro' G and C draw 96.
the diameter FK. Make CG, CF, CP continual
proportionals, and at P draw the ordinate PM to
the diameter FK; then GM will be a tangent. By
Prop. XLVI.
2 Way.
Draw
If KF be the axis, draw CG, on which defcribe
the femicircle CmG, and on the diameter KF de-
ſcribe an arch Fm to cut the former in m.
mP perp. to CG, to cut the ellipfis in M.
draw GM for the tangent. By Prop. XVIII.
3 Way.
Then
Let T be the point given; thro' T draw the 98.
diameter AB, and (Prop. LXXIX.) find its con-
jugate CD. Draw any line TS = TB, and make
TR TC. Draw RA, and SP parallel to it,
cutting the diameter AB in P. Draw the ordi-
nate PM parallel to CD; then TM is the tangent
required.
For by fimilar triangles, TS: TR :: TP : TA;
that is by conſtruction, TB : TC::TP: TA; and
by divifion, BC : TC :: AP : AT; and by divi-
fion again BC: TC:: BC-AP or CP: TC-AT
or CA; that is, CP: CA :: CA: CT; therefore
(Prop. XLVI.) TM is a tangent.
I 2
PROP.
70
THE ELLIPSI S.
Fig.
99.
100.
PRO P. LXXXV. Prob.
There is given the focus F; and alſo three points of
the ellipfis, M, N, P; to defcribe the ellipfis.
Draw NMK thro' two of the points, and NPI
alſo thro' two; and draw MF, NF, PF, to the
focus. Then make FN: FM:: NK: MK. Alfo
FN: FP: NI: PI. Then draw KI, and MH
perp. to it. And FT perp. to it from the focus;
and make FA AT:: FM: MH. Then produce
TF to B, fo that BF may be to BT in the fame
ratio. And AB will be the axis, and the middle
point C, the center. And make CS = CF, and
S will be the other focus. Whence the ellipfis
may be deſcribed by the foregoing propofitions.
Draw NG, PL, perp. to KT. Then by con-
ftruction FN: FM NK: MK :: (fimilar tri-
angles) NG: MH. And FN : FP :: NI : PI : :
Alfo FA: AT::FM: MH. There-
NG :: FP: PL:: FM: MH:: FA:
AT:: FB: BT. Which (Prop. XXIX. and Schol.)
is the property of the ellipfis, where KT is the di-
rectrix, and therefore the points M, N, P, are
in the ellipfis. Alfo fince BT: BF:: AT: AF.
Therefore AT: AF :: BT-AT : BF-AF ::
BA: SF CA: CF; which is the property of
the directrix.
NG:PL
fore FN
•
: :
PROP. LXXXVI.
Prob.
Given two conjugate diameters of an ellipfis; tę
find the axes.
Let the given diameters be HI, LM; produce
CH, and make CH, CM, HN continual propor-
tionals; let GK be a tangent at H, biffect CN in
R,
D
B
9.5
Fig 92.
H
Ꮐ
R
97.
M
I
B
H
93.
E
C
G
D
K
96
D
K
m
A
H
S
M
R
A
B
F
B
T
99
M
H
T
L
B
E
D
G
94.
DF
G
H
EL
98
B
K
A
T
100%
H
P
A
}
PIXLpa. 70.
UNIL
OF
B. I.
71
THE ELLIPSI S.
R, and draw RO perp. to it interfecting GK in O. Fig.
About the center O, with radius OC defcribe the 100.
circle CGNK, cutting GK in G, and K. Thro'
K, C, and G, C, draw AB, ED, for the two axes.
2
For by conftruction CL CHN (Geom. IV.
20. Cor. 2.) GHK; therefore (Prop. LIII.) AC,
EC are conjugates. And fince GOK is the dia-
meter of the circle; GCK is a right angle, and
therefore AC, EC are the femiaxes.
Cor. If HP, HQ be drawn parallel to EC, AC.
And there be taken CA a mean between CP and CK;
and CE a mean between CQ_and CG, you'll have the
length of the axes.
By Prop. XVI, XVII. Becauſe GHK is a tan-
gent at H.
PRO P. LXXXVII.
Prob.
To defcribe an ellipfis about a given parallelogram 10.
EGHF, to pass thro' a given point P.
Biffect the oppofite fides of the parallelogram
EF, GH; and EG, FH; by the conjugate dia-
ineters AB, ID. Draw PQ parallel to ID; and
CL²× PQ-CQ³×EL²
PQ²-EL²
take AC² =
2
CL³× PQ'—CQ³×EL²
CL²-CQ²
2
ง
and CD-
And AC, CD being had,
AL, and its equal KB will be had, from whence
the ellipfis is eaſily deſcribed.
For (Prop. XL.) AC² : CD² : : ALB : EL² ::
AQB: PQ'. That is, AC: CD² : : AC²—CL²: EL²::
AC²—CQ² : PQ². Hence AC-CL² × PQ²
AC²—CQ² × EL"; therefore AC² × PQ-EL² =
CL³×PQ³—CQˆXEL'. Whence AC is deter-
mined.
2
I 3
Again,
72
THE ELLIPSI S.
Fig.
2
2
Again, fince AC²: CD²:: AC-CL² : EL' ;
101. by divifion, ADz :CD² :: CL² : CD²—EL². And
ADz:CD²
fince AC2-CL: EL²:: AC²-CQ²: PQ²; by
divifion, AC-CL: EL :: CQ-CL: EL-
PQ. Therefore CL: CD-EL:: CQ-CL":
EL²—PQ². And by addition, CL² : CD²—EL²:
CQ² : CD²-PQ; therefore CL x CD²-PQ² =
CQ¹× CD²—EL"; whence CD² × CL2-CQ² =
CL³XPQ-CQXEL. And therefore CD is de-
102.
termined.
2
Cor. Hence AC : CD² : : CL²-CQ²: PQ²—EL'.
For AC² × PQ-EL² = CL² × PQ²—CQ² ×
EL² = CD² × CL²—CQ².
2
PROP.
LXXXVIII.
If a cylinder be cut obliquely by a plane; the fellion
will be an ellipfis.
Let AGH, RIS, be the baſes of the cylinder,
AMB the fection; AB its diameter interfecting the
axis of the cylinder KO in C; then fince AO =
OH, therefore AC CB. Take any point P, and
ſuppoſe two planes parallel to the baſe AGH, to
paſs thro' C and P, cutting the ſection in the lines
CD, PM. Thro' CD, PM, draw the planes KG,
NE. If the plane RH be fuppofed perp. to AGH
and AMB, then (Geom. V. 15.) CD, PM will be
perp. to AB, and GO, EF, to AH. And KI
CD OG, and NL PM FE.
ture of the circle, the rectangle AFH
By the na-
FE² =
PM². Then by fimilar triangles, AC: AO::AP:
AF, and AC: AO:: PB: FH; then multiplying,
AC: AO2 or CD2 :: APB: AFH or PM². There-
fore (Prop. VI. Cor. 1.) the curve AMB is an el-
lipfis.
Cor.
B. I.
73
THE ELLIPS I S.
Cor. If the plane be parallel to the baſe of the cy- Fig.
linder, the fection will be a circle.
PRO P.
LXXXIX.
102.
If an oblique cone ABD be cut by a plane EF, in 103,
fub-contrary pofition, the fection ELF will be a circle.
Let the plane ABD be drawn thro' the axis of
the cone AC, perp. to the plane of the baſe BGD.
And let another plane ELF alfo perp. to ABD, cut
the cone, fo that the angle AFE may be equal to
ABD. Then the cone is faid to be cut in fub-con-
trary pofition.
Thro' any point R in the diameter EF, draw the
plane HLK, parallel to the bafe BGD, cutting the
plane ELF, in the line LR. Alſo let another plane
ACG, cut the plane HCK, in the line OM. Then
(Geom. V. 11.) OM and CG are parallel; whence
by fimilar triangles, CD: OK:: AC: AO:: CG
or CD: OM; whence OK = OM; therefore EMK
is a circle.
Again (Geom. V. 15 and Cor.) LR is perp. to
the plane ABD, and alſo to EF and HK; and fince
the angle KFR = EBD or EHR; therefore the
triangles ERH, KRF are fimilar, and ER: HR::
KR: FR. And ERF HRK (Geom. IV. 17.)
RL; therefore ELF is alfo a circle.
Cor. If a cone be cut by a plane parallel to the
bafe, the fection (HLK) will be a circle.
This appears from the demonftration.
PRO P. XC.
If a cone ABC, be cut by a plane paffing thro' both 104.
its fides; the fection BNDK will be an ellipfis.
Draw the plane ABC thro' the axis of the cone,
and let the plane DNB paſs thro' the tangent at B,
1 4
and
74
ELLIPSIS.
THE
104.
Fig. and let BH = HD, and thro' H and any point I,
of the diameter BD, draw two planes, parallel to
the baſe of the cone BC; cutting the plane BNDK
in the lines KN, LQ, which will be parallel to
one another (by Geom. V. 11.); and the fections
FNGK, and OQPL will be two circles (by Cor. to
the laſt Prop.).
105.
Then by the nature of the circle, OIXIP =
IQ' or IL², and FH x HG = KH² or HN". By
the fimilar triangles DIP, DHG, and BIO, BHF;
DI: IP:: DH: HG, and BI: IO:: BH : HF.
And by multiplying, DIXIB: OIXIP or IQ' ::
·BH×HD : FĤXHG or HN². Therefore (Prop.
XL.) the curve BNDK is an ellipfis.
Cor. 1. The fquare of the conjugate is equal to
the rectangle of the diameters of the fruftum: KN²
= EDX BC.
For draw DE parallel to BC; then by fimilar
triangles, fince DB is biffected in H, therefore FH
-ED, and GH BC; and the rectangle FHG
to
I
2
KH' BCXED; and KN
44
BCXED.
EDXBC
BD
Cor. 2. The latus rectum of the ſection is =
For (Prop. IV. Schol.) the latus rectum is equal
KN2
BD
Cor. 3. If it be a right cone, the Square of the
tranfverfe is equal to the Square of the fide of the
fruftum, and the rectangle of the diameters of the
bafes. BD² DC²+ED×BC.
BC-ED
Let DW be perp. to BC; then WC .2
BD²
And (Geom. II. 23) BD' BC+CD2-2 BCW
BC-ED
BC'+CD'-2BCX
CD+BCXED.
Cor.
B. I.
75
THE ELLIPSI S.
Cor. 4. In a right cone, the distance of the focus S, Fig.
from the center, is equal to half the fide of the fruf- 105.
tum: HS
FB.
For (Cor. 3.) BD²—CD² = BC×ED = (Cor.
1.) KN². And dividing by 4, BH2-BF² =
HN2, and BF2 BHHN2 = (Prop. II.)
,
HS2, and BF
HS.
Cor. 5. If AI be perp. to the baſe of the cone, and 106.
AB perp. to MH the diameter of the ellipfis. And
if AN AB, and FND be drawn parallel to the
bafe CM. Then FD is the latus rectum; and DS
(parallel to MH) is = HZ the diameter at top.
Draw AL, AG parallel to MC, MH, and let R
latus rectum; then the triangles ABL, ANG
are fimilar and equal, and AG
angles MCH, DFS, GFA, are
MC: MH:: DF: DS:: GF
(fimilar triangles GFA, LAH) AF: AH :: (fimi-
lar triangles AFD, AHZ) FD: HZ; therefore
DS = HZ, and MHXFD = MC×HZ = (Cor.
1.) fquare of the conjugate =(IV. Schol.) MH x
R, therefore FDR.
=
AL. The tri-
fimilar; whence
GA or AL::
Cor. 6. In a right cone, if F be the focus, C the 107.
center of the ellipfis; AG AF. The circle GOI
parallel to baſe of the cone, will cut the plane of the
ellipfis CAH in the line ED, the directrix of the el-
lipfis.
For (Prop. XXVIII. Cor. 2. and XXIX. Schol.)
if DH be the focal tangent, DE will be the direc-
trix; where CH = CA, and AF, are fet perp. to
AM. But if they be not perp. but in any other
fituation parallel to each other, as CB and AG;
then BG will cut AM in the fame point D,
and ftill DE is the directrix. And DB will be
parallel to MN; for (Cor. 4.) CFAP or AN;
and PC (drawn thro' the middle of AN, AM)
will be parallel MN. And (XVI.) CA; CD:: CF:
CA::
76
ELLIPSI S.
THE
Fig. CA :: (by divifion) AF: AD; that is, PA: CA::
107. AG: AD; therefore DG is parallel to PC or MN.
108.
Cor. 7. In a right cone, if AG = AF, and the cir-
cle GNI be drawn parallel to the bafe. Then the dif
tance of any point P from the focus, FP; will be e-
qual to PN the distance of P from the periphery of
the circle GNI.
For DE is the directrix; draw DR the focal tan-
gent; then (Prop. XXVIII.) if the ordinate PQ
be produced to the tangent at R, then QR = FP;
but QL parallel to AG, being made equal to QR
(as in the laſt Cor.), then DL will be parallel to
the baſe HM. Whence FPQL = PB (becauſe
the plane SPQR is parallel to the plane DBL) =
SG PN.
Cor. 8. And fince there are two foci; another fuch
circle drawn the fame way below the bafe HM, will
have the fame property. And their diſtance along the
fide of the cone will be equal to the tranfverfe axis.
SCHOLIU M.
Hence appears the reaſon why the Ellipfis is
called one of the Conic Sections. For a cone cut
by any plane which paffes thro' both its fides, will
always make the figure of an ellipfis, whoſe proper-
ties have been demonſtrated in this book; except
it happen in fome particular cafes, that the two
axes of the ellipfis be equal; and then the foci co-
incide in the center; and the ellipfis becomes a cir-
cle; which therefore may be reckoned one fort of
an ellipfis.
CONIC
Fig. 101.
E
F
S
F
D
E
104.
P
R
102.
B
E
103.
UA
H
B
H
M
K
G
105.
E
K
H
G
F
G
S
B
W
C
108.
B
106. F
107.
E
F
B
A A A
M
P1. XII. pa. 76.
UNIL
OF
CY
[ 77 ]
!
CONIC SECTIONS.
BOOK II.
Of the HYPERBOLA.
IF
DEFINITION S.
DEFIN.
I.
I.
F the ends of two threads SPQ, FPQ, be Fig.
faſtened at the points S, F. And be made to
paſs thro' a ſmall bead, or pin P, and knotted to-
gether at Q. Then taking hold of Q, and draw-
ing the threads tight; if the bead be moved along
the threads; the point P will defcribe the curve
mpAPM, called an Hyperbola.
DE F. II.
}
And if the end of the long thread be fixed at F,
and that of the ſhort one at S; and the curve NBR
be deſcribed after the fame manner; that curve is
called the oppofite Hyperbola; and both curves to-
gether, MAm, NBR are called oppofite Sections, or
oppofite Hyperbolas.
DE F. III.
The two fixed points F, S, are called the Foci.
DE F. IV.
The line AB (paffing thro' the foci, when con-
tinued), contained between the two parts of the
curve, is called the tranfverfe Axis.
DEF.
78
THE HYPERBOLA.
Fig.
I.
2.
1.
DE F. V.
The middle point of AB, that is C, is called
the Center of the hyperbola, or of the oppofite fec-
tions.
DE F. VI.
If VY be drawn thro' the center C perp. to AB;
and with radius CF, and center A, an arch be
deſcribed, cutting VY in V and Y; then VY is
called the conjugate Axis.
DE F. VII.
Any line TO drawn thro' the center C, and
terminated at the oppofite fections, is called a Di-
ameter; and the extremity T (or O) its Vertex.
And the line drawn thro' the center, parallel to the
tangent at the vertex, is called its conjugate Dia-
meter.
DE F. VIII.
If any diameter OT is continued within the
curve, the part within, TP, is called the Abſciſſa.
DE F. IX.
Any line PM, drawn parallel to the tangent at
the vertex T, and terminated at the abfciffa and
curve; is called an Ordinate to that diameter TO.
And if it go quite thro' the curve, it is called a
double Ordinate.
DE F. X.
The line LI, drawn thro' the focus F, perp. to
the tranſverſe axis AB, and terminating at the
curve, is called the Parameter or Latus Rectum.
DE F. XI.
If the ends of the two axes be joined by the
lines BY, BV; and thro' the center C, two lines
CH.
B. II. :
79
THE HYPERBOLA.
I.
CH, CG, be drawn parallel to BY, BV; or (which Fig.
is the fame) if VY be placed at A perp. to BA;
and the lines CH, CG, be drawn from the center
C, thro' the ends E, D; theſe lines CH, CG, are
called the Affymptotes, of the hyperbola, or of the
oppoſite hyperbolas.
DE F. XII.
When the tranſverſe and conjugate axes are equal,
ACCV or AD, the curve is called an equilateral
Hyperbola, or right angled Hyperbola.
DE F. XIII.
A right line which meets the hyperbola in one 2.
point T, but does not cut it, as TH, is called a
Tangent to it, in that point T.
DE F. XIV.
If two oppofite hyperbola's KO, TW, be in
like manner deſcribed to the tranſverſe VY (≈ DE),
and conjugate AB; theſe are called conjugate Hy-
perbolas, with regard to the former.
PRO P. I.
The difference of the lines SP, FP, drawn from
the foci, to any point P of the curve, is equal to the
tranfverfe axis AB.
For by conſtruction PS-PF
ASAF =
AB+BS — AF = (becauſe BS = AF) AB.
Cor. Hence CF CS, or the foci are equally dif
tant from the center.
PROP.
89
THE HYPERBOLA:
Fig.
I.
I.
PROP. II.
The Square of the diftance of the focus from the
center, is equal to the fum of the fquares of the femi-
tranfverfe and femiconjugate. CF CA²+CY¹.
For make AE equal and parallel to CY, then
the radius CE = CF; and in the right angled tri-
angle CAE, CE CA+AE; that is, CF =
Cò+AE² = CA²+CY².
2
=
Cor. CFAE CA; and CF CA² =
AE CY².
PRO P. III.
The rectangle of the focal distances from either ver-
tex, is equal to the Square of the femiconjugate, FA x
SA = CY².
=
For making AE CY; by the property of the
circle, FAX AS = AE = CY¹.
2.
Cor. The rectangle of the distance of either focus
from the two vertices, is equal to the fquare of the
Semiconjugate, FAX FB = AE'
=
CY.
For SB FA, and SA = FB; whence FA ×
FB FAX SA
AE2.
PROP. IV.
3.
As the tranfverſe axis, is to the conjugate; So the
conjugate, to the latus rectum of the tranfverfe; AB:
VY VY: LI.
2
=
For (Prop. I.) SL-LF BA≈ 2CA; and
SL2CA+LF; and SL ≈ 4CA²+4CA ×
LF+LF and in the right angled triangle SLF,
SL2
B. II.
81
THE HYPERBOL A:
2
3•
SL² = SF²+LF, and fubtracting LF from Fig.
thefe two values of SL2; then 4CA+4CA × 3.
SL²
LF = SF² = 4CF²; and CF² = CA²+CA ×
=
LF. But (Prop. II.) CF CA²+CY CA
+CA × LF; therefore CY CAXLF; and
multiplying by 4, VY² = BAXLI.
2
2
2
2
Cor. 1. As the femitranfverfe, to the femiconju-
gate, fo the femiconjugate to half the latus rectum,
CA: CY: LF.
Cor. 2. As the femitranfverfe, to the distance of
the focus from the center; fo is the fame diftance, ta
the fum of the femitranfverfe and half the latus rec-
tum, CA: CF:: CF: CA+LF.
2
For (Prop. II.) CF CA2+CY
CAXLF CAX CAXFL.
2
CA² +
Cor. 3. The rectangle BFA = tranfverfe x =
latus rectum = CA x FL.
By Cor. 1. and Prop. III.
:
SCHOLIU M.
I
2
Since the tranfverfe axis is to the conjugate, as
the conjugate to the latus rectum of the tranfverfe
axis. Therefore in any other diameters, the third
proportional to any diameter and its conjugate, is
called the Latus Rectum of that Diameter. There-
fore in a right angled hyperbola, every diameter is
equal to its latus rectum.
PROP.
82
THE HYPERBOL A.
Fig.
4.
PROP. V.
From any point M in the curve, draw the lines,
MF, MS, to the foci; and the ordinate MP perp.
to the tranfverfe axis AB. Then,
As the femitranfverfe, CA:
to the distance of the focus from the center, CF ::
So the distance of the ordinate from the center, CP :
to half the fum of the lines MS, MF, or
=
2
MS+MF
2
Make SD CA; then SM = CA+DM,
and FM SM-2CA DM-CA. In the
right angled triangle SMP, SM CA+2CAX
DM+DM² = SP²+PM² CF+CP² + PM²
= CF²+2CFxCP+CP²+PM².
×
2
=
And in the right angled triangle FMP, FM² =
DM2 2DM X CA + CA FP² + PM² =
CP-CF²+PM2 CP² 2CF x CP+CF+
PM².
—
-
And fubtracting the latter equation from the
former, SM-FM 4CA×DM = 4CF×CP.
And CFXCP
CAXDM.
But fince SM DM+CA, and FM = DM -
CA; therefore SM+FM = 2DM, and DM =
SM+FM
2
; whence CFXCP CAX
=
SM+FM
2
Cor. 1. If F, S, be the foci, MP an ordinate;
then it is CA: CF :: CP: SM-CA or FM+CA.
CAXDM, and DM SM-
For CFXCP
CA = FM+CA.
!
Cor. 2. If F, S, be the foci,
MP an ordinate,
then the difference of the fquares
of the lines SM,
FM; that is, SM² — FM² = 4CFXCP.
Cor,
B. II.
83
THE HYPERBOLA.
Cor. 3. If F, S, be the foci, MP an ordinate; Fig.
then CA X SM+FM 2CF × CP.
For SM-FM²
4.
SM+MF x SM-MF =
2CA × SM+MF = 4CF × CP; And CA X
$M+FM = 2CF × CP.
SCHOLIUM.
When P falls between A and F, then FP CF
CP, and its fquare the fame as before, and the
reft of the demonſtration the fame.
PROP. VI.
If MP be an ordinate on the tranfverfe axis, it 4.
will be,
As the fquare of the tranfverfe, BA2:
Z
to fquare of the conjugate, NE ::
So rectangle of the fegments of the tranfverfe, BPA :
to Square of the ordinate, PM'.
SM+FM
Make SDCA, then DM =
And
:
2
(Prop. V.) CA: CF :: CP : DM. And by com-
pofition, CA: CA+CF or BF :: CP: CP+DM;
and alternately, CA: CP:: BF: CP+DM. And
again by compofition, CA: CA+CP or BP :: BF:
BF+CP+DM.
But BF
CP+DM
=
BC+CF SD+CF. And BF+
SM+SP.
SD+CF+CP+DM
Therefore CA: BP: BF: SM+SP.
Again, fince CA: CF:: CP: DM; by divi-
vifion CA CF-CA or AF:: CP: DM-CP;
and alternately, CA: CP:: AF: DM-CP; and
by divifion, CA: CP-CA or AP:: AF: DM
-CPAF.
K
But
84
THE HYPERBOLA.
—
Fig. But AF CF-CA SC SD; therefore
4. DMCP-AFDM-CP+SD—SC SM
}
SP; therefore,
CA: AP:: AF: SM-SP
=
and before CA: BP:: BF: SM+SP,
and multiplying, CA²: AP × BP::AF×BF:SM²
SP².
Make CN or CE perp. to AB, and equal to half
the conjugate; then (Prop. III.) AF × BF =
CN². And SM²
And SM-SP2 PM2.
SP² PM². Whence
2
CA²: APX PB:: CN2: PM or BA: NE²::
BPA: PM².
Cor. 1. As Square of the femiconjugate to the
Square of the femitranfverfe:
tances of the ordinate from the
rectangle of the dif-
vertexes: to fquare of
the ordinate. CA: CN:: BPA : PM2.
Cor. 2. As the tranfverfe axis, to the latus rec-
tum fo the rectangle of the ordinates diftances from
the vertexes, to the fquare of the ordinate. BA: lat.
rect. BPA: PM².
:
For BA lat. rect.:: BA: BA X lat. rect. ::
(Prop. IV.) BA²: NE :: (this Prop.) BPA : PM².
Cor. 3. The rectangles of the (abfciffa's or) dif-
tances of the ordinates from the vertices of the figure,
are as the fquares of theſe ordinates.
For each rectangle is to the fquare of its ordi-
nate, in a given ratio.
Cor. 4. As the Square of the femitranfverfe to
the rectangle of the focal distances from the vertex::
fo rectangle of the ordinate's diftances from the ver-
texes to the fquare of the ordinate.
Cor. 5. The whole ordinates to the tranfverfe, are
biflected by the axis. And equal ordinates are equally
diftant from the center,
For
B. II.
85
THE
HYPERBOL A.
For the rectangle of the abfciffas on the tranf- Fig.
verſe axis being in a given ratio to the ſquare of 4.
the ordinate, and that rectangle being the ſame for
the ordinates on both fides; theſe ordinates will
therefore be equal. Again, that rectangle remains
the fame, at equal diſtances on each fide the center;
and confequently the ordinates are equal at equal
diſtances.
PROP. VII.
If MP be an ordinate to the tranfverfe axis; C 4
the center, F, S the foci; and making CA : CF : :
CP: CI; then the rectangle SM X MF
CA².
CI-
Make DS CA; then (Prop. V.) CF × CP =
MS+MF
CAX
2
SM = DM+CA, and FM
= CAX CI, by conſtruction. But
(SM
(SM
2CA) DM
MS+MF
-CA; therefore
2
DM,
and CA X
But SM X
DMCA² =
DMCA× CI; whence CI = DM.
FM DM+CAX DMCA
CICA2.
Cor. 1. If CA: CF :: CP: CI; then SM — BI,
and FMAI.
For DMCI, and SM CA+CI = BI, and
FM CICA = AI.
Cor. 2. The rectangle SM x FM BIX AI.
From Cor. 1.
Cor. 3. CA: CF:: CP: CA+ FM:: CP:
SM - CA.
K 2
PROP.
86
THE HYPERBOLA.
Fig
4.
PROP.
VIII.
If F, S be the foci; C the center; MP an ordi-
nate to the tranfverfe axis; CE or CN the femicon-
jugate. Then CM³ CA - CE²+SMF.
=
2
For in the triangle SMF (Geom. II. 28.) SM²+
MF² — 2CF² + 2CM². But SM—MF² – SM²
2SMXMF+MF, and SM+MF-SM-MF
+ 2SMF = (Prop. I.) BA² + 2SMF = 4CA²
+2SMF. Therefore 4CA+ 2SMF 2CF² +
2CM². Whence CM² = 2CA² CF2+SMF=
(Prop. II.) 2CA-CA-CE2 +SMF = CA²
CE² + SMF.
:
Cor. If CA CF CP: CI; then rectangle
BPA rectangle SIF - MP².
=
2
MP2
For BPA CP2 CA² = CM²
CA² SMF CEMP² (Prop. VII.) CI²
— CA² — CEMP²
CF2
·MP² = SIF MP².
PROP.
=
(Prop. II.) CI²
IX.
If two lines SM, FM, be drawn from the foci, to
any point M in the curve; they will make equal angles
with the curve in that point.
=
Take the point m infinitely near M in the curve,
and draw Sm, Fm. And from S and F as centers,
deſcribe the ſmall arches mr, mt, thro'm; cutting
SM, FM in r and t. Then fince SM-FM
SmFm; therefore SM+Fm = Sm+FM; that
is, Sr+rM+Fm = Ft+tM+Sm; therefore take-
ing away the equal quantities Sr, Sm, as alfo Fm,
F; there remains rMtM. Therefore in the
very fmall triangles Mmr, Mmt, right angled at r
and
آپ
B. II.
87
THE
HYPERBOLA.
=
and t, we have Mr Mt, and the hypothenufe Fig.
Mm common; therefore the fides mr, mt, are e- 5-
qual, and alfo the oppofite angles, mMr, and mMt,
or mMS and mMF are equal.
Cor. 1. If KMO be drawn perp. to the tangent
at M; then the angle KMS FMO.
For they are the complements of the equal an-
gles mMS, and mMF.
Cor. 2. If MO be drawn perp. to the tangent at
M, and SMD, FM, drawn from the foci; the angle
FMO
DMO.
For DMO = KMS = FMO.
PROP.
X.
If MP be an ordinate to the tranfverfe axis, and 6.
MO perpendicular to the tangent at M; S, F the
foci. Then CA: CF:: CP: CO.
Make MG equal to MF, draw FG, which bif-
fect in D; and draw MD, which will be perp. to
FG, and biffect the angle GMF, and touch the
curve at M, by Prop. IX. and fince MO is alſo
perp. to MD; therefore GF, MO are parallel;
whence SM: GM or FM:: SO:FO. And com-
pounding and dividing, SM+MF: SM-MF::
SO+OF: SO-OF; that is, SM+MF: 2CA::
SM+MF
2CO: 2CF. And
SM+FM
2
SF+FM
2
2
2
:CA::CO:CF. And
XCA:CA::CO:CF. But(Prop, V.)
x CA CF x CP; therefore CFX CP:
CA²:: CO: CF:; CFX CO: CF²; and alternate-
ly, CA: CF:: CF x CP: CF × CO:: CP: CO.
K 3
Cora
88
THE HYPERBOLA.
Cor. 1. If CE be the femiconjugate, CA² : CE² : :
6. CP: PO,
Fig.
2
2
For CA²: CF²:: CP: CO; and by divifion,
CA²: CF² - CA CP: PO. But (Cor. Prop.
II.) CF-CA CE; whence CA: CE ::
CP: PO.
Cor. 2. BA: latus rectum:: CP: PO.
2
For CA: lat. rectum :: CA² : CA × ½ lat. rec-
tum or CE²: : CP : PO.
Cor. 3. CF2: CE :: CO: PO.
CF²
For (this Prop.) CF2: CA:
CF² CA²
CO: CP.
CO: CP.
And
CF²
(Cor. 1.) CA: CE:: CP: PO; therefore CF2:
CE: CO: PO.
Cor. 4. CA: CF:: MF: OF:: MS: OS.
::
For we had before SM+MF: 2CA:: CO: CF:
and by divifion, SM+MF2CA or SM+MF —
SM+MF: 2CA :: CO - CF or OF: CF, that
is, 2MF: 2CA or MF: CA:: OF: CF. And
from the firſt proportion by addition we have SM+
MF2CA SM+MF+2CA:: CO-CF: CO
+CF; that is, 2MF: 2MS:: OF : OS,
6.
PROP.
XI.
Let S, F be the foci, MO perp. to the tangent in
M; then MQ SOX OF -
MQ²
SMX MF.
Draw the tangent MQ, and make MG MF,
and draw FG, which is perp. to MQ, becauſe the
<GMD = FMD. Therefore the triangles SGF,
SMO are fimilar, and SO: FO :: SM : GM or
FM :: (Geom. II. 25.) SQ : QF. Whence SO×
QF = SQ × FO = SQ × SO— SF, and SQ X SF
= SOX SQ -
-QF. But SQ—QF = 2SQ ·
SQ — QF = 2SQ — SF — 2SQ — SO+FO;
there-
X
-
B. II.
89
THE HYPERBO L A.
therefore SOXSQ — QF = SO× 2SQ — SO+FO Fig.
—
=SQxSF = SQx SQ+QF, and 2QSO - SO²
+SOF = SQ²+SQF. And SQ²+SO² — 2QSQ
+SQF SOF, or SO — SQ² + SQF = SOF;
=
that is, QO+SQF SOF. But (Geom. II. 26.)
QM² = SMF
SMF SQF.
SQF. And in the right angled
triangle QMO, MO² = QO²
QO² ---- QM²
QM² = QO²
SMF SQF SOF SMF.
+ =
Cor. 1. If MQ biffect the angle SMF, and MO
be perpendicular to MQ; then SQ: QF:: SO: OF,
From the demonſtration.
2
Cor. 2. CA: CE :: rectangle SMF : MO².
SÖ
For we had SM: MF:: SO: OF; and by di-
vifion, SM- MF or 2CA: MF:
or 2CF: OF; and alternately,
CA: CF :: MF: OF, and
SO — OF,
likewife CA: CF:: MS: SO; and multiplying,
CA²: CF²:: SMF: SOF,
CA or CE :: SMF: SOF-
and CA2: CF
SMF, or MO².
CE¹
Cor. 3. Refangle SMF = BO XOA−CE
MO².
For SMF
CF
=
SOF
SOF — MO². But SOF = CO*
2
CO - CA CE² = CO+CA ×x
CO-CA-CE
SMF BOX OA
BOX AO
BOXAO
CE' MO'.
CE'. Whence
6.
PROP.
K 4
90
THE HYPERBOLA.
Fig.
7.
8.
PRO P. XII.
If SM, FM be drawn from the foci to any point
M in the curve, and SG be made equal to BA, and
FG drawn; and if MD be drawn to the middle of
FG. Then MD will touch the curve in M.
For take any point m in the line MD, and draw
Sm, Fm, Gm; then in the ifoceles triangles GmF,
GMF, Gm mF, and GM MF. Now in the
=
triangle SGm (Geom. II. 5.) §G+Gm is greater
than Sm; that is, SG+Fm is greater than Sm, and
Fm greater than Sm SG; therefore the point m
is without the curve. For if m was in the curve,
and Sm, SG of given lengths, then Fm
which now is bigger.
Sm
SG,
Cor. 1. The tangent at M biffects the angle SMF,
made by two lines drawn from M to the focus.
Cor. 2. The tangents at the ends of the tranfverfe
axis, A and B, are perp. to that axis.
PROP. XIII.
If PM be an ordinate, DCE the conjugate axis;
then if AD be drawn to the ends of the tranfverfe and
conjugate axes, and PQ parallel to it; then PM =
CQ₂ CD² = rect. DQE.
For by fimilar triangles AC: CD:: PC: CQ,
and AC: CD:: PC: CQ; and by divifion, AC²:
CD² :: PC' — AC: CQ- CD. But (Prop. VI.)
CQ²
AC: CD':: BPA or CP-CA: PM², there-
CQ-CD' DQE.
fore PM'
=
2
PROP
B. II.
91
THE HYPERBOLA:
PROP.
XIV.
Fig.
If MP be an ordinate to the tranfverfe axis, F 9.
the focus, and HMO be drawn perp. to the curve at
M, cutting the conjugate CD in H, and the tranfverfe
in O. Then will OP: OC :: CD²: CF2.
For (Prop. X.) CA²: CF² :: CP: CO; and by
divifion, CF2 -CA: CF :: CO-CP: CO, or
CD2: CF2 :: PO: CO.
CF²
Cor. 1. CD: CF2 :: MO: OH :: MP: CH::
PO: CO.
For the triangles MOP, HOC, are ſimilar.
Cor. 2. If L be the latus rectum of AB; then
As L:L+AB:: MO: OH :: OP: OC : : MP :
CH.
I
For (Cor. 1.) CD: CF:: MO: OH; but (Prop.
IV. Cor. 1.) CD² = CA × L, and (Prop. II.)
CF CA²+CD CA²+CA × L. There-
fore MO: OH :: CA× L:: CA²+CA×÷L::
L: CA+L:: L: BA+L.
I
I
2
Cor. 3. CA: CD:: CH- PM: PM:: CP:
PO: HM: MO.
For (Cor. 1.) CF2: CD:: CH: MP, &c. And
by divifion, CF-CD or CA' : CD²: : CH
MP: MP::CP: PO, &c.
PROP.
XV.
If PM be an ordinate to the tranfverfe, MQ a. 6.
tangent at M, C the center; then CQ: CA: CP,
are continually proportional.
For fince MQ biffects the angle SMF (Prop.
XII. Cor. 1.); therefore (Geom. II. 25.) SM:
MF::
92
THE HYPERBOL A:
Fig. MF:: SQ: QF. And compounding and dividing,
6. SM+MF: SM-MF:: SQ+QF: SQ-QF;
that is, SM+MF: 2AC :: 2CF: 2CQ; and al-
SM+MF
ternately, AC: QC::
2
: CF ::
X AC: ACX CF. But (Prop. V.)
SM+MF
2
SM+MF
X
2
AC CFXCP. Whence CA: CQ :: CF x CP:
CA × CF :: CP: CA, or CP : CA :: CA : CQ.
Cor. 1. If MQ be a tangent at M; C the center;
MP an ordinate; then BQ: AQ : : BP : AP.
For CP: CA :: CA: CQ; and mixtly, CP+
CA: CP-CA:: CA+CQ: CA-CQ; that is,
BP: AP :: BQ : AQ.
2
Cor. 2. If MQ be a tangent at M, CE the femi-
conjugate, MP an ordinate; and MO perp. to MQ.
Then CQ: CE: PO, are in geometrical progreffion.
For CQ: CA: CA: CP, and CQ2: CA²::
CQ²
CA² CP². And (Prop. X. Cor. 1.) CA²: CE² ::
CP: PO; and multiplying, CQ: CE² : : CA² ×
CP:CPxPO::CA CPXPO::CA-XPO:CP
XPO'. But (fame Cor.) CA² × PO = CE² × CP.
Whence CQ2: CE² : : CE² × CP : PO² × CP : CE²:
PO². And CQ : CE : : CE : PO.
2
Cor. 3. All hyperbolas defcribed on the fame tranf-
verfe axis AB, will have their tangents to interfect at
the fame point of the axis Q, when the abfciffa AP is
the fame.
For if CP, CA remain the fame; CQ is deter
mined thereby, and muſt remain the ſame.
Cor. 4. All tangents to the curve AM, interfe
the axis CA, between the center C, and vertex A.
PROP
B. II. THE HYPERBOL A.
93
PROP. XVI.
Fig.
All hyperbola's defcribed on the fame tranfverfe ax- 4.
is, will have their ordinates, ſtanding on the fame
abfciffa, in a given ratio to one another; that is, as
the respective conjugates.
For (Prop. VI.) we have BA: NE² :: BPA :
PM²; and alternately, BA2: BPA:: NE2: PM2.
Now if BA and BPA remains the fame, NE² and
PM² will be in a given ratio, and PM to NE in a
given ratio.
PROP.
XVII.
If MT be a tangent at M, and the line FD be 10.
drawn perpendicular upon it from either focus. The
interfection D will be in the circumference of a cir-
cle ADB, defcribed on the tranfverfe axis BA, as
a diameter.
Draw SM from the other focus, and produce
FD till it interfect it in G. Draw FM and CD.
Then in the right angled triangles FMD, GMD;
the angle FMD = (Cor. 1. Prop. XII.) GMD, and
the fide MD being common; therefore FD DG,
and FM MG.
MG. And FC
And FC CS, therefore CD is
Whence the triangles FCD and
FSG are fimilar. Therefore fince FDFG,
SM-MG SM-MF
parallel to SG.
will CDSG=
2
—
2.
BACA; therefore fince CD CA, the cir-
cle DAB whoſe radius is CA, will pafs thro' D.
And by a like reaſoning, if S be the other focus,
and SH be perp. to MH, the point H will be in
the circumference of the fame circle DAB.
Cor. 1. If a circle be defcribed on the tranfverfe
axis AB and a tangent at M cuts it in the points D,
H.
94
THE HYPERBOLA.
Fig. H. And FD, SH be drawn from the foci to D and
10. H. Then FD, SH are perp. to the tangent HMD.
Cor. 2. If F, S be the foci; ℃ the center; MD
a tangent at M; and if CD be drawn parallel to SM,
to cut the tangent in D. Then CD CA half the
tranfverfe.
Cor. 3. If a tangent MH interfects the circle
DAHB defcribed on the tranfverfe, in D and H;
the perpendiculars DF, HS, being drawn to it, will
paſs thro' the foci, F, S.
All theſe follow from the demonſtration of this
Prop.
Cor. 4. If CI be drawn parallel to the tangent
at M, to interfect SM, then will IM
verſe CA.
For then IM CD CA.
femitrans
PRO P. XVIII.
II.
If F, S be the foci, C the center, MH a tangent
at M. If SH, FD, be drawn perp. to the tangent
MH. Then SH × FD = BS × SA = the ſquare of
the femiconjugate axis.
x
About the tranfverfe deſcribe the circle DAHB,
to interfect SH at R, and draw CR and CD. Then
fince the << DHR is right, then RHD (Geom.
IV. 14.) is a femicircle = BHA, therefore RB =
DA, and RCB DCA. Therefore in the tri-
< =
angles SCR, FCD; the fides SC, CR are equal to
FC, CD refpectively; and the included angles at
Care equal therefore SR DF. But (Geom.
IV. 21.) HS × SR = AS × SB; that is, HS × DF
ASX SB = AFX FB (Prop. III.) fquare of
the femiconjugate.
—
PROP
G
M
Fig. 1.
K
IB
S
R
L
O
4. E
7.
p
m
.
W
H
P
K
M
5.
S
B
B
E
A
Ꮐ
A
M
P
8.
2.
B
3.
S
B.
প
S
H
E
A M
D
S
1
A
F
P
9.
M
G
10.
S
6.
F
Pl. XIII. pa.94 NIE
OF
MICH
B. II.
95
THE HYPERBOLA,
PROP.
XIX.
Fig.
If MH be a tangent at M, interfecting the axis 11.
in F; FD, SH, perpendiculars on it from the foci.
Then HM: MD :: HT : TD.
Deſcribe the circle BHAD about the tranſverſe
AB, and draw SM, FM; then the triangles SMH,
FMD are fimilar; for the angles at H and D are
right. And (Prop. IX.) SMH FMD; therefore
HM : MD :: HS: DF. And by the fimilar tri-
angles TSH, TFD, TH: TD:: HS: DF ::
(before) HM: MD.
Cor. HM: MD:: CF+CT: CF-CT:: TS:
TF.
For HM: MD :: TH: TD : : (fimilar trian-
gles) TS: TF.
PROP.
XX.
If MH be a tangent at M, SH, FD perpendiculars 12.
on it from the foci; CE the femiconjugate axis; then
the rectangle HMD rectangle SMF CE².
About the diameter BA defcribe the circle
DAHB, which will paſs through D and H (Prop.
XVII.). Thro' the center C draw MO cutting the
circle in I and O. Then (Geom. IV. 21.) HMD
= OMICM²-CI² =
CI²CM²
CM²-CA². But (Prop.
VIII.) CM² CA SMF CE; whence
=
HMD SMF-CE.
PROP. XXI.
If MH be a tangent at M; SH, FD perpendi- 12:
culars on it from the foci; CE the femiconjugate. It
will be SM: MF:: SH: CE: CE: FD².
For (Prop. IX.) the angle SMH
:
FMD; there-
fore by fimilar triangles, SM: MF:: SH: FD::
SHXFD:
96
HYPERBOLA.
THE
Fig. SH x FD: FD (Prop. XVIII.): : CE²: FD²; and
12. FM: MS:: FD: SH :: FD × SH or CE²: SH".
13.
13.
PROP. XXII.
If MP be an ordinate to the tranfverfe axis, MH
a tangent at M, C the center, AK, CG, BN perp.
to AB. Then AK: PM :: CG: BN.
For (Prop. XV.) TC: CA :: CA : CP; and
compounding, TC : TC+CA or TB:: CA: CA
+CP or BP; and fubtracting, TC: TB:: CA-
TC: BP-TB:: TA: TP. But all the trian-
gles TAK, TPM, TCG, TBN, are fimilar; whence
TA: TP:: AK: PM, and TC: TB:: CG: BN.
Therefore AK: PM :: CG: BN.
Cor. Hence TA: TP :: TC: TB.
PROP. XXIII.
If PM be an ordinate to the tranfverfe axis, MH
a tangent at M, C the center, CE the femiconjugate;
AK, CG, BN perpendiculars to BA. Then AK ×
BN PMX CG CE² = BSA BAX latus
rectum.
I
For (Prop. XXII.) AK× BN = PM × CG. Let
fall SH, FD from the foci, perp. to MH. Then
the triangles TBN, TSH, and alfo TAK, TDF
are fimilar; whence
TH: HS :: TB : BN
and TD : DF:: TA: AK
multiplying TD × TH : DF × HS : : TA × TB :
AK X BN.
But (Prop. XVII.) the points D, H, are in the
circumference of a circle defcribed on the diameter
AB. Therefore (Geom. IV. 20.) TD x TH=
ΤΑ
B. II.
97
THE HYPERBOLA.
2,
TA × TB. Whence DF x HS = AKX BN Fig.
(Prop. XVIII.) CE. But (Prop. III.) CE BSA, 13.
and alfo (Prop. IV. Cor. 1.) CE BAX lat.
rectum.
Cor. 1. FD × SHAK× BN = CE², &c.
Cor. 2. The lines NS, KS, drawn to the focus,
make a right angle NSK.
For NB: BS: SA: AK; therefore (Geom. II.
16.) the triangles NBS, SAK are fimilar, and <
BNS ASK. But < NSK
But NSK = NSA+ASK
NSA+BNS = NBA = a right angle.
Cor. 3. Hence a circle defcribed on the diameter
NK, will pass thro' the foci, S, F.
PROP. XXIV.
If S, F be the foci, MH a tangent at M, MO 14.
perp. to it, and OG perp. to SM; then MG is half
the latus rectum.
Draw SH, FD perp. to MH; then (Prop.
XVIII.) SH× FD CE² the fquare of half the
conjugate. And fince OG is perp. to MS, and OM
perp. to MH; therefore <ĜOM HMS; there-
fore the triangles SMH, FMD, and MOG are fi-
milar; whence
MS: SH:: MO: MG
MF: FD:: MO: MG.
multiplying, MF x MS: SHX FD:: MO: MG".
But (Prop. XVIII.) SH × FD = CE the fquare
X
of the femiconjugate; therefore MFX MS: CE::
MO²: MG²; and MFX MS: MO² :: CE² : MG².
But (Prop. XI. Cor. 2.) CA: CE :: MFX MS:
MO; therefore CA: CE:: CE: MG'; and
CA² CE²
CA:CE: CE: MG. Therefore (Prop. IV. Cor.
1.) MG is half the latus rectum.
•
Cor.
98
HYPERBOLA.
THE
Fig.
14.
9.
15.
Cor. 1. If MO be perp. to the tangent at M, L
latus rectum. Then AB+L: L:: rectangle SOF :
MO².
For we had SMF: MO²:: CA²: CE² or CAX
L:: CAL; and compounding, SMF+
MO: MO: CA+L: L::AB+L: L.
But (Prop. XI.) SMF+MO' = SOF; therefore,
&c.
:
Cor. 2. If MO be perp. to the curve at M, cutting
the two axes in O and H; and S, F, the foci. Then
rectangle SOF MOH.
=
For SOF: MO²:: AB+L:L::(Prop. XIV.
Cor. 2.) HO: MO:: HO× MO : MO²; therefore
SOF MOH.
Cor. 3. Therefore a circle may be defcribed from
Jome point of the conjugate CD, which will pass thro
all four points F, S, H, M.
PROP. XXV.
If FH be an ordinate at the focus, TH a tangent
at H, PM any other ordinate, continued to the tan-
gent at G. Then FM = PG.
1. FH is
FH x CN
=
= CA.
2
latus rectum, and (Prop. XXIII.)
CA X latus rectum, therefore CN
I
2
2. Again, draw AL, BZ perp. to AB, and CD
=the conjugate; then (Prop. XV.) CA: CF ::
CT: CA; and dividing, CF-CA or AF: CF::
CA- CT or TA: CA; and dividing again AF:
(CF AF or) AC:: TA: (CA-TA or) CT ::
(fimilar triangles) AL: CN or CA. Therefore AF
= AL.
3. Likewife (Prop. XV. Cor. 1.) AF FB::
TA: TB (fimilar triangles) LA or AF: BZ.
Therefore BZBF.
:
4. Alfo
B. II.
99
THE
HYPERBOLA.
4. Alfo (Prop. XVI.) CA² = CFX CT
× CF — TF = CF²
CD' CFX FT.
—
CF Fig,
2
CFT, and CF - CA or 15.
5. TF: FH:: (fim. triangles) TC CN or
CA:: TCX CF or CA: CA X CF :. CA: CF;
that is, TF: FH:: CA: CF.
6. By the laſt art. CA: CF:: TF: FH:: (fimi-
lar triangles) TP PG, And (Prop. XV. and
ICT: CA
FM}
VII. Cor. 3.) CA: CF::CP: CA + FM
:
(dividing) CP-CT or TP : FM. Therefore TP:
PG :: TP : FM; whence PG
FM.
From this procefs are drawn the following co-
rollaries.
Cor. 1. If HZ be the focal tangent, AL, CN,
CD, BZ, PG, perpendiculars on AB; then CN=
CA.
Cor. 2. AL AF, and BZ BF.
Cor. 3. CF x FT — ÇD².
Cor. 4. TF: FH:: CA: CF.
Cor. 5 TA: AF:: TP : FM.
For TA: AL (AF) :: TP : PG (FM), by fi:
milar triangles.
Cor. 6. CA: CF:: TP : FM.
2
For CA; CF : : TF : FH : ; TP : PG or FM,
L
PROP.
100
THE HYPERBOLA
Fig.
16.
17.
18.
PROP. XXVI.
If AB be the tranfverfe axis, C the center, F
the focus; and if CF, CA, CT be in geometrical
progreffion (or which is the fame, if TH be the fo-
cal tangent), and TE be perp. to TB. Then if MF
be drawn to the focus, and ME parallel to TB; it
will always be CF : CA :: FM˜: ME.
Draw the ordinate FH at the focus, and the tan-
gent TH; then (Prop, XV.) CF: CA:: CA:
CT; and therefore (Prop. XXV. Cor. 6.) CF:
CA:: FM: TP, or CF: CA:: FM: EM.
Cor. MF is to ME always in the fame given ra-
tio, wherever the point M is taken.
SCHOLI U M.
The writers on Conic Sections, call the line TE,
the Directrix.
PROP. XXVII.
Any diameter MN is biffected in the center C.
For take CP CQ, and erect the ordinates.
PM, QN; and draw NM; then (Cor. 5. Prop. VI.)
QNMP, therefore the triangles CPM, CQN
are equal and fimilar; therefore CN CM.
PROP.
XXVIII.
If DF be an hyperbola, whofe center is C, and ſe-
mitranfverfe CD, femiconjugate CA, QV parallel to
CD. Then CA³ : CD² : : CA² + CQ² : QV³.
Draw the ordinate BV; then (Prop. VI.) CA² :
CD'¹ : ´: BV² : CB- CD; and by addition, CA²:
CD::
B. II.
ΤΟΙ
THE
HYPERBOLA.
2
CD²:: CA² + BV: CB2; that is, CA: CD:: Fig.
CA² + CQ² : QV².
Cor. 1. If AM be its conjugate hyperbola, PM an
ordinate; then CP - CA¹: PM² :: CA² + CQ² :
QV².
: :
For CPCA²: PM²:: CA²: CD (this
Prop.) CA² CQ² : QV².
Cor. 2. CA² + CQ² : QV² :: CA² + Cq² : qv².
PROP. XXIX.
18.
If MC, VC be two conjugate femidiameters; and 19:
if from the ends M, V, the perpendiculars MP, VQ
be let fall on the tranfverfe AB. Then CQ=
AP X PB = CP2 CA".
For draw the tangent MT, and (Prop. XV.)
CP: CA: CA: CT; and dividing, CP: AP ::
CA: AT:: (by adding) BP : PT. Whence CP X
PT AP × PB CP² CA, and CA² =
CP CPT.
2
2
The tangent MT (def. 7.) is parallel to CV;
whence the triangles TPM and CQV are fimilar.
AC² + CQ² = CP² — CPT + CQ. And (Cor. 1.
Prop. XXVIII.) APB CA² + CQ²:: PM² :
QV¹:: (fim. triangles) TP : CQ. That is, CP
x PT: CP² + CQ CPT:: TP : CQ; and
alternately, CPT: TP :: CP² + CQ² - CPT:
+CQ
CQ²; and ſubtracting, CPT — TP' : TP² : : CP²
CPT: CQ; that is, CPTP × PT : TP² :
CPTP x CP: CQ, or PT: PC:: TP :
CQ². Whence PT x CQ = PC × TP², and
CQ² = TP × PC = AP × PB.
CA² + CQ CA² + APB.
2
=
Cor. 1. CP
Cor. 2. CP2
CQ
2
CA CP - APB.
=
I... 2
Cor.
102
HYPERBOLA.
THE
Fig.
19.
19.
3
Cor. 3. QV2 - PM² CD'.
2
:
For (laft Prop.) CA²; CD² :: CA² + CQ²:
QV²: (Cor. 1. Prop. XXVIII.) CP² - CA² or
CQ²: PM² :: (ſubtracting the antecedents) CA² :
QV² PM²; therefore CD" = QV² - PM2.
2
Cor. 4. CA: CD : : CP : QV ; : CQ : PM.
For (Prop. XXVIII. and Cor.) CA² : CD² :
(CA² + CQ² or) CP2: QV²:: (CP² — CA² or)
CQ2: PM².
PROP. XXX.
The difference of the Squares of two conjugate dia-
meters is equal to the difference of the Squares of the
CV² =CA² CD2.
two axes, CM²
For in the right
triangles CPM, CQV, CM*
CQ + QV; there-
CV² -
CV² = CPz
CP² CQ + PM² QV²
≈ CP² + PM², and CV
fore CM²
= (XXIX. Cor. 2.) CA+ PM² QV²
2
(Cor. 3.) CA²- CD².
Cor. In an equilateral hyperbola, every diameter is
equal to its conjugate.
PROP. XXXI.
19. Let MC, CV be two conjugate femidiameters; then
the rectangle of the diftances of the foci from the ver-
tex M of the diameter MC, is equal to the fquare of
the femiconjugate CV, FMX SM = CV².
Let CA, CD be the femiaxes. Then (Prop. VIII.)
CM² CA² CD + SMF, and SMF =
CM2 CA² + CD² = (Prop. XXX.) CM² +
CV2 CM² CV²,
Cor. 1. If F, S, be the foci; then FM × SM =
MC Xlat. reatum of MC.
༡
For
B. II.
103
THE
HYPERBOLA.
For CV2
I
MC x lat. rectum of MC.
Fig.
19.
Cor. 2. If MO is perp. to the tangent at M; then
CA: CD :: CV : MO.
For (Prop. XI. Cor. 2.) CA: CD2: SMF or
CV²: MO²; and CA : CD :: CV : MO.
PROP. XXXII.
If F, S, be the foci, CM, CK femidiameters con- 20.
jugate to one another, TM a tangent at any point M;
FT, CI, SH perpendiculars on the tangent TM; CD
the femiconjugate axis. Then FM: FT:: SM :
SH :: CK : CD.
For fince CF CS, FT SH 2CI. The
= =
triangles FMT, SMH, are fimilar; whence MF:
FT MS: SH: : (MF MS) 2AC (FT -
SH) 2CI :: AC: CI; that is,
AC: CI :: MF: FT, and
fimilar triangles AC: CI:: MS: SH;
therefore AC: CI:: MS x MF: SH × FT.
But (XXXI.) MS × MF = CK', and (XVIII.)
SH×FT = CD²; therefore,
AC²: CI² :: CK²: CD² :: MF2: FT².
and AC: CI :: CK :CD :: MF: FT:: MS:
SH.
::CK
Cor. MF: FT:: AC: CI:: CK: CD.
From the demonftration.
PROP.
XXXIII
All infcribed parallelograms, whofe fides are paral- 21.
lel to two conjugate diameters, are equal to the rectan-
gle of the two axes.
Let CI be perp. to the fide GH, that touches.
the curve in M; then (Cor. Prop, XXXII.) AC :
L 3
CL::
!
104
HYPERBOLA.
THE
Fig. CI: CK: CD; therefore CK x CI = AC XCD;
21. but CK X CI= KCMG ACX CD. But KCMG
is a quarter of the infcribed parallelogram; and AC
× CD is a quarter of the rectangle of the axes;
therefore the wholes are equal.
22.
PRO P. XXXIV.
Let CA be the femitranfverfe axis; CM any fe-
midiameter; AE, MT, two tangents at A and M.
Then the triangle CAE = CTM.
Draw the ordinates MP, AD, to the axis CA,
and diameter CM. Then the triangles CPM, CAE,
and alſo CAD, CTM, are fimilar. Whence CD:
CM :: CA: CT : : (XV.) CP : CA :: CM: CE.
Therefore (Geom. II. 12. Cor. 1.) the lines DP,
MA, ET, are all parallel to one another. Whence
(Geom. II. 10.) triangle AME = triangle AMT,
which taken from the triangle CAM, leaves tri-
angle CAE = triangle CMT.
Cor. 1. The lines DP, MA, ET, are parallel to
one another.
This is plain from the demonftration.
Cor. 2. The triangle AIT = triangle MIE.
This appears by taking CTIE from the equal
triangles CAE, CTM.
Cor. 3. The triangle MPT = triangle ADE =
trapezoid MPAE = trapezoid MDAŤ.
For AME
then PMT
then ADE
AMT, add the triangle AMP,
MPAE; or add AMD (AMP),
(MPT) ADMT,
Cor. 4. Triangle CDA
CPM.
For CD: CM:: CP: CA. Therefore (Geom. II.
17. Cor. 1.) CDA ≈ CPM.
PROP.
1
Fig.n.
M
F
H
14,
E
D
B
G
17
B
M
M
G
M
12.
15.
H
.0
B
E
13.
E
B
T
A
H
m
B
16.
T
E
F
P
M
I
18.
P
M
PEAO
F
D B
T
19
B
S
PI.XIV.pa. 10.4.
UNIV
OF
CH
B. II: THE HYPERBOLA.
105
PRO P. XXXV.
Fig.
Let BA be the tranfverfe, MV a diameter, MP, 23.
QL ordinates to them; AE, MT, tangents; QFL
perp. to AB. Then the triangle QIR trapezoid
IAEF.
::
Let CK be the femiconjugate to CM._Produce
QL to R. By fimilar triangles, CA : AE :: CP :
PM :: CI: IF. And by adding the antecedents,
and the confequents, CA: AE: CA + CP: AE
+ PM: CA + CI :: AE+IF. But the rec-
tangle BIA = AI × CA + CI, and rectangle BPA
= AP × CA + CP. Therefore BPA : BIA : :
AP × CA + CP : AI × CA + CI :: AP ×
AE + PM
AE+IF
2
AEIF.
: AIX
2
:: trapezoid PAEM:
The triangles MPT, QIR are fimilar; whence
triangle MPT : QIR : : MP² : QI² :: BPA :
BIA :: trapezoid PAEM: trapezoid AEIF. But
(XXXIV. Cor. 3.) MPT
QIR AEIF.
=
PAEM, therefore
For (XXXIV.) CAE CMT, fubtract both
Cor. The triangle QLF
trapezoid LMTR.
FIRL=
from CLR, and then LRAE LMTR. But
LRAEFIAE FIRL QIR
—
=
QLF; therefore LMTR = QLF.
PROP
XXXVI.
Let CK be the femiconjugate to MC; LQ an or- 23.
dinate to MC, then CM³ : CK² :: rectangle MLV:
LQ Square.
Let BA be the tranfverfe axis, produce it, and
draw KO perpen. to it, to cut CM in O. Then
(Cor. XXXV.) QLF = LMTR.
L.4
But
168
THE HYPERBOL A:
=
Fig. But (XXIX.) CP² — CA² = CN2, and (ib:
23. Cor. 4.) CP X PM CN X NK, therefore tri-
angle CPM triangle CNK. Alfo the triangle
CPM : CAE :: CP² : CA². And by divifion,
triangle CPM (CPM CAE or) PAEM::
CP²: (CPCA or) CN:: triangle CPM: tri-
CN²
angle CNO; therefore PAEM ČNO, and fub-
tracting theſe from the equal triangles CPM and
CNK, and then CAE COK, or (XXXIV.)
CTM COK.
2
=
2.
The triangle CTM: CRL :: CM²: CL²; and
CTM CRL - CTM :: CM²: CL
:
- CM³;
that is, triangle CTM: trapezoid LMTR: CM* :
rectangle VLM. But the triangles OKC, FQL,
are fimilar; therefore CK: LQ:: triangle OKC:
triangle FQL; that is, CK: LQ: CMT:
LMTR (before) CM re&angle VLM; or
CK: CM :: LQ: rectangle VLM.
Cor. 1 As the tranfverfe MV to its latus rec-
tum:: fo the reƐlangle MLV : to Square of the or-
dinate LQ².
Cor. 2. The rectangles of the distances from the
vertexes of any diameter; are as the fquares of the
ordinates, ftanding at thefe difiances.
A
Cor. 3. The diameter biffets all the double ordi-
nates. And the ordinates at equal distances from the
vertexes, are equal.
For the ratio being the fame for the ordinates
on each fide, ftanding on the fame point; thefe
ordinates must be equal. And as the terms of
the ratio remain the fame at equal diftances from
the vertex; therefore the ordinates will be equal
at equal diftances.
Cor. 4. Triangle CNO
ENKCPM, and COK
trapezoid APME, ánd
CTM.
PROP.
B. II. THE HYPERBOLA.
107
PROP.
XXXVII.
Fig.
If two right lines GI, HP be drawn parallel to 24.
two conjugate diameters DK; MV, to interfect in R; 25.
then CM²: CK :: rectangle PRH: rectangle GRI.
PN²
10²
2
For (XXVIII. Cor. 2.) PN2: IO:: CK +
CN': CK² + CO. And
2
(fig. 24.) PN2: IO²
CN2.
PN² PK + CN: CO - CN².
: :
PN RN2-PN PRH.
=
LILR GRI.
Alfo (fig. 25.) PN² : PN*
CN²: CN³ - CO², But PN2
But 102
And CO2- CN
RN' PRH. And CN CO
=RL-LG' GRI. Whence
2
CN²
IO² :: CK² +
IO
PN-
2
RL-LI
PN: PRH:: ČK + CN2: GRI. And alternate-
ly PRH GRI:: PN: CK+CN:: (Prop.
XXVIII.) CV² or CM²: CK².
Cor. Hence if more lines be drawn paraliel to HP,
GI; the rectangles of their fegments, will be as the
rectangles PRH and GRI, parallel to them. The
fame holds in the oppofite hyperbola, with the ſmall
letters.
PROP. XXXVIII.
If a line HG interfect any diameter AB in D,
a diameter FK be drawn parallel to HG.
rectangle ADB rectangle GDH:: CB': CF'.
:
and 26.
Then, 27.
Draw the diameter MV conjugate to FK, and
AI parallel to GH. Then the triangles CDE, CAI
CA X CE
are fimilar; therefore CI: CA::CE:
DC; and CI: AI:: CE : DE =
CI
AI X CE
CI
Alfo
108
HYPERBOLA
THE
Fig. Alfo (Prop. XXXVI.) CM² : CF² :: CE² — CM² :
26.
EH² =
27.
CE2 x CF2
CM2
= CD² — CA² =
CECI
CF. Hence (fig. 26.) ADB
CA2X CE2
CA CA² X
2
CI
CI2
Again CF2
'CM² X AI
× —
•
CM²:: AI: CICM², and
CF2 x CI - CF2 x CM; and CF
X CI² CM² × AI² = CF² × CM².
But rectangle GDH = EH²
ED² =
CE2XCF2
AI² XCE
-CF2
CM²
CI2
CE x CF2 x CI
2
AI2 x CE² x CM²
AI² × ×
CF2
CM² X
CI²
CE2 x CF2 x CM²
CM² X CI²
CE2-CI2
CE² x CF2
CF² =
CF2=
CI2
CF2 X
CI2
28.
Therefore rectangle ADB : GDH:: CA²: CF2
And (fig. 27.) the demonftration is the fame, on-
ly ADB CA² - CD2, and GDH ED² -
EH². Whence the ſame concluſion will follow.
Cor. The fame holds in the oppofite hyperbola, with
the small letters.
PROP.
XXXIX.
If two lines GH, LE interfect one another in D
(within or without the curve), and the femidiameters
CF, CK be drawn parallel to them; the rectangle
GDH: rectangle LDE:: CF: CK2.
Thro' D draw the femidiameter CB; then (laſt
Prop.) rectangle GDH: rect, ADB :: CF² : CB²;
and
B. II.
109
THE
HYPERBOLA:
and rectangle ADB : LDE:: CB2: CK²; there- Fig.
fore (ex equo) GDH: LDE:: CF²: CK².
And a like demonftration is had when the inter-
fection is without the curve.
Cor. 1. If any line GH cut two parallels LE,
MQ, in D and N, then HDG: EDL :: HNG:
QNM.
For EL, QM being parallel to the fame dia-
meter; theſe rectangles will be as the fquares of
the femidiameters parallel to theſe lines.
Cor. 2. If two parallel lines GH, TO, cut other
two parallels LE, MQ; their feveral rettangles
will be in the fame ratio; HDG : EDL : : HNG:
QNM :: OKT : ERL :: OST: QSM; and that
whether they interfect within or without the curve.
And in the fame, or the oppofite hyperbolas.
For each pair will be refpectively proportional
to the fquares of the femidiameters, parallel to
thefe lines.
PROP. XL.
28.
If DM be a tangent at M, and DN any line cut- 29.
ting the hyperbola (or the oppofite fections) in O and
N;
and if CF, CK be femidiameters parallel to DM,
DN. Then DM: rectangle ODN :: CF² : CK'.
For (by Prop. XXXVIII. and fig. 27.) we have
ADB: GDH CB CF. And fuppofing G,
H, to coincide in M; we ſhall have (fig. 29.) ADB:
DM²:: CB²: CF. And alternately, ADB: CB² : :
DM² : CF². And (by the fame Prop.) ADB :
CB: ODN CK; therefore DM: CF ::
ODN : CK².
: :
Cor. 1. If DM, DN be two tangents, and the fe- 30.
midiameters CF, CK be drawn parallel thereto; then
DM: DN:: CF : CK,
For
110
THE HYPERBOL A.
i
Fig.
For fuppofe O, N, to come together in one
30. point (fig. 29.) then ODN, becomes DN²; whence
DM² DN² :: CF: CK", and DM: DN:: CF:
CK.
31.
32.
33.
:
!
Cor. 2. If two parallel tangents DM, TG, meet
a third tangent NDT; then DM: DN:: TG: TN.
For if the femidiameters CK, CF be drawn pa-
rallel to DT and DM. It will be DM: DN::
FC: CK, and TG: TN:: CF: CK. Therefore
DM : DN :: TG : TN.
Cor. 3. If M, G be the points of contact of the
parallel tangents MD, GT; and N the point of con-
tact of the tangent ND. And if OE, CF be paral-
lel to DM, and QR, CK parallel to ND. Then CF2:
CK²:: DM2 DN: TG: TN: 10E:
NO² :: IQE: PQL:: GR²: LRP.
2
All this appears from this Prop. and Cor. I.
and 2.
PROP.
XLI.
If DM, DN touch the hyperbola at M and N,
and MN be drawn, and any line RL parallel to
DN; cutting DM, MN, in G and H; then GR,
GH, GL are continually proportional.
For (Cor. 3. laft Prop.) RGL: GM² : : DN² :
DM²:: (fimilar triangles) GH': GM. Therefore
RGL GH; and GR: GH: GL
Cor. If FP, parallel to DN, touch the curve in
P, and cuts DM in O; then OPOF.
For then GL will fall on FP; the points R and
Lupon P, G upon O, and H upon F. Whence
GR GL OP, and FOGH. Therefor
OP' (RGL) = OF' (GH), and OP OF.
-
PROP.
i
I
Fig. 20.
21.
R
P
N
!
B
K
H
D
S
AK
M
D
F
24.
23.
P
D
K
D
26.
B
M
R
N
22.
P
M
A
M
25.
C
K
M
F
E
B
27.
H
R
28.
F
N
T
P1.XV. pa.110
UNIV
OF
RICH
B. II. THE HYPERBOL A.
III
PROP.
XLII.
Fig.
If AB be any diameter, MT a tangent cutting it 34.
in T, and MP an ordinate. Then CT, CA, CP are
continually proportional.
:
At A and B draw the tangents AK, BH, cut-
ting MT in K and H; then fince AK, PM, BH
are parallel; BP: AP:: HM: MK :: (Cor. 2.
XL.) HB AK:: (fimilar triangles) TB: TA.
And BP AP; BP:: TB + TA: TB, or 2CP:
BP: 2CB: TB, and CP: BP :: CB: TB. And
dividing PC: BP PC:: CB: TB- CB; that
is, PC BC: BC: CT.
:
Cor. 1. If MT be a tangent at M, MP an ordi-
nate.
Then BP: PA :: BT : AT.
harmonically divided in T and A,
From the demonftration,
Cor 2. If MT be a tangent at M,
nate.
Then PA: PT :: PC: PB.
And BP is
MP an ordi-
For we had PC : PB : : BC: BT; and by divi-
fion, PC PB:: (PCBC or) PA: (PB - BT
or) PT.
:
Cor. 3. If MT be a tangent at M, MP an ordi-
Then TA: TP:: TC: TB.
nate.
For CT: CA: CA: CP, and dividing, CA-
CT: CT: CP-CA: CA, or TA: TC:: AP:
AC; and adding, TA: TC: TA + AP: TC +
AC. That is, TA: TC:: TP : TB.
Cor. 4. Two tangents drawn from the fame dou-
ble ordinate, on each fide the curve, meet in one and
the fame point of the diameter.
For the fame quantities, and the fame propor-
tion, determine both of them.
Cor.
1
112
THE HYPERBOLA.
!
Fig. Cor. 5. If AB, DS be two conjugate diameters,
34. MR parallel to AB, cutting DS in R; MG a tan-
gent at M, cutting the conjugates in T and G. Then
CR: CS :: CS: CG.
35.
Draw the ordinate PM, then by fimilar trian-
gles, CGCT :: RG: RM or CP :: (by diviſion)
CR: TP; and alternately, CG: CR :: CT: TP.
But (XXXVI.) CS : CA² :: PM²: BPA; that is,
CS: CPX CT:: CR: CPCA or CP-CP
× CT; alternately, CS: CR:: CP x CT: CP²
CR²
CPX CT or CP x TP :: CT: TP :: (before)
CG: CR:: CG x CR: CR". Therefore CS2 -
CG X CR.
Cor. 6. The fame things fuppofed as in Cor. 5. the
rectangle GRC = CS² + CR³, and rectangle RGC
= CS² + CG².
2
For CS2 CG X CR GR-
=
GR X CR
Again CS2
= CG x RG
CG.
=
CR X CR-
CR², and GR × CR = CS² + CR²:
CG X CR = CG X RGCG
CG", and RG x CG CS +
>
PROP.
XLIII.
Let AB be any diameter, CD the femiconjugate,
AM, DN, conjugate hyperbola's, PM an ordinate,
QV parallel to CD. Then CA: CD :: CA' +
CQ² : QV².
2
2
Draw the ordinate BV parallel to CA. Then
(Prop. XXXVI.) CA: CD2 :: BV: CB2
CA² ĈD²
CD2; and compounding, CA: CD²:: CA² +
BV² : CB². That is, CA²: CD² ::CQ² + CA²:
QV¹.
Cor. 1. If AM, DN be conjugate hyperbolas;
CA, CD conjugate femidiameters, PM an ordinate,
QV parallel to it. Then CP-CA² : PM² : : CQ³
+ CA²: QV².
For
B. II.
113
THE
HYPERBOLA.
For CP2
CQ² : QV².
CA: PM²:: CA: CD:: CA + Fig.
Cor. 2. If qv be parallel to QV; then CQ² +
CA²: QV²:: Cq² + CA²: qv².
2
Cor. 3. If the ordinate PM be produced to the
conjugate hyperbola at N. Then CP-CA2: PM²::
CP²
CP² + CA²: PN2.
For CP2-CA2: PM²:: CA: CD2::CA +
CP² : PN².
CD²
CA²
Cor. 4. PN² + PM: PN — PM':: CP: CA².
By compounding the terms in Cor. 3.
PROP. XLIV.
35.
If AB be any diameter, CD its femiconjugate; 34.
HT a tangent at M, cutting the parallel tangents
AK, BH, in K and H; MP an ordinate. Then
AK, CD, BH, are in continual proportion.
:
By fimilar triangles AK: PM: TA: TP ::
(XLII. Cor. 3.) TC: TB:: (fim. triangles) CG :
BH. Whence PM × CG AKX BH.
2.
X
Then CP: CT :: CP²: CP × CT or CA
(XLII.). And by diviſion, CP — CT or PT
CT: CP² - CA or BPA: CA²: : (XXXVI.)
PM² CD². And by fimilar triangles, TP : TC::
PM: CG:: PM²: PM × CG or AKx BH; there-
fore PM² CD:: PM: AK x BH.
PM² : AK × BH.
AKXBH = CD'.
Whence
Cor. 1. If AB be any diameter, CD its femicanju-
gate, PM an ordinate; AK, BH, HM, tangents at
A, B, M; then AK × BH = PM × CG = CD²
=AC Xlatus rectum of AB.
For we had before AK x BH
(by this Prop.) CD² = (Schol. IV.)
rectum.
PM x CG -
AB × latus
Cor.
114
HYPERBOLA.
THE
Fig. Cor. 2. If the two parallel tangents at the ver-
36. texes, AK, BH, be cut by the two tangents MKH,
QRN; then AK X BH = BQ × AR.
37.
×
For they are both equal to CD².
Cor. 3. The fame things fuppofed, AK : BQ::
KO: OH.
For AK: BQ : : AR : BH :: AK + AR or
KR: BQ + BH or QH:: (fimilar triangles) KO:
OH.
Cor. 4. A line drawn thro' RH, would interfect
the line drawn thro' KQ, ſomewhere in the diameter
AB produced.
This is plain from the proportions in Cor. 3.
Cor. 5. If the tangent QR was to touch the op-
pofite hyperbola (as gr), it will still be AK × BH-
Bq x Ar.
PROP.
XLV.
In the hyperbola and oppofite fections if TH, TK,
be two tangents at H and K, and if HK be drawn
thro' the points of contact, and biffected in Ọ, and if
TO be drawn; it will biffect all the parallels to
HK, as IN. That is, if HO = OK, then IP =
PN.
Thro' the center C draw FG parallel to HK;
then fince HO = OK, CFCG. And CA is a
femidiameter, and HO, OK, ordinates to it. And
fince IN is parallel to HK; IP, PN will alfo be
ordinates; and (XXXVI. Cor. 3.) IP PN; and
CGE and CFD are femiconjugates.
Cor. 1. If LM be drawn parallel to HK, which
paſſes thro' the points of contact; the parts intercept-
ed between the curve and tangent, on each fide, are
equal; LI = MN.
For
Fig: 29.
12
V k
A
D
C
H
M
K
G
32.
34.
T
M
F
R
30.
B
D
M
P
E
35.
P
R
S
K
G
31.
D
K
M
N
P
33.
R
M
n
36.
A
D
H
B
B
N
M
Pl. XVI.pa.114.
N
M
OF
B. II. THE
115
HYPERBOLA.
For fince HO OK, therefore LP
=
and IP being = PN, the remainder IL
PM, and Fig.
NM.
37.
Cor. 2. The tangent at A, is biffected in the
point A.
Cor. 3. HO, OK, and alfo IP, PN, are ordi-
nates to the diameter CA, paffing thro' T the inter-
fection of the tangents.
SCHOLI U M.
All theſe things hold equally, if one of the tan
gents be drawn to the oppofite hyperbola.
PROP. XLVI.
If TL, TM be two tangents at H and K, and
38.
HK be drawn thro' the points of contact; and from
the interfection T, any line TI be drawn, cutting
HK in D, and the curve in F and I. Then TF :
TI :: DF: DI.
Thro' F, I, draw AC, LM, parallel to HK;
then (Prop. XLV. Cor. 1.) LINM, and AF
BC; whence MI LN, and CF AB.
by fimilar triangles, AF: LI:: TF : TI.
and CF MI:: TF: TI.
=
multiplying, AF x CF :: LIX MI :: TF2: TI2.
that is, AF x AB:: LIX LN:: TF: TI².
But (Prop. XL. Cor. 3.) AF x AB :: LI ×
LN :: HA* : HL2: : (fimilar triangles) DF² : DI²,
Therefore TF² : TI² :: DF² : DI²; and TF:
TI :: DF: DI.
!
Cor. If it be TF : TI :: DF : DI, the line HK
joining the points of contact (of the tangents TH,
TK,) will pass thro' the point D.
M
SCHO-
116
THE HYPERBOLA.
Fig.
38.
39.
40.
SCHOLIU M.
If one of the tangents be drawn to the oppofite
hyperbola, 'tis equally true.
PRO P. XLVII.
If MT, AH be two tangents at any points
M
and A; and if the femidiameters CM, CA be drawn,
to interfect the tangents in H and T. Then the tri-
angle MNH = triangle ANT. And the like for
the oppofite fections.
›
Draw PM parallel to AH. Then by fimilar
triangles, CM: CH :: CP : CA : : (XLÍI.) CA :
CT. And alternately, CM. CA :: CH: CT;
whence (Geom. II. 17. Cor. 1.) the triangle CAH
triangle CMT; which fubtracted from the qua-
drilateral CANM, leaves MNH = ANT.
Cor. The triangle CAH = triangle CMT.
From the demonftration.
PROP.
1
XLVHI.
If CM, CV be any two conjugates. And MP,
QV ordinates to any other diameter AC; and CD
the conjugate of AC; then CQ CP² - CA
- APB.
This Prop. is demonftrated from Prop. XLII.
and Cor. 1. Prop. XLIII. in the fame manner as
Prop. XXIX. was from Prop. XV. and Cor, 1.
Prop. XXVIII.
And the fame corollaries will
follow, viz.
Cor. 1. CP CA² + CQ² = CA² + APB.
=
Cor. 2. CP2
CQ² = CA² = CP²
APB.
Cor.
B. II.
117
THE
HYPERBOLA.
Cor. 3. QV2PM
2
CD².
Cor. 4. CA: CD:: CP: QV:: CQ: PM.
Cor. 5. Hence the triangle CPM triangle CQV.
This is proved the fame way as in Prop. XXXVI.
PROP. XLIX.
Fig.
40.
If AH be a tangent at any point À; MC, VC, 41.
two conjugate femidiameters, cutting the tangent in Z
and H; AC, CD two conjugates. Then the rectan-
gle ZAH = CD².
To the diameter AC draw the ordinates MP,
VQ. Then by fimilar triangles CP: PM: CA:
AZ, whence CP X AZ CA x PM (Cor. 4.
laft Prop.) CQ × CD; therefore CP: CQ : : CD :
AZ.
Again, CQ: QV : : CA : AH, and CQ × AH
=CA × QV = (ib.) CP × CD. Therefore CP:
CQ : : AH :´CD :: CD : AZ, and AZ× AH÷
CD2.
Cor. If SN be drawn parallel to CD, then FN¹
- SFG
CD2.
For by fimilar triangles CA: AZ:: CF: FG
and CA: AH:: CF: FS
and multiplying, CA: AZ X AH or CD :: CF²:
FS x FG.
and compounding CA: CD:: CA² + CF²: CD'
+ SFG:: (XLIII.) ÇA² + CF2: FN²; therefore
CD² + SFG FN2, and FN-SFG
=
CD'.
M 2
PROP.
118
THE HYPERBO L' A.
Fig.
PROP. L.
42. If PT, QT be two tangents at P, Q; and PQ
drawn between the points of contact; and if any line
LD cuts PQ, and the curve in S, A, B. Then SL²:
SD²:: rectangle ALB: rectangle BDA.
Draw the femidiameters CG, CM, CF, parallel
to TP, TQ, LD, (which I have put without the
figure to avoid confufion); and draw LI parallel to
QT, to cut PQ in I. Then the triangles SLI,
SOD, are fimilar, as alfo PIL, PQT; whence
SD: SL :: DQ: LI. But the ratio of DQ to
LI is compounded of the ratio of DQ to PL, and
(PL to LI or) PT to TQ. Therefore DQ: LI::
DQ × PT :: PL x TQ Therefore SD: SL::
DQ X PT: PL X TQ. Take E a mean propor-
tional between AL and LB; and H a mean pro-
portional between BD and DA. Then ALB =
E², and ADB H².
>
Then (Prop. XL.) GC² : FC² : : LP² : ALB or
E, and FC: MC² :: BDA or H² : DQ². There-
fore extracting the roots, GC: FC:: LP: E; and
FC: MC :: H: DQ; therefore GC : MC :: PL
× H: DQ × E. But (XL. Cor. 1.) PT: QT
GC: MC :: PL × H: DQ × E. And multiply-
ing means and extremes, PT x DQ × E = QT
× PLX H. Whence H: E:: DQ × PT: PLX
TQ:: SD: SL; and H2 or ADB : E2 or ALB::
E²
SD² : SL².
Cor. Hence in any infcribed trapezium LNDO,
the diagonals LD, NO, interfeƐt in the fame point
S, where the lines PQ, KR interfect, that join the
oppofite points of contact.
For if LD cuts the tangents PL,
D, and PQ in S; we have, SL
QD in L and
SD:: ALB :
BDA::
B. II.
119
THE
HYPERBOLA.
BDA :: EE: HH. And SL: SD:: E: H; and Fig.
compounding, SL + SD: SL:: E+H:E; or E 42.
+H: E: LD : SL.
Alfo for the tangents KL, RD, it will be SL²:
SD: ALB ADB. Whence, as before, it will
be E+H: E:: LD: LS. Therefore PQ, and
KR interfect LD in the fame point S. And by
the fame reaſoning, the diagonal NO, cutting the
conjugate hyperbolas in a and b, will pafs thro' S,
the interfection of PQ, KR.
PROP.
LI.
1
If MG be any diameter, MD, GT two tangents 43.
at M and G; DTN a tangent at N, cutting the
former in D and T. And if the femidiameter CK be
drawn parallel to ND. Then the rectangle DNT =
CK2.
For let CF be the femiconjugate of MG, then
(XL. Cor. 1.)
CF: CK :: DM: DN.
and CF: CK :: TG : TN.
multiplying CF² : CK² : DM × TG : DN × TN.
But (XLIV,) CF² = DM × TG; therefore CK² =
DN X NT.
PROP.
LII.
If FG be drawn thro' the points of contact of two 44
tangents AF, AG; and AV be drawn thro' A their
interſection, parallel to GF. And if from any point in
it as V, LVO be drawn thro' the middle of FG, to
cut the curve in P and L; then will VP: VL::
OP: OL.
*
From L draw LAQR, cutting the curve in Q,
from Q draw QSP parallel to GF.
M 3
Then (Prop.
XLVI.)
120
THE HYPERBOL A.
Fig. XLVI.) LA: LR :: QA: QR, and LA X QR
44. LRXQA.
45.
The ratio of PQ to SQ is compounded of the
ratio of PQ to OR and OR to SQ. But by fi-
milar triangles PQ : OR : : LQ : LR; and ÖR:
SQ :: AR: AQ.
AR : LR × AQ.
LQ X AR — LRX
Therefore PQ : SQ :: LQ X
And by divifion, PS: SQ ::
AQ: LR × AQ. But LQX
AR-LR X AQ = LR — QR × AR— LR ×
AQ = LRX AR — AQ — QR × AR LR x
QRAR X QR = AL × QR = LR × AQ
before. Therefore PS: SQ :: LR × AQ:LR X
AQ; and therefore PS SQ, and confequently,
fince QS is an ordinate, PS is an ordinate, and P
is in the curve. Laftly, by reafon of the parallels
FG, PQ, VA, VP: VLAQ: AL:: (before)
RQ:RL::OP : OL.
PROP. LIII.
If FG join the points of contact of two tangents
FA, GA; and AV be parallel to GO. Then if
two tangents VH, VI, be drawn from any point. V in
it. Then HI drawn thro' the points of contact, will
paſs thro' O the middle of FG.
For let B, L be two points in the curve; where
VO, AO cut it. Then in refpect of the ordinate
FG, it is (by laft Prop.) VP: VL :: OP : OL. And
in reſpect of the line HI, it is (Prop. XLVI.) VP :
VL:: OP: OL; therefore O is common to both
lines FG, HI; and therefore HI paffes thro' O.
Cor. 1. If FA, GA be two tangents, and FG
drawn thro' the points of contact. Alſo HV, IV two
other tangents, and HI drawn thro' their points of
contact. Then if AV be drawn thro' the interfections
of
Fig. 37.
#
m
n
P
E
D
M
H
N
1
A
T
39.
N
P
H
42.
K
B
P
40.
M
RAHO
N
α
C
S
B
M
n
H
38.
K
41.
P
M
S
M
K
L
44.
T
F
43.
S
M
R
P1.XVII. pa.120.
UNIV
M
OF
B. II.
121
THE HYPERBOLA.
of the tangents; then will O, where FG, HI inter- Fig.
fect, be the point thro' which all lines pafs, which 45.
join the points of contact of any two tangents, which
meet in the line AV.
Cor. 2. If FA, GA be two tangents, and if AV
be drawn parallel to FG; then if any line HI be
drawn thro' the middle of FG; then the tangents
drawn from H and I will meet fome where in the
line AV.
Otherwiſe it could not be VP: VL::OP: OL,
as by Prop. LII.
Cor. 3. If IH be continued to interfect AV in
D, and the tangents DP, DL drawn; then LP paf-
fing thro' the points of contact will pass thro' O.
PRO P. LIV.
If VH, VI be two tangents, and IHD be drawn
thro' the points of contact, and if any line OPVL
be drawn thro' V, to cut the curve in P and L;
then two tangents PD, LD drawn from P and L,
will meet fome where in the line ID.
1
Let O be the interfection of VP and IH; thro'
O draw FG, fo that FOOG, and draw VAD
parallel to it. Then (LIII. Cor. 2.), the tangents
PD, LD, will meet in the line VA, as at D. Then
if DO does not pafs thro' the points of contact
H, I; let a line drawn from D to N paſs thro' the
points of contact. Then (Prop. XLVI.) we fhall
have VP: VL:: NP: NL. But HV, IV, being
tangents, it will be (by the fame Prop.) VP: VL::
OP: OL. Therefore the former proportion is
falfe, except N coincide with O. Therefore DO
paffes thro³ the points of contact H, I, Or which
M 4
is
45.
122
THE HYPERBOL A.
Fig. is the fame, the tangents PD, LD meet in the line
46.
47.
༑
IH.
Cor. If FA, GA be two tangents, and GBF
be drawn thro' the points of contact. And from any
point B in FG, the tangents BI, BH are drawn.
Then IH (drawn thro' the points of contact H, I.)
will pass thro' A.
For a line drawn from A will pafs thro' the
points of contact H, I, and therefore no other
line can paſs thro' them.
PROP.
LV.
If a line MN be drawn thro' the focus F, and
tangents MR, NR drawn from M and N. Then
the line RF drawn from the intersection of the
langents to the focus, will be perp. to the line MN.
Let C be the center, and make CF, CA, CT
continually proportional. At T erect TL perp. to
the tranſverſe axis BA. Then (Prop. XV.) a tan-
gent drawn from the end of the ordinate erected
at F, will cut the axis at T. And (LIII. Cor. 2.)
The tangents drawn from M and N, will meet
fome where in the line TL, as at R. Make SG
= BA, draw GF, which (Prop. XII.) will be
perp. to MR at D. Draw MLO parallel to BA.
The triangles SFG, MOG are fimilar, whence
SG or BA: SF:: MG: MO; that is, CA: CF::
MF MO, But (Prop. XXVI.) CF CA::
:
ML: MF. Therefore ML: MF:: MF : MO,
and ML X MO MF. But the triangles MDO,
MLR are fimilar, being right angled at D and L.
Whence MD: MO :: ML: MR; therefore MD
X
× MR = MLX MO MF". And MD: MF::
MF: MR. Whence the triangles MDF, MFR,
are fimilar, and therefore the angle MFR = <
MDF a right angle.
Cor.
B. II.
123
THE HYPERBOLA.
Cor. A circle defcribed about the diameter MR will Fig.
pass thro' the focus F, and thro' the points L, G.
For MFR and MLR are right angles, and the
triangle MGD being equal and fimilar to MFD,
will be contained in the oppofite femicircle.
PROP. LVI.
47.
If two tangents HE, HK touching the conjugate 48.
hyperbolas in R and D, cut any diameter AB in E
and K, fo that AE BK. And if FZ be a third
tangent; touching the curve at I, and cutting the
former in F and G. Then the rectangle EF x GK
= KH × RE, a given quantity.
For fince CERA is ſimilar and equal to CPKB;
therefore RE PK. And (Prop. XLIV. Cor. 3.)
PK: RF:: KG: GH, and compounding PK or
RE: RE+ RF:: KG: KG + GH; that is, RE :
EF:: KG: KH; whence EF X KG
a given quantity.
KHX RE
Cor. Producing LE and ZF to V; then VE x
KZ = KH × RE, a given quantity.
For LZ is parallel to EH, and VL to HK;
therefore the triangles VEF and GKZ are fimilar;
and VE:EF:: GK: KZ, whence VE x K2 =
EFX GK KHX RE.
PRO P. LVII.
If ABFE be an infcribed trapezium; and if the
diagonals AF, BE be biffected. The line drawn thro
the points of biflection will pass thro' the center C.
Thro' the points of contact of the oppofite fides
AE, BF, draw the line HI, and the diameter GD
parallel to it. Then (Prop. LVI.) GA × DB = a
given quantity = GEX DF. Whence GA: GE::
DF:
49
124
THE HYPERBOLA:
Fig. DF: DB. Therefore if AF, GD, EB be drawn,
49. then (Cor. 1. Lemma to Prop. LX. Ellipfis), the
line that paffes thro' the middle of AF, EB,
paffes thro³ the middle of GD. But as GD is på-
rallel to HI, the middle of GD is the center C.
50.
Cor. And if there be any four tangents to the by-
perbola, and lines (or diagonals) be drawn to the op-
pofite interfections. Then a line drawn thro' the mid-
dle of thefe, will pass thro' the center.
This is fhewn the fame way as was done in the
ellipfis.
PROP. LVIII.
If AB be any diameter; and if two lines AP, BP
be drawn thro' any point in the curve as P; and
the ordinate DF be drawn interfecting the lines in E
and G. Then DE, DF, DG are in continual pro-
portion.
Draw the tangents AN, BH, the conjugate ICL,
and ordinate PŎ. Then by fimilar triangles
AB
BH: AO: OP
and AB AN:: BO: OP
:
×
multiplying, AB2: AN X BH:: AO X OB:
OP² : : (XXXVI.) AB² : IL; therefore AN x
BH = IL².
Again (XXXVI.) AB² : ADB :: IL² : DF²;
and by fimilar triangles
AB: AD::BH: DE
and AB BD:: AN: DG
:
multiplying, AB2: ADB:: ANX BH: DG X DE:
therefore IL²: DF² :: ANX BH: DGX DE.
But ILANX BH; therefore DF2DG X
DÉ, or DE, DF, DG ÷.
Cor. If from the ends of any diameter AB, two
lines PH, PB be drawn thro' any point P of the
curve,
B. II.
125
THE HYPERBOLA.
curve, to interfect the tangents BH, AN in H and Fig.
N. The rectangle of the tangents is equal to the 50.
Square of the conjugate. AN x BH = IL².
This appears from the demonſtration.
PROP. LIX.
If C be the center, F the focus, and if CF: CA:: 51.
CA: CT, and the directrix TO be drawn perp. to
AB. And if any line MNO be drawn, and the lines
FN, FO, MFH be drawn to the focus. Then OF
biflects the angle NFH.
Draw MP, NQ perp. to TO, and ND parallel
to MF, cutting FO in D. Then (by fimilar trian-
gles) MP: NQ:: MO: NO:: MF: ND; and
alternately, MP : MF : : NQ : ND.
MP MF: CA: CF:: NQ
NQ: ND:: NQ: NF. And ND
fore< =
NFD = NDF (by the parallels) DFH.
PROP. LX.
But (XXVI.)
NF; therefore
=
NF. There-
If AB be any diameter, BG a tangent at B equal 52.
to the latus reilum. If AG be drawn cutting the
ordinate PM (produced) in D. Then PM BP X
PD.
2
For XXXVI. Cor. 1.) AB: BG: APB:
PM". And by fimilar triangles, AB : BG:: AP:
PD: AP x PB: PD x PB:: APB : PM.
Therefore PD x PB PM'.
:
Cor. Hence PM the fquare of the ordinate, is al-
ways greater than PB x BG, the rectangle of the
parameter and abfciffa; by the rectangle GF x FD,
which is fimilar to the rectangle AB x BG. And
by reafon of that excefs the curve is called an HY-
PERBOLA
For
126
THE HYPERBOLA:
Fig. For if GF be parallel to AB, the, rectangle
52. DFG is fimilar to ABG, for by the fimilar trian-
gles ABG, GFD, AB: BG:: GF: FD.
SCHOLIU M.
The rectangle under the diameter AB, and its
parameter BG, is called the Figure of that diameter.
PROP. LXI
53. If QT, PT be two tangents, and if QF, PF be
drawn from the points of contact to the focus F; then
the line TF will biffect the angle QFP.
54.
:
Make CF CA:: CA: CG, and draw the di-
rectrix DGR perp. to AB. Produce FT to K,
and draw LR, KR (Prop. L.) tangents at L, K.
Let QD, PE be perp. to DR. Then (XXVI.
Cor.) QF: PF :: QD : PE :: (fimilar triangles)
QR: PR : : (XLVI.) QN: PN; therefore
(Geom. II. 25. Cor. 1.) FN biffects the angle PFQ.
PRO P. LXII.
If CB be the femitranfverfe, BD the femiconju-
gate axis; S, F the foci, PM an ordinate. Thro'
C and D, draw CO, to cut the ordinate PM in O,
and make CICO. Then SM BI.
For (Cor. 1. Prop. V.) CB: CS :: CP; SM +
CB. Whence SM X CB + CB
SM X CB CS x CPCB.
=
CD = CS, and by fimilar triangles,
CSX CP; and
But (def. 6.)
CB: CD or
CS:: CP: CO or CI, and CS x CP CBX CI;
therefore SM x CB
— – —
CB x CI - CB; whence
by divifion SM CI
=
CB = BI.
Cor. 1. The fame things remaining, FM = AI.
For
D
M
Fig.45.
B
F
H
47.
P
46
N
S
n
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R
T
m
N
:
E
49.
B
H
H
A
F
48.
2
料
​K/B
P 9
50.
P
51.
H
N
E
N
B
D
HV
M
10
P1.XVIII. pa.126.
N
ස
OF
B. II.
127
THE
HYPERBOLA.
For FM
SM + AB BI + BA AI.
=
Cor. 2. If MS be produced to N, fo that SN =
SC, and NL be perp. to CB. Then CB — SL, BD,
and SM are continually proportional; but when P is
between S and B, it is CB + SL.
For by fimilar triangles, SM : SP:: SN or SC:
SL; whence SL × SM = SC × SP = CF × SP
=
CFX CPCS CFX CP-CS2. But we
had CF x CP or CS x CP SM X CB + CB²;
therefore SLX SM SM x CB + CB² — CS²;
and SM X CB-SL CS2 CB (II. Cor.)
=
BD². But if P is between S and B, SLX SM =
CF × CS- CP, &c.
Cor. 3. Put Z for the cofine of the angle BSM,
CD2
then will SM = CB + Zx CS®
Or when BSM is
greater than a right angle, it is CBZ X CS.
For by trigonometry, Rad. (1.): SN or SC : :
S.SNL (Z): SL = Z× SC, and CB + SL = CB
+ ZX SC.
PROP. LXIII.
Fig.
54.
If ABDC be an infcribed trapezium, and any point 55.
E be taken in the curve, from which the two lines
EN, EH are drawn parallel to any two adjoining
fides as AB, AC, to interfect the oppofite fides in
N and Q, and in H and R. Then taking the rec-
tangles upon the oppofite fides; ERXEH will always
be to EN X EQ, in a given ratio, wherever E is
placed.
Draw DFG, BPL parallel to AC; thro' C and
L draw COLS; and thro' the middle of AC, BL,
draw the diameter MK, which will biffect the pa-
rallels DF, TE; and alfo SG, OR which are ter-
minated
128
THE
HYPERBOLA.
Fig. minated at AB, CL. Then will SD
FG, OT
55. ER; and theſe would alſo be equal if SG, RH
were tangents.
56.
By reaſon of the parallels, and from the fimi-
lar triangles, HO: SD:: OC: SC :: AR: AG:
and alternately, HO: AR or EQ:: SD or FG:
AG.
Again, BP or ER PN: DG GB; and
multiplying the refpective terms, HO × ER: EQ
× PN: FG × DG: AG × GB :: (XXXIX
Cor. 2.) ER X RT or ER X EO: RAX RB of
EQ X EP. And compounding, FG × DG : AG
x GB:: HOX ER +EO × ER: EQX PN +
EQ × EP : : EH × ER : EQ × EN. But it is
evident, as long as the points A, B, D, C, are
fixed; that FGD to AGB is a given ratio, and
confequently that of REH to QEN.
Cor. 1. The fame things fuppofed, it will be EH :
EN: EO: EP:: HO: PN.
For EHXER : EQ × EN :: FG X GD: AG ×
GD :: HO × ER : EQ × PN. And alternately,
EH X ER: HO × ER :: EQ × EN : EQ × PN;
and by dividing equally, EH: HO:: EN: PN;
and by divifion, EH: EO :: EN : EP.
Cor. 2. If the points A, B, C, E, be fixed; and
D any way changed; it will still be EH EN::
HO: PN:: EO: EP, or RT: RB a given ratio.
Cor. 3. If the points A, C, coincide; MQ be-
comes a tangent. And if B, D, coincide; BN becomes
a tangent.
Cor. 4. If BC be drawn to cut EH
it be made as EH: EI::EN: EV.
ing drawn will touch the curve in B.
in I; and if
Then BV be-
For (Cor. 2.) when D is moveable, it is, EH :
EN EO EP. And if D comes to B, then
:: :
(Cor.
B. II.
THE
129
HYPERBOLA.
(Cor. 3.) BND becomes a tangent; in which cafe Fig.
H comes to I and N to V; and then it is, EH: 56.
EN: EO: EP:: EI: EV.
Cor. 5. If we make EH: EN :: EO: EP, and 55.
draw CO, BP; they will meet at L in the curve.
By Cor. 2.
• Cor. 6. If there be five points, B, L, D, C, E;
and thro' any point E, there be drawn EH, EN, cut-
ting BD, BL, DC, LC, in N, P, H and O; fo
that EH EN: EO: EP. Then CA, BA being
drawn parallel to EH, EN, they will meet at A in
the curve.
This is the reverſe of the foregoing. For if A
was out of the curve, L would not be in it.
PROP. LXIV.
If ABDC be an infcribed trapezium, and from 57.
any point E, of the curve, four lines EL, EK, EO,
EM, be drawn to the four fides of the trapezium,
in any angles whatever, equal or unequal; cutting
them in L, K, O, M. And likewife from any other.
point e, four more lines, el, ek, eo, em, be likewise
drawn respectively parallel to the former, cutting the
fides in l, k, o, m.
Then taking the rectangles of the lines falling up-
on the oppofite fides, it will be, rectangle EL × EM:
el × em: : EK × EO : ek × eo.
Thro' E, e, draw EQN, eqn, parallel to AB;
and ERH, erb, parallel to AC. Then by fimilar
triangles, ER er:: EL: el; and EH: eb:: EM:
em. And multiplying, ER X EH: er X eb:: EL
× EM: el × em.
eo.
Again, EQ: eq :: EK: ek, and EN: en:: EO:
And multiplying, EQX EN: eq xen: EK
XEO: ek x eo. But (Prop. LXIII.) ER X EH :
×
er X
130
HYPERBOL A.
THE
Fig. er X eb:: EQ × EN : eq × en.
Therefore EL X
57. EM: el xem:: EQ × EN: eq x en: :) EK × EO:
ek x eo.
58.
59.
Cor. 1. In the infcribed trapezium ABCD; if
AB, CD, produced, interfect in X; and the diago-
nals AC, BD, in Y; and YEXe be drawn. Then
EX: eX:: EY : eY.
For ſuppoſe BD, AC to become the diagonals;
and that EL, EK, EO, EM, and el, ek, eo, em,
all lie in one line Ee; and that AB, CD, inter-
fect in X, and AC, BD, in Y; and that Ee paffes
thro' X, Y; which is a particular cafe of this
Prop. Then L, /, M, m, coincide in X; and
K, k, O, o, in Y. Then the proportion EL ×
EM: el xem :: EK X EO: ek X eo, becomes EX":
eX²:: EY²: eY", and EX: eX:: EY: eY.
2
Cor. 2. If XA, XC, be two tangents, and AYC ·
paffes thro' the points of contact. It will be EX:
ex:: EY: eY.
For let B approach to A, and D to C; then
BD will coincide with AC. And it will still be
EX: eX :: EY: eY, as in Prop. XLVI.
PRO P. LXV.
If any line MN be drawn thro' the focus F, and
ZCQ be a diameter drawn parallel to it; AB the
tranfverfe axis. Then AB, QZ, MN, are conti-
nually proportional.
Draw the diameter CDL thro' the middle MN,
and the ordinate MP perp. to AB, and tangent
MTO. Then (VII. Cor. 3.) CF x CP CAX
FM+ CA, and CA X FM CFX CP-CA
= = (XV.) CF x CPCT X CP, or CA x FM
TF x CP. And by fimilar triangles CT: CO ::
TF:
B. II.
131
THE
HYPERBOLA.
TF: FM TF x CP or CA x FM: FM x Fig.
CP :: CA: CP:: CT: CA. Whence CO = CA. 59.
But (XLII. Cor. 5.) CZ CO x CG = CA
× LM; and multiplying, QZ = BA × MN.
=
Cor. 1. If F be the focus, ZQ a diameter parallel
to MF; MT a tangent at M, cutting ZQ in O;
then CO CA.
From the demonftration.
Cor. 2. If AB be the axis, F the focus, MP an
ordinate, MT a tangent. Then CA × FM = TF
* CP.
From the demonſtration.
2
I
Cor. 3. If any line MN be drawn thro' the focus
F; the rectangle MFN MN x latus rectum.
= ½
Let Llatus rectum; then (Prop. XXXVIII.)
CZ² : MFN :: CA² : AFB :: (IV. Ĉor. 3.) CA² :
CA XL:: CA: L:: CAX ML: LX ML.
But CZ CAX ML; therefore MFN÷LX
ML.
I
PROP. LXVI.
I
2
In an equilateral hyperbola, the fquare of the femi- 6ɑ:
tranfverfe is equal to the rectangle of the distances of
either vertex from the foci. CA' SAF.
For (Prop. III.) the rectangle of the diſtances of
the two foci, from either vertex, is equal to the
fquare of the femiconjugate. But (Def. XII.)
the tranfverfe and conjugate are equal; therefore
that rectangle is equal to the fquare of the femi-
tranſverſe axis,
N
PROP.
132
THE HYPERBOLA.
Fig.
60.
61.
62.
PROP. LXVII.
In an equilateral hyperbola, the rectangle of the
distances of the ordinate from the two vertices, is
equal to the fquare of the ordinate. APB PM2.
For (Prop. VI.) as fquare of the tranfverfe :
fquare of the conjugate: : fo the rectangle of the
diſtances of the ordinate, from the two vertices
BPA to fquare of the ordinate.
:
terms are equal (Def. XII.).
latter are equal.
2
But the two firſt
Therefore the two
Cor. CP² — CA² = PM², C being the center.
PRO P. LXVIII.
If AV be an equilateral hyperbola, AC the femi-
tranfverfe axis, CQ perp. to CA; QV parallel to
CA. Then CA² + CQ² = QV³.
2
For (Prop. XXVIII.) if CD be the femiconju
gate, then CA²: CD:: CA² + CQ: QV². But
ČA CD'; therefore CA+CQ² = QV².
2
Cor. The hypothenufe AQ= ordinate QV parallel
to CA.
For AQ² = CA² + CQ² = QV²; and AQ =
QV.
PROP. LXIX.
If CA be the femitranfverfe axis, CF, CH, the
affymptotes; AD the femiconjugate; PM an ordi-
nate continued to F. The rectangle HMF AD'.
By fimilar triangles, CP2: PF':: CA: AD' ::
(VI.) APB or CP-CA: PM: : (by divifion)
CP² - CP² + CA or CA: PF-
PM²; that
2
is,
B. II.
THE HYPERBOLA.
133
is, CA² AD²:: CA: PF PM'; therefore Fig.
AD PF PM HMF.
Cor. 1. If the ordinate be continued to the affymp-
totes, the parts between the curve and afymptotes are
equal. FM GH.
=
For PH PF, and PG
remainders are equal, FM
PM; therefore the
GH.
Cor. 2. If more ordinates be drawn to cut the af
fymptotes, their rectangles are all equal.
PROP. LXX.
62.
If from the point M of the curve, MK be drawn 62.
parallel to the tranfverfe axis AB, cutting the afſymp-
totes in I and K; then the rectangle KMI
CA.
For the triangles IMF, KMH and CAD are fi-
milar, whence AD: AC:: MF: IM, and AD :
AC:: MH: KM; and multiplying, AD²: AC² ::
MF × MH: IM × KM. But (LXIX.) MF ×
MH AD; therefore IM X KM AC².
×
×
Cor. 1. The rectangle IM x MH KM × MF.
For by fimilar triangles KM: MH;: IM: MF.
Cor. 2. If more lines be drawn parallel to CA ;
all their rectangles are equal to one another.
PROP. LXXI.
If two parallel lines PS, FH, cut the hyperbola, 63.
and the affymptotes; the rectangle FMH
PQS.
rectangle 64.
Thro' M and Q draw OL, IN perp. to the axis.
Then by reafon of the parallels, the triangles PQI,
FMO, are fimilar, as alfo MHL, QSN. There-
fore MO: MF:: QI: QP, and ML : MH ::
N 2
QN,
134
HYPERBOLA:
THE
Fig. QN: QS. And multiplying MOX ML: MF x
63. MH :: QI × QN : QP × QS. But (LXIX.
64. Cor. 2.) MO × ML = QI× QN; therefore FMH
= PQS.
65.
66.
Cor. 1. If the tangent DG be parallel to FH, then
the rectangle FMH DG'.
Cor. 2. If the diameter AB be parallel to HM,
then rectangle FMH CB".
PRO P. LXXII.
Any right line FMGH be drawn thro' the curves
the parts of the line between the curve and afymp-
totes, are equal; FM GH.
For (Prop. LXXI.) rectangle FMH HGF;
that is, FM X MG + FM x GH GH x GM
+ GH × MF; and FM X MG GH X GM;
whence FM = GH.
Cor. 1. Hence FG MH.
Cor. 2. If AB be a tangent at D; then AD =
DB, or the tangent is biffected in D.
For in this cafe M and G coincide in D.
Cor. 3. If a diameter CD be drawn thro' the
point of contact D, it biffects all the parallels MG,
and all the lines FH.
PRO P.
LXXIII.
If from two points in the curve, M, Q, there be
drawn the parallels MH, QN, to cut one affymp-
tote; and likewife the parallels MO, QI, to cut the
other affymptote. Then the rectangle MÒx MH =
the rectangle QI × QS.
Draw EML, PON perp. to the axis. Then
the triangles EMO, PQI are fimilar; and alfo
MLH,
B. II.
135
THE
HYPERBOL A.
MLH, QNS. Whence ME: MO:: QP: QI; Fig.
and ML : MH :: QN: QS. And multiplying, 66.
MEX ML: MO X MH: QP × QN: QIx QS.
But (LXIX. Cor. 2.) ME × ML = QP × QN
therefore MO × MH = QI × QS.
SCHOLIUM.
The Prop. holds equally true, if QS, &c. cut
the affymptote beyond C towards h.
PROP
LXXIV.
If two lines KN, FT cut one another in E; and 67.
likewife cut the curve and affymptotes; then rectan- 68.
gle LKM : DFV :: KEN: FET.
Thro' N, draw IG parallel to VD; then the tri-
angles LNG, and LED are fimilar; and alfo MNI,
MEV; whence LE: LN:: ED: NG, and ME:
MN:: EV: NI. And multiplying, LE × ME:
LN × MN::EDXEV:NGXNI.
But LEX MELK + KE × EN + NM =
LK × EN + KE × EN + LK + KE × NM =
KE× EN + LK × EM + KE × NM (KE × LK)
= KEX EN + LK × KM (fig. 67.).
Alfo by the fame reafoning, DEX VE
FET
+ DFV. Alfo (LXXII.) LNM LKM, and
GNI DFV. Whence we have
KEN + LKM: LKM:: FET + DFV: DFV.
And by divifion, KEN: LKM:: FET: DFV,
PROP.
LXXV.
All infcribed parallelograms, between the affymp- 69:
totes and the curve, are equal; ABCD = GFCH.
For (by Prop. LXXIII.) the rectangle AB X
AD = GF X GH; whence AB: FG:: GH:
N
3
AD,
136
HYPERBOLA.
THE
:
Fig. AD, or CD: CH:: CF CB; therefore (Geom.
69. III.
9. Cor. 1.) parallelogram AC = parallelo-
gram GC.
jo.
Cor. 1. The hyperbola continually approaches to
the affymptote; but can never touch it, till it be in-
finitely continued.
For EL being reciprocally as CE, EL continu-
ally diminiſhes as CE increaſes, and when CE is
infinitely great, EL will be infinitely ſmall.
Cor. 2. If a line be drawn parallel to one afſymp-
tote, it will interfect the oppofite hyperbolas in one
point, and no more.
For AD can only cut the curve in A, for the
curve AL afterwards approaching CE, muft di-
verge from AD. And the fame line continued
can never cut the oppofite hyperbola, becauſe it
lies wholly without the oppofite angle, in which the
curve is wholly contained.
Cor. 3. The affymptote may be confidered as a tan-
gent to the curve at an infinite diftance.
From Cor. 1. and Cor. 4. Prop. XV.
PROP.
LXXVI.
If the lines AB, DF, touch the hyperbola; then
the triangles CAB and CDF made by the tangents,
and affymptotes, are equal.
From the points of contact, E, G, draw EH,
GI parallel to the affymptote CB; and EO, GP,
parallel to the affymptote CF. Then (LXXII.
Cor. 2.) fince AE EB, and FG GD; there-
=
fore AH HC, and CO = OB; alfo FI = IC,
and CP CD.
www.
=
But triangle ACB to triangle FCD: CA X
CB or 2CH x 2CO: FC x CD or 2CI X 2CP ::
CH X
B. II. THE
137
HYPERBOLA.
CH × CO : CI × CP :: parallelogram CHEO : Fig.
parallelogram CIGP; but (Prop. LXXV.) CHEO 70.
CIGP; therefore triangle ACB = FCD.
Cor. 1. All infcribed triangles are equal to one ano-
ther, and double to the infcribed parallelograms.
For triangle ACB parallelogram CHOE::
2CH X 2CO
2
or 2CH x CO: CH X CO:: 2 : 1.
Cor. 2. If GI be parallel to the affymptote CB;
and GF a tangent at G; then the fubtangent IF
= CI the distance from the center.
For DF is biffected in G; and therefore CF is
biffected in I.
Cor. 3. Any two tangents to the curve LG in-
terfect one another within the angle of the affymptotes
BCF.
For if BA be a tangent at E, this interſects CF
(the tangent at an infinite diſtance) in A. And
therefore the fame BA interfects any other tangent
FD, between A and B,
Cor. 4. Hence alfo the rectangle ACB is equal to
the rectangle FCD,
Cor. 5. If AD and FB be drawn, they will be
parallel.
For CA: CD :: CF: CB, from the equal rec-
angles CAB, CDF.
PROP.
N 4
138
THE HYPERBOLA.
Fig.
LXXVII.
PROP.
71. If thro' any point F in the curve, a line FP be
drawn cutting the curve and affymptotes in P, B, M;
and from В the tangent BEA; and from E the point
of contact, EH; as likewife FQ, PR all parallel to
BC. Then 2EH = FQ + PR =
FQ+ PR = FQ-pr in the
oppofite fection.
For draw FL parallel to CA, then the triangles
MPR, FBL, are fimilar, and MP BF; there-
fore PR LB, and LC FQ; and fince AE =
EB (Prop. LXXII. Cor. 2.)
= 2
AB; therefore
CL + LB =
HE CB, and 2HE CB =
I
रं
QF + PR; or 2he = QF
QF- pr.
Cor. 1. Hence the fum of any two lines FQ, PR
(made by drawing any line BP, from the fame point
B in the affymptote), will always be the fame.
For that fum is always equal to 2EH.
Cor. 2. If BP be parallet CM; then 2EH =
FQ
PRO P. LXXVIII.
72. If AC be the femitranfverfe of the equilateral by-
perbola AM, CF its affymptote, and CH perp. to
CA, and HM parallel to CA. Then if MF be
drawn perp. to the affymptote, 2CH x HM CF
MF².
For draw the femiconjugate AD, which is equal
to AC. The triangles CAD, CHV, VFM are
fimilar, and ifoceles; whence CH = HV, VF =
MF. And VF: VM :: VH: CV, and VM x
VH = VFX CV. Then CV
2CH x HM = 2HV X HM
2HV2, and
= 2HV ×
X
HV + VM
Fig.59.
M
G
B
60.
P
N
6.2.
F
K
B
n
m
F
B
E
65.
H
G
E
2.9
67.
B
H
G
m
69.
C
H
D
F
Į
FP
C
61.
64.
F
63.
A
C B
66.
68.
E
K
70.
P
H
B
F
Ꮐ
OE
M
71.
:
P
R
Pl. XX. pa 138.
M
OF
ICH
B. II.
THE
139
HYPERBOLA.
HV + VM = 2HV² + 2HV × VM
2
—
2CV x VF+VF VF² CV + VF
—
= CF2
VF CF2 MF2.
CV² + Fig.
2
VF 72.
Cor. 1. In a right angled hyperbola; the affymp-
totes make a right angle with each other.
=
For fince CAAD, the angle ACD CDA
a right angle. And twice ACD
2.
angle
the angle of the affymptotes.
a right
Cor. 2. If SB be drawn perp. to the afymptote
from the focus; CB = CA.
For the triangles CAD, CBS, are ifoceles, and
fimilar; and CD CS; therefore CB CA.
=
z
PRÒP.
LXXIX.
If thro' the given point G in the curve, the line 73.
GA be drawn parallel to the affymptote CH. And
any line GL be drawn cutting the curve and affymp-
tote in F and L; and if FK be drawn parallel to the
afymptote CB. Then KL will always be equal to
the given line AG.
Produce FG to D; then the triangles DGA,
FLK are fimilar and equal, for the reſpective ſides
are parallel; and (Prop. LXXII.) GD = FL;
therefore AG KL.
Cor. If from F in the curve, FK be
rallel to the affymptote CB; then FK
part of the other affymptote.
drawn pa-
AD, the
PROP.
140
THE HYPERBOL A.
Fig.
74.
PROP. LXXX.
If CA, CD be the femiaxes; the afymptote CF,
biffes the line DA, drawn between the ends of the
two axes; and all lines HK parallel to it, contained
between the conjugate hyperbolas DH, AK.
For draw AE perp. to AC, and equal to CD;
and draw DE; then DCAE is a rectangled paral-
lelogram. But (Def. XI.) the affymptote CF
paffes thro' E, and (Geom. III. 2.) the diagonals
CE, AD biffect one another in O; therefore DO
= AO.
Again, fince DA, HK are parallel to the af
fymptote CG; therefore (Prop. LXXV.) CI × IK
= CO×OA = CO × OD = (LXXV.) CI × IH
therefore IKIH.
Cor. 1. If DA joining the ends of the two axes,
cut the affymptote in O; then 4CO DA² = CA
+ CD2.
For ACD is a right angle, therefore (Geom. II.
20. Cor. 6.) OC OD OA, and CO² = DO
=AO², and 4CO² = 4DO² = 4ÃO² = DA² =
CA² + CD2.
2
Cor. 2. The rectangle of the fides of the infcribed
parallelogram, is equal to the fourth part of the fun
of the fquares of the two femiaxes.
For draw AP parallel to CE, then COAP is
an infcribed parallelogram, and CO × CP or CO
the rectangle of the fides; and that is
CA² + CD2
4
PROP
B. II.
141
THE HYPERBOLA.
LXXXI.
Fig.
PROP.
If AB be any diameter; CG, CL, affymptotes; 75.
GH a tangent at B. Then GH is equal to the con
jugate diameter of AB.
For let KF be a tangent at A, CD the femi-
conjugate. Then, becauſe (LXXV. Cor. 3.) the
affymptote FG is a tangent at an infinite diftance,
therefore (Prop. XXXIV.) AK × BH = CD²;
but the triangles CBH, CAK are equal and fimi-
lar, and AK = BH; therefore BH2 CD²,´and
BH = CD, and GH or KF = 2CD the conjugate
of AB.
Cor. 1. If CG be an affymptote, BG a tangent at
B. Then CB, BG are femiconjugates to each other.
Cor. 2. Hence the further any diameter AB is from
the axis; the greater it is, and the greater its conju-
gate GH.
PROP.
LXXXII.
If F be the focus, and it be CF: CA:: CA: 76.
CT, and TE be perp. to CA; and if MF be drawn
from any point of the curve to the focus, and MN
parallel to the afymptote CO. Then MN - MF.
Draw ME parallel to CA, and AL perp. to it,
then (Def. 6.) CL CF, and the triangles MEN,
CAL are fimilar, and MN: ME:: CL or CF:
CA :: (XXVI.) FM : ME. Therefore FM =
MN.
PROP.
142
THE HYPERBOLA:
Fig.
77.
78.
\
PROP. LXXXIII.
If F be the focus, and FP perp. to CA, and from
M any point in the curve, MP be drawn parallel to
either affymptote, to cut FI in P. Then FM-MP
latus rectum; or FM + MP, when Mis on the
other fide of I.
I
4/4
For if CF, CA, CT be in geom. progreffion,
and TN perp. to CA. Then (LXXXII.) FM =
MN, and FI = IQ = (by the parallels) PN;
therefore NM - MP = PN = FI; that is, FM
MP-FI.
PRO P. LXXXIV.
If CF, CB be affymptotes, and DA be drawn pa-
rallel to the asymptote CB, cutting the conjugate by-
perbolas in D and A; then CD, CA drawn from
the center to the interſections, will be conjugate dia-
meters.
=
Thro' A, D draw the tangents FG, FL. Then
(LXXX.) fince AO = DO, and (LXXII. Cor. 2.)
FAFG; therefore AO or OD CG or
LC, and DA = CG or CL.
CG or CL. Whence (Geom. I.
5. Cor. 3.) CD is parallel to AG or FG. And
likewife CA is parallel to LD or LF; and there-
före (Def. 7.) ĈA, CD are conjugate diameters.
Cor. 1. Hence it appears that the two conjugate di-
ameters NA, PD, belong to the conjugate hyperbolas
AM, DH; and alfo to the oppofite hyperbolas KN,
PQ. And that all thefe four conjugate hyperbolas,
make but, as it were, one figure BAMHDÍKNPQ
•
Cor. 2. Hence also the conjugate hyperbola DH
paſſes thro' the ends D, of all the conjugate diameters
PD
B. II.
143
THE
HYPERBOLA.
PD; as well as the hyperbola AM paſſes thro' all the Fig.
ends A, of the first diameters NA.
Cor. 3. In an equilateral hyperbola, every diameter
78.
is equal to its conjugate.
For then (fig. 73.) CA
CD, and therefore
CO is perp. to DA, and to all the parallels HK.
Or (in fig. 77.) CO is perp. to DA, and therefore
CD = CA.
PROP. LXXXV..
If LKE be drawn parallel to the affymptote CG, 79.
and any line DV be drawn, cutting this line, the
curve, and affymptotes in E, F, T, D, V; and the
tangent GBQ be drawn parallel to DV, and CB the
diameter.
Then BG: CG:: rectangle FET: 2KE × EV,
::
Draw LP parallel to GQ. Then (LXXVI.
Cor. 1.) 4CL x LK GC x CQ. And by fi-
=
milar triangles CL: LP: CQ: QG :: CQ ×
CG or 4CL × LK: QG × CG. Whence CL x
CG X QG = LP x 4CL X LK, and CG x QG=
4LP x LKCG X 2BG.
× ×
—
LK ×
Then (LXXIV.) LKM : DFV :: KEN : FET.
But KM = EN infinitely; therefore LK: KE::
DFV: FET. And LK× FET = KEX DFV =
KE X BG². And KE × BG² × CG =
FET X CG, or KE x BG X 2LP X LK
FET × CG. And 2KE X BG x LP =
FET, and BG: CG:: FET: 2KE X LP
× EV.
X
Cor. 1. CG X BG 2LP x LK.
From the demonftration.
LK X
CG X
or 2KE
Cor. 2. If more parallels be drawn it will be
FET fet: KEV: kev,
Cor.
$44
HYPERBOLA.
THE
Fig.
79.
80.
Cor. 3. If two parallels meet one line LE; then
FET: fot:: KE : KO.
Cor. 4. Hence alfo fet: fot: kev: Kov.
:
PROP. LXXXVI.
If FBE be an hyperbola between the affymptotes
Cf, CL; and it be CG: CH :: CK: CL; and lines
or ordinates be drawn parallel to the other affymptote
Cf. Then the area GHDF area KLEB.
The infcribed rectangles being equal (Prop.
LXXV.), their fides are reciprocally proportional,
EL: BK:: (CK: CL:: (hypoth.) CG: CH: :)
DH: FG; therefore if any lines CG, CH, CK,
CL are proportional; the ordinates GF, HD, KB,
LE will alfo be proportional, and in the fame ra-
tio inverfly. GL
Divide GL into an infinite number of equal
parts at I, P, G, R, &c. in geometrial progreffion;
and draw the ordinates IM, PN, QO, RS, &c.
parallel to GF or Cf; let the fame be continued to
L, and let r to s, be the ratio of the terms of the
progreffion. Whence we ſhall have
i
r:s::CG: CI:: CI: CP:: CP: CQ:: CQ:
CR, &c. and (Propor. XXII.) r : s : : GI : IP ::
IP : PQ : : PQ: QR, &c. and by what went before
s:r::IM: PN:: PN: QO: : QO : RS, &c.
and multiplying the two laft,
:
rs: rs :: GI × IM : IP × PN: IPX PN : PQ
× QO:: PQ X QO: QR × RS, &c. therefore the
infinitely ſmall parallelograms are equal; that is,
GIM IPNPQO QRS, &c. KVA =
VTW = TXZ = XYa, &c.
Now, in the progreffion CG, CI, CP, CQ, &c.
if we take any equidiftant terms CG, CH, and CK,
CL (Propor. XXI.) thefe fhall be proportional;
that is, CG: CH :: CK : CL. And the terms
CH,
B. II.
145
THE HYPERBOL Á.
CH, CL, being equidiftant from CG, CK; there Fig.
is the fame number of parallelograms between G 80.
and H, as between K and L. And theſe paralle-
lograms being all equal, the fum of all between G
and H, is equal to the fum of all between K and
L. But by the method of indiviſibles, the fum of
all the parallelograms GIM, IPN, &c. compoſe
the area GHDF; and the fum of all the parallelo-
grams KA, VW, &c. compofe the area KLEB;
therefore the area GHDF area KLEB.
Cor. 1. If CF, CD be drawn, and alfo DH, FG 81.
parallel to the asymptote Cf. Then the fector CFD
= area FGHD.
ww
For fince. rectangle CGF CHD, triangle
CGF CHD; take away CGq common, then
CqF GqDH. Add FqD common, then CqF +
FqD or ſector CFD = GqDH + FqD = FGHD.
Cor. 2. If it be CG: CH:: CK: CL; then feltor
CFD fector CBE.
For CFD FGHD, and CBE = BKLE.
Cor. 3. If CK: CL :: Cd : Cf. And BK, EL
be drawn parallel to Cf; and Dd, Ff to CL; then
fector CBE = CDF = area BKLE = DdfF.
For CK: CL: Cd: Cf:: CG: CH. Whence
fector CFD = DdfF = FGHD, &c.
Cor. 4. Hence the ſpace DdfF Space DHGF.
For the parallelogram dG added to each makes
the fame ſpace CƒFDH.
Cor. 5. If CG, CH, CK, CL are in geometrical
progreffion; then fector CFD, CDB, CBE are equal;
and also the Spaces FGHD, DHKB, BKLE.
Cor. 6. If CL be the nth term in geometrical pro-
greſſion, in the ratio of CG to CH; then feltor CFE
=21 × CFD; and FGLE = 2 → 1× FGHD.
— n- .XFGHD.
For
146
HYPERBOLA.
THE
Fig.
For if CH is the fecond term, then CFD = 1×
81. CFD. If CK be the third term, then CFB =
CFD + CDB = 2 x CFD.
If CL be the fourth
term, CFE = FCD + DCB + BCE = 3 × CFD,
and ſo on. And the like for the areas FGHD,
FGKB, &c.
Cor. 7. If CG, CH, CK, CL, are proportionals;
then LE, KB, HD, GF are also proportionals, and
in the fame ratio, or FG, DH, BK, EL are pro-
portionals, in the inverse ratio.
From the demonftration.
SCHOLIU M.
Hence if CG = 1, and CH, CK, CL, &c. be
any numbers; the (fectors or) hyperbolic Spaces GFDH,
GFBK, GFEL, &c. are analogous to the logarithms
of theſe numbers.
For (Cor. 6.) whilſt the numbers CG, CH, CK,
CL, &c. proceed in geometrical progreffion, the
correfpondent (fectors or) ſpaces, proceed in arith-
metical progreffion; and therefore (from the na-
ture of logarithms) are proportional to the loga-
rithms of thefe numbers.
82.
Make CGCN 1. Then by computation
from infinite feries, &c. if the number CH be
10, the ſpace GFDH or log. 10. will be found
2.30258509.
If the number CK be 100, the
fpace GFBK or log. 100 will be 4.60517018. If
the number CL be 1000; the ſpace GFEL or
log. 1000, will be 6.90775527. And if the hy-
perbolic ſpaces correfponding to the intermediate
numbers be likewife computed; they will be the
logs. of theſe numbers. And theſe logarithms, be-
cauſe they are computed from the hyperbolic areas,
are called hyperbolic Logarithms.
PROP,
Fig.52.
A
m
54.
CH
L B
B
P.
M
T
F
S P
h
56.
I
PN
B
57.
CM H
m
h D
K
E
k
B
C H
M
P
E
53.
K
T
55.
9
E
R
G
58.
Pl. XIX. pa. 130.
UN
во
OF
K
b
Fig. 72.
B
M
H
F
B
G
73.
N
D
74
T
E
A
H
N
Q
n
M
76.
12:
B
N
K
C
75.
G
•D
L
H
I
78.
M
F
M80.
GIPQR H
BAWZ
CGH K
81.
N
E
टेसे
E
P
Z.
N
83.
C
G
P
m
77.
TA
G
B
79.
M
82.
D
B
E
P1.XXI.p.146.
B. II.
147
THE HYPERBOLA.
Fig.
PROP. LXXXVII.
If two hyperbolas QE, FD, have the fame af- 83.
fymptotes CP, CH; if HDE be drawn parallel to
the afymptote CP; then HD is to HE every where
in the given ratio of CN to CP.
For the infcribed rectangles are CG x GF, and
CG X GQ. Therefore (LXXV.) CH x HD =
CG X GF, and CH× HE = CGX GQ. Whence
CGXGQ.
GF : GQ :: CGFN : CGQP : : CH × HD : CH
× HE :: HD: HE.
Cor. 1. The area FGHD: area QGHE:: ordi-
nate GF: ordinate GQ or as infcribed parallelo-
gram CF infcribed parallelogram CQ.
:
For the ordinate HD to HE is every where as
CN to CP. Therefore if the baſe GH be divided
into an infinite number of equal parts, and ordi-
nates HDE erected. It will be GF: GQ:: HD:
HE: . fum of all the HD. fum of all the
HE: area FGHD: area QGHE.
Cor. 2. If CG = GS = 1, and GF 84.
•43429448; then the area FGHD is the common lo-
garithm, or Brig's logarithm of the number CH.
For the hyp. logarithm is to the common loga-
rithm, as 2.30258509 to 1, or as I to .43429448,
or as GS to GF; that is, (Cor. 1.) as the area
GSEH to GFDH.
SCHOLIU M.
=
Hence Brig's logarithms will be had by comput-
ing the hyperbolic areas GFDH, whofe infcribed
parallelogram is 43429448. Thus if CG 1,
CH 10, the area FGHD = 1; if CH = 100,
the area FGHD = 2. If CH = 1000, FGHD
3, and fo on; and the like for the intermedi-
Q
ate
148
THE HYPERBOLA.
Fig. ate numbers. Likewife fince the area at G is no-
84. thing; therefore the log. of CG or 1 is o; and for
numbers lefs than CG the area lies on the other fide
of SG; and therefore the logarithms of numbers
leſs than 1, are negative. And fince the area SGCB
is infinite, therefore the logarithm of o, is an infi-
nite negative.
85.
It is plain likewife that logarithms may be of as
many different forms as you pleaſe. For if CG be
I, then the log. GSEH will be as the infcribed
parallelogram CRSG. Therefore as that paralle-
logram is greater or leſs, the logarithms of numbers
will be proportionally greater or lefs. And there-
fore in different fpecies of logarithms, the loga-
rithms of particular numbers will be in a given
ratio.
On the contrary, if you have the logarithms of
numbers, the hyperbolic area may be computed.
For it will be as .43429, &c. to the infcribed paral-
lelogram CGSR:: fo log. of CH, to the area
GHES or ſector CSE.
PROP.
LXXXVIII.
If AN be an equilateral hyperbola, AM any hy-
perbola with the fame tranfverfe axis.
The area
ANP of the equilateral hyperbola: area AMP of the
other hyperbola: as the femitranfverfe CA: to ſe-
miconjugate CD.
·CA2
For if CE be the femiconjugate of AN, then
CECA. And (Prop. LXVII.) CP² — CA² =
PN. And (Prop. VI ) CA²: CD² : : CP²
or PN2 PM. And CA: CD:: PN: PM.
Therefore if the abfciffa AP be divided into an in-
finite number of equal parts and ordinates erected;
it will always be CA: CD:: PN: PM :: fum of
all the PN: fum of all the PM:: area ANP :
area AMP.
Cor.
B. II.
,149
THE
HYPERBOLA.
Cor. i. The areas of two hyperbolas on the fame Fig.
tranfverfe, and having the fame abfciffa, AP; are 85.
as their conjugates.
Cor. 2. As femitranfverfe CA
Semitranfverfe 86.
CB: CANP the area of the hyperbola AN: CBMP
the area of the hyperbola BM, on the fame abfciffa CP.
By reaſoning as before CD: CD² + CP² ::
CA²: PN² :: CB2: PM²; and CA: CB : : PN :
PM all the PN: all the PM :: area CANP :
::
area CBMP.
Cor. 3. Therefore if the area of an equilateral by-
perbola be known; the area of any other hyperbola
may be found.
PROP. LXXXIX.
}
If NAM be a hyperbolic conoid, made by revolv- 87.
ing round the tranfverfe axis. If C be the center,
BA the tranfverfe; then 2CA + AP : CA +
AP:: circumfcribing cylinder: folidity of the conoid
NAM.
3
Let femitranfverfe CA
= c, AP = x, PM = y, p =
VI.) tt: ec : : 2tx + xx :
21x + xx. And pyy
ptt
t, femiconjugate CD
3.1416. Then (Prop.
yy, and yy =
tt
X
CC
= X2tx + xx, for the cir
CC
cle NM. Therefore if AP be divided into an in-
finite number of equal parts, according to the
method of indivifibles; and planes drawn thro' all
thefe points, all theſe circular planes will make
up the folid. Now (Arith. Infinites Prop. II.)
the fum of all the 2tx is -
2txx
2
= txx, and (ib.
Prop. III.) the fem of all the xx is =
O 2
; there-
3
fore
150
THE HYPERBOLA.
Fig.
ptt
87. fore the fum of all the pyy = × txx +
|
cc
3
pttx
CC
I
× tx + xx. And the circumfcribing cylin-
3
derpyyx =
pttx
сс
CC'
X 2txxx; therefore the cy-
I
xx
linder to the folid :: 2txx + xx : tx + 3×× : :
2t + x: t +
3
x.
I
Cor. 1. The folidity of the hyperboloid to a cone
of the fame baſe and altitude :: BP : BP + .AC.
For the cone is of the cylinder; and it is, cy-
linder: folid: 2t + x : t + x; therefore cone:
folid 2t+x: 3t+x.
:
3
I
3
Cor. 2. If AP = AC, the conoid NAM =
the circumfcribing cylinder.
9
If AP = 3'AC, the conoid
If AP = AB, the conoid
=
the cylinder.
12
2
the cylinder.
5
4
=
9
If AP = 2AB, the conoid the cylinder, &c.
Cor. 3. The hyperbolic conoid NAM, is equal to
the fruftum of the affymptotic cone EIAKF, leffened
by the cylinder GIAKH.
For (Prop. LXIX) the rectangle EMF = AK;
therefore (Geom. IV. 38.) the annulus or ring
contained between the two circles NM and EF is
equal to the circle IK. Therefore if AP be divid-
ed into an infinite number of equal parts; all the
rings compofing the hollow folid NIAKM = all
the circles GH, compofing the cylinder GIKH.
Therefore, fruftum EIKF cylinder GIKH =
fruftum EIKF - folid NIAKM'; that is,
conoid NAM.
hyp.
Cor.
B. II. THE HYPERBOLA.
151
Cor. 4. The hollow folid NIAKM inclofing, the Fig.
cylinder GIKH.
hyp, conoid is
PROP.
XC. Prob.
To defcribe a hyperbola by continued motion.
1 Way.
Draw the line DB, and let G be a fixed point;
fix the two rulers together IK, KL, fo as to make
the given angle IKL; let there be another ruler
SL, which moves about the fixed point L in the
leg KL, as a center. Then cauſe the fide KL to
move along the line DB, and coincide with it;
whilft the ruler SL flides by the fixed point (or
pin) G. Then the lines SL and KI will crofs one
another; and their interfection F, will defcribe the
hyperbola FOG.
For take SG FL, and draw SC parallel to
IK meeting DB in C. Then G is a fixed point in
the curve, and FK is parallel to SC; and GL
paffes thro' L, fo that KL is always given. There-
fore this is a property of the hyperbola, (by
Prop. LXXIX.) whence the curve is a hyper-
bola whofe affymptotes are SC, LC; and center C.
2 Way.
87.
88.
Let FL, GH be affymptotes, AB a diameter. 89.
Let AG be parallel to the affymptote FL; take
HK = CG, make the angle CHM = GCF.
Then move the right line HK along the affymp-
tote HG, whilft the right line AM (moveable
about A) paffes thro' the point K; then the in-
terfection (of HM, AM), M will defcribe the hy-
perbolas MB, Am.
For the triangles KHM, KGA are fimilar; whence
KH: HM :: KG: GA; that is, CG: HM ::
CH GA; whence CH x HM = CG × GA;
therefore (LXXV.) M is in the hyperbola.
О3
SCHO-
152
THE HYPERBOLA:
Fig.
89.
90.
91.
92.
SCHOLIU M.
The hyperbola may alſo be deſcribed as directed
in Def. 1, and 2.
PROP.
XCI. Prob.
To describe an hyperbole, by finding many points
in the curve.
I Way.
Let BA be the axis; S, F the foci. Take any
point N beyond F, and with extent AN, and one
foot in F, defcribe an arch M. Then with extent
NB, and one foot in S, croſs the former arch at M.
Thus find as many points as neceffary, and draw
the hyperbola thro' them.
By this conftruction, SM-FM AB; there-
fore (Prop. 1.) the point M is in the curve.
2 Way.
Let AB, DG be two diameters conjugate to one
another. Divide CA into any number of equal
parts at L, L, &c. draw DA, and parallel to it,
draw LI, LI, &c. draw CF CB, and perp. to
it. Then thro' I, I, &c. draw MN, MN, &c.
parallel to AB; and fo that IM or IN be equal to
LF. Then M, M, &c. and N, N, &c. will be
oppofite hyperbolas.
2
2
For by fimilar triangles CD: CA²:: CI:
CL: (compounding) CD + CI: CA² + CL2
or IN² or IM. For CA + CL² = CF² + CL²
LF2. Therefore M and N are in the curves,
by Prop. XLIII.
3 Way.
Let CA or CB be the femitranfverfe, CD the
femiconjugate; produce CD at pleaſure, and draw
DE
B. II.
153
THE
HYPERBOLA.
DE parallel to CB; draw DA, and from any Fig.
point P in the axis, draw PF parallel to AD, to 92.
cut CD in F. With radius CF, and center C de-
ſcribe a circle to cut DE in G. At P, draw the
ordinate PM, perp. to CP, and equal to DG. Then
M will be in the curve.
For by fimilar triangles CA: CD:: CP2:
CF2 (by divifion) CP2 CA²: CF² — CD2.
But CP CA² BPA; and CF — CD² =
CG2
CD²:
CD
=
DGPM²; therefore CA:
BPA PM. Whence (Prop. VI.) the
point M is in the curve.
PROP. XCII. Prob.
Given the afſymptotes FG, MN, and the point B 93.
in the curve, to defcribe the hyperbola.
iWay.
Thro' B draw feveral right lines to cut the af
fymptotes, as ABN, DBĚ, GBL, &c. Then
make NT AB, ES = DB, LI = GB, &c.
Then the points T, S, I, are in the curve.
Likewife thro' any point ſo found as T, you may
draw other lines as HTQ, KTO, &c. and make
PQ = HT, KT = OR, &c. then R, Q, &c.
are alſo in the curve.
This proceſs is plain from Prop. LXXII.
2 Way.
Thro' the given point B, draw BA, BD paral- 94.
lel to the affymptotes. Then take any length CG,
and draw GH parallel to CD, and fo that CG:
CA:: CD: GH; then H is in the curve, and fo
for more points.
I
Or take the equal parts CA, AE, EG, GI,
&c. And make EFAB, GH = ÷ AB, IK
AB, and ſo on. Do the fame in the affymp-
04
3
tote
154
THE HYPERBOLA.
Fig. tote CD; then thro' all the points f, B, F, H, K,
94. draw the hyperbola required.
95.
95.
96.
PROP.
XCIII. Prob.
A right line DE in the hyperbola being given; to
find its diameter, and the center.
Draw HI parallel to DE; biffect DE, HI, in F
and G; thro' which draw GFAB; then AB is the
diameter. Biffect AB in C, then C is the center.
For the diameter biffects all the whole ordinates,
and is itſelf biffected in the center.
PROP. XCIV. Prob.
Any diameter AB in a hyperbola, being given; to
find its conjugate.
Draw any line KI parallel to AB, and biffect it
in L; thro' L and the center C, draw LC, for
the conjugate, which will be terminated by the
conjugate hyperbolas, if they are there; otherwife
its magnitude will be determined by Prop. XXXVI.
PRO P. XCV. Prob.
Any diameter AB being given; to draw an ordi-
nate to it, from a given point H in the curve.
Thro' the center C, draw the diameter HK; and
from K, draw KL parallel to AB. From L draw
LH for the double ordinate.
For fince HC = CK, and CO is parallel to KL,
therefore HO = OL.
:
Or thus,
Draw HI parallel to AB, biffect IH in G; thro'
G and the center C, draw GC, and HO parallel
to it for the ordinate.
For
R
N
Fig.84
G
B
E
S
M
P
90
M
N
85
IM
E
D
B
A
86.
C
P
D
C
P
H
D
K
88.
B
A F
F
F
D
F
B
89.
91.
MMMB
E
B
92.
冬
​n n n
95. D
H
94.
93.
D
B
B
F
G
H
K
A
E
G
D
R
NH
K
L
PIXXII pa:154
UNIL
OF
MICH
B. II:
155
THE HYPERBOLA.
For GC, biffecting IH, is the pofition of the Fig.
conjugate, and HO being parallel to it is an ordi- 96.
nate to CO.
:
PRO P. XCVI. Prob.
To find the two axes of a given hyperbola. 97.
Draw two parallels thro' the figure, FK, ML;
thro' the middle of them draw the diameter AB;
biffect it in C for the center.
On the center C defcribe an arch, of a circle,
to cut the curve in two points G, H. Draw GH
and biffect it in O. Thro' O and the center C, draw
CO for the axis. And CD perp. to it for the
conjugate.
For AB paffing thro' C is a diameter, and fince
CG CH; therefore CO paffing thro' the middle
of GH, will be perp. to it, and being perp. to the
ordinate GO, it is the axis.
PROP. XCVII. Prob.
A hyperbola being given; to find two conjugate dia- 98.
meters, making a given angle with one another.
About any diameter AB, defcribe the arch of a
circle ADB, to contain an angle equal to the given
one, or its fupplement. To the point F where it
cuts the hyperbola, draw BF, AF; which biffect
in O and L; and from the center C, draw CO,
CHL. Then CH and CO (terminating at the
conjugal hyperbola,) will be the two axes.
For fince ALAF, and AC
is parallel to BF. And fince FO=
parallel to AF; whence LCO
angle.
2
I
2
AB, CL
FB, CO is
OFL
the given
SCHOLIU M.
and BE be
If CE be the femiconjugate of AC,
drawn and HG parallel to it, cutting CO in G;
then
156
HYPERBOL A.
THE
Լ
Fig. then CG is the conjugate, and G a point in the
98. conjugate hyperbola. For by Prop. LXXXIV.
BE, GH being both parallel to the affymptote;
CH, CG are conjugates, as well as CB, CE.
99.
100.
ΙΟΙ.
PRO P. XCVIII. Prob.
Given the affymptotes and a point in the curve; to
find two conjugate diameters.
From the given point B, draw BHD parallel to
the affymptote GC, make HD HB.
=
Draw
DCE, and make CE = CD, then DE is the con-
jugate of AB.
For draw BF parallel to CD, then the triangles
CDH, FBH are fimilar; and they are equal, be-
caufe BH HD, and therefore CH = HF; and
confequently GB = BF. And therefore FG is a
tangent at B (LXXII. Cor. 2.). And fo FG is
equal to the conjugate of CB (LXXXI.); whence
DE equal and parallel to FG, is its conjugate.
Cor. Hence if two conjugate diameters be given,
and the angle they make with one another, be given;
the affymptotes may be drawn. Which is done by
drawing CG parallel to DB, and CF parallel to EB.
PROP. XCIX.
Prob.
To draw a tangent to any point M in the hyperbola.
I Way.
Draw the femidiameter MC, and find its conju-
gate CD (XCIV.); then from M, draw MT pa-
rallel to CD, which will touch the curve in M.
2 Way.
Let BA be any diameter, draw MP an ordinate
to it; and make CP, CA, CT, continual propor
tionals
B. II.
157
THE HYPERBOLA.
tionals; then draw MT which will be a tangent Fig
at M. By Prop. XLII.
3 Way.
Draw the lines MF, MS, from the point M to
the focus's; make ME FF, draw FE, which
biffect in D, then MD drawn will be a tangent at M.
Or produce SM to G, ſo that MG = MF. Draw
FG, and MT parallel to it, for the tangent at M.
For in the ifoceles triangle EMF, ED = DF;
therefore EMD = FMD, whence (Prop. IX.)
MD is a tangent And fince MD is parallel to
GF, and MF MG; therefore EMD = EGF,
and EMF 2EGF, and EMF = EGF = EMD,
therefore MD is a tangent.
Ι
2
4 Way.
101.
Let CH, CK be the affymptotes; from the point 102;
M draw MD parallel to one affymptote CH. Make
DFDC; and draw FM which will be a tangent
at M.
For produce FM to G; then fince FD = DC;
therefore (by the parallels CG, DM) FM
whence (LXXII. Cor. 2.) MF is a tangent.
PROP. C. Prob.
MG;
From a given point without the hyperbola, to draw
a tangent to it.
I Way.
Let T be the point given. Thro' T draw the 101.
diameter BA. Make CT, CA, CP, continual
proportionals. At P draw the ordinate PM, to
the diameter AB. From M to T draw MT for
the tangent. By Prop. XLII.
2 Way.
Let T be the point given. Thro' T draw the
diameter AB; and find its conjugate CD (XCIV);
draw
103.
158
HYPERBOL A.
THE
Fig. draw any line TS-
line TS TB, and make TR = TC;
103. draw AR, and SP parallel to it, cutting the dia-
meter AB in P. Draw the ordinate PM parallel to
CD; then TM is the tangent.
104.
For by fimilar triangles TS: TR::TP: TA;
that is, by conſtruction, TB: TC :: TP : TA.
Therefore (XLII. Cor. 3.) TM is a tangent.
PROP. CI. Prob.
The focus F being given, and three points in the
curve; to defcribe the hyperbola.
Thro' the two points N, M, draw NMK, and
alſo thro' N, P, draw NIP. Draw MF, NF, PF
to the focus. And make FN: FM :: NK : MK,
and FN : FP :: NI: PI. Thro' K and I, draw
KG; draw FT, MH perp. to it. Take the points
A, B, fo that FM: MH:: FA: AT:: BF: BT;
then BA is the axis, and C the middle point, is
the center. Make BS AF, and S is the other
focus; and the hyperbola may be deſcribed by
Prop. XCI.
Draw NG, PL perp. to KT. Then by con-
ftruction, FN: FM :: NK: MK :: (ſimilar trian-
gles) NG: MH, whence FN: NG :: FM: MH.
Alſo FN FP: : NI : PI :: NG: PL, whence
FN NG :: FP: PL. Therefore FN NG ::
:
FP: PL:: FM: MH::(conftruction) FA : AT::
BF BT. And this (Prop. XXVI.) is the pro-
:
perty of the hyperbola, where TG is the direc-
trix. For, fince AT: AF:: BT: BF:: (com-
pounding) BT+AT. BF + AF:: BA: SF::
CA: CF; therefore (XXVI. Schol.) TG is the
directrix.
PROP.
B. II.
159
THE HYPERBOL A.
Fig.
PRO P. CII. Prob.
Given two conjugate diameters of a hyperbola; 105.
CH, CL, to find the two axes.
Join the ends of the diameters with the line
LH, which biffect in O, draw the affymptote COI.
Draw CF parallel to LH, and CF is the other af
fymptote. Draw CP to biffect the angle HCF,
and CQ perp. to it.
Then the axes lie in the
lines CP, CQ
=
Make IO CO, and thro' H, L, draw IK,
IG. Alfo draw HP, LQ perp. to CP, CQ. And
take CA a mean proportional between CK and CP;
alfo CE a mean between CG, CQ. Then CA, CE
are the femiaxes.
For (LXXX.) fince LO OH, CO is one af-
fymptote, and CF (parallel to LH) is the other;
and fince CO OI, therefore (LXXII. Cor. 2.)
IK, IG are tangents. And fince CK: CA: CP
are in geometrical progreffion; therefore (Prop.
XV.) CA is the axis, and it biffects the angle HCF.
And fince CG: CE:: CE: CQ, and CE perp. to
CA; CE is the other axis.
PROP. CIII. Prob.
To defcribe a hyperbola about a given parallelo- 106.
gram EGHF, to pass thro' a given point P.
Let the conjugate diameters AB, ID biffect the
oppofite fides of the parallelogram; draw PQ pa-
rallel to ID; and take AC²
CQ² × EL²
CL² × PQ²
EL-PQ
€Q¹× EL”—CL³× PQ²
CL² ~ CQ-
And CD2 =
For
160
HYPERBOLA.
THE
Fig. For (Prop. XXXVI.) CA² : CD³ :: ALB : EL^: ;
106. AQB: PQ. That is, AC: CD :: CL-CA2:
2
EL²:: CQ - CA: PQ. Hence CL2- CA2
× PQ² = CQ²
CA² x EL*; therefore CA² ×
2.
2
EL² — PQ² = CQ2 × EL - CL× PQ, which
CQ²
determines AC.
Again, fince AC: CD::
CD²
24
CL-CA: EL;
by compounding, AC: CD :: CL: CD + EL.
And fince CL CA: EL :: CQ² CA² :
PQ2; and (by divifion) CL2 CA: EL:: CL²
CQ: EL² - PQ. Therefore CL: CD +
EL²:: CL-CQ² : EL² - PQ. And (dividing)
CL² : CD² + EL²::CQ²: CD² + PQ. Whence
CLª × CD² + PQ² = CQ² × CD² + EL²; and
CD² x CL² - CQ² = = CQ² × EL CL² × PQ'
2
which determines CD.
2
Cor. AC²: CD :: CL-CQ² : EL² — PQ².
For AC x EL-PQ² = CQ² × EL² — CL
× PQ = CD² × CL² — €Q².
PROP.
CIV.
107. If a cone MDN be cut by a plane, parallel to any
plane DXY within the cone, which paſſes thro' the
vertex D; the fection LAT will be an hyperbola.
Let the plane produced cut the oppofite cone
BDp in Btl, biffect AB in C, and parallel to the
bafe MN, draw PTQ, CGKF, ptq, whofe fections,
with the cone will be circles; and the interfections
with the plane of the hyperbola, will be LT, lt.
Draw PQ, pq perp. to LT, it; and CK a tangent
to the circle GKF, then the rectangle PIQ = LI²,
and rectangle piqli, and rectangle GCF =
CK'. The plane PDQ interfecting the other planes,
will
B
G
Fig. 96.
E
102.
13
M
F
L
97.
F
H
99.
H
105.
H
B
H
?
C
P
{
F
I
0
F
100/
103
T
98.
E
H
D
E
101.
M
D
SB
P
M
Gr
104.
B
TAF
R
H
XM
106.
E
I
P
D
BIQ
K
PI.XXIII:pa.160.
UNIV
OF
ル
​FICH
B. II.
161
THE HYPERBOLA.
will make the fimilar triangles AIP, ACG; and Fig.
alfo BIQ, BCF. Whence
AI: IP :: AC : CG
and BI: IQ :: BC: CF.
And multiplying,
AI X BI: PI × QI:: AC x BC: CG x CF.
that is, AIB LI² :: AC²: CK².
Therefore (Prop. XXXVI.) the fection LAT is a
hyperbola.
Cor. 1. The fame plane cutting the oppofite cone,
the fection will be the oppofite hyperbola tВl, equal and
fimilar to TAL.
For Ai ip: AC: CG, and Bi: iq:: BC: CF.
Whence AiB : piq:: AC× BC: CG × CF; that is,
AiB : il² :: AC²: CK²; therefore when Bi = AI;
then li LI, and the curves are both alike.
Cor. 2. The fquare of the conjugate, 4CK² = BH
X AE, the rectangle of the diameters of the fruftum.
For the circles BH, AE being parallel to the
bafe; fince AC AB, therefore CG
I
2
I
2
BH,
and CF AE; therefore CK' or GC x CF =
=
BHX
AE.
2
Cor. 3. The latus rectum =
BH X AE
AB
×
For it is a third proportional to the tranſverſe
and conjugate.
Cor. 4. The hyperbola is right angled when CK´=
CA, or when BH × AE = AB².
107.
Cor. 5. Thro' the vertex of the cone D, draw the 108.
plane DXY parallel to the plane of the hyperbola
ALT; and thro' X and Y, draw the planes DMX,
DMY, touching the cone in the lines DX, DY; and
interfecting the plane of the hyperbola in the lines NC,
OCo; then theſe lines NCn, OCo, are the asymptotes
of the oppofite hyperbolas LAT, IBt; and YO or
XN, the femiconjugate; and the angle NCO = an-
gle XDY; and MD paſſes thro' the center C.
For
162
THE HYPERBOLA.
Fig. For thro' A draw the circle AGES parallel
108. to the baſe of the cone, which will touch the plane
of the hyperbola in A. This will cut the planes
DMN, DMO, in the lines sr, Gw. And then
the lines sr, rA, Aw, wG are all tangents to the
circle; therefore rsrA, and AwwG. Like-
wife rA = Aw, becauſe the parallels thereto NI
= IO. And fince sr, Gw, are parallel to XN,
YO; therefore XNsrrA AwwG =
YO; and therefore XN² or rA² rectangle LNT,
by the property of the circle; and confequently
(LXXI. Cor. 1.) CN, CO, are affymptotes; and
(LXXXI.) rw is the conjugate, and rA
YO is the femiconjugate.
109.
=
XN or
Draw PK parallel to AB; then by Prop. XVI.
Cor. 2. ellipfis, which alfo agrees with the circle,
QZ : PZ : : QM: PM :: (by diviſion) QM
QZ or MZ: PM-PZ. And alternately, QZ :
MZ:: PZ: PM - PZ:: (by addition) QZ +
PZ or PQ: (MZ + PMPZ) 2PM. And al-
ternately, QZ: PQ:: MZ: 2PM. But by fimi-
lar triangles,
QZ: PQ:: ZD: PK.
and MZ: PM:: ZD: PF.
or MZ: 2PM:: ZD: 2PF.
▾
whence ZD: PK :: ZD: 2PF.
Therefore PK 2PF, whence KP is biffected in
F, and confequently AB (parallel to it) is biffected
in C, and fo C is the center.
Laftly, fince CN, CO, are parallel to DX, DY;
the angle NCO = angle XDY.
Cor. 6. If AI be perp. to the baſe of the cone,
and AB perp. to MH the tranfverfe of the hyperbo-
la. And if AN AB, and FND be drawn parallel
to the bafe PQ. Then FD is equal to the latus rec-
tum. And DS (parallel to MH) = HZ the dia-
meter at the vertex H.
Draw
B. II.
163
THE
HYPERBOLA.
Draw AL, AG parallel to PQ, MH, and put Fig.
R latus rectum. Then the triangles ABL, 109.
ANG, are ſimilar and equal, whence AG = AL.
The triangles MCH, DFS, GFA, are fimilar;
whence MC: MH:: DF: DS:: GF GA or
AL: (by the ſimilar triangles GFA, LAH) AF :
AH (triangles AFD, AHZ fimilar) FD: HZ.
Whence DS HZ. And MH x FD - MC x X
HZ (Cor. 2.) fquare of the conjugate
Schol.) MH x R; therefore FD = R.
(IV.
Cor. 7. In a right cone, the Square of the tranf- 110.
verfe BA BE AEX BH, the Square of the
fide of the fruftum lefs the rectangle of the diame-
ters of the bafes.
Let BF be perp. to the baſe AE, then FE =
BH+AE
BE2
2
; and (Geom. II. 23.) AB² = AE² +
2AE X
BH + AE
2
-BE-AEX BH.
Cor. 8. In a right cone, the Square of the fide of
the fruftum is equal to the fum of the fquares of the
tranſverſe and conjugate, BE² — AB² + ÂE² + BH².
Cor. 9. In a right cone, the distance of the focus
from the center, is equal to half the fide of the fruf-
tum, BE.
I
2
For if S be the focus, O the center; C half the
conjugate. Then (Prop. II.) OS = OB² + C² =
AB² + 4 AEX BH
2
I
I
BE or AH.
2
(Cor. 8.) BE², and OS
Cor. 10. In a right cone, if F be the focus, C III.
the center, and AG AF. The circle GOI paral-
lel to the base of the cone RS, will cut the plane of
the hyperbola FAH in the line ED the directrix of
the hyperbola.
In the plane of the hyperbola, let FH, AL, be
perp. to AF, and HD a tangent at H; then (XXVI.
P
and
164
THE HYPERBOLA.
Fig. and Sch.) D is the point thro' which the directrix
III. paffes. But (XXV. Cor. 2) AL AF
II2.
AG by
conſtruction. Alſo (Cor. 9.) CF
Alfo (Cor. 9.) CF APAN;
=
and therefore CP drawn thro' the middle of AN,
AM will be parallel to MN or to RS the baſe of
the cone. Now (Prop. XV.) CF: CA :: CA:
CD:: (by divifion) AF: AD; that is, AP: CA::
AG: AD; therefore DG is parallel to CP, or the
baſe of the cone. And on the contrary, if DG is
parallel to RS, DE will be the directrix.
Cor. 11. In a right cone, if F be the focus, and
AG = AF, and the circle GI be drawn parallel to
the baſe of the cone; then the distance of any point P
from the focus as FP, will be equal to the distance of
P from the periphery of the circle GI.
=
Let PQ be an ordinate to AQ; draw QH, QS
parallel to AG, DG. Then will (Geom. V. 12.
Cor.) QHSG diftance of P from the peri-
phery of the circle GI; becauſe the plane SQP is
parallel to the circle GI. By Cor. 10, DE is the
directrix; therefore if DR be the focal tangent,
and if the ordinate QP be continued to the tangent
at R; then (Prop. XXV.) QR = FP, and AĽ =
AF AG. And by fimilar triangles, AL: QR::
=
(DA : DQ : :) AG : QH; that is, AG: FP ::
AG: SG; therefore FPSG.
SCHOLIU M.
Hence we have the reaſon why the Hyperbola is
called one of the Conic Sections. For if two op-
pofite cones be cut by any plane, it will always
make the figure of an hyperbola in both the cones,
as is demonſtrated in this propofition. The pro-
perties of which have been demonftrated at large
in this Book.
CONIC
M
P
P
Fig. 107
B
H
109.
H
B
I
E
>
112.
E
S
M
K
H
S
E
K
B
108.
C
N
F
E
A
P
S
110.
M
B
H
F
C
P
111.
D
E
H
R
n
2
PLXXV.pa 164.
pa.164.
UN
OF
[165]
CONIC SECTIONS.
BOOK III.
Of the PARABOLA.
I
DEFINITION S.
DEFIN.
I.
1.
F one end of a thread equal in length to CH, Fig.
be fixed at the point F, and the other end fixed
at H, the end of the fquare DCH. And if the
fide CD of the ſquare be moved along the right
line BD, and always coincides with it. Then if
the ſtring FGH be always kept tight, and cloſe to
the fide GH of the fquare. The point or pin G
(where it leaves the fquare) will defcribe a curve
MRALGK called a Parabola.
DE F. II.
The fixed point F, is called the Focus.
DE F. III.
The right line BD, is called the Directrix.
DE F. IV.
If the line BN be drawn thro' the focus F,
perpendicular to BD; then AN is called the Axis
of the Parabola; and A the Vertex.
P 2
DEF.
166
THE PARABOL A.
Fig.
I.
2.
3.
DE F. V.
A line drawn thro' the focus F, perpendicular to
the axis, as LR is called the Parameter or Latus
Rectum.
DE F. VI.
Any line drawn within the curve, parallel to
the axis, as GH is called a Diameter.
And the
point G where it cuts the curve, is the Vertex.
DE F. VII.`
A right line drawn from any diameter to the
curve, and parallel to the tangent at the vertex,
as PM, is called an Ordinate. If it go quite thro'
the curve, it is called a double Ordinate.
DE F. VIII.
The part of the diameter between the vertex
and ordinate, as GP is called the Abfciffa.
DE F. IX.
A right line meeting the curve in one point G,
but does not cut it, is called a Tangent in that
point.
PROP. I.
If BD be the directrix, G any point in the curve,
the line GD drawn to the directrix, parallel to the
axis, is equal to the line GF drawn from the fame.
point G to the focus; GD GF.
For HG + GF = length of the ftring = HD,
take away GH from both, and then GD GF.
Cor. 1. The distances of the focus, and of the di-
rectrix from the vertex, are equal. AB AF.
For when D is at B, G will be at A; confe-
quently AB AF.
Cor.
B. III.
167
THE
PARABOLA.
Cor. 2. If GP be an ordinate to the axis; then Fig.
AP + AF FG.
3.
For AP + AF
BPGD.
Cor. 3. FGFP —
FP half the latus rectum.
PROP. II.
I
The distance of the focus from the vertex is the 1.
latus rectum : AF = 4 LR = ½ LF.
I
I
For when the pin G comes to L, then LF =
FB (Prop. I. Cor. 1.) 2FA, and AFFL.
For the fame reafon FA FR, therefore FA =
=
LR.
1
== 2
SCHOLI U M.
As the latus rectum to the axis is four times the
diſtance of the vertex A from the focus F. So
in any other diameter GH, four times the diſtance
of its vertex from the focus, or AFG is called its
Latus Rectum.
PROP
III.
The Square of any ordinate to the axis, is equal to
the rectangle of the latus rectum and abfciffa PM"
= LR X AP.
For (I. Cor. 2.) MF = AF + AP = (Prop. II.)
AP+LR, and FP AP AF AP
=
I
4
LR. And in the right angled triangle MFP,
MP² MF2-FP MF +FP X MF-FP
= 2AP X LR AP XLR.
=
2
=
Cor. 1. If F is the focus, MP AP X 4AF.
Cor. 2. The abfciffas are as the fquares of their
ordinates.
AP: AQ :: PM² : QN².
P 3
For
4.
168
THE PARABOL A.
Fig.
4.
5.
5.
For AP AQ:: AP x LR: AQ x LR::
PM²: QN².
2
Cor. 3. The latus rectum is a third proportional to
the abfciffa and ordinate. AP: PM: LR
PRO P. IV.
As the latus rectum to the fum of any two ordi-
nates : : fo their difference, to the difference of the
abfciffa. Lat. rect.: CD :: ND : PQ.
Let L latus rectum, then (Prop. III.) LX
AP PM²; and L x AQ = NQ².
And by
fubtraction, L x AQ-L x APNQ-PM2;
therefore L : NQ + PM : : NQ — PM : AQ-
AP; that is, L: DC :: ND : PQ.
Cor. 1. If MD be the axis, NC an ordinate to it.
Then the rectangle NDC MD x parameter.
Cor. 2. The rectangle NDC is every where as MD.
PROP.. V.
If a line AN be drawn from the vertex to cut the
curve in M, and the ordinate QO in N; and the
ordinate PM be drawn. Then the rectangle PM ×
QN = AQ × parameter.
Let L latus rectum, then (Prop. III) QO² =
LXAQ; and (III. Cor. 2.) PM: QO or L X
AQ:: AP AQ:: (fimilar triangles) PM: QN;
and PM² × QN = PM X L X AQ, or PM x
×
QN = LX AQ.
Cor. 1. PM X QN QO².
Cor. 2. Any line befides the axis, paſſing thro' the
vertex A, will cut the curve again in fome point M.
For
I
B. III.
169
THE
PARABO L A.
For QN QA
latus rectum : PM the ordi- Fig.
nate where it cuts the curve.
PROP. VI.
6.
If from any point C in the axis, the line CF be 6.
drawn, cutting the curve in D and F, and the or-
dinates DB, FQ be drawn. Then CA² = ABX AQ.
For (III. Cor. 2.) AB: AQ : : BD² : QF² : :
(fimilar triangles) CA + AB2: CA+AQ; whence
AB × CA² + 2CA × AQ + AQ² = AQ ×
CA² + 2CA × AB + AB³, and AB× CA² + AQ²
= AQ × CA² + AB²; whence AQ — AB× CĄ
= AQ² × AB — AB² × AQ = AQ — AB× AQ
× AB. Therefore CA AQ× AB.
2
PRO P. VII.
If AB be perpen. to the axis AE, EG an ordi- 7.
nate, and AG drawn. Then any line BF being drawn
parallel to the axis AE, cutting the curve in M, and
AG in E. Then BM, BD, BF are continually pro-
portional.
For (III. Cor. 2.) AP: AE :: PM² or CD² :
EG (fimilar triangles) AC AE; therefore
AP: AC AC: AE; that is, BM: BD:: BD:
. BF.
PRO P. VIII.
If from any point M in the curve, MF be drawn 8.
to the focus, and MD be drawn parallel to the axis
cutting the directrix BD in D; and DF be drawn.
Then the line MG drawn thro' the middle of DF
will touch the curve in M.
From any other point m in the line MG, draw
mF to the focus, and md perp. to BD. Then fince
P 4
MF =
ני
170
PARABOLA.
THE
Fig. MF MD, and mFmD, the triangles FMD and
8. FmD are ifoceles, and FG GD; but the angle
Ddm being right Dm or mF is greater than dm;
therefore the point m, by the generation of the pa-
rabola, is without the curve.
9.
1
1
Cor. 1. If the tangent MD be continued to cut
the axis at T; the distance of the focus and interſec-
tion, is equal to the distance of the focus and point
of contact. FT
FM.
For MD being parallel to FT, the < GTF =
GMD GMF; therefore FT
FM.
Cor. 2. The tangent MG biffels the angle DMF
made by the line perp. to the direttrix, and that drawn
to the focus.
í
Cor. 3. A line drawn from any point M to the
focus, and another parallel to the axis, make equal
angles with the curve; FMm OMr.
PROP. IX.
If the tangent at M, cut the axis AB at T;
then the distance of the interfection from the vertex,
AT AP, the abfciffa to the ordinate PM.
From T draw any line TF cutting the curve in
D and F, and draw the ordinates DB, FQ; then
(Prop. VI.) TA' AB X AQ.
TA²
Now let the
points D, F, approach to M, and coincide with it;
then B, Q coincide with P, and TDF becomes the
tangent TM; in which cafe TA2 APX AP =
AP2, and TA = AP.
TA²
Cor. 1. The fubtangent PT is double the abſciſſa AP.
Cor. 2. The tangent at the vertex A, is perp. to
the axis AB.
For,
B. III:
171
THE
PARABOL A.
For, from the generation, the ordinates are perp. Fig.
to the axis, or parallel to the directrix.
Cor. 3. All parabolas having the fame abfciffa,
will have their tangents meeting in the fame point of
the axis.
For if AP be given, AT equal to it, will be
given for them all.
PROP. X.
9.
If two parabolas be defcribed on the fame axis; 10.
the ordinates will be proportional, or in a given ratio
to one another.
For let the common abfciffa be AB, the ordi-
nates BC, BD. 1, L the parameters. Then (Prop.
III.) 7× AB = BC², and L × AB = BD²; there-
fore BC BD :: √l× AB: √L × AB :: √/l:
VL, which is a given ratio.
:
Cor. The correfpondent ordinates of two parabolas
defcribed on the fame axis, are in the fubduplicate
ratio of the parameters.
PROP. XI.
If F be the focus, MO perp. to the tangent MT; 11.
then the fubnormal (or distance of the ordinate from
the point of interfection) PO= the latus rectum.
1
2
Let L latus rectum, then (IX. Cor. 1.) PT =
2AP; and by firnilar triangles TP or 2AP: PM ::
PM: PO, and 2 APX PO PM (Prop. III.)
LX AP; therefore 2PO L or PO = L.
Cor. 1. If F be the focus,
gent MT: Then FT
MO perp. to the tan-
FM- FO.
For (Prop. II.) AFL, and (I. Cor. 2.) FM
AP+ AF AP+L, and FP AP-
= =
=
AF=
172
PARABOL A.
THE
Fig. AF
AP —
&
+
L. And FO AP —÷L÷
11. ÷ L = AP + L = FM = (VIII. Cor. 1.) FT.
Cor. 2. A circle defcribed with the radius FM,
and center F, will pass thro' T, M, O.
Cor. 3. The angle MFO is double the angle MTO.
Cor. 4. If OG be perp. to MF; then MG =
half the latus rectum.
For fince FMFO, the < FMO =<FOM;
therefore the right angled triangles MOP, MOG
are fimilar and equal; and MG = OP = ½ L.
I
Cor. 5. Hence MG OP 2AF = para-
= ½
meter.
!
PROP. XII.
12. If MT be a tangent at any point M, AG a tangent
at the vertex; then if FG be perp. to MT from the
focus, it will pass thro' the interfellion G.
For if BD be the directrix, and MD parallel to
the axis; then (Prop. VIII.) fince MF = MD,
MG drawn to the middle of FD is perp. to it, and
FG GD; and (I. Cor. 1.) fince FA = AB;
therefore AG is parallel to DB, or perp. to AF,
and is therefore a tangent at F. Therefore the
tangent AG, the tangent MT, and the perp. FG
all interfect in one point G.
I
Cor. 1. If MP be an ordinate, AG a tangent at
A, MT a tangent at M; then AG' AP X la-
tus rectum AP X AF.
I
4
For (IX. Cor. 1.) fince TA = ; TP, GA =
MP; therefore GAMP (Prop. III.)
AP ×
I
X
parameter.
Cor. 2. If MT, AG be tangents; then AG =
half the ordinate MP.
Cor.
B. III.
173
THE PARABOL A.
Cor. 3. If FG be perp. on the tangent MT from Fig.
the focus. Then FA, FG, FM, are in continual 12.
proportion.
For in the right angled triangle FGT, FA: FG::
FG FT or FM.
PROP.
XIII.
If F be the focus, MO perp. to the curve at M. 11.
Then MO² = FO × latus rectum.
1
2
For (Prop. XI.) PO parameter, and (XI.
Cor. 1.) MF = FO. But (Geom. II. 23.) MO²
MF+FO 2FO x FP 2FO2 - 2FO
XFP2FOX FO-FP2FO X PO FO
=
=
X parameter.
PROP.
XIV.
If FH be an ordinate at the focus; THG a tan-
gent at H, or the focal tangent; PM an ordinate,
continued to the tangent at G. Then TP FM =
PG.
I
4
=
Draw the tangent at the vertex AL. Then (IX.)
TA AF(Prop. I. Cor. 1.) parameter, and
(def. V.) FH = parameter; therefore TF
FH, and confequently by fimilar triangles TP =
PG; but TPTA + AP AF + AP =
(I. Cor. 2.) FM; and therefore PGFM.
Cor. If TH be the focal
=
I
tangent, AL a tangent
at the vertex; then TA AL AF = para-
= = —
meter; and TF = FH =
I.
z
parameter.
PROP.
13.
174
THE PARABOL A;
Fig.
14.
PROP. XV.
If AR be the axis, ML any diameter; AE, MT,
two tangents at A and M.
MYE.
Then the tritangle AYT
then (IX. Cor. 1.) TA
I
I
= PM or EA
2
2
Draw the ordinate MP,
=TP; therefore AY
EY; therefore the triangles AYT, EYM are fi-
milar and equal.
Cor. 1. If AD be parallel to MT; the triangle
MPT ADE = rectangle MPAE DPAM =
DATM.
This is evident by infpecting the figure.
Cor. 2. If QIF be perp. to AB, and QRL parallel
to MT; then the triangle QIR = rectangle IAEF.
For the triangles MPT, QIR are fimilar; whence
MPT : QIR : : MP² : QI² :: (III. Cor. 2.) PA:
IA: rectangle PAEM rectangle IAEF; but
MPT MPAE; therefore QIR IAEF.
Cor. 3. The triangle QLF
The triangle QLF = parallelogram
LMTR.
For QIR
IAEF, add to both LRIF, then
LQFLRAE LRTM.
Cor. 4. After the fame manner triangle iqR =
rectangle iAEf; and qfL. TRLM.
:
For the triangles MPT and qiR are ſimilar;
whence MPT qiR:: MP²: qi¹ :: AP: Ai::
PAEM: ¡AEƒ; but MPT = PAEM; therefore
qiR = iAEƒ.
Then fince TAY MEY, add to both AifMY,
and then TiƒM = AifE = iqR. Take ifLR from
both; then TRLM = qfL.
Cor. 5. Any ordinate Qq is biffeted by the diameter
ML.
For
B. III.
175
THE PARABO L A.
For (Cor. 3. and 4.) triangle LQF = LMTR Fig.
Lqf; therefore thefe triangles LQF, Laf, are 14.
fimilar and equal, whence LQLq.
PROP. XVI.
In any diameter MK, the abfciffas are as the fquares 15.
of the ordinates; ML: MK:: LQ': KG".
For draw the tangent MT to cut the axis AS in
T. Then (XV. Cor. 3.) triangle QFL = = paral-
lelogram LMTR, and GBK = KSTM. But
QL²: GK² :: triangle QLF : GKB :: Par. LMTR:
KMTS :: LM : KM.
Cor. Any diameter interfects the curve only in one
point, viz. the vertex.
For the abfciffas and ordinates continually increaf-
ing, the curve recedes more and more from the
diameter.
PROP. XVII.
In any diameter ML, the fquare of any ordinate, 16.
is equal to the rectangle of the parameter and ab-
Scilla; LQ² = 4FM × ML.
From M draw the tangent MT cutting the axis
in T, and draw AK parallel to it, from the princi-
pal vertex, and draw the ordinate MP to the axis.
Then (III. Cor. 1.) 4AF × AP PM², and MK
= AT = (IX.) APPT. In the right angled
triangle TPM, MT or AK' = MP² + PT =
AK²
4AF x AP + 4AP' = 4AP X AF + AP =
×
×
(II. Schol.) 4AP × FM = 4MK x FM; that is,
AK² = MK × 4FM.
But (XVI.) AK² : QL² : : MK: ML::4FMX
MK : 4FM × ML, but AK² = 4FM × MK;
therefore QL² = 4FM × ML.
Cor.
176
PARABOL A.
THE
Fig. · Cor. 1. The parameter is a third proportional to
16. the abfciffa and ordinate.
في
17.
Cor. 2. If MP be perp. to the axis AP, then the
latus rectum of the diameter ML, exceeds the latus
rectum of the axis, by 4AP.
For 4FM (I. Cor. 2.) 4AF + 4AP, and 4AF
is the latus rectum of the axis.
Cor. 3. If MC is perp. to the tangent MT, then
TC is half the latus rectum of the diameter ML.
For TP : TM : TC÷, or 2MK : AK
rameter.
z pa-
Cor. 4. The line GD, perp. to the directrix BD, is
4 of the latus rectum of the diameter GH.
By Prop. I.
SCHOLIU M.
Since the fquare of the ordinate is always equal
to the rectangle of the abfciffa and latus rectum;
for that reaſon the curve is called a Parabola.
PRO P. XVIII.
In any diameter DP, the whole ordinate paffing
thro' the focus F, is the latus rectum to that diameter,
MN latus rectum of DP; and its abfciffa DP is
I
4
the latus rectum.
4
÷
L.
Draw the tangent DT, then DP = TF = (VIII.
Cor. 1.) DF (II. Schol.)
=
But (XVII.) NP or PM²
DP 4DP²; therefore PM
NML.
=
lat. rectum
LX DP = 4DP x
2DP = L, and
=
2
Cor. 1. If an ordinate of any diameter paſs thro'
the focus, as MN. The abfciffa is equal to the dif-
tance of the focus from the vertex: DP DF.
Cor
D
Fig.1.
K
H
4.
L
A
•
A
P
7.B
JM
B
A
R
M
R
N
D
m
5.
2.
M
3.
H
B
B
JB
6.
B
G
9
A
F
B
M
E
F
M
F
7
M
12.
B
H
A
1.3.
P
M
B
14.
I
10.
M
Pl.XXV.pa176.
UNIL
OF
MICH
B. III.
177
THE
PARABO L A.
I
2
I
X z
17.
Cor. 2. If any line MN be drawn thro' the focus Fig.
F, the rectangle NFM = NM × parameter..
For NP² DP x NM, becauſe NM is the la-
tus rectum; and DT or PF² = DO² + TO² =
4AF × AO + 4AO² = AO × 4AF + 4AO –
(becauſe AO TA, and 4AF + 4AO ≈ lat. rec-
tum of DP) TAX MN. But rectangle NFM —
NP2 PF2DP X NM-TAX NM
NM.
PROP. XIX.
24
AFX
If ML be any diameter, HP an ordinate to it; 18.
and HL perp. to it. Then HL = MP × latus
rectum of the axis.
Let BO be the directrix, continue PM to O,
draw the tangents MT, AD, alfo draw FDO from
the focus F. Then (VIII. Cor. 1.) TF = FM,
and FD perp. to MT; alfo (XII.) AD perp, to FT.
Then (XVII.) HP² = 4MF × MP. The trian-
gles HPL, DTA, DTF are ſimilar; whence HP² :
HL² :: DT² : DA² : : DF² : FA² : : TF or MF:
FA: 4MF x MP: 4AF x MP; but HP² =
4MF × MP; therefore HL² = 4AF × MP.
~
Cor. 1. The Squares of the perpendiculars HL, are
as the abfciffas MP; or HL² : bľ : : MP : Mp.
Cor. 2. If from the vertices M, H, of two dia- 19.
meters MP, HG; two ordinates be drawn MN and
HP; then the abfciffas MP, HN, will be equal.
For if L parameter; then MG LX HN,
and HL2 LX MP; but MG HL; therefore
LX HN = L x MP, and MP = HN.
=
PROP.
178
THE PARABOL A.
Fig.
20.
14.
21.
1
PROP. XX.
If from any point M a tangent be drawn to cut
any diameter AP in T; and the ordinate MP be
drawn; then the diftance of the interfection from the
vertex, AT; is equal to the abfciffa AP.
Draw the diameter MN, and draw MG, AL,
perp. to AP, MN; and draw AN parallel to MT,
and AN will be an ordinate to MN. Then (XIX.
Cor. 2.) AP = MN
= MN (by parallel lines) AT.
Cor. 1. The fubtangent PT is double the abfciffa AP.
Cor. 2. The tangents from each end of the double
ordinate Mm, meet in the fame point of the diameter T.
For the abfciffa of the diameter drawn thro' m,
will be equal to AP = AT.
Cor. 3. Hence if AR, ML are any diameters be-
fides the axis; and AE, MT, tangents; the trian-
gle AYTMYE.
For (this Prop.) ME = DM = AT; and the
triangles AYT, MYE are fimilar, and therefore
equal.
PROP. XXI.
If BD be the directrix, GL any diameter, ML an
ordinate, MT a tangent, MK perp. to GL. Then
the Square of the tangent MT", is to the Square of
the ordinate, ML as the distance of the point of
contact from the directrix, MD; to the latus rectum
of GL.
2
:
1
For (Geom. II. 22.) MT² = ML² + TL² f
2TLK. But (XVII.) ML² = 4FG × GL, and
(XX.) TL² = 4GL", and 2TLK = 4GLX LK;
whence MT24FG X GL + 4GL² + 4GL
× LK
B. III.
179
THE PARABOLA.
× LK = 4GL X FG+ GL + LK
×
4GL Fig.
x FK. Therefore MT: ML² :: 4GL × FK : 21.
4FG X GL:: FK: FG.
Cor. 1. If the diameter GL becomes the axis AC; 16.
then the fquare of the tangent MT², to the ſquare of
the ordinate MP2: AF + AP, to AF; that is, as
MP²
latus rectum of MK
to latus rectum of the axis.
For if F be the focus, AF (fig. 16.) = AB
(fig. 21.)
Cor. 2. The Square of the tangent MT2, is to the 21.
Square of the perp. MK:: as the diſtance of the point
of contact and directrix MD, to parameter of the
axis AB.
2
2
I
For (XIX.) MK² = 4AB × GL. And MT²:
ML²:: FK: FG, and ML2: MK:: 4FG × GL:
4AB × GL : : FG: AB. And ex equo, MT² :
MK2 FK: AB.
Cor. 3. ML2: MK:: FG: AB.
2
For ML²: MK² : : 4FG × GL: 4AB × GL::
FG: AB.
Cor. 4. LK² = 4HG × GL.
24
=
For LK ML-MK4FGX GL-4AB
× GL = 4HG × GL.
X
Cor. 5. MT rectangle 4FK × GL.
If AG be
PROP. XXII.
any diameter, AB a tangent, AF perp. 22.
to the axis. BD parallel to AG.
Then AF
1
For let L
BD × latus re&um of the axis.
1
latus rectum of AG, latus
rectum of the axis. Draw the ordinate GD.
Then (XVII.) L× AG = GD²; that is, LX BD
=AB. But (XXI. Cor. 1.) AB²: AF² :: L:/::
Q
LX
180
THE PARABOL A.
Fig. LXBD: 1x BD. But LX BD AB'; therefore
22. /X BD
lx
23.
AF²,
Cor. 1. The line BD is every where as the fquare
of the baſe AF.
Cor. 2. AF: BD :: FH: FD.
For 1x BD AF², and (IV. Cor. 1.) AF ×
FH/X DF; whence AF: BD::/: AF::FH:
DF.
PROP. XXIII.
If FG, HQ be ordinates to the axis AQ; BFT
a tangent at F; FL parallel to AQ; and QH pro-
duced to the tangent at B. Then BH: HL::HL:
FM.
Put p for the parameter, then fince TG 2AG,
by fimilar triangles, 2AG: GF::FL: LB::px
LF: px LB; then p x LF x GF = p × 2AĞ ×
LB = (III.) 2FG² × LB, and 2FG × LB = p ×
LF (IV. Cor. 1.) HL x LC, which reduced
to a proportion, BL: LH :: LC : 2FG or FM;
and by divifion, BL- LH: LH:: LC-FM:
FM; that is, BH: LH:: KC or LH: FM.
Cor. 1. FM XLB = px FL = HL`× LC.
Cor. 2. If BF be a tangent at F, FL a diameter,
BLC perp. to FL; then BL² = rectangle HBC.
For fince BH: HL::HL: FM. By compound-
ing BH: (BH + HL) BL :: HL: (HL + FM)
HK. And by addition, BH: BL::(BH + HL)
BL: (BL+HK) BC.
PROP.
B. III. THE PARABO L A.
181
PROP. XXIV.
Fig.
If AD, PQ, be two diameters; AC a tangent at 24.
A interfecting QP in C; and if any point F be taken
in AD, and CL be made equal to AF; and the lines
PF, AL, drawn; they will interfect at S, a point
in the curve.
Draw BS parallel to AD, and the ordinates GS,
DP. Then fince AF CL, PL + AF = PC
AD. By fimilar triangles, AF : PL:: AS:
SL: AB BC; and compounding AF: AF +
PL or AD: AB: AB + BC or AC. Again by
fimilar triangles BS: CL or AF:: AB: AČ. And
multiplying, BS: AD:: AB: AC; that is, AG:
AD :: GS²: DP. Therefore (XVI.) as P is in
the curve, S is in the curve alſo.
Cor. Lines drawn from A and P thro' any point
S in the curve, will cut off AF in the diameter AD,
equal to CL the part from the tangent AC.
PROP.
XXV.
If any diameter interfect two parallel lines; the 25.
rectangle of the fegments of the lines, will be as the
Segments of the diameter. DBE : FCG :: AB: AC.
=
Draw the diameter PL of the parallels DE, FG,
which will be parallel to AC; and draw the ordi-
nate AR. Then the rectangle DBE HE' -
HB² - HE² — RA² = (XVII.) LX PH-LX
PR, L beingthe latus rectum of PL. Likewife rectan-
gle FCG LG LC LX PL-LX PR.
2
=
Therefore DBE: FCG:: LX PH - PR: L x
PL-PR:: RH: RL :: AB: AC.
Q 2
Cor.
1
182
PARABOLA.
THE
Fig. Cor. 1. If DE be an ordinate to any diameter PL,
25. and E be cut by any other diameter AC. Then the
reliangle DBE = AB X parameter of PL.
26.
27.
t
Cor. 2. If XQ be a tangent at P; then rectangle
FCG : FQ² :: AC : AQ.
For then D and E coincide with P, and HE va-
niſhes.
Cor. 3. If any line FG be cut by two diameter's
AC, NO; the rectangles of the parts, are as the
fegments of thefe diameters. FCG: FOG:: AC: NO.
For (Cor. 1.) FCG ACXL; and FOG =
NO X L; therefore FCG. FOG:: ACXL: NO
XL:: AC: NO.
Cor. 4. If two parallel lines, DE, FG, be cut by
two diameters AB, NO; the rectangles of the parts
will be as the fegments of the respective diameters ;
DBE: FOG :: AB: NO.
For DBE
FCG :: AB AC; and FCG:
FOG:: AC: NO. Therefore ex equo, DBE :
FOG: AB: NO.
Cor. 5. And if PQ be a tangent at P; then the
rectangle FOG: PQ²:: NO: AQ.
PRÒ P. Xxvi.
If two parallel lines DE, FG, interfect any other
two parallels, HL, MN, in B and O. The rectangle
DBE : HBL: FOG: MON.
Thro' the interfections B, O, draw the diame-
ters AB, VO. Then (XXV. Cor. 4.) DBE :
FOG:: AB: VO:: HBL: MON.
Cor. 1. If any line FG cut two parallels HL,
MN, in O and I; then rectangle HIL: MON::
FIG: FOG.
This
H
Fig.15.
F
T
M
16.
T
M
R
B
G
C
K
A
L
18 M
F
P
1
19.
L
H
P
IC.
1
F
B
D
B
H
22.
G
A
D
21.
H
F
K C
N
L
17.
D
P
20.
J
M
Ꮐ
P
23.
F
G
M
B
I
? K
¿
T
M
C
P
24.
B
P
26.
25
R
E
G
H
B
D
E
G
G
L
F
D
M
Pl. XXVI: pa.182.
UNIL
OF
MIC
CH
B. III.
183
THE PARABOLA.
This is plain by fuppofing DE to fall upon FG. Fig.
Cor. 2. If BD, BL be tangents at D and L, and 28.
KF, RN, parallel to them. Then GOF: ION ::
DR² BD²
DR2: NRÍ:: BD2: BL:: GKF: KL:: parame-
ter of the diameter paſſing thro' D: parameter of the
diameter thro' L.
For if DE, HL (fig. 27.) become tangents;
then D and E coincide, as alfo H and L; and
DBE, HBL become BD² and BL²; and the like
for the reſt.
Alfo BD, and BL are equal to the ordinates
drawn to the diameters at D and L, from the point
where the diameter from B cuts the curve. And
if d and be the parameters, H the common ab-
fciffa, then BD² : BL²: d× H:/× H: :d : l.
PROP. XXVII.
If DM, DN be two tangents, MN the line join- 29.
ing the points of contact; and if any line GL be
drawn parallel to either tangent DN, cutting the
curve in R and L, and MN in H. Then GR, GH,
GL are continually proportional.
For (XXVI. Cor. 3.) RGL GM:: DN:
DM² :: (fimilar triangles) GH: GM². There-
fore RGL GH².
PROP.
XXVIII.
If HT, KT, be two tangents, HK the line drawn 30.
thro' the points of contact. Then the line drawn
from the interfection T, thro' O the middle of HK,
will biflect all the lines within the curve, parallel to
HK viz. if HO = OK, then IP = PN.
By Cor. 2. Prop. XX. The tangents HT, KT
meet in the diameter belonging to the ordinate
Q3
HK.
184
THE PARABO L A.
Fig. HK. Therefore fince HOOK, TO is the dia-
30. meter.
And fince IN is parallel to HK, IP and
PN are ordinates, and IP = PN, by Cor. 5.
Prop. XV.
31.
Cor. 1. If LM be drawn parallel to HK joining
the points of contact; the parts intercepted between
the curve and tangent on each fide are equal, LI =
NM.
For fince HO = OK, by fimilar triangles LP
= PM; take away the equals IP = PN; then the
remainder LI MN.
Cor. 2. The tangent at A the vertex, is biſſected
in that point.
1
Cor. 3. HO, OK, and alſo IP, PN, are ordi-
nates to the diameter paffing thro' the interfection T
of the tangents.
PROP.
XXIX.
If TH, TK be two tangents, and HK the line
drawn thro' the points of contact. And from the in-
terfection T, any line TI be drawn cutting HK in D,
and the curve in F and I. Then TF: TI :: DF:
DI.
Thro' F and I draw AC, LM parallel to HK;
then (XXVIII. Cor. 1.) LI= NM, and AF =
BC, whence MI LN, and CF AB.
By fimilar triangles AF: LI:: TF: TI,
and CF MI:: TF: TI.
and multiplying, AFX CF:LIXMI :: TF: TI.
that is, AFX AB: LIX LN:: TF2: TI.
But (XXVI. Cor. 2.) AF × AB : LI × LN : :
HA²: HL2: : (parallel lines) DF: DI. There-
fore TF2 TI:: DF2: DI'. And TF: TI:: DF:
DI.
Cor.
B. III.
185
THE PARABOL A.
Cor. If it be TF : TI :: DF : DI. The line HK Fig.
which joins the points of contact, will pass thro' the 31.
point D.
PROP.
XXX.
If AF, AG, be two tangents; FG a line drawn 32.
thro' the points of contact; and thro' the interfection
of the tangents, AV be drawn parallel to GF. Then
from any point V, if there be drawn VL thro' O the
middle of FG, to cut the curve in P and L. Then
will VP : VL:: OP: OL.
Draw AQRL, and thro' Q draw PSQ parallel
to FG. Then (XXIX.) AQ: AL:: RQ: RL,
whence AL X RQ = RLX AQ.
:
By fimilar triangles, PQ OR :: LQ: LR;
and OR SQ:: AR: AQ. Whence by com-
pounding PQ: SQ :: LQ × AR : LR × AQ.
And by divifion PS: SQ:: LQ × AR — LR ×
AQ: LR X AQ.
But LQX AR-LR X AQ= LR + QR X
AR- LRX AQ = QR × AR + LR × QR =
AL × QR = (before) RL × AQ; therefore PS
=SQ. And confequently as Q is in the curve,
P is in the curve. And by reaſon of the parallels,
OR, PQ, VA; VP: VL :: AQ : AL : : (XXIX.)
QR: RL:: OP: OL.
PROP. XXXI.
Let AO be a diameter; FA, GA, two tangents;
FG an ordinate. Let two other tangents VH, VI,
be drawn from any point V, in the line VA being
parallel to FG. I fay HI drawn thro' the points of
contact, will pass thro' the point O, the middle of FG.
For thro' O and V draw VPOL cutting the
curve in P and L. Then fince VA is parallel to
Q4
the
33:
186
THE PARABOLA..
Fig. the ordinate FG; therefore (XXX.) it is, VP:
33. VL
VL :: OP : OL. And fince the tangents VH,
VI meet in V; therefore (XXIX.) VP : VL ::
OP: OL. Therefore the point O is common to
both FG and HI; and fo HI paffes thro' O.
34.
Cor. 1. If A, V, be the concourfe of the tangents
FA, GA, and HV, IV; then the line, (joining the
points of contact of any two tangents meeting in AV)
will pass thro' O.
Cor. 2. If FO, GO, be ordinates to the diameter
AO; and any line HI be drawn thro' O; then the
tangents HV, IV, will meet fomewhere in the line
AV drawn parallel to FG.
PROP. XXXII.
If VH, VI, be two tangents at H and I; and
IHD be drawn thro' the points of contact. And if
any line VPCL be drawn thro' V; and the tangents
PD, LD be drawn; they will meet fomewhere in the
line ID.
Draw FG thro' O, where VI and IH interfect;
fo that GO may be OF; and draw VAD paral-
lel to FG. Then (XXXI. Cor. 2.) the tangents
PD, LD, will meet fomewhere in the line VA, as
at D. Then if DO does not paſs thro' the points
of contact H, I; let a line drawn from D thro'
N, paſs thro' the points of contact. Then (XXIX.)
we ſhall have VP: VL:: NP: NL. But HV,
IV, being tangents, we have VP VL OP:
OL. Therefore the former proportion cannot be
true, except N and O coincide. Therefore DO
paffes thro' the points of contact H, and I. That
is, the tangents PD, LD, meet in the line IH..
Cor. If HV, IV be two tangents, and IHD be
drawn thro' the points of contact, I, H. And from
any
B. III.
187.
THE
PARABOL A
any point D, the tangents DP, DL are drawn; then Fig.
LP drawn thro' the points of contact L, P, will paſs 34.
thro' V.
For a line drawn thro' V will pafs thro' the points
of contact L, P, and therefore no other can.
PROP.
XXXIII.
M and N; and 35.
The angle MFN
If MD, ND be two tangents at
MF, NF be drawn to the focus F.
=twice the angle MDN.
Draw the axis TAFB.
Produce MD to cut the
axis in T; and let ND cut it in G. Then (XI.
Cor. 3.) the < MFB = 2 < MTB, and <NFB
= 2 <NGB; therefore MFB + BFN, that is,
MFN2MTG + 2NGA = 2DTG + 2DGT
=2MDG (Geom. II. 1.)
Cor. 1. If MD, ND, be two tangents, and the 36.
line MN joining the points of contact, paſs thro' the
focus F. Then MDN is a right angle.
For when MFN is a right line, the < MFB +
BFN is equal to two right angles; and therefore
MDN is one.
Cor. 2. If MD, ND be two tangents, and the
line MN joining the points of contact, paffes thro' the
focus F. Then MD, ND, meet fome where in the
directrix GE.
For the focal tangent being drawn will cut the
axis in G, ſo that AF AG (by Prop. IX.), which
is the point through which the directrix is drawn
(I. Cor. 1.), and (XXXI. Cor. 2.) the tangents
MD, ND will meet fome where in GE.
Cor. 3. If MD, ND be two tangents, and MN
paffes thro' the focus F. Then the line DF, drawn
from the interfection of the tangents to the focus, is
perp. to MN.
For
188
THE PARABOL A.
Fig.
For if GE be the directrix; and ME parallel to
36. the axis. Then (Prop. I.) ME MF, and (VIII.
Cor. 2.) <DME DMF; therefore the triangles
MDE, MDF are fimilar and equal, for MD is
common to both, whence ED = DF, and <MFD
MED a right angle.
37.
38.
PROP.
XXXIV.
If AB be any diameter, and any line AP be drawn
from A, thro' any point P of the curve; and thro
P, the line NP be drawn parallel to AB. Then if
the ordinate DF be drawn to interfect these lines in
E and G; DE, DF, DG will be continually propor-
tional.
Draw the ordinate PO; then by fimilar triangles,
OP: DE :: AO: AD: : (XVI.) OP: DF2, and
OP X DF2 = DEX OP², or DF2 DE XOP or
DG X DE.
Cor. As AD: DE:: OP: latus re&um of AB.
For DEX OPDF AD X parameter.
PRO P. XXXV.
If any line MN cut the directrix TO in O, and
from the points M, N, where it cuts the curve, MF,
NF be drawn to the focus F; and MF be produced
to H. Then OF drawn to the focus, will biffect the
angle NFH.
Draw MP, NQ perp. to TO; and ND parallel
to MF. Then by fimilar triangles, MP: NQ::
MO: NO:: MF: ND. But (Prop. I.) MP =
MF; therefore ND NQ NF. Whence the
= =
angle NFD NDF DFH, by the parallels
ND, FH.
Cor.
B. III. THE PARABOL A.
189
Cor. Hence if the angle NFH be biffected by the Fig.
line FO drawn from the focus; it will cut MN at 38.
◇ in the directrix.
PROP. XXXVI.
If QT, PT be two tangents. And if QF, PF 39.
be drawn from the points of contact to the focus.
Then the line TF will biffelt the angle PFQ.
Let DR be the directrix; draw QPR, and produce
TF to K; and draw LR, KR, tangents at L and
K (XXXII. and Cor.); let QD, PË, be perp. to
MR; then by fimilar triangles, QD: PE :: QR :
PR : : (XXIX.) QN: PN. That is, (Prop. I.)
QF: PF :: QN: PN. Therefore (Geom. II. 25.
Cor. 1.) FN biffects the angle QFP.
PROP.
XXXVII.
If a trapezium ABDC be infcribed in a parabola, 40.
and any point E be taken in the curve, from which
two lines EN, EH, are drawn parallel to any two
adjoining fides, as AB, AC, to interfect the oppofite
fides in N, Q, and H, R. Then taking the reƐtan-
gles on the oppofite fides; it will always be ER ×
EH to EN X EQ, in a given ratio.
Draw DFG, BLP, parallel to AC; thro' L
and C, draw COLS. Thro' the middle of AC,
BL, draw the diameter MK, which will biffect the
parallels DF, TE; and alfo SG, OR, terminated
at AB, CL. Then (XXVIII. Cor. 1.) will SD =
FG, and OT = ER, which would alſo be ſo, if
SG, RH were tangents.
By fimilar triangles and parallel lines, HO: SD::
OC: SC :: AR AG. And alternately, HO:
:
AR or EQ :: SD or FG: AG. Again, BP or
ER :
190
PARABOLA.
THE
Fig. ER: PN:: DG: GB. And multiplying, HO x
40. ER: EQ × PN:: FG x DG AG X GB ::
(Prop. XXVI.) ER × RT or ER × EO: RA ×
RB or EQ × EP. And by compofition, FG ×
DG: AGX GB :: HO × ER + EO × ER : EQ
× PN + EQ × EP :: EH × ER: EQ × EN.
And it is evident, that as long as the points A,
B, D, C, are fixed; that FGD to AGB is a given
ratio, and therefore that of REH to QEN.
41.
40.
Cor. 1. The fame things fuppofed; it will be,
EH EN:: EO : EP : : HO: PN.
×
And dividing all
For EHXER: EQ × EN :: FGX GD: AG X
GB: HO × ER: EQ x PN.
the terms, EH: EN :: HO
ing, EH: EN : : EO : EP.
:
PN.
And divid-
Cor. 2. If the points, A, B, C, E, are fixed, and
D variable at pleasure; it will still be EH to EN,
and HO to PN, in the given ratio of EO to EP or.
of RT to RB.
Cor. 3. If the points A, B, coincide; AG then
becomes a tangent. And if C and D alfo coincide;
DH becomes a tangent. And then Q, N, coincide.
Cor. 4. If BC be drawn to cut EH in I, and if
it be EH: EI: :EN: EV.
EN: EV. Then if BV be drawn
it will touch the curve in B.
}
For (Cor. 2.) if D be moveable, it will be EH:
EN: EO EP, wherever D is. Therefore fup-
poſe D to move round till it coincide with B; then
BND will become a tangent at B, and N will fall
upon V; that is, BV will be a tangent.
Cor. 5. If you make EH: EN :: EO: EP, and
draw OC, BP, they will meet at L in the curve.
For fuppofing L a point in the curve, it will be
by (Cor. 2.) EH: EN :: EO: EP; therefore L
cannot be out of the curve.
Cor.
Fig.27
F
M
D
30
V
T
E
I
H
7277
P
G
K
I
28.
D
R
K
H
31
H
M
29.
32.
B
S
F
33-
F
G
G
F
34.
L
T
36.
E
G
37
N
G
P
M
B
E..
D
M
T
B
35.
N
G
H
A
38.
M
P
PL.XXVII.pa.100..
M
N
OF
.
B. III.
191
THE PARABOLA.
Cor. 6. If there be five points B, L, D, C, E; Fig.
and thro' any point E, there be drawn EH, EN, cut- 40.
ting BD, BL, DC, LC, in N, P, H, O, ſo that
EH: EN :: EO : EP. Then CA, BA being drawn
parallel to EH, EN; they will meet at A in the curve.
This is the reverfe of the foregoing Cor. For
if A was out of the curve, L would not be in it.
Cor. 7. If A and B coincide, then AB becomes a
tangent. And the like for any other two points.
PROP. XXXVIII.
If a trapezium ABDC be infcribed in a parabola, 42.
and from any point E of the curve, four lines EL,
EK, EO, EM, be drawn to the four fides of the
trapezium, in any angles whatever, equal or unequal,
and cutting them in L, K, O, M. And likewife
from any other point e, four more lines be alfo drawn
to the fides of the trapezium, refpectively parallel to
the former, cutting them in l, k, 0, m. Then taking
the rectangles of the oppofite fides; the rectangle EL
× EM: el x em:: EK × EO : ek × eo.
Thro' E, e, draw QEN, qen, parallel to AB;
and EHR, ehr, parallel to AC. Then by fimilar
triangles, ER: er :: EL: el; and EH: eb: :
EM: em.
: em. And by multiplying ER × EH :'er ×
eb :: EL X EM: el X em.
eo.
Again, EQ eq:: EK: ek, and EN: en:: EO:
And multiplying, EQX EN: eq × en :: EK
'x EO: ek X eo. But (Prop. XXXVII.) ER ×
EH: er × eb :: EQ × EN : eg × en. Therefore
EL X EM: el x em:: (EQ X EN: eq × en: :)
EK X EO: ek X eo.
Cor. 1. If ABCD be an infcribed trapezium; and
if AB, CD, produced interfect in X; and the diago-
nals AC, BD in Y. And XeYE be drawn. Then
EX:
43.
192
THE PARABOL A.
Fig. EX: eX:: EY: eY. And if E be at an infinite
43. distance, then eX = eY.
44.
For fuppofe BD, AC, to become diagonals
(fig. 42.); and that EL, EK, EO, EM, and el,
ek, eo, em, all lie in one line Ee; and that AB,
CD, interfect in X, and AC, BD, in Y; and that
Ee paffes thro' XY; which is a particular cafe of
this Prop. Then L, 1, M, m, coincide in X;
and K, k, 0, 0, in Y. Then the proportion, EL
X EM: el X em:: EK X EO: ek x eo, becomes
EX²: eX²:: EY: eY; and EX: eX:: EY: eY.
!
2
But if eE be infinite, then EX = EY, and eX = eY.
Cor. 2. If XA, XC, be two tangents, and AYC
paffes thro' the points of contact.
eX: EY: eY.
Then will EX:
For let B approach to A, and D to C; then
BD will coincide with AC. And it will still be
EX: eX:: EY: eY.
PROP. XXXIX.
In any diameter, as its latus rectum, to the fum of
any two ordinates: fo their difference, to the differ-
ence of the abfciffa. Latus rectum of AQ: ND ::
DC MD.
Let Llatus rectum; then (XVII.) L X AP
= PM², and LX AQ = QN²; and fubtracting
one from the other, L x AQ-LX APQN²
PM² = QN + PM × QN PM; that is,
L× PQ = ND × DC. And L: ND :: DC: PQ.
1
Cor. If NC be an ordinate to any diameter AQ;
MD any other diameter. Then the rectangle NDC
= MD x parameter of AQ
PROP.
B. III. THE PARABOL A.
193
PROP. XL.
Fig.
If PM be an ordinate to any diameter AP, and 45-
AM be drawn cutting any other ordinate QO in N.
Then PM × QN = AQ × parameter of AQ.
Let L latus rectum; then (XVII.) QO² =
LX AQ, and (XVI.) PM² : QO2 or L x AQ ::
AP: AQ :: (fimilar triangles) PM: QN:: PM²:
PM X QN; therefore LX AQ = PM × QN.
Cor. 1. PM X QN = QO².
Cor. 2. Every line paſſing thro' A, except the dia-
meter AP, will meet the curve again in fome other
point M.
For QN : QA :: L: PM the ordinate where it
meets the curve again.
PROP. XLI.
If AQ be any diameter, and any line CF be drawn 46.
cutting it in C, and the curve in D, and F. And
if the ordinates DB, FQ be drawn. Then CA =
AB X AQ
2
2
For (XVI.) AB : AQ : : BD² : QF¹ : : (fimilar
triangles) CB² or CA + AB" : CQ² or CA + AQ ;
whence ABX CA² + 2CA × AQ + AQ² = AQ
×CA² + 2CA× AB+ AB², and ABXCA² + AQ²
AQ × CA² + AB². Therefore AQ — AB ×
CA² = AQ² × AB- AB² x AQ = AQ — AB
× AQ X AB. Therefore CA AQ × AB.
X
2
Cor. Any line that is not a diameter, drawn within".
the curve, will interfect it in two points.
For
194
PARABOL A.
THE
Fig.
For fince DF is not a diameter, and therefore
46. not parallel to AQ, it will interfect AQ ſomewhere
as at C. Therefore by this Prop. it is AB: AC::
AC: AQ, the abfciffa belonging to the ordinate
paffing thro' the other interfection F.
47.
48.
PROP. XLII.
If AB be any diameter, EG an ordinate, AB a
tangent at the vertex; and if AG be drawn. Then
any line BF drawn parallel to the diameter AE, cut-
ting the curve in M, and the line AG in D; BM,
BD, BF will be continually proportional.
2
For drawing the ordinate MP, and DC parallel
to it. Then (XVI.) AP: AE :: PM or CD² :
EG:: (fimilar triangles) AC²: AE². Therefore AP :
AC:: AC: AE; that is, BM : BD : BF.
PROP. XLIII.
If FG, HQ, be ordinates to any diameter, AQ;
BFT a tangent at F; FL. a diameter paffing thro
F. Then if QH be produced to the tangent FB;
then BH, HL, FM are in continual proportion.
Let p be the parameter of AQ; then (XX.) TG
2AG, and by fimilar triangles, 2AG: GF::
FL: LB::px LF: :p x LB. Then px LF
× GF =p × LB × 2AG = (XVII.) 2FG² × LB,
and 2FG × LB = px LF = (XXXIX. Cor.) HL
x LC. Whence BL: LH:: LC: 2FG or FM;
and by divifion, BL-LH or BH: LH:: LC-
FM or LH: FM.
X
×
1
Cor. 1. BL x FMpx FL = HLC.
Cor. 2. If FL be any diameter; FB a tangent at
F, cutting the ordinate QH in B. Then BL2 =
rectangle HBC.
For
B. III.
195
THE PARABOL A.
For draw the diameter AQ; then BH: HL:: Fig.
HL: FM, and (by adding) BH : (BH + HL) 48.
BL::HL: (HL + FM) HK. And (adding a-
gain) BH: BL:: (BH + HL) BL: (BL + HK)
BC.
PROP. XLIV.
If AC be a tangent at A, AH any line drawn 49.
within the parabola, DF any diameter cutting the
tangent in B, and the line AH in F. Then AF:
FH: BD: DF.
2
AG, and p X
ढे
Draw the diameter GLO to the ordinate AG or
GH, and draw HC parallel to DB, put platus
rectum. Then (XVII.) px GL =
4GL 4AG AH. But AG AH; there-
=
fore GO HC
= HC (XX. Cor. 1.) 2GL, and
HC = 4GL, and p x 4GL or px HC AH.
Therefore p: AH:: AH: HC: : (fimilar trian-
gles) AF FB; and alternately AH: FB::p:
AF: (XXXIX.) FH: FD :: AH - FH FB
-FD; that is, FH: FD:: AF: DB; or AF:
FH: DB: DF.
Cor. 1. If LG be any diameter, AH an ordinate,
AC a tangent at A, HC parallel to GL; then AH'
= HC x parameter of LG.
Cor. 2. The fame things fuppofed, HC = 4GL.
Cor. 3. The fame things ftill fuppofed, AF² =
BD × parameter of LG.
For AF: FH: BD: DF. And compounding,
AF: AH:: BD: BF. And by fimilar triangles,
AF: AH :: BF: CH. And multiplying, AF:
AH² :: BD : CH. Whence CH X AF BD
× AH² = BD × HC × p; therefore AF² =
px BD.
R
=
PROP.
196
THE PARABOL A.
:
Fig.
50.
51.
PRO P. XLV.
If FT be a tangent at F, LI parallel to it.
LB, FC, ID, ordinates to any diameter AD. Then
LB + ID=2FC, when both L and I are on the
fame fide of AD.
Draw the diameter FK which will biffect LI in
G; and thro' L, draw the diameter LP. Then by
fimilar triangles, fince IG GL, IK = KP =
FN. And by parallel lines, PD = NC = LB.
Therefore adding equals to equals, IK + LB =
FN+ NC FC. And doubling, IP + 2LB or
ID + LB = 2FC.
Cor. If L be on the other fide of AD; then ID
LB 2FC.
For then fubtracting equals from equals, it will
be IK LB FN NC FC; and by
doubling, IP2LB or ID- LB2FC.
PRO P. XLVI.
If AF be the axis, F the focus; FD perp. to AF.
Draw any line FM, and MD parallel to AF. Then
FM MD = FI, half the parameter of the axis;
when D is without the curve.
=
For make AC AF, and draw the directrix
CE. Then (Prop. I.) FM ME = MD+ DE
= MD + CF =(Prop. II.) MD + FI.
Cor. If D is within the curve, Fm + md = FI
half the parameter.
For Fm me = de dm FI- dm. And
Fm+dmFI.
PROP
D
E
Fig.39
V
D
41.
M
P
A
G
A
E
R
EX
EX
my
F
RT
40.
A
M
42.
B
R
ko
K
45.
E
43.
B
44.
P
D
46.
B
P
C
47.
B
P
G
B
N
48
F
G
M
PI.XXVIII.pa.196*
OF
པི;, བྷ
IV
•
B. III.
197
THE PARABOL A.
Fig.
PROP. XLVII.
If any circle cut a parabola in four points B, C, 52.
D, E; and from the points of interfection, ordinates
be drawn to the axis AI; the fum of the ordinates on
one fide of the axis, is equal to the fum of the ordi-
nates on the other; CG + BI EF + DH.
Draw the lines BC, DE, joining the points by
two and two, let them meet in N. Let PT, QS
be the diameters of BC, DE; from the vertices of
which let the tangents PR, QR, be drawn, which
are parallel to BC, DE. Then (XXVI. Cor. 2.)
RP²: RQ²:: BNC END. But (Geom. IV. 21.
:
Cor. 3.) BNC END; therefore RP² = RQ",
and RP RQ.
RQ. Therefore drawing PQ, and
biffecting it in O, RO will be perp. to PO or OQ,
confequently RO is the axis. But (XLV.) BI +
CG2PO2Q0 = FE + HD.
And if C fall on the other fide of AI, then
(XLV. Cor.) BI — CG = 2PO = 2QO
2QO = HD
+FE, and BI CG + HD + FE.
Cor. 1. If E and D coincide in Q, then HD +
FE = 20Q
For then HD + FE becomes OQ + OQ or
20Q.
Cor. 2. If C, E and D, coincide in Q, then BI
= 30Q.
For then HD +FE+CG becomes OQ+ 0Q
+ OQ.
Cor. 3. If C falls in A the principal vertex; then
BIFE + HD; or BI = 20Q, if E and D
coincide.
R 2
PROP
198
THE PARABOL A.
Fig.
XLVIII.
PROP.
53. If CF, AG be two diameters; and from any point
B in CF, any line BE be drawn cutting the line AC
in D, the diameter AG in G, and the curve in L
and E. Then the rectangle LDE = rectangle BDG.
54..
For by fimilar triangles, BD : DG :: BC: AG ::
(XXV. Cor. 3.) LBE: LGE; and BD x LGE
= DG X LBE; that is, BD x LGE = DG X
LB X BG + GE, or BD x GE x LD + DG =
DG X LB x BG + DG x LB X GE; and BD x
GE X LD + BDX GEX DGDG X LB X GE
DG X LB x BG. That is, BD X LD × GE
+ LD X DG × GE = DG × LB × BG, or BG X
LD X GE=DG × LB x BG; therefore LD x
GE = BL × DG; and adding LD × DG, LD ×
DG + LD × GE = LD × DG + BL × DG;
that is, LDX DE BD x DG.
×
Cor. The rectangle LD x GE = rectangle LBX
DG.
PRO P. XLIX.
If AB be any diameter; LBN an ordinate; LC,
NC, two tangents at L and N; DG any line paz
rallel to the diameter AB, cutting the tangents in D
and O, and the curve in F. Then FG rectangle
OFD.
For (Prop. XLIV.) OF: FG:: LG: GN::
FG: FD. Therefore FG² = OF × FD.
PROP.
B. III. THE
THE PARABOL A.
199.
PROP. L.
Fig.
If PG, AO be any two diameters, PO an ordinate 55.
to AO. And if any line AF be drawn from A, to cut
PO, PG in N and L, and the curve in F; then
AL' AN X AF.
:
For by reafon of the parallels, LG, AD; AF:
AL: DF: DG: (XXXIV.) DE: DF ::
(Geom. II. 13. Cor.) OP: ON: (by parallel
lines) AL: AN; and AF × AN = AL².
PROP.
LI.
If LC, NC, DF, be three tangents, at L, N and 56.
A; then will DA: AF:: LD: DC:: CF: FN.
Thro' D, A, C, and F, draw the diameters DG,
AH, CI, FM. Draw LA, AN ordinates to DG,
FM; fince theſe ordinates are biffected by their di-
ameters, we ſhall have LG = GH, NM = MH,
and alfo LI IN; that is, LH + HI = NM +
MI = MH + MI = 2MI + HI, and LH or 2GH
= 2MI, and MI = GH = LG; and GI = HM
= MN.
By reaſon of the parallels, DA : AF :: GH:
HM:: or LG: GI:: (fimilar triangles) LD : DC.
For the fame reafon, DA: AF:: GH: HM :: or
MI: MN :: CF : FN.
Cor. 1. If the diameter CRI cut the tangent DF
in B; then BFDA.
For GHIM; therefore by reafon of the pa-
rallels, DA = BF.
Cor. 2. If RI be the axis, and LN an ordinate to
it; LC, NC, tangents at L and N, DAF a tangent
at any point A; then LD = CF, and DC = FN.
R 3
For
200
THE PARABOL A.
Fig. For in that cafe CI is perp. to LN, and CL=
56. CN, and the angle NCILCI; therefore the
right angled triangles LCI, NCI, LDG, NFM
are fimilar; whence CF: IM:: FN: NM:: LD:
LG or IM; therefore CF = LD, and confequent-
ly FN CD.
57.
58.
59.
PRO P. LII.
All parabolas are fimilar figures.
Two figures are fimilar, when fimilar right lined
figures can always be infcribed in them, being
drawn in a like fituation.
=
Take BC: bc:: AL: al; then BC x al be
x AL, and BC x al X AL bc x AL'. But
(Prop. III.) BC X AL DL, and be x al = dl².
×
Whence DL²: dl² :: BC X AL : bc X al :: BC X
AL X al or be x AL: bc X al :: AL: al².
Whence AL: al:: DL: dl; therefore the trian-
gles ALD, ald are fimilar, and confequently the
curves DBA, dba are fimilar.
bc
Cor. Hence two parabolas, that have the fame pa-
rameter and abfciffa, are both equal and fimilar.
PROP. LIII.
If a parabola CAD, and ellipfis AGB, be de-
fcribed, having the fame vertex and focus A and F.
Then as the tranfverfe of the ellipfis: to the distance
of the focus from the further vertex fo the latus
rectum of the parabola to the latus rectum of the
ellipfis. AB BF:: CD: GH.
For (Book I. Pr. VI. Cor. 2.) BA: GH:: BFX
FA: GF'. But (Prop. II.) AF
But (Prop. II.) AF = CD; and
GF' = GH". Therefore BA GH:
व
: xX
BF X
CD :
B. III.
201
THE
PARABO L A.
4
CD: GH:: BF × CD: GH; therefore BA Fig.
× GH² = BF × CD × GH; and BA × GH = 59.
BF X CD.
Cor. If a parabola and ellipfis has the fame focus
and latus rectum. The distances of the vertices from
the focus will be in the fame ratio. AB : BF :. AF:
FO.
For BA GH:: BF x FAGH'.
:
But GH
= 40F; therefore BF × FA : 40F² : : BA :
40F :: BA ×OF: 4OF²; therefore BF × FA =
BA X OF.
PROP. LIV.
Two equal parabolas upon the fame axis, but hav- 60.
ing different vertices, approach infinitely near one ano-
ther, but never meet.
FG
Let P be the parameter of the two parabolas
DAL, FBC. Then (Prop. III.) px AB + BG =
DG2, and px BGFG'; and fubtracting, px.
AB DG²
rectangle DFC. Now
fince px AB is a given quantity, and fince the or-
dinate FG continually increaſes with the axis BG,
FG and FC at laft become infinitely great, and
therefore DF is infinitely little.
Cor. In the parabola the ordinate continually in-
creafes, as the abfciffa increaſes; but in a lefs ratio;
and when the curve is infinitely extended, the abfciffa
becomes infinitely greater than the ordinate.
For px AG = DG', whence p: DG:: DG :
AG, ſo that the greater AG or DG is, the leſs is
the ratio of p to DG or of DG to AG; and when
AG becomes infinite, then p x AG is infinite
therefore DG is infinite, and DG is infinite; and
is therefore infinitely greater than p; and confe-
quently infinitely lefs than AG.
R 4
PROP.
202
THE PARABOL A.
Fig.
61
PROP. LV.
उ
The area of a parabola FAG is the area of the
circumfcribing parallelogram, or AB x FG.
3
Draw AD, DG parallel to BG, AB; divide AD
into an infinite number of equal parts, and thro'
all the points C, fuppofe lines, as ĈN, to be drawn
parallel to AB. Then according to the method of
indiviſibles, all the lines CN, compofe the area
ANGD. Put AP
Put AP, PNy, r parameter;
then by the nature of the curve (Prop. XVII.) rx
» = CN, and the fun of all the
yy, and x =
CN or of all the
infinites) is X
yy
3
yy
r
yy
(by Prop. III. Arithmetic of
area ANC; and putting x for
xy = ÷ AC × CN, and
2, the area ANC =
F
उ
=
I
3
2
confequently the area ADG AD X DG.
Whence the area ANGBAD x DG or AB
X BG. And the area FAG AB × FG.
Otherwife,
2
T
X
3
Suppofe AB divided into an infinite number of
equal parts, thro' which draw the ordinates as PM.
Then fince rx yy, we have y or PM = êx =
=
rxx. And (Prop. V. Arith. Infinites) the
fum of all the PM's is =
3
√r x xž
2
3
= = √rx x x
3
2
2
yx for the area APM.
Whence the area ABF
2
3
ABX BF, and the area FAG = AB × FG.
2
उ
3
Cor. The area of the complement ANGD = ÷ AB
X BG.
PROP.
P
B
Fig.49.
52
F H
N
B
H
D
B
55
F
N
G
E
58.
50.
E
Le c
A 51.
F
G
M
K
B
53.
E
G
B
54.
56.
DA
F
A A
H
57.
29.
60.
TL.XXIX. pa.202.
M
OF
B. III. THE PARABOLA.
203
PROP. LVI.
Fig.
A parabolic conoid is
1
Z
its circumfcribing cylinder.
61.
Let AB be the axis, and fuppofe it divided into
an infinite number of equal parts; thro' which let
circular planes be drawn as MN, then all theſe
circles make up the folid. Let AP, PM =
y, parameter=r, c = 3.1416.
Then (XVII.)
rx
yy, and crx
cyy circle MN.
And
(Prop. II. Arith. of Infinites) the ſum of all the
crx
crx =
2
=
x x = ½ crxxcxy²
I
2
AB X cir-
cle FG.
2
Cor. The folidity of the fruftum FMNG is equal
to a cylinder whofe hight is PB the hight of the fruf
tum; and bafe, half the fum of the bafes of the
circle FG+ circle MN
fruftum =
=
2
X PB.
Let F fruftum, PB = h, FBb, then F
cbb
2
xx+b
cyy
2
x. But (XVI.) x:yy: x +
(cbbx
cyyx =
h: bb, and bbx ➡yyx + yyb, therefore 2F
+ cbbb―cyyx =) cyyx + cyyh + cbbb
cbb + cyy
cyybcbbb, and F-
xb.
2
LVII.
PROP.
The content of a parabolic spindle, or of the folid 62.
made by the parabola revolving round the ordinate BG,
8
is the circumfcribing cylinder.
15
=
=
Let BG = a, AB b, GPx, PM y, r
= parameter, c = 3.1416. Then (IV. Cor. I.) PM
xr BG + BP X PG, or ry=2ax
XX, and
y =
204
THE PARABOLA.
1
Fig.
62. y =
XX
4аaxx
and cyy=
4ax3 + x4
C
доро
2ax
j
circle PM.
Then if GB be divided into an infinite number
of equal parts, and circles as PM drawn through
them, the fum of all theſe conftitute the folid.
But (Arith. Inf. Prop. III.) the fum of all the
4aa
44
4aa
rr
XX
3rr
x³, and (ib. IV.) the ſum of all the
X3.
a
ry
x4, and (V.) the fum of all
24
X5
rr 5rr
There-
rr
fore the content of the folid GPMX: +aax³
rr
3
—ax4 + 3x³. And when x becomes BG or az
5
then the whole folid =
C
X
rr
8
15
C
X
× 4 as — as + as =
rr
=
as; but rb aa, and rrbbat. There-
8
fore, folid ABG = —×1/rrbba = — cbba =
circle AB × GB.
61.
15
15
8
15
Cor. The fegment of a parabolic spindle ANOB
is equal to a cylinder, whofe bafe is eight times the
greater bafe AB, together with three times the leffer
baſe NO, and alſo four circles which are mean pro-
portionals between the two baſes ; and the hight — the
hight of the fruftum.
For let AP x,
3.1416, NO = t.
yy
T
I
15
PN=y, AB = b, c
And rx yy;
then x
CN, and the area deſcribed by CN revolving
about BO is cX OC²
CX ON² = cbe -CX
2cbyy
** = 2cbx — CXX
CXX =
cy4
Therefore if
rr
AC
B. III.
THE
205
PARABOL A.
AC be divided into an infinite number of equal Fig.
parts, and planes drawn thro' them, the fum of all 61.
the circular annuli compofing the folid ACN=
2cby3
3r
cys
5rr
•
This taken from the cylinder ACOB
or cbby, leaves cbby
2cby cys
cxxy = cbby
cbby —
здо
2
I
+ =cbby -
5rr
2— — cbyx+
I
3
$
— cby × b — t + — cz
3
8
x b
t
11
1 cbby +
4
15
cbyt + 2 cgtt.
3
5
63-
15
PRO P. LVIII. Prob.
To defcribe a parabola by continued motion.
I Way.
Having a diameter, the latus rectum, and the
angle the tangent makes with the diameter. Pro-
duce the diameter PA, till AG be equal to the la-
tus rectum; draw the tangent AL, and make the
angle AGD GAL BAC. Then if the an-
= =
=
gle BAC be moved about the fixed point A, whilft
the leg AB carries with it the end D of the ruler
DM, along the line GD, and DM keeps always
parallel to GA. Then the interfection of DM
with the leg AC, at M, will defcribe the parabola
AMF.
—
Draw the ordinate MP parallel to AL, the lines
PM, GD are equal, as being equally inclined to
the parallels GP, DM. The triangles AGD, APM
are fimilar; for MPA GALAGD, and fince
GAL BAC or DAM, take away DAL, then
GAD LAM AMP. Whence AG: GD::
PM: AP; that is, latus rectum: PM :: PM: AP,
and AP × latus rectum = PM'; therefore (XVII.)
AMF is a parabola.
—
If AP is the axis, then the angles GAL, AGD,
DAM, APM, are right angles. Then it appears
at
206
THE PARABOL A.
Fig. at once, that DL or AG: AL or PM: AL or
63. PM: LM or AP.
64.
55.
2 Way.
Let AP be a diameter, AC a tangent at A. Pro-
duce the diameter AP, and make the ifoceles tri-
angle HBL, where BL is parallel to AC, and draw
HLIQ; draw HF parallel to AC, and equal to
the latus rectum; alfo draw ID parallel to AP.
Then if the figure FHBLQ be fo moved that HB
always flides along the diameter HA, and HQ
moves the line ID along CA, always parallel to
itſelf, whilft the point F turns the line GK about
the fixed point A. Then the interfection M, of the
lines AK, ID, will defcribe the parabola AOM.
For draw the ordinate PM. Then fince HB -
BL; therefore HA AI = PM. And the trian-
gles AHF, APM being fimilar, it will be AH or
PM: HF:: AP: PM. Whence HF × AP =
PM²; therefore (XVII.) the point M is in the pa-
rabola.
SCHOLIUM.
A parabola may likewiſe be defcribed by Def. 1.
PROP. LIX. Prob.
The axis and parameter being given, to deſcribe a
parabola; by finding feveral points in it.
I Way.
the
Let AC be the axis, make AF AB
latus rectum, and draw the directrix BD perp. to
BC, and F is the focus.
From any point G (above B,) as a center, with
the diſtance GF, crofs DH at D and H; with the
fame diſtance, and from the centers D and F, de-
ſcribe two arches cutting one another in K. Like-
wife from H and F, defcribe, two arches cutting
-
each
B. III. THE PARABOL A.
207
each other in L. Then K, L are two points of the Fig.
curve. And fo for more points.
For FGDK is an equilateral parallelogram, and
KD KF. Therefore (Prop. I.) K is in the curve;
and L likewiſe.
2 Way.
Let AC be the axis, make AF
tus rectum. Draw KCL perp. to AC.
center F with the radius BC, cut KL,
L, which are two points of the curve.
like for more points.
=
4
65.
AB la- 65.
From the
in K and
And the
For fince BC= FK FL; therefore (I. Cor. 2.)
K and L are in the curve.
SCHOLIU M.
When the parameter is not given, draw an or-
dinate; and find a third proportional to the ab-
fciffa and ordinate, which will be the parameter.
The reft as before.
3 Way.
Let AE be the axis, F the focus; let AC = CS 66.
= latus rectum, and CS perp. to AE, draw alfo
AD perp. to AE. Divide CS into any number
of equal parts, and AC into as many equal
parts as is the fquare of CS. Then from the
fcale CS, fet along AD, 1, 2, 3, 4, &c. parts;
thro' all which points draw parallels to AE; and
on theſe parallels, from the fcale AC, fet off 1, 4,
9, 16, &c. parts, viz. all the fquares of the former
numbers, having their origin at AD. And thro'
the points where they terminate, draw a regular
curve for the parabola, as AiG3S5.
For let CS n parts, and AC nn parts. There-
fore the parts in AC to the parts in CS, will be as
n to I. Now (Prop. II.).
FGCS n parts of the fcale CS;
4
2
and AF ACnn parts of the fcale AC =
n parts of the fcale CS.
There
208
PARA BOLA.
THE
Fig. Therefore FG:
66. ought by Prop. II.
67.
68.
:: 2
FA-n: 1n: 2:1, as it
And fince the ordinates 11,2G,
33, &c. are as the fquares of the abſciſſa A1, A2,
A3, &c. therefore (III. Cor. 2.) all the points I,
G, 3, S, &c. are in the curve of the parabola.
PROP. LX. Prob.
Given any diameter, and the tangent at the ver-
tex; to defcribe the parabola, by finding a number of
points in it.
I Way.
Let AB be the diameter, AL a tangent; and if
the parameter be given, produce PA to G, ſo that
AG may be equal to the parameter; biffect GA in
C. Draw AF perp. to AP. Take any point O,
and with radius OG, and center O, cut the lines
AP, AF in P and F. Thro' P draw MM paral-
lel to AL, and make PM, PM = AF. Then
M, M, will be two points in the curve.
For GP is the diameter of a circle whofe ordi-
nate is AF, and therefore GA X AP
AF²; that
PM²;
And thus
is by conſtruction, AP x latus rectum
therefore (XVII.) M is in the parabola.
as many points as you will may be found, and
then the curve drawn thro' theſe points.
2 Way.
If the diameter AP, tangent AL, a point M
in the curve, be given. Thro' M draw ML paral-
lel to AP, cutting the tangent AL in L. Prolong
PA upwards, and make AF LB. Draw AB
and MF, their interfection N, will be in the curve
of the parabola. And thus any number of points
may be found.
This conſtruction is evident from Prop. XXIV.
Ś CHO-
B. III. THE PARABOL A.
209
SCHOLIU M.
The parameter is found, by ſeeking a third pro-
portional to the abſciſſa and ordinate when drawn.
rallel to AP.
3 Way.
Fig.
68.
Let AP be the diameter, GAL the tangent. 69.
Make AG to the parameter, and draw FG pa-
Make GF AL, and draw LM pa-
rallel to AP. Thro' A draw FM to interfect LM
in M, a point in the parabola; and fo for more
points.
For draw FH parallel to GA, and the ordinate
PM, and draw HL. Then LA FG = HA.
And the triangles APM, AHF are fimilar, and
AP: PM: AH: HF; that is, by conftruc-
tion AP: PM:: AL or PM: AG the latus rectum.
Therefore (XVII. Cor. 1.) M is in the curve of
the parabola.
4 Way.
I
T
Let AB be the diameter, AL a tangent. Pro- 70.
duce BA, and make AE equal to the latus rec-
tum. Thro' E draw ED perp. to EB, cutting
the tangent AL in L. Make the angle LAF =
LAE, and AF AE. Thro' F draw DHC pa-
rallel to EB; then to the axis DC, and focus F,
and directrix DE, defcribe a parabola, by Prop. LIX.
For biffect DF in H, then (I. Cor. 1.) H is the
vertex, HC the axis; and fince AF AE, and
angle FALLAE; therefore (Prop. I. and VIII.
Cor. 2.) AH is a parabola.
SCHOLIU M.
If the abfciffa and an ordinate are given.
The parameter is found, by feeking a third pro-
portional to the abfciffa and ordinate.
The reft as before.
PROP.
210
THE PARABOLA.
Fig.
LXI.
Prob.
71.
72.
PROP.
A right line DE being given in a parabola; 'to
find its diameter, the axis, parameter, and focus.
Draw HI parallel to the given line DE. Biffect
DE, HI, in O and G, thro' which draw AOG for
the diameter.
Draw HF perp. to AG and biffect it in B; and
draw MB parallel to AG, for the axis.
Make MB: HB:: HB: parameter of the axis.
Then the parameter let from M to S gives the
focus S.
4
For fince DE, HI are biffected by AG; there-
fore (XV. Cor. 5.) DE, HI are ordinates to the
diameter AG. And fince MB is a diameter (be-
ing parallel to AG), and biffects HF at right an-
gles; therefore (IX. Cor. 2.) MB is the axis, and
HB, BF, ordinates. The reft is plain from Prop.
XVII. Cor. 1. and Prop. II.
SCHOLIU M.
The parameter of the diameter AG is found in
the fame manner, by faying as AG: GH:: GH:
parameter fought.
PROP.
LXII. Prob.
Any diameter being given in a parabola; to draw
an ordinate from a given point in the curve.
1 Way.
Let AB be the diameter, H the point given;
thro' H draw HC parallel to AB; and at the fame
diftance from AB, on the other fide, draw DL pa-
rallel to AB, to cut the curve in L; draw HL,
which will be a double ordinate to AB.
For
B. III.
211
THE PARABOL A.
For the parallels being equidiftant, makes HF Fig.
= FL.
2 Way.
From H draw any line HK cutting AB in G;
make GK GH; thro' K draw KL parallel to
AB, cutting the curve in L. Draw HL, which
will be a double ordinate to the diameter AB.
For HG = GK; therefore by reaſon of the pa-
rallels, HF is FL.
PROP. LXIII. Prob.
72.
To find a diameter in a parabola, that makes a 73.
given angle with its ordinates.
Draw any diameter DO (LXI), and make the
angle ODC equal to the angle given, ſo that DC
cut the curve in C. Biffect DC in B, and draw
the diameter AB parallel to DO, for that required.
For DB and BC are ordinates to the diameter
AB, and the angle DBA ODB the angle given.
= =
Cor. There will be two diameters which anfwer
this Prob and equally diftant from the axis; except the
given angle be right, and then the axis is the diame-
ter fought.
PROP.
LXIV. Prob.
To any given point M of a parabola, to draw a 74.
tangent.
1 Way.
Find any diameter AB
draw MC parallel to AB.
the diameter MC (LXII.);
rallel to DC, fo is MT the
This appears by Def. VII.
S
(LXI.), and thro' M
Find an ordinate DC to
thro' M draw MT pa-
tangent fought.
3 Way,
212
THE PARABOL A.
Fig.
75.
76.
77.
2 Way.
Draw any diameter AB, from M draw an ordi-
nate to it MB (LXII.); make AT
draw TM the tangent at M.
This is plain from Prop. XX.
3 Way.
AB, and
Draw MF from the given point to the focus;
and draw ME parallel to the axis AB. Make ME
= MF, and draw FE. Then draw MT parallel
to FE, for the tangent in M.
For in the ifoceles triangle EMF, <E=<F;
and fince MD is parallel to EF, < DMF = MFE
= MEF = TMË. Therefore (VIII. Cor. 3.) MT
is a tangent at M.
4 Way.
From the given point M, draw MG parallel to
the axis AB; and make MG MF. Draw GF
and biffect it in D; draw MD for the tangent at M.
By Prop. VIII. and Cors.
5 Way..
Having the directrix CG and the focus F; draw
MF from the given point to the focus; draw FT
perp. to MF, to cut the directrix in T. Then
draw MT, which will touch the curve in M.
By Prop. XXXIII. Cor. 3.
78.
PROP.
LXV. Prob.
From a point given without the parabola, to draw
a tangent to it.
1:
I Way.
Thro' the given point T, draw the diameter AB;
and make AB = AT. From any point D in the
curve,
Fig.61
F
G
B
H
62.
M
d
IG
63.
B
I
B
G..
7
M
P
G
D
D:
B
H
C
A
F
65.
4
Ο
64.
•
d b
P
M
K
70.
67.
L
N
N
m
P
M
i
E
I D
71,
H
D
C
H
B
Р
321
G
66.
F
F
E
F
L
b
M
69.
P
m
68.
M
M
S
I
H
A
72.
K
P1.XXX.pa.212
UNIL
OF
MIC
B. III. THE PARABOLA.
213
curve, draw the ordinate DC (LXII.); and thro' B Fig.
draw MN parallel to DC. Then MT, NT, drawn, 78.
will be tangents at M and N.
This appears by Prop. XX.
2 Way.
From the given point T draw any line cutting 79.
the curve in H and C. Find FL a mean propor-
tional between TH and TC. Thro' L draw the
diameter FL cutting the parabola in F, the point
of contact.
This is plain by Prop. XLIII. Cor. 2.
¨. PRO P.
LXVI.
Prob.
Given the focus F of a parabola, and three points 80.
in the curve M, N, P; to defcribe the parabola.
I Way.
}
Thro' N, M, and alfo N, P, draw the lines NK,
NI. And draw MF, NF, PF, to the focus.
Then make FN: FM :: NK: MK; and FN :
FP: NI: PI. Thro' K and I, draw the line KI,
and FT perp. to it from the focus.
A, for the vertex. Then with the
directrix IK, deſcribe a parabola thro' the point M
(Prop. LIX.), which will paſs thro' the points N, P.
Biffect FT in
focus F, and
Draw NG, PL, MH perp. to KT. Then by
conftruction, FN: FM:: NK: MK:: (fimilar
triangles) NG: MH. Alfo FN: FP: : NI: PI
(fimilar triangles) NG: PL. Therefore MF, NF,
PF are to one another, as MH, NG, PL. Confe-
quently, fince MF MH; therefore NF NG,
and PF PL. And all the points M, A, P, N,
(Prop. I.) are in the parabola.
2 Way.
From N thro' M and P, draw the lines NO, NB; 81.
thro' F draw NH, alfo draw MF, PF.
S 2
Draw FO
biffecting
214
PARABOL A.
THE
Fig. biffecting the angle MFH, cutting NM in O. Alfo
81. draw FB biffecting the angle PFH, to cut NP in
B. Thro' O and B draw the line OB. Then with
the focus F and directrix OB, defcribe a parabola
thro' the point M, which will alfo pafs thro' the
points N, P.
82.
For (Prop. XXXV. Cor.) if N, M, be two
points in the parabola; FO will cut NM in the
point O of the directrix OB. And if N, P, be
two points in the parabola; FB will cut NP in the
point B of the directrix OB. Therefore the direc-
trix and focus remaining the fame, all the three
points M, N, P, are in the fame parabola,
SCHOLI U M.
After a like manner an ellipfis or hyperbola whofe
focus is given, may be defcribed thro' three given
points; by the help of Prop. LXIII. B. I. and
Prop. LIX. B. II.
PRO P. LXVII. Prob.
To defcribe a parabola thro' three given points, and
whofe diameters fhall make a given angle with the or-
dinates.
Thro' two of the points M, N, draw the right
line MN, and thro' the point P, draw the diame-
ter PF, making the angle PFM angle given.
Thro' G the middle of MN, draw the diameter AG
parallel to PF. Then make, as rectangle MFN:
MG PF: AG. Then make as AG: GM::
GM : AH the parameter of AG. Laftly, defcribe
a parabola with the diameter AG, and latus rec-
tum AH, this will pafs thro' the points M, P, N,
as required.
For fince MFN: MGN:: PF: AG; therefore
(XXV. Cor. 3.) the points M, P, N, are in the
parabola.
B. III.
215
THE PARABOLA.
parabola. And < AGM = PFM angle Fig.
given.
PROP.
LXVIII. Prob.
82.
To draw a tangent to a parabola parallel to a given 83.
line CD.
Draw the directrix AB; thro' the focus F, draw
BH perp. to the given line CD, to cut the direc-
trix in B. Draw BP parallel to the axis AF. Bif-
fect BF in G, and draw GP parallel to CD, cut-
ting BP in P, and GP will touch the parabola in P.
For draw PF; then the triangles BPG and FPG
are fimilar and equal. For the angles BGP and
FGP are right angles, being equal GHD. And
BG GF, and PG common, therefore PB PF,
whence (Prop. I.) P is in the parabola, and (VIII.)
PG is a tangent.
PRO P. LXIX.
Prob.
About a given line as a diameter, to deſcribe a co-
84.
nic fection, to pass thro' a given point P; and fo as the
ordinates may make a given angle with the diameter.
Let AB be the diameter, from P draw PQ mak-
ing the given angle with AB. Biffect AB in C,
and draw CD parallel to PQ. Then if Q fall be-
tween A and B, make as rectangle AQB: QP ::
AC: CD². And with the femidiameters AC, CD,
deſcribe the ellipfis ADPB.
But if Q fall below B, with the femidiameters
AC, CD, deſcribe the hyperbola BPF.
Laftly, if AB do not terminate, but is infinitely
extended towards B; make as AQ: QP:: QP:
L, the latus rectum of AQ. Then with the dia-
meter AQ, and latus rectum L, defcribe the pa-
rabola AP, for that required.
S. 3
AH
216
PARABOLA.
THE
Fig.
84.
85.
All this process is plain from the properties of
theſe figures before laid down.
PRO P. LXX. Prob.
To defcribe a conic fection to pass thro' five given
86. points, A, B, C, D, E.
Draw lines between the points A, C; and be-
tween the points B, D; to interfect in R. And
thro' the fifth point E, draw EK, EH parallel to
BD, AC; meeting them in G and Q. Then in
the lines EK, EH, find the points K, H, fo that
rectangle AGC: KGE:: ARC: BRD :: EQH :
BQD; taking QH, GK, towards fuch parts as the
problem requires.
and draw LN.
And draw OP
But if LN, OP
parabola.
Biffect KE, BD in L, and N,
Alfo biffect AC, EH, in P and O.
to interfect LN in F the center.
are parallel, the curve will be a
Produce FP, FO (if neceffary) to I and M, ſo
that IF FM, and that PA: PA- OE: FI
-FP2: OFFP2, to this add FP², and we have
IF² and confequently IF or FM, and IM the dia-
meter to the ordinates AP, EO. And the ſection
may be drawn by Prop. LXIX.
For fince IFFP: FO- FP :: PA2:
PA-OE², by compounding IF² — FP² : IF²
FO: PA: OE²;
:
that is, IPM: IOM : : PA² : OE; (therefore by
the properties of the conic fections) IM is the dia-
meter of the ordinates PA, OE. The reft appears
by Prop. XLIII. Cor. 2. B. I.
The fame demonftration ferves for the hyperbo-
la, mutatis mutandis; and by Prop. XXXIX. Cor. 2.
B. II.
Or
Fig. 73.
D
B
C
M
76.
82.
74.
M
T
F
B
M
T
75.
B
T
77.
A
M
78.
B
E
B.
K
H
TI
L
G
79.
F
C
P
M
H
80.
N
81.
B
H
B
G
P
M
F
H
P
83
84
A
B
P
I
85.
K
E
E
P
R N
B
M
N
IM
D
86.
F
N
B
PIXX.
n
N/
OF
RICH
M
B. III. THE PARABOL A.
217
Or thus by points.
Fig.
Draw AB, BC, DA, CD, forming the infcribed 87
trapezium ABCD, (or if A or C be infinitely dif
tant, draw them in given directions). From E draw
EF, EG, parallel to AB, AD. In EF, EG, take
H and N, either both the fame way, or both the
contrary ways, from F and G, in refpect of E;
and fo that EF: EG:: EH: EN. And draw BH,
DN, to interfect in M a point of the curve. And
thus innumerable points may be found, and the
curve drawn thro' them.
If BH, DN be parallel, then M is infinitely dif-
tant; or they may be parallel to the affymptotes
of a hyperbola, or to the diameters of a parabola.
The practice is expeditiously performed thus;
in EF and EG, take as many equal parts as you
will, at 1, 2, 3, 4, &c. fo that the magnitude of
theſe in EF, to thofe in EG, may be as EF to
EG. Then thro' all the points in EF, draw lines
from B; and thro' all the points in EN, draw lines
from D, and where the correfpondent lines inter-
fect, will be ſo many points of the curve.
For (Prop. LXX. Cor. 5. B. I.) fince EF: EG:
EH: EN, then BH, DN, drawn will meet at M
in the curve. And the like for the hyperbola
(Prop. LXIII. Cor. 5. B. II.). And for the para-
bola (XXXVII. Cor. 5. B. III.)
Cor. 1. If fewer points be given, provided others
be found to make up five; the curve will be de-
fcribed in the fame manner.
may
Cor. 2. If four points be given, and the curve is
to paſs thro' theſe points, and touch a right line given
in pofition, in one of thefe points. It is only fuppofing
two points to coincide in one; and then the line drawn
thro' them, becomes a tangent to that point. And
then the conftruction will be the fame as before.
S 4
Cor.
218
PARABOLA.
THE
Fig. Cor. 3. Likewife if the curve is to pass thro' three
87. points, and touch two right lines, given in pofition,
in two of theſe points. The conſtruction will be the
fame, by fuppofing, two and two of thefe points to
coincide.
88.
PROP. LXXI. Prob.
To draw a conic fection thro' four given points B,
C, D, E, to touch a right line HI given by pofition.
Draw lines thro' two and two as CE, DB, meet-
ing in G, and cutting the tangent in I and H.
Take GK a mean proportional between GE and
GC; IP a mean between IC and IE; GN a mean
between GB and GD; and HL a mean between
HD and HB, and HQ a mean between HO and
HF. Then in the tangent HI, find the point A,
fo that HA: AI :: GK × HL: IP × GN; and A
will be the point of contact. Therefore by Prop.
LXX, thro' the five points A, B, C, D, E, de-
ſcribe the conic fection, and it is done.
For (XLIII. Cor. 1. B. I. &c.) rectangle BGD :
EGC:: DHB: OHF; that is, GN: GK:: HL2:
HQ, and GN: GK:: HL: HQ; whence GNX
HQ = GK × HL. Again XLIV. Cor. 3. B. I.
&c.) AH': AI :: OHF: CIE :: HQ: IP²; and
AH: AI:: HQ: IP :: HQ X GN: IPXGN::
GK × HL: IP × GN.
According to the various fituation of the points
C, D, E, B, and tangent HI; an ellipfis, hyper-
bola or parabola will be deſcribed.
PROP
B. III.
219
THE PARABOL A.
PROP. LXXII. Prob.
Fig.
To defcribe a conic fellion thro' three given points, 89.
B, C, D; to touch two right lines given by pofition,
HI, NK.
Thro' two of the points B, D, draw BD meet-
ing the tangents in H and K.
And thro' two
others of the points, C, D, draw CD, alfo meet-
ing the tangents in I and L. Take four lines T,
V, X, Y, fo that TT rectangle BHD, VV =
BKD, XX = CID, YY = CLD. Then in the
lines HD and ID, find the points R, S, fo that
HR: KR:: T: V, and IS: LS:: X: Y. Draw
RS cutting the tangents in A and E, the points
of contact. Then thro' the five points A, B, C,
D, E, draw a conic fection by Prop. LXX.
Draw IF parallel to KN, meeting the curve in
O and F. And make IZ = OIF. Then fup-
pofing A, E, the points of contact, (XLIV. Cor. 3.)
rectangle OIF or IZ² : LE² : : CID or XX : CLD
or YY. And IZ: LE :: X: Y :: (conſtruction)
IS: LS. Therefore the points S, E, Z, are in
one ftreight line. Again let the tangents meet in
G, then rectangle OIF or IZ: IA² : : GE² : GA².
And IZ: IA:: GE: GA; therefore the points E,
Z, A are in one right line. And by the fame rea-
foning the points E, R, A, are in one right line.
Therefore the points A, E, are in the right line
RS. That is, the points of contact lie in the right
line RS, as determined in the conftruction.
PROP.
220
THE PARA BOL A.
PRO P. LXXIII.
Prob.
Fig.
90.
To draw a conic fection thro' two given points D,
E; to touch three right lines, TG, TH, CH, given
in pofition.
Thro' the given points D, E, draw the line FQ,
to cut the tangents in G, Q, and P. And let XX
= EQD, YY = DGE, and ZZ = EPD. And
make, 'as FQ : FG :: X: Y. And GK : PK
Y: Z. And from F draw FH to the interſection
of the tangents TH, CH. And make FI: FH::
IM: MH. Thro' K, M, draw a line interfecting
the tangents AT, CH, in A and C the points of
contact. Thro' F and A, draw a line interfecting
the tangent TH in B the point of contact. Then
thro' A, B, C, D, E, draw a conic fection, by
Prop. LXX.
2,
For fince GK: PK :: Y: Z, therefore GK²:
PK Y or DGE: ZZ or EPD; therefore
(Prop. LIV. B. I.) K is in the line joining the points
of contact A and C. Alfo fince FQ: FG :: X:
Y, and FQ: FG EQD: DGE; therefore
(LIV. B. I.) as FQ cuts the tangents TG, TQ,
in G and Q, F will be in the line AB, joining the
points of contact A and B. Alfo for the tangents
TG, HC, IM² : MH² :: LIN: LHN. And for
the tangents TG, TH; FI: FH:: LIN: LHN.
Therefore IM²: MH' :: FI² : FH". And IM :
MH :: FI: FH; therefore by conftruction, the
point M is in the line AC. Whence both K and
M being in the line AC, determines the points of
contact A and C; and FA drawn determines the
point of contact B. Whence the five points A, B,
E, C, D, are given.
PROP.
B. III.
221
THE PARABOLA.
PRO P.
!
LXXIV.
Prob.
Fig.
To defcribe a conic fection thro' a given point E, 91.
to touch four right lines, OT, TH, HR, RO, given
in pofition.
Draw the diagonals TR, HO, to interfect in
K. Thro' K and the given point E, draw FQ;
cutting the tangents in G, P, Q. Then find the
point D fo that KG: KP² :: DGE: DPE; then
D is another point of the curve. Therefore by
the laſt Prop. draw the curve thro' the two points
D, E; to touch the lines TO, TH, HR.
Or thus; find the point F, fo that FQ² : FG²::
EQD EGD. Then draw FH, and find the point
M, fo that FI: FH:: MI: MH. Thro' K and M
draw CA; and thro' F and A, draw FB. Thro'
B and K, draw BS. Then A, B, C, S are points
of contact; thro' which and the point E, draw the
fection, by Prop. LXX.
For (Prop. LIV. Cor. B. I.) the lines joining the
oppofite points of contact, pals thro' K. And fince
KG: KP²:: DGE: DPE, (LIV. B. I.) D will
be in the curve. And fince FQ cuts the tangents
TO, TH, and it is FQ² : FG² : : EQD : EGD;
therefore (Prop. ib.) F is in the line joining the
points of contact A, B. Alfo fuppofing M in the
line AC, drawn thro' the points of contact of the
tangents TO, HR; then (LIV. B. I.) MI2:
MI²
MH LIN: LHN:: (for the tangents TG,
TH) FI²: FH. And MI: MH:: FI: FH. On
the contrary, fuppofing MI: MH:: FI: FH, then
M will be in the line AC. Therefore a line drawn
thro' M, K, will pafs thro' the points of contact
A, C. And a line from F thro' A, paffes thro' the
point of contact B; and laſtly, a line drawn thro'
B and K (LIV. Cor. B. I.), will paſs thro' the other
point of contact S.
2
PROP.
222
THE PARABOLA.
Fig.
92.
PROP. LXXV. Prob.
To defcribe a conic ſection, to touch five lines given
in pofition, GT, TH, HR, RO, OG.
Take any four of the tangents forming a quadri-
lateral as IGTHI, leaving out the tangent OR; draw
the diagonals IT, GH to interfect in L. Then
leave out the tangent OG, and draw the diagonals
of the quadrilateral KTHRK, to interfect in M;
draw LM, to cut the tangents in A and C, the
points of contact. Go thus round the figure; leav-
ing out fucceffively the tangents GT, TH, HR;
drawing the diagonals of the trapezia DHROD,
ZROGZ, FOGTF; interfecting in N, P, Q.
Then MN, NP, PQ, QL being drawn, will in-
terſect the tangents in the points of contact, B and
S, C and E, S and A, E and B. Then thro' the
points A, B, C, S, E, deſcribe a conic ſection by
Prop. LXX.
For (Prop. LIV. Cor. B. I.) the interfection of
the diagonals of the trapezium IGTH, which is
L. And alſo of the trapezium KTHR, which is
M, are in the line AC, drawn thro' the points of
contact A and C. Likewife the interfection of
the diagonals of the trapezia KTHR, and DHRO;
that is, M and N, are in the line BS, drawn thro'
the points of contact B and S. Thofe of DHRO
and ZROG, N and P, are in CE. Thoſe of ZROG
and FOGT, (P and Q), are in SA. Thoſe of
FOGT and IGTH (Q and L), are in the line EB.
Therefore A, B, C, S, E, are the five points of
contact.
When the center or focus is given, it is equiva-
lent to two lines or points; and an affymptote to
an hyperbola, to two tangents.
Many
:
Fig.87
B
m
G
3
1
91.
M
G
M
K
89.
F
E
R
T
N
H
B
YH
K
G
B
88.
E
с
KP
A
F
H
M
K
B
90.
E
E
S
92.
R
M.
A
D
T
B
PI.XXXII.pa.222.
UNIL
OF
B. III.
223
THE PARABOL A.
Many more problems might be added; but what Fig.
I have here given, are abundantly fufficient for an 92.
introduction.
PROP. LXXVI.
If a cone ABC be cut by a plane LDG parallel to 93.
the fide of the cone AB, the fection LDG will be a
parabola.
Draw the plane ED parallel to the baſe BGC.
Then fince AB is parallel to the plane LDG; a
plane touching the cone in the line AB, will be
parallel to LDG; whence arch BL BG, and
BI paffing thro' the center of the bafe is perp. to
LG, and LIIG. And by the nature of the
circle, the rectangle BIC IL².
IL². From the vertex
D draw DI. Then by fimilar triangles AE: ED::
ED X DI
DI: IC =
becomes
;
AE
BI X ED X DI
AE
therefore BI × IC = IL²,
ED2
= IL², or
AE × DI =
IL². But AE and ED are given quantities; there-
fore make AE: ED :: ED: L; then
ED2
AE
=L,
and L X DI IL. Therefore (Prop. XVII.)
the curve LDG is a parabola, and L the parameter.
Cor. 1. The latus rectum
ED2
AE'
or a third pro-
portional to AE and ED. And if F be the focus,
DF =
ED2
4AE
By Prop. II.
Cor. 2. Let DP be perp. to AB, then in a right
cone, 2PE the latus rectum; and DF
I
2
PE.
For if AN be perp. to ED, it biffects it, and the
triangles AEN, DEP are fimilar; and ÁE: EN
or
224
THE PARABOL A.
Fig.
93.
or ED :: ED : EP =
ED²
2AE'
ED2
and 2EP =
AE
latus rectum.
94.
95.
96.
Cor. 3. If AF be perp. to the bafe of the cone, and
AP perp. to DI the axis of the parabola; and if
ANAP, and HNG drawn parallel to the bafe
BC; then HG is equal to the latus rectum. And HA
(on the oppofite fide) = ED the diameter at the ver-
tex D.
Draw AL parallel to BC, then the triangles AHN,
APL are fimilar and equal, and AH AL = ED.
And by fimilar triangles (AED, AHG) AE: ED::
ED²
AH or ED: HG =
AE =(Cor. 1.) latus rectum.
Cor. 4. If MVN be a right cone, F the focus of
the parabola ARZ. And if AG be made equal to
AF; the circle GI, parallel to the base of the cone,
will cut the plane of the parabola in the line ED the
directrix of the parabola.
For (Prop. I. Cor. 1.) if DE be the directrix,
then AD AF (by conſtruction) AG. And
fince ANAK, the triangles ANK, AGD are
fimilar, and DG parallel to NK; that is, GI pa-
rallel to MN, will cut the parabola in the line DE.
Cor. 5. In a right cone VMH, if F be the focus,
and AG = AF, and the circle GNI be drawn pa-
rallel to the bafe. Then the distance of any point P
of the parabola, from the focus; that is PF, will be
equal to PN, the distance of P from the periphery of
the circle GNI.
For let PQ be an ordinate, and draw QS paral-
lel to the baſe of the cone; then fince the plane
SQP is parallel to the bafe, SG will be equal to
PN, becauſe in a right cone the fegments of the
fides contained between parallel circles are equal.
And
Fig. 93.
94.
AA
95
G
AA
96.
P1.XXXIII. the End.
B. III.
225
THE PARABOLA.
And fince AQ = AS, and AD = AG, DE being Fig.
the directrix; therefore DQ SG. But (Prop. I.) 96.
FPDQ = SG PN.
SCHOLIU M.
From this Prop. the reaſon appears, why the
Parabola is called one of the conic fections: and
that is, becauſe a cone cut by a plane parallel
to one of its fides, will always form the curve call-
ed a parabola, as we have fhewn in this propofi-
tion. The properties of which curve we have de-
monftrated at large in this Book; and of the other
fections in the foregoing books; and that from the
mechanical deſcription thereof upon a plane, with-
out regard to the cone or any other body what-
ever; without analytical calculations; without har-
monical ratios; but by the pure fyathetic method
of demonftration.
FINI S.
THE
NATURE AND PROPERTIES
OF
CURVE LINES.
BOOK I.
OF THE
CONCHOID, CISSOID, CYCLOID, QUADRA-
TRIX, LOGARITHMIC CURVE; the SPI-
RAL of ARCHIMEDES, the LOGARITH-
MIC SPIRAL, the HYPERBOLIC SPIRAL.
BOOK
II.
Of CURVE LINES in general, and their
Affections.
Ex his principiis via ad majora Sternitur.
NEWT. QUAD. CURV.
[ i ]
THE
PREFACE.
THE enfuing Treatife, which is concerning curves
of higher kinds, I have given here, as a proper fe-
quel to the Conic Sections. And to proceed gradually, 1
have in the first Book given account of many of the
common fort of curves, as the Conchoid, Ciffoid, Cy-
cloid, &c. and ſeveral forts of Spiral Lines, which
will oblige thefe people that want to be acquainted
with the nature of thefe curves. And as the reader
will often meet with the mention of thefe curves,
therefore he ought to know the nature of them, and
their effential properties; and this is what every one
ought to be acquainted with, that expect to make any
great advances in this branch of science. And in
handling thefe, I have taken the shortest way of de-
monftration I could find, whether geometrical or alge-
braical. But I have purpofely omitted fuch properties
as could not be demonftrated without having recourfe
to fluxions. And for this reafon alfo I have faid no-
thing about feveral other curves of like fort.
In the fecond Book I have treated of fuch general
properties, as belong to all curves univerfally. In
Sect. 1, are laid down feveral theorems about the
transformation of curves. By the help of some of
thefe theorems, the area of one curve being given, or
perhaps the length; the area of the other may be de-
termined. By others, their lengths, furfaces, or fo-
lidities may be found. And this by common geometry
without the help of fluxions.
T 2
In
iv
THE PREFACE.
In the ſecond ſection are laid down ſeveral propo-
Jitions, relating to all forts of curves in general, their
nature and properties. By these are determined the
pofition of the axis, the ordinates, and their affymp-
totes, if they have any; and what curves have afflymp-
totes, and what not.
In this part I have given the
learner a taſk of the higher geometry; and here is a
large field, for any one that has a mind to exercife his
invention therein.
In the third fection you have feveral propofitions
concerning curve lines, drawn from fluxions, and
which are ferviceable for investigating the nature of
fuch curves. When the reader is acquainted with all
thefe things as an introduction, he may proceed with
pleasure to read other books on the fame fubject, and
particularly that inimitable piece of Sir J. Newton's,
entitled, An Enumeration of Lines of the third Or-
der. And therefore I hope, what is here delivered,
will be grateful to all lovers of geometry, and parti-
cularly to fuch as defire to make any confiderable pro-
grefs therein.
W. Emerſon.
BOOK
[ 1 ]
BOOK I.
SECT. I.
Of the CONCHOI D.
A
DEFINITION S.
DEFIN. I.
I.
Q is a ſtreight ruler in which the points A, Fig.
B are fixed. Then if this ruler is fo mov-
ed that it always paffes thro' the fixed point P,
whilſt the point B flides along the right line CD.
The curve NAFG defcribed by the point A, is
called a Conchoid, or the Conchoid of Nichomedes.
DE F. II.
If B be between A and P, the curve AFG is
called the Superior Conchoid.
DE F. III.
If a is between P and B, the curve afg is called
the Inferior Conchoid.
DE F. IV.
If PT is perp. to CD, and if Ba in the inferior 2.
conchoid, be greater than TP; the curve aPƒZPg,
is called the Nodated Conchoid.
DE F. V.
The fixed point P is called the Pole or Center of I.
the conchoid.
T 3
DEF.
2
B. I.
THE CONCHOID.
Fig.
I.
DE F. VI.
The line CD is called the Directrix of the con-
choid.
PROP. I.
I.
1.
All lines as AB comprehended between the curve and
directrix, are equal, which paſs thro' the pole P.
This is evident from the generation.
Cor. If BTF be perp. to the directrix CD, any
line AB, comprehended between the curve and direc-
trix, and palling thro' P, is equal to the perpendicular
FT.
PROP. II.
The directrix of the conchoid CD is an affymptote
to the curve; or approaches infinitely near, but never
touches it.
From any point A of the curve, draw AP to
the pole, and AH perp. to CT. Then the trian-
gles PBT, ABH are fimilar, and BP: PT :: AB:
AB × PT
AH=
PB
Then fince the diſtance of
the point A from CD; that is, AH is always reci-
procally as PB; AH continually diminiſhes as PB
increaſes, and when PB is infinitely great, AH will
be infinitely ſmall. Therefore CD is an affymp-
tote to the curve.
And the fame reaſoning may be applied to the
inferior conchoid afg.
PROP.
Sect. I. THE CONCHO I D.
PROP.
III.
Fig.
If from any point A of the curve, AH, FTP be
3.
drawn perpend. to the directrix HT; and AL paral-
lel to it. And if the quadrant FGK be deſcribed
cutting AL in G. Then PL X LG TLX LA.
For the triangles AHB, TLG, are fimilar and
equal; for the angles at H and L are right. And
AB TF TG, and AHTL; therefore
BH GL. Alfo the triangles AHB, ALP, are
fimilar; whence AH: HB:: PL: AL; that is,
TL: GL: PL: AL; therefore PL X LG =
TL X AL.
Cor. 1. The fame things fuppofed, AG = BT,
and GL HB.
For AL HT, and GL = HB; therefore by
fubtraction, AG
BT.
Cor. 2. This Prop. holds good alfo in the inferior
concboid.
This appears by applying the fame demonftra-
tion to the inferior conchoid, denoted by fmall
letters.
Cor. In the inferior conchoid, gl bh.
PROP. IV.
In the fuperior conchoid, if PTF and AH be
perp. to the directrix HT, and AL parallel to it.
TL: TFAH':: PL: AL.
For by Prop. III. TL: GL :: PL: AL; and
TL²: GL² : : PL² : AL". But GL²
But GL2
GT2
TF2 AH; therefore TL: TF
24
TL2
AH²:: PL : AL².
I 4
Cor.
4.
3.
THE CONCHOID.
B. I.
Fig. Cor. Hence putting FT = a, PT = b, axis TH
3. = x, ordinate HA = y; then the equation of the
b+ y√ aa—yy
fuperior conchoid is
y
x= AL, or
being reduced y++ 2by³ + bb-aa + xx × yy —
·
2aaby
aabb.
PROP. V.
In the inferior conchoid, if pft and
the directrix bt, and al parallel to it.
ab² : : Pľ : al.
For (Pr. III.
ab be perp. to
Then tl: tf²
Cor. 2.) tl: gl: : Pl: al. And tľ:
But gl bbba” — ab¹ — tf²
ab²; therefore tľ: tf - ab²:: Pl : al.
gl: P: al.
:: =
Cor. 1. Hence putting tf = a, tPb, axis th
=x, ordinate bay, the equation of the inferior
conchoid will be
by ✔aa
y
yy
= x = al. And
when reduced, y4 — 2by³ + bb aa + xx × yy +
2aaby aabb.
Cor. 2. The fame equation holds for the nodated
conchoid, all the difference being, that a is here greater
than b.
SECT.
[ 5 ]
SECT. II.
Of the CISSOID.
L
DEFINITION S.
DE F. I.
5.
ET ABG be a circle, C the center, QED a Fig.
fquare; alfo P, F two fixed points in the lines
AP, ED, ſo that GP radius CG EF = FD.
Alfo let CD produced ad infinitum, be perp. to CP.
Let the fquare QED be fo moved, that the fide
EQ may always pafs thro' the point P, whilft the
end D ſlides along the right line CD. Then the
point F (the middle of ED) will defcribe the curve
GBF, called the Ciffoid, or the Ciffoid of Diocles.
DE F. II.
The circle ABG is called the Generating Circle.
DE F. III.
The line CD is called the Directrix.
PROP. I.
If AB and BG be quadrants, the ciffoid paffes thro'
the point B.
For when the fquare PED comes into the pofi-
tion PCB, then the angles C and E being right,
EF falls upon CB, and F upon B, cutting off the
quadrant GB.
5.
PROP.
6
B. I.
THE CIS SO I D.
Fig.
6.
PRO P. II.
If from any point F of the curve, FL be drawn
parallel to the directrix CD, to cut the circle in M.
Then LM, LG, LF, are in geometrical progreffion.
For draw GH parallel to CD, and FH parallel to
AG. Then the triangles CPR, ERD are fimilar and
equal. For the oppofite angles at R are equal, the
angles at C and E are right, and CP ED;
therefore CR = ER, and confequently ER + RD
CD. Let LF or GH = x, LG or FH = y,
AC a. Then FOy-a,OD/FD FO
Ї
a
Vaa
yy.
yy + zay aa
And OF+FD OF + OH =y.
The right angled triangles FOD, ERD are fimilar,
OF: : ED : SER
whence OD: {PD
FD
RD
And by com-
pounding, OD : OF + FD : : ED : ER + RD or
CD; that is, √2ay — yyy :: 2a: x + OD or
x+√2ay—yy; therefore x2ay-yy + 2ay—
jy2ay, and x/2ay — yyyy, but by nature
of the circle, LM = 2ay-yy, therefore LM
(√/2ay—yy): LG (y) :: LG (y) :LF (x).
Cor. 1. Hence OD — LM, the ordinate of the
circle.
For OD was = √2ay -yy, and LM -
√2ay-yy.
Cor. 2. If GH = ∞, HF = y, CG = a, the
equation of the curve is 2axxyxx = y³.
For x/2ay-yyyy, and by fquaring, 2ayxx
-yyxxy; and dividing, 2axx - yxx — y³.
Cor.
Sect. II.
7
THE CIS SOI D.
Cor. 3. The four lines AL, LM, LG, LF, are
in continued proportion.
For by the nature of the circle AL: LM:: LM:
LG:: (by this Prop.) LG: LF.
Cor. 4. Produce FL to N, and draw NC, then
will CDFN.
For LF
CO, and NL LM (Cor. 1.)
QD, therefore NF CD.
Cor. 5. FH: AL:: GH: FH.
For y 2a-y: xx:yy, by Cor. 2.
PRO P. III.
Fig.
6.
If thro' any point of the curve F, LF be drawn 7.
perp. to the diameter AG, cutting the circle in M;
and the line GFV be drawn, cutting the circle in V;
then arch AV ≈ arch GM.
For (Prop. II.) LM: LG:: LG : LF :: (fimi-
lar triangles) GZ: ZV. And LM² : LG² : : GZ² :
ZV²; that is, ALG LG:: GZ: AZG. And
: LG²
dividing the antecedents by LG, and the con-
fequents by GZ, then AL: LG :: GZ : AZ. And
compounding, AL + LG LG:: GZ + AZ:
:
AZ; that is, AG: LG:: AG: AZ; therefore
LG AZ, and LM = ZV, and arch GM = AV.
Cor. 1. Arch AM = arch GV.
Cor. 2. If two lines AM, GV move about the
points A, G, ſo that their interfection be always in
the directrix CB; and if ML be perp. to AG; then
the interfection of GV and ML, at F, will be in the
ciffoid.
For by that means the arches AV and MG will
be equal; and F will be in the curve by this Prop.
Cor. 3. The lines AO, GO, DO are all equal.
For
8
THE CISS OID. B. L
;
For the triangle AOG is ifoceles, and fince AC
7. CG, therefore AO = OD.
Fig.
Cor. 4. The line AMID = GV.
For AOOD, and by reaſon of the equal
arches VB, BM, IO is QM; and ID = AM
or GV.
Cor. 5. The lines OI, OV, OF, OM are equal.
Becauſe the arches VB,
CL, are equal.
BM, and the lines ZC,
Cor. 6. The line GF
AI or MD.
8.
For the triangles AZI, GLF are fimilar and equal.
PROP. IV.
If GS be perp. to the diameter AG, and from any
point F in the curve, FN be drawn parallel to GS,
cutting the circle in N, and ANS be drawn. Then NF
= GS.
Draw GFV, and produce NF to M. Then fince
arch NG arch GM (Prop. III.) arch AV;
therefore the angle NAG AGV; whence AS is
parallel to GV. And fince NF is parallel to SG;
therefore NFGS is a parallelogram, and NF = GS.
Cor. 1. The line IZ
LF.
For the triangles AZI and GLF are fimilar and
equal.
arch AV,
Cor. 2. If NF be drawn perp. to AG, thro' any
point F in the curve, and GF be drawn. Then NF,
GF, LF, are continually proportional.
For draw NG, then fince arch NG
add GV, then NGV GVA
whence NGV or NGF is a right angle. Therefore
in the right angled triangle NGF, where GL is
perp. to NF, it is NF: GF :: GF: LF.
a femicircle;
Cor.
Sect. II.
9
THE CIS SOI D.
Cor. 3. GS, GF, LF, are continually propor- Fig.
tional.
PROP. V.
8.
If AE be perpendicular to AG the diameter of the 9,
generating circle; then AE is an asymptote to the
curve.
For let GH = x, HF = y, AG = 2a. Then
(Cor. 2. Prop. II.) the equation of the curve is
y³
2axx —yxx — y³, and xx = za—y'
24 ·y'
where EF =
24-y. So that as xx continually increaſes, and y³
cannot exceed za'; therefore 2a -y or EF conti-
nually decreaſes; and when xx becomes infinitely
great, 24-y or EF becomes infinitely fmall; and
3
when y = 2a,then xx =
24
3
infinitely. Therefore
C
AE is an affymptote.
PRO P. VI.
Thro' any point F of the ciffoid, draw GH, cut-
ting the circle in D, and the affymptote in H. Then
GF = DH.
Thro' D and F draw BD, LM parallel to AH.
Then (Prop. III.) arch AD = arch GM. Whence
GL AB, and by reafon of the parallels AH,
BD, LF; GF = DH.
Cor. The line GD line FH; or the line within
the circle is the line without the ciffoid.
:
ΙΟ
PROP.
10
B. I.
THE CIS SO I D.
.
Fig.
PROP.
VII.
8.
8.
Thro' any point F, draw NF perp. to AG, and
GS parallel to it, alfo draw ANS. Then GN: GF::
GL: LF :: AG: GS.
For it is fhewn in Cor. 2. Prop. IV. that NGF
is a right angled triangle, and GL perp to NF.
Therefore GN: GF :: GL: LF: (by fimilar
triangles) AZ ZI:: AG: GS.
Cor. AN: NG:: NG: GF:: GL: LF.
Still by fimilar triangles.
PROP.
VIII.
Thro' any point F, draw NF perp. to AG, and
GS parallel to it, alfo draw ANS. Then GA: GL::
GS2 : GF².
For (Prop. IV. Cor. 3.) GS: GF :: GF: LF;
and GS²: GF :: GS: LF :: (fimilar triangles)
AG: GL.
PROP. IX.
All ciffoids are fimilar figures.
This is plain, becaufe the abfciffe and ordi-
nates of feveral ciffoids, will be in the fame ratio:
when either of them is in a given ratio, to the di-
ameter of its generating circle.
SECT.
F
N
A
h
T
C
H
A
3.
H
·K ·
Ν
B
Fig.1.
G
P
$
2.
C
T
4.
b
h
k.
t
F
a
f
P
N
A
6.
E
E
N
5.
B
C
G
P
7.
8.
A
Z
M
A
H
B
I
M
B
D
C
M
G
Curves.
10.
G
H
D
A
9.
E
F
H
Pl.I. pa.10.
NIL
OF
[ 1 ]
SECT. III.
Of the CYCLOID.
I'
DEFINITION S.
DE F. I.
F a circle DEBL be made to roll along the right Fig.
line AF, until the fixed point D, which at firſt 11.
touched the line at A, comes to touch it again at
F; that point D will defcribe the curve ADGF,
called a Cycloid or Trochoid.
DE F. II.
The circle DEBL is called the Generating Circle.
DE F. III.
The line DB perpendicular to the middle of AF,
is the Axis of the cycloid.
PROP. I.
:
The bafe of the cycloid AF is equal to the circum- IF,
ference DEBL of the generating circle.
For all the particles of the circle are fucceffively
applied to the right line AF; and therefore the
fum of all, or the whole circle is equal to that line.
PROP. II.
Draw PM parallel to the bafe BF, then the line 12.
PM is equal to the arch DP
For let the circle DPE come to the place LME;
then by the circular motion about C, the defcribing
point
12
B. I.
THE
CYCLO I D.
Fig. point D will have moved thro' the arch DP. And
12. by the progreffive motion along BF, the point H
is come to E, fo that BE arch BH, which is
equal the arch DP or LM. But EM is parallel
to BP, becaufe the arch LM DP. Whence
BPME is a parallelogram. And PM BE =
arch BH arch DP.
13.
14.
-
Cor. The ordinate of the cycloid is equal to the
fum of the arch and its fine. AM DP+ AP.
PROP. III.
If AM be an ordinate cutting the circle in P, MT
a tangent at M. MT is parallel to the cord PD.
For whilft the cord BP or EM is defcribing the
infinitely ſmall part of the curve at M, the point
E may be confidered as fixed for a moment, and
therefore the ſmall arch at M will be perp. to EM.
But by the demonſtration of the laſt Prop. BPME
is a parallelogram, and EM parallel to BP. But
the angle DPB in a femicircle is alfo right; there-
fore DP is parallel to the particle of the curve at
M, or to the tangent MT.
Cor. 1. If ME be parallel to PB, then ME is
perp. to the curve in M.
Cor. 2. If MT be a tangent, and the cord DP be
drawn, then AP: AD:: AM: AT.
For DP, TM being parallel, the triangles ADP,
ATM are fimilar.
PRO P. IV,
If the ordinate GDM be drawn; the length of the
arch of the cycloid AM is double the correspondent
cord AD, in the generating circle.
Draw the ordinate Enp infinitely near GDM,
draw the cord Aon, and Dr perp. to it.
Then
the
Sect. III.
13
THE CYCLOID.
the triangle oDn is ifoceles; and An biffects the Fig.
angle DnE. For the angle AnD AEn EnA 14.
=noD, by reafon of the parallels; that is, onD
-
2
noD; and nDoD, whence nr no. But
(Prop. III.) the tangent pT is parallel to the cord
An; therefore Monp is a parallelogram, and Mp
= no 2nr; that is, the difference of the arches
Ap, AM, is twice the difference of the cords.
An, AD. And as this always happens till they
vaniſh at A; therefore the whole arch Ap is twice
the cord An, and the arch AM twice the cord
AD.
Cor. 1. The length of the femicycloid AMF is dou-
ble the diameter AB, of the generating circle.
Cor. 2. The length of the whole cycloidal curve is
four times the diameter of the generating circle.
Cor. 3. If GM is an ordinate, then AG, arch
AM, and 4AB, are in continual proportion.
For AM
2AD, and AM' = 4AD² = 4AG
× AB, and AG : AM :: AM : 4AB.
Cor. 4. AG, arch AM, and 2AF are in continu-
al proportion.
For (Cor. 2.) 4AB 2AF.
PROP. V.
If GM be an ordinate, AG the abfcifja; then AB: 15.
AG : : AF² : AM².
For (Prop. IV. and Cor. 1.) AF 2AB, and
AM 2AD; therefore AF4AB, and AM²
=
= 4AD. And AF': AM': 4AB' : 4AD' ; ;
AB²: AD: AB: AG.
Cor. The abfciffa AG is directly as the fquare of
the arch AM².
U
For
14
B. I.
THE CYCLO I D.
Fig.
15. × AG :
Therefore AM
16.
For AF²: AM² ::AB: AG :: 4AB² : 4AB
: (IV. Cor. 1.) AF² : 4AB × AG.
4AB X AG. And therefore
This alfo might be inferred from
AM' is as AG.
Cor. 3. Prop. IV.
PROP.
VI.
If ME be perp. to the bafe BF; then AB × ME
AB X MF
I
4
MF2.
For (IV. Cor. 3.) 4AB × AG = AM².
AG AB- BG, and AM AF
=
But
MF; there-
2
fore 4AB X AB BG AF FM = (IV.
Cor. 1.) 2AB — FM
FM; that is, 4AB' - 4AB X
BG4AB2 4AB X FM + FM. And 4AB
X FM 4AB × BG = FM².
X
FM². Whence AB ×
FM = AB X BG + FM²; or ABX BG = AB
× FM
MF
I
4
4
=
FM2; that is, AB X ME AB X
MF².
PROP. VII.
6.
If ME be perp. to the baſe BF, MG parallel to it;
then EF arch BD
For GM or BE
EF BF BE
the fine GD.
(II. Cor.) AD + DG, and
(Pr. I.) ADB
BEADB
AD
DG = DB
DG.
17.
PROP. VIII.
If GDM be an ordinate, AN parallel to it, and
NM parallel to AB; the external cycloidal Space
ANM is equal to the circular fegment AGD.
For draw the tangent MT and cord AD, and
ump parallel to NM. Then (Prop. III. Cor. 2.)
AG:
Sect. III.
15
THE CYCLOID.
AG: GD :: TG: GM:: (fimilar triangles) pm: Fig.
Mm; therefore AG X Mm GD X pm; that is, 17.
the area NMpnarea GgdD. But thefe are the
correſponding elementa of the cycloid and circle;
therefore by the method of indivifibles, the ſum
of all the NMpn, or the area ANM fum of all
the GgdD or the area AGD, of the circle.
Cor. 1. The area of the internal cycloidal space is
equal to the rectangle AG x GM- the circular area
AGD.
×
For area AGM AGX GM- ANMAG
=
X GM
area AGD.
Cor. 2. The area of the whole cycloidal Space
AMFB is equal to thrice the area of the femicircle
ADB.
For (Cor. 1.) area AMFB = rectangle AB × BF
— area ADB = (Prop.I.) AB X femicircumference
ADB area ADB, = 4 areas ADB — area ADB
3 times the area ADB.
Cor. 3. The whole cycloidal Space AMFB is the
circumfcribing parallelogram.
For the external ſpace is equal to the femicircle
ADB = AB X arch ADB ABX BF.
÷ =
Cor. 4. The whole cycloidal ſpace without the circle,
ADBFM is double the femicircle ADB.
U 2
SECT.
[ 16 ]
SECT:
IV.
Of the QUADRA TRIX.
Fig. A
18.
18.
DEFINITION S.
DE F. I.
CBL is the quadrant of a circle, and ſuppoſe
the radius CL to move uniformly round the
center C, from A to B; and at the fame time, the
line FG to move uniformly along AC, from A to
C, continuing always parallel to CB; and fo that
both motions begin and end together, at A and
B. Then the interfection N of the lines CL., FG,
during the motion, will defcribe the curve AND,
called the Quadratrix, or Dinoftratus's Quadratrix.
DE F. II.
The circle ALB is called the Generating Circle.
DE F. III.
The radius AC is called the Axis, and CD the Bafe.
PROP. I.
Thro' any point of the curve N, draw the radius
CNL, and NF parallel to CB. Then AF: AC ::
angle ACN : angle ACD.
For by the generation, AF: AC: arch AL:
arch AB:: angle ACL : angle ACB.
Cor. 1. AF: FC:: angle ACN: angle NCD,
Cor. 2. The abfciffe AF, are as the angles at the
center ACN, or as the arches which meafure them.
PROP.
Sect. IV. THE QUADRATRIX.
17
PROP.
II.
Fig.
The bafe is a third proportional to the quadrant and 19.
the axis. ALB: AC:: AC: CD.
:
Thro' N infinitely near D, draw CL; and draw
NF parallel to CD; and NI, LS perp. to it. Then
(Prop. I.) AC : AF:: angle ACB: angle ACL : :
arch ALB arch AL. And by divifion, AC :
CF or NI:: arch ALB : arch LB. And alternate-
ly, ALB: AC or CB:: LB: NI. But when L
coincides with B, S alfo coincides with B; and N
and I, with D. Therefore ALB CB :: LB or
LS: NI:: (fimilar triangles) CS or CB : CI or CD,
That is, ALB : CB or CA:: CB or CA; CD.
PROP. III.
::
With the radius equal to the bafe CD defcribe the 20
quadrant DNH. Then will this quadrant DNH be
equal to the axis AC,
Deſcribe the quadrant ALB; then (Prop. II.)
ALB: AÇ :: AC:CD :: (fimilar ſectors) ALB
DNH; therefore AC
DNH,
IV.
PROP.
Thro any point N of the curve, draw the radius 19.
CL; and FN parallel to the bafe; then CD: CA::
AF; arch AL.
For (Prop. II.) CD: CA: CA: ALB::
(Prop. I. Cor. 2.) AF: arch AL.
U 3
PROP
18
B. I.
THE QUADRATRIX.
Fig.
21.
£2.
PRO P. V.
Upon the bafe CD, defcribe the quadrant DPH.
Thro' any point N of the curve draw NF perp. to AC,
and CN cutting the infcribed quadrant in P. Then
AF arch HP.
Draw the quadrant ALB; then (Prop. IV.) AF:
arch AL:: CD: CA:: (fimilar fectors) arch HP:
arch AL. Therefore AF arch HP.
Cor. FC arch PD.
=
For (Prop. III.) AC — HPD, and AF = HP;
therefore FC PD.
PROP. VI.
If the radius CA be continued to Q, ſo that AQ
= AC; and the femicircle BAK be defcribed. And
make AQ: AF: AK: AL. And draw CL, and
FN parallel to CK, cutting CL in N; then will N
be in the curve of the quadratrix continued without
the circle.
For the part AN is conftructed by the fame law
as the part AD within the circle.
Cor. 1. The quadratrix DANn is continued ad in-
finitum, by finding a fufficient number of points N, n,
&c. by the fame rule.
Cor. 2. The line QX, erected perpendicular to the
line QAC, is an affymptote to the curve.
For fince AF: AL : : AQ : AK, when F is
infinitely near Q, L will be infinitely near K, and
the interfection N will be infinitely diftant. Whence
the curve AN, and the line QX, continually ap-
proach to one another; and yet never meet, but
at an infinite diſtance.
SECT.
A
Fig.u.
D
E
L
T
13.
A
A
C
B
D
12.
G
M
B
F
B
E
F
M
E
C
B
E
F
A
15
B
T
M
14.
F
16.
T
17.
C
M
B
E
F
A
18-
19.
F
G
F
ID SB
L
F
H
21.
D B
X
n
D B
}
K
B
20.
H
N
D
B
22.
N
m
:
F
D B
FIII.pa.18.
UNIV
OF
MICH
[19]
SECT.
V.
Of the LOGARITHMIC CURVE.
DEFINITION S.
DE F. I.
F in the right line AE, the parts AB, AC, AD, Fig.
AE, &c. be taken in arithmetic progreffion, 23.
from the fixed point A; and at the points A, B,
C, D, E, &c. the ordinates AF, BG, CH, DI,
EK, &c. be drawn parallel to one another, and in
geometrical progreffion. And if theſe ordinates
are fuppofed infinite in number and infinitely near
to one another; then the curve line FGHIK drawn
thro' the extremities of all the ordinates, is called
the Logarithmic Curve.
DE F. II.
The line AE is called the Axis of the curve.
PROP. I.
The axis AE is an asymptote to the curve.
Let the curve be continued backwards from A,
and the ordinates AF, bg, ch, &c. continually de-
creafing, and let Aen x AB. Then fince AF,
bg, ch, &c. are continual proportionals; therefore it
will be as AF: ek :: AF": bg". Now as n increaſes,
the ratio of AF" to bg", and that of AF to ek, will
continually increaſe; and fince AF is given, ek
continually decreaſes. And when n is infinite, AF
will be infinitely greater than bg", and confequent-
ly ek infinitely leſs than AF; therefore Ae is an af-
fymptote.
PROP
U 4
247
20
THE LOG. CURVE. B. I.
Fig.
PROP. II.
25. If TE is a tangent to the curve at E, BE an or-
dinate: then the fubtangent TB is every where the
fame, wherever the point E is taken.
26.
Suppoſe the ordinates BE, CF, DG equidif
tant and infinitely near each other; draw the tan-
gent FV, and draw En, Fr, parallel to AD.
Then fince BC = CD, it will be BE: CF :: CF:
DG (by divifion) CF- BE: DGCF ::
nF: rG. And alternately BE: nF :: CF: rG.
But the triangles FnE and EBT are fimilar, as
alfo the triangles GrF, and FCV. Therefore BT:
nE: BE: nF:: CF: rG:: CV: rF, but nE=
rF; therefore BT CV, and fo it will be, every
where.
PROP. III.
If AF, BG, DI, EK, be four ordinates, and AB
the distance of two of them be equal to DE, the dif
tance of the other two; and thro' the points G, F,
and K, I, the lines GFT, KIV be drawn, cutting
the axis in T and V. Then will AT DV.
Draw FS, IL parallel to AE. Then fince AB
-DE, therefore AD BE, and AF, DI, BG,
EK are four terms in the fcale of continual pro-
portionals, of which the first two, and the laft two
are equidiftant; and therefore it will be AF: DI::
BG: EK. And AF: BG :: DI: EK; and AF:
BG AF :: DI: EK - DI, or AF: GS ::
DI: LK. But the triangles TAF, FSG, are fi-
milar, as alfo the triangles VDI, ILK. Therefore
TA: FS: AF: SG:: DI: LK:: VD: IL, but
FS=IL, therefore TA = VD.
PROP.
Sect. V, THE LOG. CURVE.
21
PROP. IV.
Fig.
The Space BELG, comprehended between any two 27.
ordinates EL, BG is equal to TE × SL, the rectan-
gle of the fubtangent and the difference of the ordinates.
For let DI be infinitely near EL, and draw Ir
parallel to AE. Then LT being a tangent, the
triangles LTE and LIr are fimilar, whence LE :
ET :: Lr: rI; therefore LE × rI = ET × Lr.
But LEX rl area DELI, therefore ET x Lr
area DELI. But (Prop. II.) ET is a given
quantity, whence the fum of all the DELI or the
area BÉLGET x fum of all the Lr, or ET
× LS.
Cor. 1. The area of the whole curve infinitely pro-
duced towards A, is equal to the rectangle of the ſub-
tangent and ordinate, LE × ET.
For then LS becomes equal to LE, by Prop. I.
Cor. 2. The whole space infinite towards A, is
double the triangle TEL.
Cor. 3. The logarithetic spaces infinitely long, are
as the bounding ordinates.
Cor. 4. The space between any two ordinates, is to
the space between any other two; as the difference of
the first two; to the difference of the last two.
PROP. V.
The folid made by the infinitely long Space FLEA 27.
revolving about the axis AE, is equal to half a cylin-
der of the fame bafe, and hight equal to the fubtangent;
TE
× circle EL.
2
For let EL, DI, be two ordinates infinitely near,
draw Ir parallel to AE. Then the line LT being
a tan-
22
THE LOG. CURVE. B. I.
Fig. a tangent, the triangles TEL, IrL are fimilar;
27. therefore LE : ET : : Lr : rI, and LE×rI = Lr
× ET, and LE² × rI = LE × Lr X ET. Put
LE=y, ET=t, c=3.1416; then rI X yy =
×
Lrx ty, and rI X cyy Lr X cty; but rIx cyy
folid LIDE; whence Lr x cty folid LIDĚ.
Therefore if LE be divided into an infinite number
of equal parts equal to Lr = 1, then the fum of
all the cty
fum of all the folids LIDE. But
(Arithm. Inf. Prop. I1.) the fum of all the y is
yy
and the fum of all the cty =
etyy
2
- fum of all
22
the folids LIDE the whole folid LEAF infi-
nitely extended towards A. But cyy area of
the circle whofe radius is EL or y; therefore the
whole folid LEAF = X area of the circle
whofe radius is LE.
Cor. 1. The infinitely long folid is to a cone of the
fame bafe, and hight the fubtangent; as 3 to 2.
Cor. 2. The folid LEBG, contained between two
ordinates, is half the cylinder, whofe hight is the fub-
tangent, and baſe the difference of the circles whoſe
radii are LE, GB.
2
For the folid LEAF = tx circle LE, and
GBAFt X circle GB; therefore the difference
LEBGtX: circle LE
PROP.
circle GB.
VI.
23.
If
Iƒ AF = 1, BG = a, AE = x× AB, EK
y ; then the equation of the curve is y = a*.
For let AB, AC, AD, AE, &c. be arithmetic
proportionals, then AF, BG, CH, &c. are in
geometrical progreffion. And we ſhall have AF* :
BG* :: AF : EK; that is, 1: BG* :: 1 : EK, or 1 :
a*:; 1:y, and y = a*.
PROP.
Sect. V. THE LOG. CURVE.
23
PROP. VII.
Fig.
If the ordinate AF, BG, CH, &c. reprefent num- 23.
bers in geometrical progreffion, and be placed at equal
diftances; the abfciffas AB, AC, AD, &c. being in
arithmetic progreffion, will reprefent the logarithms of
theſe numbers.
That the abfciffas are in arithmetical progreffion,
when the ordinates are in geometrial progreffion,
appears from Def. 1. And from the nature of lo-
garithms, if they be in arithmetical progreffion,
the numbers anſwering will be in geometrical pro-
greffion. Therefore the abfciffas in this curve are
the logarithms of the ordinates belonging thereto.
Cor. Hence logarithms may be of as many different
forms as we pleafe. But as a multiplicity of ways
would be ufelefs, Mathematicians confine themselves to
two forts only. 1. When AF, BG, CH, DI, &c.
are 1, 10, 100, &c. and AB, AC, AD, &c. are
1, 2, 3, &c, and thefe are called Brigs, or the com-
mon logarithms, and the fubtangent is, 4342944 8
&c. 2. Napier's or the hyperbolic logarithms, where
the fubtangent is 1.
PROP.
VIII.
28.
If FG, QS be two logarithmic curves, AF
PQ = 1, BG = HS = any other ordinate; GT, 29.
SV, tangents; then will AB: PH : : TB : VH.
Let the equal ordinates LI, NK, be placed in-
finitely near BG, HS; then fince they are equal,
they have a like poſition upon their abfciffas; that
is, LB: NH:: AB: PH. Draw In, Kr, paral-
lel to AB, PH. Then the triangles Gln, GTB,
ás alſo SKr, SVH, are fimilar; whence BT: nl : :
BG:
24
B. Ia
THE LOG. CURVE.
Fig. BG: nG: HS: S: HV: rK; and alternately,
28. BT: HV:: nI or LB: rK or NH:: AB: PH.
29.
28..
23.
Cor. 1. The logarithms of equal numbers are pro-
portional to the fubtangents of the curves they belong to.
For AB, PH are the logarithms of the equal
numbers BG, HS.
Cor. 2. In a number extremely near 1; its exceſs
above 1, is to its logarithm; as 1, to the fubtangent.
For let LI be 1, and BG infinitely near it;
then by fimilar triangles, Gn: nl or LB its log. : :
IL or I: LT the fubtangent,
Cor. 3. The distances of two equal ordinates in
two different curves, are as the fubtangents of the
curves.
PROP. IX.
EK
If DI, EK be two ordinates; then DE = log.: DI
For AE = log. : EK, and AD = log. : DI,
AD = log. : EK-log.: DI= (by
whence AE
EK
the nature of logarithms) log.: DI;
EK
= log. :DI, for that curve.
Cor. If BG, CH, and DI,
that is, DE
EK, be any ordi-
CH
EK
nates; then BC: DE:: log.:
:
BG
log・・ DI' by
any table of logarithms.
CH
For in this fig. BC= log. :
BG'
and DE = log. :
EK
CH
DI; and (VIII. Cor. 1.) BC: DE:: log. BG:
:
EK
log: DI'
by any table.
SECT.
[ 25 ]
SECT.
VI.
Of the SPIRAL of Archimedes.
IF
DEFINITION S.
DE F. I.
the radius CA be moved uniformly round the Fig.
center C, defcribing the circle ABDA; and at 30.
the fame time a point is fuppofed to move uniform-
ly along that radius from C to A, ſo that both mo-
tions begin and end together; the curve CFGA
defcribed by that point, is called the Spiral of Ar-
chimedes.
And if the radius (produced) move round a fe-
cond time, while the defcribing point continues its
motion, the curve defcribed, is the Second Spiral;
and ſo on.
DE F. II.
The circle ABDA is called the Generating Circle.
PROP. I.
If F be any point in the curve, then will CF: 30.
radius CA: : arch AB: circumference ABDA.
For fince the motion of the radius, and the mov-
ing point are both uniform, and the radius moves
over AB, while the point moves over CF; alſo
the radius moves over ABDA, whilft the point
moves over CA. Therefore by the laws of uni-
form motion, CF : CA :: arch AB: arch ABDA.
Cor. 1. If the angles about C be taken in arith-
metic progreffion, the distances of the moving point
from ℃ will alſo be in arithmetic progreffion.
Cor.
26
ARCHI M. SPIRAL: B. I.
Fig. Cor. 2. The angles about the center, are as the
30. diftances of the moving point from the center.
Cor. 3. The arches of the circle ABD, are as the
distances of the moving point from the center.
Cor. 4. If radius CA =r, circumference ABDA
c, any arch ABD = z, and its ordinate CP=
y; then the equation of the curve is, y =
rz
31.
PRO P. II.
To any point P of the curve, draw CPD from the
center, and CT perp. to it; and let the radius CA:
arch ABD CP: CT. Draw PT, and it will
touch the curve in P.
=
: :
Put c circumference ABDA, r = radius CA,
z=arch ABD, y = CP. Draw Cm infinitely
near CD, cutting the fpiral in n, and draw Pr perp.
to Cm. Then (Prop. 1. Cor. 3.) arch ABD: CP::
arch ABDm: Cn. And by divifion, arch ABD:
CP :: arch Dm Cn CP or nr; that is, z:y:
:
y
Dm: nr= × Dm; and by ſimilar ſectors, CD
Z
(r) : CP (y) : : Dm : Prx Dm. But if PT
be a tangent, the triangles nPr and PTC are fimi-
lar; therefore nr (~/× Dm): rP (× Dm)
PC (y): CT =
zy
Whencer: z:: y: CT.
Z:
:
Cor. If CT be perp. to CP, and the arch POL
be defcribed with the radius CP. And if CT be
made equal to the arch POL; then TP will be a tan-
gent at P.
For
i
Sect. VI.
27
ARCHI M. SPIRA L:
For by ſimilar ſectors, CD (r): CP (y) :: arch Fig.
J2
ABD (z): arch POL =
CT by this Prop.
31.
p
III.
PROP.
Let F be any point in the curve; with radius CF, 32;
defcribe the arch FG; then AC: CF:: circumfe-
rence ABD : arch FG.
For let CA=r, circumference
c, arch AB
z, arch GF = v, CF = y. Then (Prop. I.)
circumference (c): radius (r) :: arch AB (z): CF
(y); and by fimilar fectors, CF (y) : FG (v) : :
CA (r) : arch AB (z). And multiplying, cy: rv: :
rz: zy :: r :y.
r:y. And taking the product of the
extremes and means, cyyrrv, and rr: yy:: c : v.
Cor. Hence another equation of the curve is
yy =
rrv
C
PROP IV.
H
I
3
The Spiral Space CPNC is equal to the Sector 33.
CNQL, whofe radius is CN, and arch NQL.
Draw CIS infinitely near CPQ, and Pr perp. to
CS. Then CQ² or CN² : CP²: : fector CQS :
CP2
× CQS.
ſector of the ſpiral CPI = CNX CQS.
There-
fore the fum of all the fectors CPI = fum of all
CQS
•
the CNX CP¹. But if the arch NQL be divid-
ed into an infinite number of equal parts, equal to
SQ, and lines drawn from the center; all the fec-
tors formed thereby will be equal. And all the
CP's will be in arithmetic progreffion. Therefore
(Arith. infin. Prop. III.) the fum of all the CP¹ =
CN'
28
B. I.
ARCHI M. SPIRAL.
Fig. CNX by the number of them. And the fum of
3
33.
all the
34.
CQS
CN2
× CP² = CQS x the number of
उ
them. Whence the fum of all the CPI or the fpi-
ral ſpace CPNCCQS x number of fectors
CQS = the whole fector CLQN.
I
3
I
3
Cor. The whole ſpiral Space CPNA ≈ ¦ the ge-
nerating circle ABDA.
For then CN becomes equal to CA.
PROP. V.
If the area of the first circle be A, the first, fecond,
third, fourth, &c. Spiral ſpace, will be —A,— A,
12A, 37 A, &c.
19
3
3
3
3
It muſt be confidered that in defcribing the fe-
cond ſpiral CFGALMN, the firſt ſpiral ſpace is
twice deſcribed, becauſe there are two revolutions.
Alfo in defcribing the third fpiral, the firſt ſpiral
ſpace is thrice defcribed, and the fecond twice.
Likewife in defcribing the fourth fpiral, the firſt
fpiral ſpace is defcribed four times; the fecond
thrice, the third twice; and fo on. Likewife
in deſcribing the ſecond, third, fourth, ſpiral; the
correſponding circle is twice, thrice, or four times
deſcribed, reſpectively; by reaſon there are fo ma-
ny revolutions.
There-
Sect. VI.
29
ARCHIM. SPIRA L.
Therefore let them be denoted as follows.
orders fimple/circles' ratio of
fpiral areas the circles
Iſpaces
C = 9A
I
a
A |A
A
2
b
B
B
4A
3
C.
C
4
d
D
5
ё
E
254 |
&c.
D= 16A
E=25A
Then by Prop. IV. and what is now mentioned,
we ſhall have
3
a
a+b=
a + b + c
I
3
I
A
3 X 2B.
။
× 3C.
a + b + c + d
× 4D.
a + b + c + d + e =
× 5E,
&c.
Then fubtracting each term from the following
one, we ſhall have, the fimple fpiral ſpaces, viz.
1
a — A
b
C
11
3 2
3
3.C
2
3
ا 1000
1.1.
I
I
B
A =
A
A =
A.
3
3
3
27
B =
A.
A
19 A.
3
3
3
3
D
3
64
27
D-
C
A-
A
37 A.
3
3
3
3
3
e = SE
3
125
4D =
3
64
61
A
A
A.
3
3
d =
11
w.
3
Cor. 1. If a, b, c, d, &c. be the first, second,
third, fourth, &c. ſpiral ſpace, and A, B, C, D, &c.
the correspondent circles. Then will a, b, c, d, &c.
be reſpectively equal to 1A, 1B, 1C, 27, D, &c.
3
X
12 27
37
Cor.
Fig.
34.
30
SPIRAL. B. I.`
ARCHI M.
Fig.
Cor. 2. The area of the nth ſpiral, will be — A × :
34. 3n Xn-I+I:
Take the feveral differences as follows,
उ
I
I
A
I dif.
3
2 dif. 3 dif
7.
2 A
2
A
2 A
3
3
19 A
4. A
2 A
3
6 A
2 A
4
37 A
3
6r
5
AA
8 A
3
Then by Prop. II. of the differential method, the
nth ſpiral ſpace is = A +
n
I
n
× 2A +
I
I
1
+6×
I
× 2A = ¦ A: X1+6×.
X
= ÷ A × : 1 + 6 × 7=-=-=¹×
1
3
n
2
X
2
n
2
X
I
2
2
n
I+
AX:1+6x
2
I
I }
انن
=AX:1 + 3n x n − 1 =
Cor. 3. If P be the nth circle.
ral ſpace is equal to
I
X
I 2 N2
X
2
A + nA x n − 1.
nA×n — I
Then the nth fpi-
-I+ip.
n x n − 1 +
nn
For nnAP, and A=
P
nn
Cor. 4. The fpiral Spaces contained between every
two fucceeding circles, are 2A, 4A, 6A, 8A, &c.
For theſe are the firft differences of the whole
fpaces.
SECT.
K
24
K
Fig 23
I
h
F
A
H
F
26
G
E
An
L
2,5
F
F
S
K
A
B
C D
E
T
AV
D B
E
A
TV
BCD
27
28
A A
F
A
T B
DE
A T
LB
30
D
P
G
B
33
B
12
m
D
31
n
P
ว
C
L
A
O
B
K
34
M
D
D
B
L
H
P
K
29
B
G
C
A
D
C
32
NH
Pl. III.pȧ.30.
UNIV
OF
MICH
Sect. VII.
[ 31 ]
Of the
SECT.
the LOGARITHMIC SPIRAL.
VII.
DEFINITION S.
DE F. I.
IFC, te that it may every where cut the
F the curve FGDA be drawn about the center Fig.
C, fo that it may every where cut the radii CF, 35.
CD, CG, CA, &c. in one and the fame angle;
that curve is called the Logarithmic or Proportional
Spiral.
DE F. II.
The lines CF, CD, CG, &c. are called Ordinates.
PROP. I.
If any number of ordinates CF, CK, CD, CA, 36.
&c. be drawn, making equal angles at the center C.
Then thefe ordinates fhall be in continual proportion.
Suppoſe an infinite number of ordinate CF, CG,
CH, CI, &c. to be drawn, making equal angles
at the center. Then the triangles CFG, CGH,
CHI, CIK, &c. are fimilar; for the angles at G,
H, I, K, are equal (by def. I.), and the angles at
the center are all equal by conſtruction. Whence
CF : CG:: CG: CH:: CH: CI:: CI;
CK, &c. continued as far as CA. But fince the
angles FCK, KCD, DCA are equal; as many
continual proportionals may be interpofed between
KC and DC, and alſo between DC and CA, as
there are between FC and KC. Therefore CF, CK,
CD, CA are equidiftant terms. But in a ſeries of
terms in continual proportion; the equidiftant terms
are alfo in continual proportion (Propor. Prop. 21.)
Therefore CF: CK:: CK: CD:: CD: CA.
X. 2
Cor.
32
B. I.
THE LOG. SPIRAL
Fig.
Cor. 1. If the angles at the center be in arithme-
36. tic progreffion, the ordinates will be in geometrical
progreffion.
Cor. 2. The Spiral makes an infinite number of re-
volutions, before it comes to the center C.
For theſe proportional ordinates are continued ad
infinitum downwards.
Cor. 3. Let any radius CA cut the ſpiral in its
Jeveral revolutions; then the distances of the feveral
points of interfection from the center, will be in geo-
metrical progreffion, and that in the ratio of CA to CB.
This appears by Cor. 1. for each revolution
makes four right angles.
PROP. II.
37. If CD, CE, CF, CA, &c. be in geometrical progref
fion, and CD = 1; then will the angles DCE, DCF,
DCA, &c. be the logarithms of CE, CF, CA, &c.
38.
For fince CD, CE, CF, CA are in geometrical
progreffion; the angles DCE, ECF, FCA, are
equal, and the angles DCE, DCF, DCA, in arith-
metic progreffion. But it is the property of loga-
rithms to be in arithmetic progreffion, when the
numbers anſwering are in geometrical progreffion.
Therefore theſe angles are the logarithms of the or-
dinates.
Cor. Let CF, CG, CA, be
angle FCG angle FCA:: log.
For let CD = 1.
:
any ordinates; then
CG
:
CA
CF: log. CF
Then angle DCA = log. CA,
and angle DCG = log. CG,
alfo angle DCF = log. CF,
and
Sect. VII.
33
THE LOG. SPIRAL.
log.: CG-log. : CF,
and by fubtraction,
< DCG - DCF = log. : CG
CG
log. CF
Alfo
=
that is, angle FCG = log. :
DCA-DCF log.: CA — log.: CF,
that is, angle FCA = log. :
CA
CF
Therefore in any fyftem of logarithms it will be,
CG
:
CA
angle FCG: angle FCA:: log.: CF: log. CF
PROP. III.
Fig.
38.
If CM is any ordinate, and CT perp. to it; and 39.
if the angle CMT be made equal to the angle of the
Spiral. Then MT is a tangent àt M.
For the tangent MT, and the infinitely fmall
part of the curve at M, coincide.
Cor. TM is to CM every where in the fame given
ratio.
PROP. IV.
If MT be a tangent at M, CT perp. to the or- 39.
dinate CM; then will the length of the whole spiral
MDGC be equal to the tangent TM.
Take the point n infinitely near M, and draw
nr perp. to CM. Then in the right angled trian-
gle nMr, all the angles are given, therefore Mr is
to Mn in a given ratio, fuppofe as r to s, then Mn
11
S
S
× Mr. And all the Mn = X all the Mr,
r
S
that is, the length of the curve = × CM. But
the triangles nMr and TMC are fimilar, whence
X 3
TM ;
34
B. I.
THE LOG. SPIRAL.
:
Fig. TM: CM: nM: Mr::s:r; therefore TM =
39. S
LO.
41.
× CM = length of the whole curve.
PRO P. V.
Let mD be a tangent at m, and CD perp. to mD;
CM, Cm, ordinates. Then mD: mC::CM Cm:
Mm the length of the fpiral contained between the
ordinates.
Draw CT, Ct perp. to CM, Cm, produce mD
to t, and draw the tangent MT. Then from the
S
laſt Prop. TM = — CM, and tm
S
TM tm
Х
x CM
r
= Cm, and
r
Cm. But (Prop. IV.)
TM curve MGC, and tm curve MGC; there-
fore TM-tm the part of the curve Mm;
therefore Mm =
r
x CM-Cm, and CM -- Cm :
Mmrs: Cm : tm :: (fimilar triangles) D:
mC.
Cor. Hence the parts of the curve are as the dif-
ferences of the ordinates including them.
PROP. VI.
If CM be an ordinate, MT a tangent, CT perp.
to CM. Then the area of the ſpiral MDC, -de-
fcribed by a radius making an infinite number of revo-
lutions, is half the triangle TCM.
Draw the ordinate nC infinitely near MC, and
nr perp. to CM. Then fince all the angles of the
triangle nMr are given, Mr is to nr in a given
ratio, fuppofe as r to g; then nr =
11
r
× Mr. And
the
Sect. VII.
36
THE LOG. SPIRAL.
the area of the triangle MCʼn =
MC X nr
2
2
Fig.
X
2r 41.
an infinite number
Then the fum of
MC x Mr. Divide MC into
of equal parts equal to Mr.
all the MC, nC, &c. to C, is equal to half fo
many times the greateſt MC; that is, (as the num-
MC
ber of them is denoted by MC,) equal to X
2
MC2
MC or
And the fum of all the
q
XMCX.
2
2r
q
MC2
9
Mr = X
or x MC².
Therefore the
2
4r
27
ſum of all the triangles MCn, or the whole area of
9
the fpiral ſpace = × MC². But by the fimilar
4r
triangles Mar, MTC; MC: TC:: Mr: nr::
MC x q
:q, therefore TC xr = MC × q, and r
=TC, and
q × MC
4r
=TC. Whence the whole
area of the fpiral fpace MDC = TC x MC =
triangle TCM.
Cor. 1. The Spiral Space MDFGM, comprehended
between the part of the ordinate GM, and the curve
TC
4MC
MDFG, is equal to X MC²
For the fpiral-fpace MDGC =
- GC.
q
× MC², and
4r
the ſpace GC =
9
47
ence, viz. MDFG =
× GC², therefore the differ-
9
TC
× MC² — GC² :
=
4r
4MC
× MC² GC2.
Cor. 2. If MC, mC be two ordinates;
tao tangents; CT, ct, perp. to CM, Cm.
X 4
MT, mt,
Then the
40.
area
36
THE LOG. SPIRAL. B. I.
Fig. area MCm, comprehended between thefe ordinates, and
40. the Spiral; is equal to half the difference of the tri-
angles TCM and tCm; or equal to
CM²
Cm².
TC
4MC
This follows from the Prop. and Cor. 1.
SCHOLIU M.
X
The area mentioned in this Prop. is that made
by infinite revolutions of the radius; where every
fucceeding revolution covers a part of the former.
And fo the fame parts of the area are defcribed
over and over. And therefore the area mentioned
in Cor. 1. is the true area of the whole ſpiral; and
is that contained between the curve and the part
of the ordinate intercepted by the curve,
PRO P. VII.
42. In the Spiral CAB, if the ordinate CA = 1, and
the circle ADE be defcribed with the radius CA;
and any other ordinate CB be drawn. Then arch
AD =
log. of CB.
For fince AC1, therefore (Prop. II.) the
angle ACD is the logarithm of the ordinate CB.
But the angles at C, being proportional to the
arches of the circle; therefore the arch AD will
alfo be the logarithm of CB, according to fome
fpecies or other of logarithms.
Cor. 1. From the point A where it cuts the circle,
draw the tangent AT, and CT perp. to CA. Then
if the fubtangent CT be .43429448, then AD is
Brig's logarithm of the number CB.
For by the nature of logarithms (demonſtrated
Cor. 2. Prop. VIII. Sect. V.), the exceſs of a num-
ber (extremely near 1) above 1: is to its logarithm:;
I:
Sect. VII.
37
THE LOG. SPIRAL
I: fubtangent. Therefore draw the ordinate Crn Fig.
infinitely near CA. Then the triangles nAr, ATC, 42.
are fimilar; whence nr : Ar :: AC: CT. But nr
is the excefs of Cn above 1, and Ar is its loga-
rithm; and AC is 1, and CT the fubtangent
-43429448; and (Prop. VII. Cor. Sect. V.) this
is the fubtangent to Brig's logarithms. Therefore
Ar is Brig's logarithm of Cn, and confequently
AD is Brig's logarithm of CB.
Cor. 2. If the fubtangent CT be 1; then AD is
Napier's logarithm of CB. In other cafes AD will
belong to fome other fpecies of logarithms.
For (Cor. Prop. VII. Sect. V.) 1 is the fubtan-
gent of Napier's logarithms.
SECT.
[ 38 ]
B. II.
SECT.
VIII.
Of the HYPERBOLIC SPIRAL.
Fig.
43
44.
D
DEFINITION S.
DE F. I.
RAW any line CG; and from C as a center,
defcribe as many arches of circles as you
pleaſe from CG, as VE, GF, gf, &c. all equal
to one another. Then the curve paffing thro' all
the points which terminate theſe arches, as FEƒdC
is called the Hyperbolic or Reciprocal Spiral.
DE F. II.
Any right line drawn from the center to the
curve, as CF, is called an Ordinate.
DE F.
III.
The line CG is called the Axis.
PROP. I.
If CA be perp. to the axis GC, and equal to any
of the arches VE, GF, &c. and AB be parallel to
CG. Then AB is an affymptote to the curve.
=
Thro' F draw DB parallel to CA, then fince
DBCA GF, therefore BF GF DF =
the difference between the arch and the fine, but
when the radius CG is encreafed infinitely, that dif
ference FB is infinitely diminiſhed, and becomes leſs
than any given line. Therefore AB is an affymptote.
Cor. The distance of the affymptote from the axis,
is equal to any arch GF, between the axis and curve.
PROP.
Se&t. VIII. THE HYPER. SPIRAL
39
PROP.
II.
Fig.
Thro' any point F of the curve, draw the arch GF 45:
and the ordinate CF; and draw any other ordinate
CH, cutting the arch GF in L. Then will CF ×
arch GF CH X arch GL.
Thro' H defcribe the arch HI.
Then the fec-
tors CHI, CLG, are fimilar; whence CH or CI:
CG: arch IH or GF: arch GL.
Whence CHX
arch GL = CG X arch GF.
Cor. 1. CI × arch IH = CG × arch IN.
X
For CI: CG :: IN: GF or IH.
Cor. 2. If radius CG r, CA or arch GF-
a, ordinate CH y, arch GL = z.
equation of the curve is arzy.
PROP. III.
Then the
If HT be a tangent at any point H, and CT be 46.
drawn perp. to CH. Then the fubtangent CT is
equal to the arch IH.
With any radius CG defcribe the arch GFL,
cutting the ordinate CH in L, and draw Chl in-
finitely near CHL. Then (Prop. II.) CG x GF
= CHXGL = Cr x Gl=CH-rb x GL + L/
= (becauſe rh, Ll, are infinitely ſmall) CH × GL
+CH X L/GL X rb. Whence CH × LI =
GL X rb. But the ſectors CH, CL/ are fimilar,
whence CH: Hb:: CL: L, and CL × Hb =
CH × L/ = GL X rh. Alſo the triangles CHT
and brH are fimilar; therefore rh: bH :: HC :
CT, and CT × rb = HC × Hb; and CL × CT
× rb = CH x CL x Hb CH X GLX rb =
(Prop. II.) CLX GFX rb. Therefore CT
(Defin. 1.) IH.
GF
Cor.
40
B. I.
THE HYPER. SPIRAL.
Fig. Cor. 1. The fubtangent CT = CA the distance of
46. the affymptote, and therefore is always the fame.
45.
25.
47.
Cor. 2. The angle CHT continually increafes, in
approaching to the center.
For CT is given, and CH continually decreaſes.
Cor. Hence the curve makes an infinite number of
revolutions about the center.
For in the logarithmic fpiral, where the angle
at H is perpetually the fame, the curve makes an
infinite number of revolutions. And much more
does it fo in this, where the angle at H continually
increaſes, and makes it keep without the logarith-
mic fpiral.
PROP. IV.
If CF, CH, be any two ordinates, thefe ordinates
are reciprocally as the angles they make with the axis
CG. CF: CH :: <GCH : <GCF.
For (Prop. II.) CF × GF = CH × GL. There-
fore CF CH:: arch GL: arch GF: ≤GCL :
<GCF.
PROP. V
The length of the ſpiral between any two ordinates,
is equal to the length of the logarithmic curve between
two ordinates of equal lengths with the former; when
the fubtangents of both curves are the fame.
Let HDC be the ſpiral, CEK the logarithmic
curve; let the ordinate CH, Cn and DG, CF be
infinitely near, and CH DG, and Cn CF, in'
=
the two curves. With the radius Cn deſcribe the
arch nm; alfo draw Fr parallel to AD. Then
fince the fubtangent VD = CT, alfo DG =
CH. Therefore the right angled triangles CTH
and
Sect. VIII.
41
THE HYPER: SPIRAL.
and DVG are equal and fimilar, whence the angle Fig.
CHT = DGV. And fince CCF, therefore 47-
CH- Cn or mH DG — CF or rG, and fince
<H=<G, therefore the triangles mHn and
FGr are equal and fimilar; and Hn GF; that
is, the curve Hn contained between the ordinates
CH, Cn, is equal to the curve GF contained between
the ordinates DG, CF equal to the former; and fo
it will always be. Therefore the fum of all the
Hn, or the ſpiral contained between any two ordi-
nates, is equal to the fum of all the GF, or the log.
curve contained between two ordinates equal to the
former.
Cor. The fpiral making an infinite number of re-
volutions, is alſo infinite in length.
For the logarithmic curve, and its affymptote
is infinite in length; and confequently the ſpiral,
which is equal to it, is the fame.
PROP. VI.
The area of the whole fpiral FLDC, making in-
CF X CT
finite revolutions, is equal to
2
tangle of the ordinate aod fubtangent.
balf the rec-
Draw the ordinate Cr infinitely near CF, and rn
perp. to CF.
And by fimilar triangles Fn: nr::
CF CT the fubtangent. Whence CT x Fn =
CF nr; but the triangle FCr=
CFX nr
; there-
2
CTX Fn
fore the triangle FCr =
And the fum
2
48.
lutions
of all the triangles FCr defcribed in all the revo-
CT x all the Fn. That is, the whole
area of the fpiral ſpace =
CT × FC.
Cor.
42
B. I.
THE HYPER. SPIRAL.
Fig.
Cor. 1. The Spiral Space FLNDF contained be-
48. tween the curve, and part of the ordinate DF, is
equal to
CT × DF.
2
For the whole ſpiral ſpace FLNDC = ÷ CT ×
CF, and the ſpace DSC, which is covered by the
other, is CT × CD. And their difference
CTX CF CD CT × DF.
2
Cor. 2. Any part of the fpiral Space contained be-
tween two ordinates CF, CL, is equal to CT ×
CF
CL.
2
I
=
х
For CTX CF = ſpace FLNDC, and ½ CT
× CL = LNDC. And their difference CT
X CF-CL.
Cor. 3. The fpiral space contained between any two
ordinates, is half the space in the logarithmic curve
contained between two ordinates, equal to the former;
when the fubtangents of both curves are the fame.
For (Prop. IV. Sect. V.) the logarithmic ſpace
is equal to the rectangle of the fubtangent and dif-
ference of the ordinates, and in the ſpiral it is
equal to half that rectangle.
I might here have given more properties of all
theſe curves, and likewiſe have fhewn the nature
and properties of fome other forts of curves, as
the Catenary, the Epicycloid, &c. But as this cannot
eafily be done without having recourfe to the me-
thod of Fluxions; I forbear purfuing them any fur-
ther, as the making uſe of any other method, would
be carrying the reader too far about.
BOOK
Fig 35
D
36.
D
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E
F.
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D
C
H
37-
A
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F
A
t
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39.
n M
F
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до
D
38.
D
41.
M
42
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D
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كـ
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G
45
F
44
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47
m
D
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48
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In
40.
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43.
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A
PL.IV pa. 4-
OF
MIC
[ 43 ]
воок
II.
Of CURVE LINES in General; and
their Affections.
C
URVE lines may be conceived to be gene-
rated theſe three ways. First, they may be
made by the ſection of a folid and a plane. Se-
condly, they may be defined by an equation, ex-
preffing their nature, or fome principal property
thereof. Thirdly, they may be conceived to be
defcribed in plano by local motion.
1. Firſt, thoſe who confider curve lines as ge-
nerated in the ſurface of a ſolid by a plane cutting
it; find it proper and neceffary to confider the
рго-
perties of that folid; and to demonftrate from
thence what muſt be the nature of a figure arifing
from ſuch a ſection. Accordingly the ancient geo-
metricians confidered the conic fections, as made
by a plane cutting a cone.
2. In the fecond cafe, equations, which exprefs
the nature of curve lines, are not properly defini-
tions of theſe curves, but certain arithmetical com-
putations, grounded on fome property of thefe
curves, and which (duly managed) may lead us to
the knowledge of thefe properties. For theſe
equations, being analytical expreffions, will (like
the ſcience itself) lead us in a retrograde order from
the equations themſelves, to the properties on which
they are founded. For it is well known in analy-
tics, that by confidering the things fought, as if
they were known; at laft, by equations expreffing
their conditions and relations, we arrive at the real
knowledge of them. Juft thus in geometry, by
denoting
44
CURVE LINES. B. It
denoting the conditions of a curve by an equation
expreffing its nature, we at laſt arrive at fome pro-
perty of that curve, upon which that equation is
founded. Therefore it is plain, that equations are
not true definitions of a curve, but only artificial
expreffions, by the help whereof, and the given con-
ditions, we are able to trace the firft, moft fimple,
and defining property of fuch curve. An equation
then is nothing elſe but a defcription, by which a
figure is rendered more eaſy to the conception.
When the nature of a curve is expreffed by an
equation, we must find out by what motion of
points, interfection of lines, or by what defcription
or conſtruction, a curve is generated, which ſhall
have the conditions expreffed in that equation. An
equation then pre-fuppofes a curve deſcribed ac-
cording to fome law or rule, upon which that equa-
tion is founded.
3. Thirdly, The deſcription of a curve in plano,
is the moſt true and natural definition of a curve,
and the proper original thereof, and from which
all other properties may be determined. Now this
conftruction may be effected various ways, as by
the interfection of lines, the motion of points,
fimple or compound after feveral manners, which
will generate feveral forts of curves. Thus a cir-
cle is deſcribed by a line revolving round a fixed
center; an ellipfis is defcribed by a flexible line
moving round two fixed centers; a cycloid is de-
ſcribed by a point in the circumference of a circle,
whilſt it turns round along a right line; and fo of
others.
The ancients admitted only right lines, the circle,
and the conic ſections into geometry, excluding all
other curves as not geometrical. But modern geo-
meters admit all curves into geometry, that can any
way be expreffed by equations. Curves are beft
divided into kinds by the dimenfions of the equa-
tions
CURVE LINES.
45
tions expreffing their nature. By help of theſe
equations, we gain a better idea of fuch figures,
and more eafily comprehend the nature of them.
DEFINITION S.
DE F. I.
A Curve Line is that which is deſcribed by a Fig.
moving point which continually changes its direc- 49.
tion, as ABC; whilft a right line always goes on in
the fame direction, as AD.
DE F. II.
If BM be any curve line, and any right line AD
being drawn of an indefinite length, and having
its origin in A; and if any part AP be cut off, and
from P, the right line PM be drawn to the curve,
making a given angle with AP; and this be every
where done; then the line AP is called the Ab-
fciffa, and PM the Ordinate, of that curve.
DE F. III.
A line of any Order is denominated from the
higheſt power of the equuation, expreffing the na-
ture thereof; which powers are made of the ab-
fciffa, or the ordinate or both.
Thus if x abfciffa; y = ordinate; a, b, c,
given quantities; if ax = by + cc, this denotes a
line of the first order, or a right line. If yyay
bx+xx, this denotes a line of the fecond order,
or a curve of the first order. And fo does this
equation ay + xy bx. And this ayy
ayy bxy
bxy-ccx.
Again, y³ 3axy bxx, denotes a line of the
third order, or a curve of the fecond order.
fo on to higher orders, according to the dimen-
fions of x or y, or both of them in one term.
DE F. IV.
And
If x be the abfciffa, and y the ordinate of any
curve; then if the equation expreffing the nature
Y
of
50.
46
CURVE LINES.
B. II.
Fig. of the curve, be made up of x and y and given
50. quantities, and x and y denote ftreight lines, then
that curve is called a Geometrical or Algebraic Curve.
51.
52.
51.
DE F. V.
But if the nature of a curve be expreffed by an
equation, wherein one of the indetermined quan-
tities x or y, repreſents a curve line, then the curve
denoted by fuch an equation, is called a Tranſcen-
dent or Mechanical Curve.
DE F. VI.
Exponential Curves, are thofe where the equa-
tions expreffing the nature thereof confiſt of terms
whoſe indices are indetermined or variable quan-
tities made of x or y. Thus a*y, is an expo-
nential curve, as alfo y+axy.
DE F. VII.
Curves that turn round a fixed point or center,
are called Spirals or Radial Curves, as ADE. And
any line CE, drawn from the center C to the curve,
is called an Ordinate.
DE F. VIII.
A tangent to a curve, is a line drawn thro' a
point of the curve without entering into it; fo that
being produced, it leaves the curve on both fides.
DE F. IX.
The Subtangent is the line PT, or the diftance of
the ordinate from the interfection of the tangent
and abfciffa.
perp.
And in fpirals, it is the line CT, which is
to the ordinate CE, and meets the tangent at T.
DEF.
CURVELINES.
47
DE F. X.
The Subnormal is the line PQ, which is the dif
tance of the ordinate from that point of the ab-
fciffa, where a perpendicular to the curve (or tan-
gent) interfects it.
Fig.
52.
And in the ſpiral, it is the line CQ, perp. to 51.
CE, and cutting the line EQ (which is perp. to
the curve) in Q.
DE F. XI.
The Axis of a figure is a line paffing thro' the
center, which is fome remarkable point of the
curve; if it biffects the ordinates, it is called a
Diameter.
DE F. XII.
An Affymptote is a right line which continually
approaches the curve, but never touches it, till it
be at an infinite diftance.
SCHOLI U M.
x
All curves are more fimple or compounded, ac-
cording as their equations confift of fewer or more
terms. But the equations are the more comprehen-
five the more terms they contain; and become uni-
verfal when all the powers and products of the ab-
fciffa and ordinate, are expreffed therein. To give
fome inſtances, let abfciffa; y ordinate;
a, b, c, d, &c. known quantities, which may be
either affirmative or negative. Then a line of the
firft order will be denoted thus, y + ax + b = o.
A line of the ſecond order thus, y² + ax + b ×
y + cxx + dx eo. Lines of the third order
thus, y³ + ax + b × yy + cx² + dx + e × y +
ƒx³ + gx² + bx +ko. And fo on; and thefe
equations are univerfal, and take in all cafes of
thefe feveral orders. And in theſe ſeveral equa-
+
Y 2
tions,
48
B. II.
CURVE LINE S.
1
Fig. tions, any of the quantities a, b, c, &c. may
51. be wanting; and then the curve changes into a
more fimple one of the fame order.
When theſe coefficients are multiplied out at
length, the number of terms in the equations of
the ſeveral orders, will be in the firſt 3; in the fe-
cond 6; in the third 10; in the fourth 15, &c.
which are thus generated, the firſt 1+2 = 3;
the fecond 1+2+3 = 6; the third = 1 +
2 + 3 + 4 = 10; the fourth = 1 + 2 + 3 +
4+5=15, and ſo on.
I
i
SECT.
CURVE LINES.
49
SECT. I.
The generation of Curves, by which the
length of any given Curve is determin-
ed; the use of tangents in determining
their areas, furfaces or folidities.
PROP. I.
If AOM be any given curve, AP its axis, MT Fig.
a tangent at M. And BC be fuch a curve, that 53.
MDC be drawn parallel to AP; it may be, AD:
tangent MT:: AN a given line : CD.
Then the Space ABCD will be equal to the rectan-
gle of AN and the curve line AOM.
Take the point m infinitely near M, and draw
mc parallel to MC. Produce the ordinate PM to
s, and draw rC parallel to sM. Then the trian-
gles MPT, Msm are fimilar, whence Ms: Mm ::
MP or AD: MT (conftruction) AN: CD.
And therefore CD X Ms Mm X AN; that is,
the rectangle CDdr or CDdc rectangle Mm x
AN. Hence the fum of all the rectangles CDdc
:
fum of all the rectangles Mm × AN; that is,
the area ABCD = rectangle AN× curve line AOM.
Cor. 1. Hence if any tangent MT be known, the
correſponding ordinate CD will be known.
Cor. 2. And if the area ABCD can be had, the
length of the curve AOM will also be bad.
Y 3
PROP.
50.
B. II.
CURVE LINES.
Fig.
54.
PRO P. II.
Let AM be any curve, AP the axis, PM the or-
dinate perp. to it; and let the curve KZ be of fuch
a nature, that taking any point M in the curve ÂM,
and drawing the line MB perp. to the curve in M,
cutting the axis in B, and producing MP to Z; it
may be always PM: MB:: arch AM: PZ. Then
the area AKZP will be equal to half the fquare of
the arch AM.
Take the point m infinitely near M, and draw
mB. Alfo draw ms parallel to MZ, and mr perp.
to it. Then the triangles Mmr, MPB are fimilar,
and mr : Mm :: MP: MB:: (conſtruction) arch
AM: PZ. Then Mm x arch AM = mr × PZ
X
= PpsZ. And the fum of all the Mm X AM =
fum of all the PpsZ — area AKZP.
Divide AM into an infinite number of equal
parts each equal to Mm = 1. Then all the arches
beginning at A, conftitute a ſeries of quantities in
arithmetic progreffion, whofe number is AM, and
the greateſt alfo AM. Therefore the fum of all
the AM (or AM x Mm)
AM
AM²
XAM =
2
2
by Prop. II. of the Arithm. of Infinites. There-
AM²
fore
area AKZP.
2
PROP.
Sect. I.
52
CURVE LINE S.
Fig.
PROP. III.
24
Let AIB be any curve, AD the axis, BD the or- 55、
dinate perp. to AD. And let KL, CE be two curves,
fo related to this, that producing BD to E and L, and
taking a given right line KN, it may be DL DE
+ KN2, and that the rectangle KN x BD = area
ADEC. Then the rectangle KN × curve AIB =
area ADLK.
Take the point b infinitely near B, and draw
bdel parallel to BDEL; and draw Bm, En, Lr pa-
rallel to AD. And put AD = x, DE = y, DL
=v, curve AIB = 2, BD w, and KN = a,
then by conſtruction vv = yy + aa; and aw =
ADEC; that is, a X BD ADEC, and a x bd
AdeC, and their difference a X mb DdeE
DE × Dď = 3 × Dd = y × Bm, and fquaring,
ᎠᎴ
x
× mb² = y² × Bm²; that is, vv yyXmb² = y²
× Bm; and tranfpofing, vv × mb² = yy x
mb² + Bm² yy x Bb2, and extracting the root
v × mb = y × Bb, and av × mb = ay × Bb, or
(becaufe a X mb − y x Bm) uy X Bm = ay . x BỎ,
a ×
y× ×
or v× Bm = a × Bb; that is, DL × Dd = a × Bb,
and the fum of all the DL x Dd = a × fum of
all the Bb; that is, the area ADLK = a × curve
AIB.
!
x
X
x
Cor. Hence if the areas of the curves, ADLK,
ADEC be known; the length of the curve AIB will
be known.
PROP.
Y A
52
B. II.
CURVE LINES.
Fig.
56.
PROP.
IV.
If aefg be any fort of Spiral, af an ordinate, ta
perp. to it, tf a tangent, HI a given line; and it be
made as af: at:: HI: ql; and aq be made equal to
af, and ql laid from q to l, in a line perp. to aq, and
if this be every where done, and the curve mlk
drawn And if the curve EFG be of fuch a na-
ture, that the ordinate CF — af, and BE = ae,
&c. and that BC x HI area mpgl. Then will
the length of the curve EF = length of the ſpiral ef.
And the area BEFC twice the area aef.
-
For fuppofe ag infinitely near af, and DG infi-
nitely near CF. Draw FN parallel to BD. Then
fince CF af, and DG ag, their difference
NGng. Put HI = a. Then fince pmkr = a
× BD, and pmlq = a × BC, therefore the differ-
ence qlkra × CD; that is, ql × qr or ng × ql
≈ax CD. But the triangles aft, ngƒ are ſimilar,
whence ng : nf:: af: at:: (conſtruction) a a: gl;
therefore a X nf = ng × gl = a X CD, therefore
nf CD = NF. Whence the triangles nfg and
NFG are equal and fimilar, and FG = fg, and all
the FG all the fg; or the curve EFG equal to
the curve efg. And fince af CF, and ag = DG;
therefore the triangle afg half the trapezium
CDGF. And the fum of all the triangles afg
\
=
half the fum of all the trapezia. CG; that is,
´the area aef= half the area EBCF.
PROP.
Sect. I.
53
CURVE LINES.
Fig.
PRO P. V.
Let AMC be any curve; AB the axis; PM, CB, 57.
ordinates, MT a tangent. Let the parallelogram
PMDB be infcribed in the curve. Then will PMDB
be the greatest poffible infcribed parallelogram, when
the length PB is equal to the fubtangent PT.
Take the point m infinitely near M, and draw
mp, md parallel to MP, MD. Then we have another
parallelogram mpBd, infcribed in the curve. Every
quantity is at its greateft, when by varying the
terms of it, the increaſe is equal to the decreaſe.
The triangles TPM and Mrm are fimilar, whence
TP: PM:: Mr: rm. Therefore PM x Mr
TP Xrm. Therefore if PB or MD be put equal
to TP, we fhall have PM × Mr MDx rm =
rDxrm. That is, the parallelogram Mp par.
mD. But when the parallelogram MPBD is changed
into the parallelogram mpbd. The former is de-
creafed by the tpace Mp, and increaſed by the ſpace
mD Mp. Therefore, fince the infinitely finall
increafe, is equal to the infinitely fmall decreaſe;
the parallelogram PMDB is a maximum.
=
=
Cor. By a like reaſoning, the greatest cylinder
PMDB, that can be infcribed in the folid, generated
by the curve AMC, revolving about its axis AB;
is when the length PB is equal to half the fubtangent
PT.
=
=
=
For let c 3.1416, then fince PM x Mr =
TP X rm; let PB PT, then TP 2PB, and
PM x Mr = 2PB x rm; multiply both by
cx PM, then c X PM² × Mr = 2cx PM × rm×
PB. But cx PM' x Mr folid defcribed by
PMmp; and zc × PM × rm × PB
2c
ſcribed by DMmd.
folid de-
Therefore
$
54
CURVE LINE S. B. II.
1
Fig.
Therefore, the increaſe of the infcribed cylinder
57. being equal to the decreafe, the cylinder muſt be
a maximum.
58.
59.
58.
59.
58.
PRO P. VI.
In any curve, if the points M, m, be taken infi-
nitely near together; and the ordinates MP, mp,
drawn; and alfo Mr, mn, parallel to the axis AP.
Then either of the parallelograms MPpr, or nPpm,
may be taken for the Space MPpm.
And in fpirals, if M, m be infinitely near, and
AM, Am drawn; and the small arches Mr, mn,
defcribed from the center A, which will be perp. to
AM, Am. Then either of the triangles AMr, Amn,
may be taken for the Space AMm.
For fince the hypothenufe Mm is infinitely fmall,
the fides Mn, nm, or Mr, rm, will much more be
infinitely ſmall. But the parallelogram MPpr is
to the parallelogram Mrmn, as PM to Mn; and
as Mn is infinitely less than MP; the parallelogram
Mrmn is infinitely lefs than MPpr; and much
more is the triangle Mnm, or Mrm infinitely leſs
than the ſpace MPpm; and therefore vanishes in
refpect of it, confequently the par. MPpr ſpace
MPpm par. nPpm.
In like manner in the fpiral, the triangle Mmn
or Mrm is infinitely leſs than AMr or Amn. And
therefore AMr or Amn, may be taken for AMm,
the difference being infinitely finall, or nothing, in
reſpect of any of them.
Cor. 1. If a curve APM be divided into an infi-
nite number of parts by the ordinates MP, mp, &c.
infinitely near one another; either the fum of the cir-
cumfcribing parallelograms, as Ppmn; or the fum of
the infcribed parallelograms, as PprM; may be taken
for the area of the curve.
And
Sect. I.
55
CURVE LINES.
And in fpirals, either the fum of the circumfcrib- Fig.
ing triangles Amn, or of the infcribed triangles AMr, 59.
may be taken for the area.
For when the ordinates are parallel, the fum of
the circumfcribing and infcribed parallelograms,
differ from one another only by one of the paral-
lelograms nPpm, which is infinitely less than the
area APM, as Pp is infinitely lefs than AP. And
much more will the difference between the curve
and either fum, be infinitely finall.
For the ſame reaſon in fpirals, the fum of the
circumfcribed, and the fum of the infcribed trian-
gles, differ from the curve by leſs than the infinite-
ly ſmall triangle AMm.
Cor. 2. The infinitely fmall part Mm of the curve
AM may be looked on as a right line.
Becauſe its deviation from a right line muſt be
infinitely ſmall in that ſhort ſpace.
PROP. VII.
Let AM be any curve, AP the abfciffa, PM an 60.
ordinate, AT a tangent at the vertex. And let AN
be fuch a curve, that drawing the tangent MT to
any point M, to cut AT in T, and producing MP,
make PN AT. And this be every where done.
Then drawing AM, the area of the curve ANP is
twice the fegment ABMA.
Take m infinitely near M, and draw mpn parallel
to MPN; alſo draw AQg parallel to MT, then
QM = AT = (by conſtruction) PN. Therefore
the parallelogram QMmq = trapezium PNnp
(Prop. VI.). But the triangle AMm (contained
between the fame parallels AQq, TMm) = half
QMmq half PNnp. Therefore the fum of all
the triangles AMm fum of all the trapezia
PNnp; that is, the fegment ABMA half the
area ANP.
I
2
PROP.
56
B. II.
CURVE LINES.
*
1
Fig.
61.
PROP. VIII.
If AM be any curve, AP the abfciffa, PM an or-
dinate, BC parallel to it, MT a tangent at M. And
if BD be a curve of fuch a nature, that CD, drawn
thro' M, parallel to AP, may be always equal to the
Subtangent TP. Then the area of the curve BCD,
is equal to the area of the curve APM.
Take m infinitely near M, and draw mp parallel
to MP, and md parallel to AP. Then the trian-
gles TPM, mrM are fimilar, whence Mr: rm ::
MP: PT or CD; therefore MP x rm = CD x
Mr; that is, the trapezium MPpm trapezium
CDdc. And fince there are as many trapezia in
the ſpace CBD, as in the ſpace APM; and theſe
trapezia being always equal, their fums must be
equal. That is, the area of the figure BCD will
be equal to the area of the figure APM.
PROP. IX.
62. If PQM be a spiral or other curve, AM an or-
dinate, AT perp. to it, MT a tangent. And if the
curve ABC be of fuch a nature, that drawing AF,
and AG perp. to it; and with the radius AM and
center A, defcribe the arch MN, and draw NB pas
rallel to AG and equal to the fubtangent AT. Then
the area of the curve ANB is twice the area of the
Spiral PQMA, beginning at A.
Take m infinitely near M, and draw the ordinate
Am, and with the center A and radius Am, defcribe
the arch mrn, and draw nb parallel NB or AG. Then
the triangles Mrn, MAT, are fimilar, whence Mr:
rm:: MA: AT or NB by conftruction. Therefore
MA× rm = NB × Mr, or NB x Nn; that is,
twice
Sect. I.
57
CURVE
LINE S.
twice the triangle MAm BNnb. And fince this Fig.
is fo every where, therefore the whole area NAB 62.
twice the area APQMA.
PROP.
X.
If AFM be any curve, AP the axis, PM an or- 63.
dinate, MC perpendicular to the curve in M. And
if the curve BOD be of fuch a nature, that produc-
ing MP to D, PD may be equal to the fubnormal
CP. Then the area ABODP
PM².
I
I
2
Let the point m be infinitely near M, and draw
mpd parallel to MD. And draw Mr parallel to
AP. Then the triangles Mmr and MCP are fimi-
lar, whence mr: Mr:: CP or PD: MP. There-
fore mr X MP = Mr x PD.
Mr x PD. Now if PM be di-
vided into an infinite number of equal parts each
equal to mr = 1; then all the PM will compofe
a ſeries of terms in arithmetic progreffion, whoſe
number and greateft term is PM. Therefore
the ſum of all the progreffion (Arith. Infinites
Prop. II.) is PM². And fince mr × MP = Mr
× PD or Pp x PD = trapezium PpdD; therefore
the fum of all mr x MP fum of all the PpdD;
that is, PM² = area APDÓB.
I
2
I
2
PROP. XI.
If AMB be any curve, AP the axis, PM an or- 64:
dinate, MT a tangent. And KZ be fuch a curve,
that producing MP to Z, it may be, TP PM ::
any right line a : PZ. Then the area APZK = 0
X PM.
2
Let m be a point of the curve infinitely near M,
and draw mpz parallel to MP, and Mr parallel to
AP. Then the triangles Mrm, TPM are fimilar,
whence Mr: mr :: TP: PM :: (conſtruction) a:
PZ.
58
CURVE LINE S. B. II.
Fig. PZ. Then Mr × PZ = a X mr; that is, the
64. trapezium PpzZ ax mr. And the fum of all
the PpzZ or the area AKZP = ax fum of all
the mr a x PM.
65.
PROP. XII.
Let AM be any curve, AP the abfciffa, PM an
ordinate, MT a tangent. And let the curves BD,
and GF be fo related, that drawing CN parallel to
PM, and producing MP to D, and drawing MNF
parallel to AP; it may always be, PT: PM ::
NF: PD. Then the area APDB is equal to the
area CGFN.
Take m infinitely near M, and draw mpd paral-
lel to MP, and mnf parallel to AP. Then the
triangles Mmr, TMP, are fimilar, whence Mr:
mr :: TP PM :: NF: PD. Therefore Mr X ×
PD = mrx NF; that is, Pp × PD = Nn × NF,
or which is the fame thing, the trapezium PpdD
trapezium NFfn. And as this is fo every where;
the fum of all the trapeziums PpdD ſum of all
the trapeziums NFfn, or the area ABDP area
CGFN.
Cor. Hence if any two of the curves AMP,
ABDP and CGFN, be given; the third may be
found.
PROP.
A
Fig. 49
52.
M
A
T
A
P
b
B
m
T
55.
D
AN
D
K
57.
m
E
ne
N
M
50.
ni
B
A
p
P
D
C
53.
B
T
SLM
Vre
d
D
A
Př
B
<
M
m
Pp
58.
T
61.
M IC
A PP B
a
n
M
Im
59.
D
122
T
E
}
D
51.
M
mn
56.
54.
E
K
F
IN
A B
C D
H
I
60.
P
M
n
n
m
62.
Pl.V. pa. 58.
N
IK
OF
Sect. I. CURVE LINE S.
59
PROP.
XIII.
Fig.
66.
Let AMB be any curve, AP its axis, PM an
ordinate; and let the curve ANE be of fuch a na-
ture, that producing PM to N,
be parallel to the chord AM.
ND parallel to MP, AP; the
equal to the area APM.
the tangent NT may
Then drawing AD,
area ADN will be
Let m be a point infinitely near M, and draw
pmn parallel to PM, and dn, NS, Mr parallel to
AP. Then the triangles Nns, TNP, and AMP
are fimilar, whence AP: PM :: TP: PN:: Ns:
Therefore AP X sn PM x Ns; that is,
×
DN x sn = PM x Mr, or the trapezium DNnd
trapezium PMmp. And the fum of all the
trapeziums DNnd fum of all the trapeziums
PMmp; that is, the area ADN area APM.
sn.
PROP. XIV.
If AMB be any curve, AP the axis, PM an or-
dinate; and let KZL be fuck a curve, that taking
any right line r, and continuing the ordinate MP to
Z, it may be, area APM: PM : :r : PZ. Then
√
the area APZK = 2r X area APM.
Take m infinitely near M, and draw mpz paral-
lel to MP. And put z = area APM fum of
the trapezia PpmM or PM x Pp, AP = x, PM =
y, PZ v; then /≈:y: gi: V =
× Pp or PZ × Pp =
ry
ry
and v
√x XPp = √ XPM× Pp
✓ trapezium PMmp. Let w = trapezium
1%
X
PMmp.
67.
60
B. II.
CURVE LINE S.
:
Fig. PMmp, and divide the area APM or z into an in-
67. finite number of trapeziums equal to w 1. Then
all theſe areas denoted by z conftitute a ſeries of
quantities in arithmetic progreffion, and the fum
68.
I
of all the roots (by Prop. V. Arithm. of In-
I
finites) is
m + I
I
X fo many times the greateſt term,
I
where m =—1, the greateſt term and num-
2)
ber of terms z, therefore the fum of all the
I
is ====
X
I
I
=2%. And therefore fince
X w. Therefore the ſum of all
PZ x Pp = √x X w.
the PZX Pp = rx fum of all the
of all the
1
W
or rx fum
√x = rx 2√/z; that is, the area
APZK = r × 2✔✅ area APM.
PROP. XV.
Let AMB be any curve, AD the axis, PM an or-
dinate, MT a tangent. And let the curves EX and
GY be fuch, that continuing MP to X, and draw-
ing MQY parallel to AP; it be TP : PM :: QY:
PX. Then the area APXE area DGQV.
Take the point m infinitely near M, and draw
mpx parallel to MP, and may parallel to MQY.
Then the triangles Mrm, TPM are fimilar, whence
Mr: rm: TP PM: (conftruction) QY: PX;
therefore Mr X PX = rmx QY, or Pp x PX =
Qq × QY; that is, the trapezium PXxp trape-
zium
Sect. I.
61
CURVE LINES.
zium QYyq. And the fum of all the trapezia PXxp Fig.
= fum of all the trapezia QYyq; that is, the area 68.
APXE = area DGYQ.
PROP. XVI.
Suppofe AMB any curve, DM an ordinate; DT 69:
perpendicular to it, MT a tangent. And let the
curves EN, and GH be of fuch a nature, that producing
DM to N, and with radius DM and center D, de-
Scribing the arch MrQ, and drawing QH and DG
perp. to DQ. Then making TD : DM:: DM ×
QH DN². Then the area DGHQ = twice the
area EDN.
Take the point m infinitely near M, and draw
drmn. From the center D, with radius DN de-
ſcribe the ſmall arch Ns; and with the radius
Dm, the arch mq, and draw qb parallel to QH.
Then the triangles Mrm and TDM are fimilar,
whence Mr: rm :: TD : DM :: (conſtruction)
DM × QH : DN²; therefore Mr x DN² = rm ×
DM × QH. But the fectors DMr, DNs are fi-
milar, whence DM: Mr:: DN: Ns, and DM ×
Ns DN x Mr, and DN x DM x Ns DN2
× Mr = rmx DM x QH; and dividing, DN ×
Ns = rm x QH = Qq x QH; that is, twice the
triangle DN = trapezium QHbq. Therefore the
fum of all theſe trapezia or the area DGHQ=
twice the fum of all the triangles, or twice the area
DEN.
N
PROP.
62
CURVE LINE S. B. II.
Fig.
70.
51.
•
PROP. XVII.
Let DOM be any curve, DA a tangent at D, DM
an ordinate, DT perp. to it, MT a tangent. And
let AGF be fuch a curve, that producing DM to F.
And taking a = any given line, let DT: 2a::
DM²: DF². Then the area ADF a x DM.
Take the point m infinitely near M, and draw
Dmf; alfo draw Mr perp. to Df, and fn perp. to
DF. Then the triangles MDT, mrM are fimilar,
and therefore DT: DM:: Mr: mr, whence DT
× mr = DM × Mr. But DM × Mr : DF × nf : :
triangle DMm: triangle DFf:: DM: DF::
(conftruction) DT: 2a. Therefore 2a x DM X
Mr DT x DFX nf, or 2a X DT X mr = DT
X DFX nf, whence 2a X mr = DF × nf = twice
the triangle DFf. And a X mr = DFf; therefore
ax all the mrfum of all the triangles DFf;
that is, a x DM area DAF.
PROP.
XVIII.
If AM be any curve, AP the abfciffa, PM an or-
dinate, BC parallel to it, MT a tangent at M. And
if BD be fuch a carve, that drawing MCD parallel
to AP, CD may be always equal to the fubtangent TP.
And both curves APM and BCD revolve about the
axis APB. Then the folid defcribed by BCD, will
be double the folid defcribed by APM.
Let be a point of the curve, infinitely near
m
M, and draw mp parallel to MP, and md parallel
to AP. Then the triangles mrM, TPM are fimi-
lar; whence mr: Mr:: TP: PM; that is, Pp:
Cc:: CD: PM; therefore Pp x PM Cc x CD.
Put circumference defcribed by M or D; then
↑
=
the
Sect. I.
63
CURVE LINES.
the area deſcribed by Cc is p x Cc, and the area Fig.
deſcribed by PM =
p × PM
2
And the folid de-
ſcribed by CcdD is p x Cc x CD, and the folid de-
sp
ſcribed by PpmM is
p× Ppx PM
2
px Cc x CD
2
61.
Therefore the folid defcribed by CcdD is double-
the folid deſcribed by PpmM. And the fum of all
the CcdD or the folid CBD is double the fum of
all the PpmM or the folid APM.
PROP.
XIX.
Let AFM be any curve, AP the axis, PM an or- 63.
dinate, MC perpendicular to the curve in M. And
if BOD be fuch a curve, that producing MP to D,
PD may be equal to MC. Then if the curve AFM
revolve about the axis AP, the furface defcribed by
AFM will to the area APDB, as the circumference
of a circle to the radius.
Take Mm infinitely fmall, and draw mpd parak
lel to MP, and Mr parallel to AP. Then the
triangles MPC and Mrm are fimilar; whence Mr:
Mm:: MP: MC; therefore Mr x MC = Mm ×
MP. Let p = circumference defcribed by M;
then the area defcribed by Mm, will be p x Mm.
And the area PpdD = Pp × PD = Mr × MC.
x
Therefore the ſurface defcribed by Mm: area PpdD::
Mm × p: Mr x MC or Mm × MP :: p: MP : :
circumference: radius. And the fum of all the
furfaces Mm: fum of all the PpdD; that is, the
furface deſcribed by AFM: area APDB:: cir-
cumference: radius.
Z 2
PROP.
64
B. II.
CURVE LINES.
Fig.
61.
PROP.
XX.
If AM be any curve, AP the axis, PM an or-
dinate, BC parallel to it, MT a tangent at M. If
MD be drawn parallel to AB; and the curve BD be
fuch, that CD be always equal to the tangent MT.
Then the furface defcribed by the curve AM revolving
about the axis AP, is to the area BCD, as the cir-
cumference of a circle, to the radius.
:
Take the part Mm infinitely fmall, and draw mp
parallel to MP, and md parallel to MD; and let
pcircumference to the radius PM. The trian-
gles mrM, TPM are fimilar, whence Mr: Mm ::
MP MT or CD; therefore Mr X CD = Min ×
MP. Then the ſurface deſcribed by Mm = Mm
Xp, and the area CcdD Cc x CD = Mr × CD.
= Mm x MP. Therefore the furface deſcribed
by Mm area CcdD:: Mm x p: Mm x MP ::p:
MP. And the fum of all the ſurfaces Mm:
fum of all the areas CcdD::p: MP; that is, the
furface deſcribed by the curve AM area of the
curve CBD : circumference: radius.
SECT.
[ 65 ]
SECT. II.
Of the Axes, Ordinates and Affymptotes
of Curves, their nature and proper-
ties in general; their affections de-
duced from the equations of the curves
the product of the fegments of paral-
lel lines; generation of curves by
Shadows.
i
PRO P. I.
If the abfciffa of a curve be reprefented by x, and Fig.
the ordinate by y; and the nature of the curve be ex-
preffed by any equation between x and y. Then if any
particular value he affigned to x, there will be as ma-
ny values for y, as is the index of its highest power
in the equation, and no more..
Suppoſe any equation, as y³ aay + xy²
axxo, and put bx, then the equation will
become y³-aay + by² abbo, which is a
cubic equation with the unknown quantity y, and
therefore by the principles of algebra, this equation
has three roots, which are the three values of y
correfponding to that value of x denoted by b.
And affigning any other value for x, as fuppofe c;
then there will be produced another cubic equation,
having three roots or values of y correfponding to
c; and ſo on for any other value of x. And the
fame reaſoning holds good for any other equation
how high foever it be, that there are as many va-
Y
X 3
lues
66
LINE S. B. II.
CURVE
Fig. lues of y, as is the dimenfion of the higheſt power
of it. And there can be no more, becauſe there
are no more roots of the equation, than the high-
eſt power of the unknown quantity has dimenſions.
Cor. 1. In an equation between x and y, expreffing
the nature of a curve; for any particular value of y;
there are as many values of x, as the highest power of
it has dimenfions in the equation.
For in the former equation, put by, then
b³ — aab + bbx — axxo, a quadratric equation
for the unknown quantity x, which equation has
two roots or values of x. And the like for other
equations.
Cor. 2. Among the feveral values of y, for any de-
terminate value of x; thofe that are affirmative are
drawn on the upper fide of the abfciffa to the curve;
and those that are negative, are drawn from the ab-
fciffa downwards to the curve.
And among the values of x, for any particular
value of y, if those that are affirmative, lie from
the origin of the abfciffa to the ordinate, on the right;
then those that are negative, will lie from the origin of
the abfiffa to the ordinate, on the left hand.
For affirmative and negative quantities muſt lie
contrary ways from any fixed point or given line.
Cor. 3. When fuch a value of x is affigned, that
in the refulting equation, fome of the roots are im-
poffible. This denotes that fuch ordinates as are re-
prefented by these impoffible roots, don't reach the curve;
and in that cafe there is no fuch ordinate.
For impoffible roots in an equation always fhew
the impoffibility of the thing propofed.
Cor. 4. Hence by affigning feveral values of x,
you may determine as many values of y as you pleaſe,
and by ibat means find as many points of the curve
as
Sect. II:
67
CURVE LINE S.
as you please; which being drawn thro' all thefe Fig.
points will be the LOCUS of that equation.
Cor. 5. In affigning any value for x, if one root or
value of y be o, then the curve cuts the abfciffa in
that point. But if y be infinite, then the ordinate to
that point, is an affymptote to the curve, which runs
out ad infinitum.
Cor. 6. If x be made infinite, and then fome value
of y is o, this fhews that x or the abfciffa is an af
Symptote to fome branch of the curve, which runs on
infinitely.
:
PRO P. II.
If a curve is defined by an equation fhewing the re- 71.
lation of the abfciffa (x), and ordinate (y); a right
line may cut that curve in as many points, as the high-
eft power of y has dimenfions in that equation, and not
in more points.
Let RAMN be a curve defined by the equation
y? acy + by²
ax¹² = 0. It is fhewn in the
laft Prop. that for any determined value of AP or
x, there will be three values of y, viz. PM, PN,
and PR, becauſe the equation has three roots or
values of y. Therefore the right line NR cuts the
curve in the three points N, M, R. But there
being no more roots or values of y, it cannot cuţ
the curve in more points. It may cut it in lefs,
when ſome of the roots are impoffible. As if PM
was to move into the poſition QO, where it
touches the curve, at O, there the two points M,
N, coincide; but beyond Q, the ordinate cannot
touch that part of the curve at all, and then the
two roots PM, PN vaniſh.
Cor. 1. Hence one root alone cannot vanish out of
the equation, or one ordinate vanifb out of the curve;
but two must vanish together.
24
For
68
B. II.
CURVE
LINE S.
Fig. For first the two ordinates PM, PN, coincide
71. with QO; and then vaniſh together, and are no
where found.
Cor. 2. In a line of the third, fifth, &c. or of
any order whofe dimenfion is an odd number; the ab-
fciffa, or any right line not parallel to an aſſymptote, will
cut the curve at least in one point.
For if the equation of the curve be odd, it will
at leaſt have one real root.
Cor. 3. If a right line cut a line of the third
order in two points, it will neceffarily cut it in three.
For if the equation of the curve has two roots
real, it will have three.
Cor. 4. If a right line touch a line of the third
order; it will neceffarily cut the curve in fome point.
For the point of contact muſt be taken for two
points which coincide together.
Cor. 5. If a right line touch a line of the third
order in the point of contrary flexure; it will cut the
curve there, and in no other point.
For three points muſt be fuppofed to coincide
there.
Cor. 6. If a right line cut a curve in a double
point of the curve, this muſt be taken for two points.
And a line that touches one arch of the curve in that
point; cuts the other in the fame point.
PROP.
III.
Every geometrical curve being continued, either re-
turns into itfelf, or goes on to an infinite diftance.
Any geometrical curve may be confidered as de-
fcribed by the continual motion of a point; and
this
Sect. II,
69
CURVE LINES.
this being effected by fome certain law, the motion Fig.
muſt needs continue ad infinitum. And therefore 71.
the ſpace run thro', that is, the line itſelf, muſt
either return back or proceed on infinitely, by
reafon of the regular motion of the defcribing
point.
Cor. 1. If any curve has two infinite branches or
legs, they join one another either at a finite, or at
least at an infinite diſtance.
For the curve being deſcribed by the continual
motion of a point, it muſt needs pafs out of one
branch into another.
PROP IV.
Let BM be any curve, AP the abfciffa, PM an 72.
ordinate. If instead of MP, MQ and the parallels
thereto be made ordinates; the equation between AQ
and QM, denoting the nature of the curve, fhall not
rife to higher dimenfions, than the equation between
AP and PM, did before.
Let AP, PM y, AQ=v, QM = w.
Then fince the lines MP, MQ, QP are given in
pofition, their ratio to one another is given; there-
fore let PM rw, and QP = sw; then AP =
v + sw, PM = rw, So now we fhall have the
equation of the curve made up of v+ sw, and
rw, which before was made up of x and y. It is
plain if + sw be fubftituted for x, and rw for
y, it will arife to no higher powers of v and w,
than it did before of x and y, becauſe v and w
are but of one dimenfion; for v + sw raiſed to any
power, can confift of terms of no higher power
than the like power of x. So that now the curve
will be defined by an equation where the higheſt
powers of v and w, are the fame as the higheſt
powers of x and y in the firſt equation
Cor.
70
B. II.
CURVE LINĖS
Fig.
Cor. If the abfciffa begin at any other point C,
72. Still the equation between CQ and QM will rife but
to the fame hight as that between AP and PM.
73.
For the abfciffa is now increaſed or leffened by
fome given quantity as a; therefore if we put
v = CQ, then AP = a+v+sw; the equation
ftill contains only one dimenfion of v and w, and
therefore can riſe to no higher power than before.
PROP. V.
An affymptote may interfect the curve in So many
points abating two, as the equation of the curve has
dimenfions, and no more.
A right line may interfect a curve in as many
points as the equation of the curve has dimenfions,
and no more, by Prop. II. Suppofe then that
DAE is a line of the third order, whofe affymp-
tote is CB. And let the right line DFE cut it in
the three points D, L, E; and the affymptote in
F. Let the line DE revolve about the pole F,
fo as that the point D may approach towards C; by
this motion the points D, E of interfection, will
move further and further from the point F, till
at laft when the line DE falls upon the affymptote
CB, the points of interfection with the curve, D
and E are removed to an infinite diftance; and
there remains only the point A, in which DE
or the affymptote CB, can cut the curve. And in
the fame manner it is fhewn in any other cafe,
that two points of interfection go off to an infinite
diſtance. And therefore the number of points in
which an affymptote can cut the curve, will want
two, of the number of dimenfions, in the equation
of the curve.
Cor. In a conic fection, which is the fecond gender
of lines, the afymptote does not cut the curve at all.
In
Sect. II.
71
CURVE
LINE S.
In the third gender it can only cut it in one point. In Fig.
the fourth gender in two points, and no more.
So forward.
PROP.
VI.
And 73.
Lines of the third, fifth, feventh, order, &c. or any 74.
odd number, have at least two infinite legs or branches,
running contrary ways.
Suppofe NAM a curve of fuch an order; then
fince this is an equation of fome odd power, there-
fore from the nature of equations in algebra, this
equation muſt at leaſt have one real root, whatever
the value of AP or x be; therefore fuppofe AP
infinite towards P, then PM will be a root of the
equation, and have a real value, and M will be a
point of the curve. And the branch AM will be
infinite. Again, let AB be x, the equation
ftill will have one real root BN, and N will be
another point of the curve; and the branch AN
will be infinite. So there will be the two branches
AM and AN both infinite. And perhaps there
may be more branches infinite upon other confi-
derations.
=
Cor. 1. Hence the roots PM or BN may be infi-
nite; in which cafe the curve runs off infinitely from
the abfciffa BP. Or they may be finite; in which
cafe AP or AB may be an affymptote to the curve.
Cor. 2. In lines of the fecond, fourth, fixth, or
any even number of dimenfions; the roots of the equa-
tion may be all impoffible; and then the figure may re-
turn into itself, and be contained within certain
bounds.
PROP.
72
CURVE LINES. B. II.
Fig.
75.
PRO P. VII.
The affections of a curve may be known from the
equation of that curve.
≈
Let the abfciffa APx, the ordinate PM = y;
then the affections of the curve may be traced out,
by putting or y, fucceffivelyo, and infinity,
and equal to any of the given coefficients, affirma-
tive or negative, &c. By fuch equations as refult,
all the points of the curve will be determined, and
where it cuts the axis, the pofition of the affymp-
totes, if it have any, and ſo on.
For example, let ay
+ x X xx,
b + x
a x × yy = b
AB; then yý =
a
b + x
a x
X
xyy = x² + bx², 05
where a = AD, b =
Xxx, and y = ± x √
This ſhows that any ordinate may be ei-
ther affirmative or negative, and may be taken both
above and below the abfciffa, as PM, Pm, and
equal to one another. Put xo, then y = 0,
this fhows that the curve cuts the axis at A the
beginning of the abfciffa. Put xa AD, then
b + a
b + a
y = a √
=a
a-
a
=
infinity; there-
fore the ordinates erected at D will be infinite, or
EDF will be an affymptote to the curve. If x be
b+x
greater than a, then ✔ will be the fquare
a
X
root of a negative quantity; which fhews that no
part of the curve can run beyond the line or af
fymptote EF. If x lie the contrary way; that is,
√
b
a + x²
now if
if x be negative, then y =
x be leſs than b, then y will have two equal values,
which
Sect. II.
73
CURVE
LINES.
which are equal to one another; and that which Fig.
was the affirmative one before, is the negative one 75.
now; and what was negative before is now affir-
mative, cauſed by changing ± √ to +
But here ſuppoſe xb, then y = √
+ x
√.
b
b
a + b
= 0; and therefore the curve will paſs thro' the
point B, and be a tangent to the ordinate. If x be
x
b.
greater than b, then y = a+
x
+ x √ x
becomes an
impoffible quantity, fo that the curve goes no fur-
ther to that hand than to D. If we make yo;
that is, x✓
b X
a + x
here the roots are x =
o; then xx x b + x = 0,
b, x = c, x = o. There-
fore the curve paffes twice thro' the point A, and
once thro' the point B ; and therefore A is a double
point of the curve.
Suppoſe bo, then the points B and A coin-
cide, and the nodus or part of the curve between
A and B vaniſhes. Here y = × √
X
a
X
and
X
a — x × yy = ׳; and this is an equation to the
ciffoid, by Cor. II. Prop. I. Sect. II. Book I.
which is plain by comparing their equations toge-
ther; x repreſenting there what y repreſents here
and y repreſenting there what x does here; and 2a
there the fame as a here.
Put
X
xx, becomes
infinity, then ax × yy = b + xx
xyyx³, or yyxx, and x =
-yy, an impoffible equation; and therefore x
cannot be infinite.
b+x
Putyinfinity, then +x✅✔
infinity,
b + x
and xx X
a
infinity; therefore fince x is
a
finite.
74
B. II.
CURVE
LINES.
−
Fig, finite, we fhall have a xo, or x = a = AD,
and DE an infinite ordinate, or an affymptote, as
before.
75.
76.
If we ſuppoſe ab; this makes no alteration
in the nature of the curve.
PROP.
VIII.
In any curve, if tangents be drawn to all points of
the curve; and if they always cut the abfcifla at a
finite diftance from the beginning of it; that curve
has an affymptote; otherwife not.
Let AmM be any curve, AP the axis, not be-
ing an affymptote, or parallel to one. To all the
points m, M, fuppofe tangents to be drawn as mt,
MT, cutting the abfciffa in t and T. Now ifm
be any point of the curve near the vertex, mt its
tangent, and if At be a finite quantity; and if m
be ſuppoſed to move further and further from the
vertex, to M, and that the diſtance AT of the
tangent MT is ftill finite; and if M be removed
to an infinite diſtance, the length AT ftill conti-
nues of a finite magnitude. Then the tangent
MT will be an affymptote to the curve.
For an
affymptote is nothing elfe than a tangent drawn to
a point of a curve infinitely diftant. But if T be
at an infinite diftance, then fuch a tangent is no
where to be found, and therefore in fuch a cafe
the curve has no affymptote.
For example, let AM be a parabola; then if
AM be infinite, AP is infinite; and AT AP is
alfo infinite. Therefore AM has no affymptote.
If AM be an hyperbola, and M at an infinite
diſtance; then the tangent MT cuts the axis AP
in the center T; then AT is half the tranſverſe
axis; confequently MT is an afymptote.
Cor.
T
T
Fig. 63. A B
m
M
F
64.
K
A
B
65.
ERR
MA
P
P
D
Z
~ x
Z
P
M
D
d
72
N
C
66.
H
d D
E
B
68.
P
K
A
67.
G
\A.
P
M
/L
m
B
70.
f
E
E
N
T
M
P
X
A
B
M
In
D
B7
Q
G
69.
hH
1
M
A
71.
R
M
P
1
B
A C
72.
73.
E
C
B
B
N
74.
1
ļ
M
P
B
j
75.
M
P
D
m
F
76.
M
T
t A
P
PL.VI. pa. 74.
OF
CH₂
Şect. II.
CURVE
75
LINES.
Cor. Hence to find the affymptote to a curve, is Fig.
the fame thing as finding the tangent to a point of 76.
it, at an infinite diftance.
PROP. IX.
If the equation of a curve be y = a + bx + cx² 77.
+ dx³, &c. to fen. And n be an even number; then
the curve has two infinite legs both on the fame fide
of the abfciffa AP.
For when x is infinite, then y = fx, all the
other terms vanifhing; fo that if yfx", or y =
fxt, or y = fxº, &c. and fuppofe x infinite, then
whether AP be+xorx, ftill y = + fx²,
or y = + fx, or y = + ƒ×³, &c. that is, the
value of y will always be affirmative, when ʼn is
even; and therefore the curve is always on the
fame fide of the abfciffa.
=
=
:
For example, if a, b, c, d be = o, and n = 4,
then y fx4, and APx, then PM fx4, and
if Ap =
x, then again pm = + fx+, and there-
fore both the parts AM, Am, infinitely extended,
lie on the fame fide of the abfciffa AP. This
curve is a kind of parabola.
PRO P. X.
If the equation of the curve be y = a + bx + cx² 78.
+ dx³, &c. to fx. And n be an odd number. Then
the curve has two infinite legs, running contrary ways,
and on different fides of the abfciffa.
For when x is infinite, all the terms of x vaniſh
except the greateſt fx"; therefore y fx. But if
n = 1, 3, 5, 7, &c. and the abfciffa -
then Pm or the ordinate, y = -fx, or y fx³,
ory =—-fx³, &c. that is, if x be negative, and
=
X,
n an
76
CURVE LINES. B. II.
Fig. n an odd number, then y = fx", whence the
78. ordinate lies on the contrary fide of what it does
when it is, for then y+fx". And the
branches of the curve AM, Am run contrary ways,
becauſe x is contrary to + x; that is, the ab-
fciffas lie different ways.
79.
For example, lety. Then if AP = x, then
y = fx3 = PM; but if Apx, then y =-
fx3 pm, on the contrary fide of AP. So the
curve MAm has two infinite branches running
contrary ways, and on different fides of AP. This
curve is the cubical parabola.
PROP. XI.
If the equation of the curve bey =
an + I
xn
; this
curve has four infinite legs contained between two af-
fymptotes; which lie in the oppofite angles DAČ,
BAE, when n is an odd number; but in the adjacent
angles, DAC, DAB, when n is even.
Let AP, PM, fuppofe x affirmative,
a^2 + 1
then PM =
; if x beo, then PM is in-
Xn
finite, and the curve runs on infinitely towards D;
therefore DA is an affymptote. And if x be infi
nite, then PM o, and the curve runs on infi-
nitely towards C, and AC is an affymptote.
Let x be negative, or AQ = x, then if n is an
a² + 1
odd number, y = -
xn
; therefore y or QN
lies on the contrary fide of BC, and the curve is
contained in the angle BAE. Let xo, y is in-
finite, and the curve runs on infinitely toward E.
And
Sect. II.
77
CURVE
LINES.
And if x = - infinity, then y = o, and the curve Fig.
runs on ad infinitum towards B.
But whilft x is negative, if n be an even num-
ber, then y = +
n + I
a
;
and therefore y or Qm
xn
lies on the fame fide of BC as at firft. And as x
iso or infinity, y will be infinite or nothing;
therefore the curve runs infinitely towards B and
D, and is contained in the angle BAD. So that
there are four infinite branches of the curve, con-
tained between the two affymptotes BC, DE; either
in the oppofite angles, when n is odd; or in the
adjacent angles, when ʼn is even. And here one of
the affymptotes AC is the abfciffa of the curve.
Theſe are called Hyperbolic Curves.
aa
X
Cor. If y = or xy =¸aa, then the curve is
the common hyperbola, and fince n = 1, therefore the
oppofite fections are contained in the oppofite angles
DAC, BAE; between the afymptotes BC, DE, as
is commonly known.
PROP. XII.
79.
If FM be any curve, FD the diameter; and if 80.
an ordinate MD be drawn to cut the curve in as ma-
my points as it has dimenfions; then the sum of the
ordinates on one fide FD is equal to the fum of the
ordinates on the other fide; taken in the fame line MG.
Let the nature of the curve be denoted by this
equation; y" + ax + b x 302 - +
cxx + dx + ex y22 &c. o, (fee Schol. to
the definitions), and this general equation referred
to fome abfciffa AP, and ordinate PM. Produce
AP to C till AC =
b
and draw AB parallel to
,
a
A 2
MP
78
. B. II.
CURVE
LINESLINES.
Fig.
b
MP and =
80.
Thro' C, B, draw CFBD, and
n
apo + b
take BD for a new abfciffa, and DM its ordinate
and let BD v, DM = 2. Take :p :: BD
1
(v): AP = pvx. By fimilar triangles, CA:
AB:: CP: PD; that is,
and PM - MD
b
b
b
a
ñ
::po+
: PD
22.
a
PD = %
n
apv + b
apv + b
,
or y = z
n
And involving,
n
zn
yn
zn.
Zr
apv + b xxx − 1 + n − 1 × apv + b
Xz"-2 &c. and y-I =
apo + b
N
Zn-I
n · IX
22-2
=
&c.
X 2"-2 &c. y-2
ſubſtitute theſe values of y, and x in the equation
I
3 + ax + b × yn − 1 + &c. = o, and we have
2n — apv + b × 2-1 +2 - 1 × apv + b²
X 22-2 &c.
+ apu + bx 27-1
X 27 - 2 &c.
apv + b
n I X
n
+ cppvv x z²-2 &c.
Here the ſecond term of the equation vaniſhes,
and therefore the fum of the affirmative roots or
ordinates is equal to the fum of the negative ones,
by the nature of equations; that is, the ſum
of the ordinates on one fide is equal to the fum
on the other fide of BD, and BD is the diameter
of the figure.
Cor. 1. In curves of the first gender, or the conic
fections; the ordinates are bijected by the diameter.
Cor.
Sect. II.
79
CURVE LINE S.
Cor. 2. In curves of the fecond gender; the di- Fig.
ameter cuts all the ordinates; fo that the sum of two 80.
parts on one fide is equal to the third part on the
other fide.
Cor.
3.
The line which cuts two parallels, that
the fum of the parts on one fide be equal to the fum
of the parts on the other; will cut all other paral-
lels in the fame manner.
PROP. XIII.
If a right line FN cut a curve of any order, in 81.
the points M, m, N, &c. and its affymptotes in F,
C, G. Then the parts FM, GN, lying the fame
way from their affymptotes to the curve, will be equal
to the part or parts Cm, lying the contrary way
from their affymptotes to the curve.
Let AP be the diameter to the ordinate PM,
and let the curve be a line of the third order.
Then (Prop. XII.) PM + Pm PG. Now be-
cauſe the curve coincides with the affymptotes at
an infinite diſtance; therefore M, F; and m, C;
and N, G coincide. But at an infinite diftance it
is alſo (Prop. XII.) PM + Pm = PN; that is,
PF + PC PG. But if this be fo at an infinite
diſtance, it will be the fame at any other diſtance;
as parallel lines are cut in the fame proportion;
therefore it will be in general PC + PF = PG,
but Pm + PM PN; therefore taking equals
from equals, Pm - PC + PM — PF = PN
PG; that is, Cm- FM GN, or Cm FM
+ GN.
=
And the fame is eafily fhewn of curves of
higher orders.
A a 2
PROP...
80
B. II.
CURVE LINES.
Fig.
82.
PROP. XIV.
If ANB be a curve of the first order, AB a
right line cutting the curve in A and B; MN ano-
ther right line, cutting the curve alſo in M and N,
and the other in P. Then the rectangle APB, will
be to the rectangle MPN in a given ratio, having
any parallel pofition.
Let AP be the abfciffa, PM and PN ordinates,
and let AP = x, PM or PN either of them y.
Then the equation of the curve (Schol. Def.) in
general, is y²+ ax + b × y + cxx + dx = 0;
in which the determinate quantity e is o, becauſe
the abfciffa begins at the curve. Let yo, then
cxx + dx = 0, becauſe all the terms vanifh where
y
is found, then ċx —— d, and x =
d
AB;
C
/
d
therefore AB
AP or PB =
x, or ra-
C
d
dx
دی
But by
C
ther x + and AP X PB = xx +
the nature of equations, the laſt term in which the
unknown quantity is wanting, is the product of all
the roots; that is, cxx + dx = PM x PN.
dx
'Therefore AP x PB: PM x PN: : xx +
I
C
cxx + dx :: 1: c; but whilft the paralleliſm of
the lines AB, MN continues, the quantity cre-
mains the fame; and confequently the proportion
of thefe rectangles remains the fame.
PROP.
Sect. II. CURVE
CURVE LINES.
81
PROP. XV.
Fig:
Let AMN be a line of the third or any other or-
der; and let the right line AD cut it in the three points
A, B, D; and the line MN in three others M, m,
N; and let P be their point of interfection. Then
will PA x PB x BD be to PM x Pm x PN in a
given ratio; whilst they are in any pofition parallel
to AD, MN.
Let the abfciffa APx, and any of the lines
PM, Pm, PN. Then (Schol. definitions)
the nature of the curve is denoted by this equa-
tion, y³ + ax + b × y² + cx² + dx + e x y +
fx3
ƒx³ + 8x² + bxo, where the determinate
quantity k iso, becauſe the abfciffa begins at
the curve. Let yo, and all the quantities
will vaniſh where y is found, therefore fx³ + 8x
+bxo. In which cafe, we ſhall have x =
+1
√ 88
b
4ff f
2f
-g
that is, AB =
2f
gg h
4ff ƒ
gg
b
and AD =
+
therefore BP =
2f
Aff.
f
AP — AB = x +
дра
gg
b
Aff f
88
and PD =
2f
it
g
AD
AP=
2f
+ √
b
Aff
f
g
√ 88
b
2f
ther x +
X DP = xxx +
g
x +
2f
gg
4ff F
A a
3
bol
2f
4ff f
g
gg
+
2f
x, or ra-
Therefore AP × BP
b
4ff →→→→ ƒ
X
b
Σ
X X
83.
82
CURVE LINE S. B. II.
b
Fig. gg
83. 4ff f
gx
x x xx + +++= xx
b
4ff 4ff
gx² bx
xx + + =x³ +
f f F + F
But by the
nature of equations, the laſt term is equal to the
product of all the roots; that is, MP × mP × NP
= ƒx³ + gx² + bx. Therefore AP x BPX DP :
8x2 bx
MP × mP × NP : : ׳ +7 +7 : fx² + 8x² +
bx 1 f. But whilft the lines AD, MN, have any
:
other places parallel to the former; the quantity
f remains the fame; and therefore the ratio of AP
× BP X DP to MP X mP × NP.
Otherwife thus,
Let R be the beginning of the abfciffa, which
cuts the curve in the three points A, B, D; and put
RPx, and let the ordinate MP cut the curve
in the three points M, m, N. Then in the gene-
ral equation, when y vanishes, we fhall have f³ |
gx² + bx + k = 0, and ≈³ |-
x³-
&
f
24
x² +
b
f
k
Fx + F
are RA, RB,
made up of the
RB, x RD,
Whence x³ +
= o, the roots of which equation
RD; therefore the equation is
three quantities ≈
RA, x
multiplied into one another.
g
b
k
f
xx +
F* + F
X
RA X X
X
X3
RB X
RD RP
RP-RA X RP RB X RP RD
= AP × BP × PD. But in any equation, the
laft term is equal to the product of all the roots.
Therefore fx + gx² + bx +k MP x mP X
NP. Therefore the product of AP x BP X
- PD product of MP x mP x NP; or AP ×
BP
Sect. II. 83
CURVE CURVE
LINES.
xx
BP × PD: MP×mP × NP :: ׳ + x +
x3
k
+7: fx³ + 8xx + bx + k : : 1 : f.
b Fig.
F* 83.
It is evident, the fame way of reaſoning may
be applied to a curve of any gender, and there-
fore when the angle of inclination APM is given;
if the right lines AP, PM; or RP, PM, cut a
curve line of any order in as many points as it has
dimenfions; it will always be, that the product of
the fegments of the firft line, between P and the
curve, is to the product of the fegments of the
fecond line likewife between P and the curve, in
a conftant ratio.
Cor. 1. If the line AP cut the curve only in one 84.
point A; then if you make AF =
g
and CF
2f"
perpendicular to it, and =
√4fb
gg and draw
2f
PC; then will PM × Pm × PN: AP× PC:: ƒ: 1.
For f׳ + gx² + bx =
fx xxx
g
f× × ×x + 1/4 x +
b
F
o, (where ko, becauſe a is in the curve), which
has only one root x, becauſe AP cuts the curve
only in A; and the quantity xx +
g
tains two impoffible roots, viz. * =
b
+1
3x + 7, con.
g
f
88
or
4ff
is greater than
Therefore CF is a real
gg
4ff F³
4fb greater than gg.
b
b
therefore
f
tity. And PC² =
A a 4
PF² + CF² = ≈ +
x
quan-
2
+
&
2f
Afk
84
CURVE
B. II.
LINES.
gx
gg
Fig. 4fb = = xx + x + 2 + 1/65R
— gg
4fb - 88
xx +
84.
Aff
gx
b
F + F
4ff Aff
f
And fx x PC²ƒ× AP × PC²
ƒx³ + gx² + bx.
Therefore fx AP x PC =
PM x Pm x PN; and AP X PC²: PM × Pm ×
PN : : 1 : f.
Cor. 2. And fince AF and FC are compofed of
given quantities, the point C is fixed; whatever be
the angle APM.
PROP.
XVI.
1
A line of any order may have as many aſſymptotes,
as it has dimenfions; and not more.
I
For let the curve be of the nth order of lines,
ſo that its equation is y+ax+b × 32-1 &c.
yn
=0. This equation will have n roots, and the
root being extracted will appear in this form, y =
Ax² + Bx²-1 &c. Now if by increafing x ad in-
finitum, if y be a real quantity the curve will have
an infinite leg, and confequently may have an af-
fymptote. And as there are n fuch roots there
may be as many branches and affymptotes, for
the fame reafon; but there can be no more, as
there are no more roots of the equation.
PROP. XVII.
The properties of curves of a fuperior order, agree
likewife with thofe of all inferior orders.
An equation of a curve of the third order is
denoted thus, y³ + ax + b × yy +
×3 +ƒx³ +8x² + bx + k = o.
cxx + dx + e
If f, g, h, k,
be
Se&t. II.
85
CURVE LINE S.
be ſuppoſede, then the equation will become y²+ Fig.
ax + bxy + cxx + dx + eo, which is one
of a lower order. Now the properties of the curve
belonging the firſt equation, muft hold good of
the curve belonging the fecond equation, with all
the quantities that remain; and therefore its pro-
perties are included in the former. For all the
difference is, that fome line or lines become o and
vaniſh, and others become infinite, fome coincide,
others become equal; likewife fome points coin-
cide, and others are removed to an infinite dif
tance; yet under theſe circumſtances, the general
properties ſtill hold good with the remaining
quantities; fo that whatever is demonftrated ge-
nerally of any order, holds true of the inferior
orders. And on the contrary, there is hardly any
property of the inferior orders, but there is fome
fimilar to it, in the fuperior ones.
For as in the conic fections, if two parallel lines
are drawn terminating at the fection; the right
line that biffects theſe will biffect all other paral-
lels thereto; and is therefore called the Diameter of
the figure, and the biffected lines, Ordinates; the
interfection thereof with the curve, the vertex; and
the interfection of all the diameters, the center; and
that diameter, the Axis, which is perpendicular to
the ordinates. So likewife in higher curves, if
two parallel lines are drawn, cutting the curve in
a proper number of points; the right line that
cuts theſe parallels fo, that the fum of the parts
on one fide the line, to the curve; be equal to
the fum of the parts on the other fide, it will cut
all other parallels in the fame manner, which cut
the curve in as many points; then theſe parts may
be called Ordinates; and the line fo cutting them,
the Diameter; the interfection of the diameter and
curve, the Vertex; the interfection of two diame-
ters, the Center; the diameter perpendicular to the
ordinates,
86
B. II.
CURVE LINE S.
Fig. ordinates, if there be any, the Axis. And when
all the diameters concur in one point, that's the
general Center.
Again, the conic hyperbola being a line of the
fecond gender, has two affymptotes; fo likewiſe
that of the third gender may have three; that of
the fourth four, and fo on; and they can have no
more. And as the parts of any right,line, between
the curve and its affymptotes are equal; fo like-
wife in curves of the third fort of lines, if any line
be drawn cutting the curve and its affymptotes in
three points; the fum of the two parts of it,
drawn the fame way from the affymptotes to the
curve; will be equal to the third part drawn the
contrary way from the third affymptote to the curve.
And fo of higher curves.
Alfo in the conic fections which are not para-
bolical. As the fquare of the ordinate or the
rectangle of the parts of it on each fide the dia-
meter, to the rectangle of the parts of the dia-
meter, terminating at the vertexes; as the latus
rectum, to the diſtance of the vertices, or tranf-
verfe diameter. So in non-parabolic curves of the
fecond gender, the folid under the three ordinates,
is to the folid under the three abfciffe or the
diſtances to the three vertices; in a certain given
ratio. In which ratio, if you take three lines pro-
portional to the three diameters, each to each;
then theſe three lines may be called each of them
the Latus Rectum, and thefe diameters the Tranf
verfe Diameters. And in the common parabola
which has but one vertex for one diameter, the
rectangle of the ordinates is equal to the rectangle
of the abfciffa and latus rectum. So in curves of
the ſecond kind which have but two vertexes for
the fame diameter. The folid under the three
ordinates, is equal to the folid under the two ab-
fciffæ
Sect. II.
87
CURVE LINES.
fciffe, and a given right line, which therefore Fig.
may be called the Latus Rectum.
Laftly, fince in the conic fections, where two
parallel lines terminating at the curve both ways;
are cut by two other parallels likewiſe terminating
at the curve; make the rectangle of the parts of
one of the firft, to the rectangle of the parts of
one of the ſecond lines, paffing thro' the fame
point of divifion as the rectangle of the parts of
the fecond of the former, to the rectangle of the
parts of the fecond of the latter two, paffing alfo
thro' the common point of their divifion. So when
four fuch lines are drawn in a curve of the fecond
kind, and each meeting it in three points. The
folid under the parts of the first line, will be to
that under the parts of the third; as the folid under
the parts of the fecond line, to that under the
parts of the fourth.
The legs of curves of the fecond and higher
genders, juft as thofe of the firft, running on in-
finitely; will be either of the hyperbolic kind, hav-
ing fome afymptote; or cf the parabolic kind,
having no affymptote. Theſe things appear plain-
ly from the foregoing propofitions; and therefore
it is fufficiently confirmed, that the properties of
curves of a fuperior order include thofe of an in-
ferior order.
PROP. XVIII. Prob.
To investigate the equation of a curve AMK, 85,
baving given the relation of the abfciffas to one ano-·
ther, and alfo the relation of the ordinates to one
another.
Let the abfciffa AD, and its ordinate DL be
given; and let AD = a, DL = b, any other ab-
fciffa AP, its ordinate PM = y. And let x
be
88
CURVE LINE S. B. II.
Fig. be denoted by a multiplied or divided by the va-
85. riable quantity v, or fome power or powers there-
of. And likewiſe let y be denoted by b multiplied
or divided by fome powers of v, according to the
nature of the figure. Then expunge v and you'll
have the equation of the curve.
Ex. 1.
Let AP vax, and PMvvby; then
alo
XX
aa
= VV =
whence
aa'
y=xx; and the curve
is a parabola convex towards AD.
Ex. 2.
Suppoſe AP vax, and PM
then v =
b
y;
V
X
b
; and xy ab; an equation
a
تو
to an hyperbola between the affymptotes.
Let AP
Ex. 3.
vax, and PM
Y
X
√vab
vubb; then v=
and y =
bbxxx
bx
; an equation to the ellipfis.
aa
Ex. 4.
XTA
Ι
am
Let AP va, PM = y=vno. Then v
12
whence bmx"= y; a figure of the
parabolic kind.
Ex
Sect. II. CURVE LINE S.
89
Ex. 5..
Suppoſe AP = x = va + v²a + via, and PM
= y = vb + v'bv'b; then is v+v² + v³ =
Fig.
85.
| ४
Let AP = xva, PM y =
y
and ay
bx, a right line.
Ex. 6.
bv
Then
༩༧- རྗ
| s
; whence J
за
I
x
V =
a
bx
goût
and confequently y =
за
x
ра
as and ifr1, then ya bx, and this is
the log. curve.
=
PROP. XIX.
☺
Prob.
To find the radius of curvature in the vertex of a
curve, where the abfciffa is perpendicular to it.
Let R be the radius of curvature, x the ab-
fciffa of the curve, y its ordinate. Then inſtead
of y in the equation of the curve, put √2Rx;
and throwing out fuch quantities as are infinitely
lefs than others, and then reducing the equation,
R will be had.
For let AGM be any curve, AG an infinitely 86.
fmall part of the curve, AF its abfciffa, GF its
ordinate perp: thereto, C the center of the circle
which paſſes thro' A and G, then CA is the radius
of the circle. And fince the part AG is common
to both the curve and the circle, CA will be the
radius of curvature required. Put RCA, AF
or BG, GF or BA y, then by the na-
ture
90
CURVE LINE S. B. II.
Fig. ture of the circle 2Rx -xxyy, but xx being
86. infinitely leſs than 2Rx, we fhall have yy 2Rx,
and y√2Rx. Therefore if 2Rx be put for
y in the equation of the curve, you'll have R re-
quired.
3.
Ex. I.
Let the curve be an ellipfis, 2t the tranſverſe,
2c the conjugate, the latus rectum. Then 2tx-
xx : yy :: 2t : 1, and 2tyy = 2tlx lxx; that is,
(by ſubſtitution) 2t x 2Rx = 2tlxlxx (be-
cauſe lxx is infinitely leſs than 2tlx) 2tlx; whence
2R, and R, or half the latus rectum.
Ex. 2.
=
Let AM be the ciffoid, whofe equation (by
Cor. 2. Pr. 2. Sect. II. B. I.) is y³ 2axx
yxx. For y put2Rx; then 2Rx/2Rx = 2axx
yxx (becauſe 2axx is infinitely greater than
yxx) 2axx. And fquaring, 8R³×³ = 4aax4. And
2R³ = aax = o, becaufe x is infinitely fmall.
Therefore the radius of curvature in the vertex is o.
Ex. 3
PL
Let the curve.be the conchoid, whofe equation
(by Prop. IV. B. I. Sect. I.) is TL X AL
✔TF²
is, TF - FL X AL =
AH; that
PF- FL × √TF² —
ting TF = a, TP = b,
TF — FL
TFFL
that is, put-
FP = d, FL = x, AL
x
➡y, the equation of the curve is a -x × y =
d. xX √2ax
XX.
And in the vertex, throw-
ing out the infinitely fmall quantities, ay d
√2ax, and aayy2ddax; that is,
4
2ddax. And aRdd, whence R =
dius of curvature in the vertex.
=
aa X 2Rx =
dd
a
>
the ra-
Cor.
Sect. II.
91
CURVE LINES.
87.
Cor. If the ordinates make not a right angle with Fig.
the diameter, but any oblique angle GFH, put s =
fine of the angle GFH, AF = x, FGy, R =
radius of curvature; then instead of y in the equa-
tion of the curve, put √2Rsx, and rejecting the in
finitely small quantities as before, and reducing the
equation, R will be found.
For let AFH be the diameter of the curve, AC
the radius of curvature, ABD a tangent at A, GB
parallel to the abfciffa AF, GD perp. to AD;
and let GD = v. Then if the arch AG be fup-
pofed infinitely fmall, AG = GF, and the part
AG of the curve will coincide with the circle.
But by the nature of the circle, AG² = DG ×
AL; that is, yy = 2Rv. But in the infinitely
fmall triangle DGB, DG (v): GB or AF (x)::
s.DBG or GFH (s): radius (1); therefore vsx;
whence yy2Rv
2Rsx, and y = 2Rsx.
Example.
Let DGA be a parabola, HAF a diameter, FG 88.
an ordinare, the latus rectum, of the diameter
HF; then if AF = x, FGy, ss. angle
GFH, we ſhall have lxyy 2Rsx, by the rule,
whence R = This may be denoted otherwiſe,
25
for Is :: AT : AP, and s =
AP
AT
√4DP² + PA, and if a = la
AP
AP
✔TP² + PA²
PA²
AP
tus rectum of the axis TP, s =
√4DP² + a × DP
DP X a
a
4DP² + a × DP
a + 4DP
a
4
I
H
8 + DP
That is, if S be the focus, s
92
B. II.
CURVE
LINES.
!
I
DS. Whence the radius of curvature in A
SA
== 1× √ SA
√SA
2
DS
the nature of the curve.
4SA
4DS
by
a
za, 1/
Fig.
88.
And if P falls upon S, then PA = ½ a, and
=a, whence Ra/2.
PROP. XX.
The Shadow of any given curve line, projected from
a lucid point, upon an infinite plane, will be a curve
of the fame gender, with the given one.
For the fhadow of any point will be in the right
line drawn thro' the light and that point. And any
right line in the given curve is projected into a
right line upon the plane. For imagine a plane
drawn thro' the light, and thro' a right line in the
given curve, and that this plane is extended to the
plane of the fhadow, it will cut this plane in a
right line, which will be the fhadow of the given
line. Therefore right lines that interfect one ano-
ther will be projected into lines or fhadows that
interfect one another. If any line cuts the curve,
its fhadow will cut the fhadow of the curve. If
any line touches the curve, the fhadows of the
line and curve will touch one another. If one part
of the curve interfect another, their fhadows will
alfo interfect. So that all points that are coincident
in the figure, will be coincident in the fhadow.
And all points at a diſtance in the figure,
will be diftant from one another in the fhadow.
Now let a right line be drawn in the figure to cut
it in as many points, as is the dimenfion of the
figure; and it cannot cut it in more. Then the
fhadow of this line will alſo be a right line, and
will
m
Fig. 77.
M
:
78.
M
2
A
80.
A
M
F
B
P
m
M
82.
G
B
R
N
I
M
85.
:
A
BA
G
M
A
P D
86. c
M
B
79.
D
M
B.
P
A
P
m
81.
F
M
772
83.
M
84
H
DB A
D
B
87
S
P
C
F
M
G
Hy
L
88.
F
с
m
PIVII. p.92.
N
OF
BICH
Sect. II.
93
CURVE LINE Ś.
will cut the ſhadow of the curve in as many points, Fig.
and no more. Therefore this fhadow is a curve of
the ſame dimenfion, or of the fame gender with
that given.
Cor. 1. Hence the fhadow of the circle, or any
conic fection, will be a conic fection. The fhadow of
lines of the third order, will be lines of the third or-
der, and fo on ad infinitum.
Cor. 2. If any part of the curve be higher (og
further from the plane) than the light is; that part
of the curve can cast no fhadow upon the plane ;
and the shadow of the curve will run on infinitely to-
wards that hand. And therefore whatever there is
in that part of the curve above the light, will be quite
loft in the fhadow, and no where be found.
Bþ
SECT.
94
B. II.
CURVE LINES.
SE C T. III.
Some fluxionary Propofitions neceffary
for inveftigating the nature of Curves.
Fig. H
89.
AVING handled feveral things of this na-
ture in my book of Fluxions, I intend not
to repeat them here; but fhall only take notice of
fuch as have been omitted there, or but flightly
mentioned. And this I the rather do, becauſe
many properties of curves of high genders are
far more eafily inveſtigated by the method of
fluxions, than by common algebra.
PROP. I.
Let AM be any curve, AP the abfciffa, PM an
ordinate perp. to it, MC the radius of curvature in
M, CF parallel to AP meeting the ordinate MPF
in F. Then the fecond fluxion of the ordinate, the
fluxion of the curve, and the line MT, are continu-
ally proportional; fuppofing the fluxion of the axis to
be constant.
Let M, m, r be three points in the curve, M.
and ms parallel to AP. Produce Mm to n, then
fince Momr, therefore Mm mn; then rn=
≈
fluxion of om, or the fecond fluxion of the ordi-
nate. Draw mF, then the triangles mrn, MFm
are fimilar, for nrm alternate angle mMF,
and mnr Mmo MmF; and therefore nr:
rm :: Mm: MF, but rm Mm, therefore nr:
Mm :: Mm : MF.
-
<
-
Cor.
Sect. III.
95
CURVE LINE S.
Cor. 1. Hence, the fluxion of the axis Mo to Fig.
fluxion of the curve Mm:: MF: MC, the radius of 89.
curvature in M.
Cor. 2. Hence the radius of curvature in any
curve is easily found; by first finding MF by this
Prop. And then MC by Cor. 1.
Example.
Let AP, PM y, AM, and let the
equation of the curve be 2ax + xxzz, this put
into fluxions, is 2ax + 2xx2zż, or ax + xx
=zz; then z =
x², and ÿì
2
a + x
ZZ
2ax + xx
a + x
*, and x² + y²
= x²
Z
a + x
a + x
x²
x²
2
ZZ
2ax + xx
2ax + xx
*
Then ÿ=
aax²
2ax + xx
ax × ax + xx
ax × a + x x x
2ax + xx
and y
ax
√2ax+xx
Then
Min
A X
a + x
(j):
·* (*) ::
2ax + xx
ax x x a + 2
2ax + xx
2
a + x x ✰
X
ZZ
a + x
Then :
१२.
a + x
a
*) :: MF
the radius of curvature.
And in the vertex where x is 0, the radius of
curvaturea. This curve is called the Catenary.
Z
Z
* (*):
3
M'N
a + x
MF
Z.
a
a + x
Z
a
x): MC
=
Rb 2
PROP
C
96
CURVE LINES. B. II.
Fig.
90.
Ey I.
PRO P. II.
If PF, QD, RC be three equidiftant ordinates in
any curve and infinitely near; and nDb be a tangent
at D. Then the fubtenfe nF or bC, will be equal to
half the fecond fluxion of the ordinate.
Db, and the
For draw FDa; then fince nD
fubtenfes nF, b℃ parallel, and the curvature
FDC the fame throughout; therefore nFbC.
Again, in the triangles nDF, bDa, fince nD =
Dỗ, and all the angles equal; therefore nF = ab
= bc.
C. Therefore 26C aC, and bC = aC.
Let FL and DO be parallel to
and let PFy, then LDỳ
OC or aC =
=
I
2
the axis PR, and
I
Oa, and Oa --
ÿ, and bC or nF = ½ aC = ½ ÿ.
2
Cor. If PF, QD, RC be three equidiftant ordi-
nates, and FD be drawn, cutting RC in a. Then
Ca fecond fluxion of the ordinate.
For aC
2bC=j, by the demonſtration.
PROP. III.
In any curve whofe ordinates are drawn from a
fixed point C. Let CM, Cm, Cn be three ordinates
infinitely near, fo that the perpendiculars Mo, mp
may be equal. And if Mm be drawn interfecting Cn
And if Cmy, mp = x, mn z. Then
in s.
the line ns = ÿ
ZZ
y
For draw mr between s and C, making the an-
gle smr = mCn; then the triangles Cmn, mrs will
be fimilar, for the angles at m and C are equal by
conftruction, and msC or msr = mnC; therefore
Cm
Sect. III.
97
CURVE LINE S.
Cm (y): mr (ż) : : mr (ż) : rs =
ZZ
y
Fig.
Draw mq 91.
parallel to Mo. Then the angle pmq = mCn =
smr, whence the angle rmp smq=mMo, and
Momp. Therefore the triangles Mmo, mrp are
fimilar and equal, whence pr = mo. Therefore
ÿ = np mo = np
rs; therefore sn = ÿÿ —
rp = rn.
ZZ
y
But snrn
Cor. In any curve whofe ordinates are referred
to a fixed point; if CM, Cm, be two ordinates,
and mt a tangent at m; then the fubtence Mt =
yÿ
2y
ZZ
For Mt is half of sn =
2
N
ZZ
ÿ
2y
PROP. IV,
If ŷ, ÿ, ÿ, &c. denote the first, second, third, 92.
&c. moment, increment, or fluxion of y; and if P,
Q, R, S, T, &c. be feveral equidiftant ordinates in
a curve, being infinitely near one another. And let
the first P be called y. Then will
y = ÿ
q = ÿ + ÿj
ÿ
r = ÿ + 2ÿ + ✯
متر
s =ÿ + 3ÿ + 3ÿ + ÿÿ
&c:
For by the nature of curvature q = p + p ;
that is, q = y + ÿ, r = q + q, s = r + r, &c.
Therefore taking the fluxions of thefe equations,
Bb3
98
B. II.
CURVE
LINES.
Fig.
92.
q =ÿ + j, r = q + q,
&c. Whencer = q + q
s
j = r + ¿, q =
x,
j + j
=
ÿ + ÿ +
ÿ + ÿ +ÿ+ ÿ
=ÿ
= j +
2ÿ + Ÿ‚ s = r + ï =
†ÿ,
q
+ q + q + q =
q +
+ 2ÿ + ÿ + ÿ = j +
.
2q + q = ÿ + ÿÿ + 2ÿ
†ÿ
3ÿ + 3ÿ †ÿ, &c.
Cor. Hence the fluxion of the nth ordinate P, Q₂
R, S, &c. is j + n − 1 •ÿ +
n
I . N - 2
IN
n
I. N
2. n 3
+
j + &c.
2.3
V
93:
94.
93.
94.
PROP.
Let M be the point of contrary flexion in any curve
AM; and if curve AM = z, ordinate PM or
CM = y, Mox. Then in curves referred to an
axis, ÿ = o, and in fpirals yÿ = ", fuppofing &
Z", x
to be given.
For at M the point of contrary flexion, three
points n, M, m, infinitely near one another, lie
in a right line. Therefore in curves referred to an
axis, if Mo or x be given, om or y will be given,
confequently ÿ= 0.
And in fpirals let CT be perp. to the tangent
MT; then fince TM is a tangent at all the points,
n, M, m. Therefore at M, CT is a conftant
quantity; let CT=t, then by fimilar triangles
yx
ż : * : : y : CT ==t, and yx
t, and yx = tà, in
fluxions, jx = tz, x and t being conftant quanti-
y x
ties. Therefore yx, and jxz = yż, But
Sect. III.
99
CURVE LINE S..
ż² = x² + y²; therefore żż = ÿÿ, and z = Fig.
ÿÿ
Z
Therefore yxż = yx ×
yy
yÿ
and z =
Z
and 94.
ZZ
żż = yÿ.
у.
Cor. Hence to find the point of contrary flexion,
let x be given, and put ÿjo, in curves referred
to an axis; or ÿ
11
22
in fpirals.
y
Ex. I.
Let the equation of the curve b3 × ya
11
-5
;
2.
X
a
then
X
a;
y -
a =
63
and y
a
जात
a
X
a
whence y = ½ ×
5
b√b
2
15
लाल
X
ba/b
a
b√b
2
and ya+
xx, and ÿ
ÿ =
therefore x-a
forex = a.
142
46√
13-14
XX a x x x = 0;
= o, and x
a = 0. There-
Ex. 2.
Let ABM be a fort of ſpiral, RFB its gene- 95.
rating circle, radius CD, CM y, arch RD
= v. And let the nature of it be denoted by the
equation vyy
a³.
The ſectors CDd, and CMo are fimilar, whence
CM (y) : Mo (x) : : CD (r) : Dd
by the equation of the curve yy
rx
=
= v, and
y
+ 2vÿÿ = 0,
zvý
rx
rx
and v =
y
, and y =
= (be-
20
ryyx
cauſe v =
vy
20773
>
whence ÿ =
Bb 4
- ryyx
Q3
and
,
дзё
100
. B. II:
CURVE
LINESLINE S.
Fig.
95.
yÿ =
- ry¹ýx
goz y 4 x ²
a³
296
Alſo ż² = x² + ÿ² = x²
rry4x2
+
Whence
r²y4x²
406
296
= x² +
rry4x²
and
490
27²y4 = 4a+y4. And r2y4 4a6, and ry² =
2a3; whence y = √√√2a³. And if r = a, then
y = r√2.
PROP. VI.
If x, y, z, &c. repreſent any right lines; and in
any curve, if x × y × %, &c. = à given quantity ;
x
Y
then +
y
X
of
Z
&c. = 0.
22 / 22
ä
For let xyz= a, then the fluxion will be zyć
+xzỷ + xyż = o ; divide by the given equation;
-o;
Zyx x Zỷ XYZ
+
then +
1.8
O
X
; that is,
+
ZYX XZY XJZ
XYZ
y
+
z'
&c. = 9.
Cor. If xyz &c. a, a given quantity; then
+
my NZ
+- $30.
y
Z
0.
&
ગેસ
PROP.
Sect. III. CURVE LINE S.
ΙΟΙ
1
PROP. VII.
If x, y, z, v, u, w,
lines, and in any curve if
quantity; then
W
W
&c. = 0.
|
Fig.
&c. reprefent any right 95,
xyz &c.
vuw &c.
+
+
22 / 22
&c.
Pla
C
any given
И
น
xyz
Let
=a, then
XYZ
+
VU W
บนซ
xyzv
xyzu
xyzw
v²uw
vu²w
VUW²
xyz xyż
+
บนซ
VUW
&c.o. And dividing all by
xyx
X
ż
ข
a, we fhall have
+ +
VU W
X
W
&c. = 0.
น
70
Cor. If
xym &c.
x
then
{
vur &c.
my
+
&c.
-
X
D
nů
ཝ །©
&c. = 0.
www
PROP.
102
CURVE LINES. B. II.
Fig.
96.
PROP. VIII.
And
If from a given point P the right lines PA,
Pa, be drawn, cutting any curve in as many points
A, B, C, and a, b, c. as it has dimenfions.
the tangents AK, BL, CM; and ak, bl, cm, be
drawn, cutting any line PE, in the points K, L, M,
I
I
I
and k, l, m. Then will PK + PL + PM &c. =
I
Pk
I
I
+Pl+Pm &c. wherever the points A, B, C,
and a, b, c, fall.
Let AC, ac be fuppofed to move parallel to
themſelves, and then the point of interfection P
will move along the right line PE. Then fince
(Prop. XV. Sect. II.) PA x PB x PC:: Pa x
Pb x Pc f : 1, a given ratio. Let APx,
: :
PBy, PC z, and PA = v, Pb = u, Pc
=
w; then xyz=fvuw, or
(Prop. VII.)
and +
+
18
al
X y
+
& IN.
Z
Ў
Y
+
· 22 | 22
xyz
f. Therefore
VUW
И
W
O,
V
И
W
W
+
+
But fince
ข
U
W
12
AP is moved with a parallel motion, it is evident
the fluxion of AP is to the fluxion of EP, as AP
to the fubtangent PK; whence, putting EPs,
x
we have
and for the fame reaſon
Ў
11
X
PK'
J
PL'
11, and 24
Z
; alfo
PM
S
И
W
ט
Pk u
Pľ
W
S
S
S
Whence
+
+
+
Pm
PK
PL
PM
Pk
Pl
Sect. III.
103
CURVE LINE S.
s
I
I
I
I
==
Pk P/
Fig.
+P; and PK + PL + PM = 2+1+96
Ι
Pm
Pm³
But if any of the points K, L, M, &c. fall on
the contrary fide of P, then the correfpondent lines
PK, PL, PM, &c. muſt be made negative.
I
I
Cor. 1. If any line PC cut the curve in the points
A, B, C, and tangents drawn; then will PK + PL
+
I
PM
&c, be always equal to an invariable quantity.
Cor. 2. And bence if a right line PE cut the
curve in D, E, I, &c. as many points as it has di-
menfions; then will
I
PI
I
+ PE
I
PK
I
I
I
+ PL + PM = PD +
For in this cafe k, l, m, coincide with D, I, E.
Cor. 3. If any of the points K, L, M, &c. fall
on the contrary fide of P; you must take
I
PL'
I &c.
PM'
I
PK'
For lying the contrary way, they must have con-
trary figns.
PROP.
104
CURVE LINE S. B. II,
Fig.
PRO P. IX.
97. If the line AB be drawn thro' a given point P
parallel to the affymptote CM; and the tangents
AK, BL drawn, cutting any line DE (drawn alfo
thro' P, and cutting the curve in three points D, I,
I
I
E; and the affymptote in M). Then = +
I
PM PI
98.
I
1
I
PE PD
PK - PL
I
Ι
I
I
For (by Prop. VII.) PM + PK + PL = PI+
PE
I
PD?
becauſe D lies the contrary way from
P, and AB meets the curve and affymptote CM at
an infinite diftance where they touch. Then by
I
I
I
I
I
I
PD PK
PK-PL
tranſpoſing, PM =PI+ PE-PD
Cor. Hence the distance of the afymptote may be
found, by finding the point M. And if
I
PM
be = 0,
then the affymptote is at an infinite diftance, and the
curve will be of the parabolical kind.
PROP,
X.
To find the affymptotes of a curve.
This Prob. may be folved by the laſt Prop. and
its Corol, or otherwife as follows. Let APx,
PM y, TM a tangent at M, and TP the fub-
tangent. Find the value of the fubtangent
yx
за
by the equation of the curve, then putting x =
infinity in the value of the fubtangent, and alfo in
the
Sect. III.
CURVE
105
LINE S.
the equation of the curve, and you will get TP Fig.
and PM, which gives the poſition of the affymp- 98.
tote TM, and TP-x, gives the interfection T.
For the triangles Mrm and TPM are fimilar,
y
y x
and rm (j) : Mr(x)::MP():TP = And
fuppofing M at an infinite diftance x will be infi-
nite, from whence both TP and PM will be had,
and confequently the pofition of TM the affymptote.
Ex. I.
Let the equation of the curve DM be xyy =
@3; in fluxions y²+ 2xyy
2xyy
yy
2
o, and ✰
2xy
y
y x
Therefore PT=
2X.
У
And PM } =
a3
✓ Then making
X
Then making x infinite,
PT = — 2× = — infinity, and PM =
2X
Ja
X
o; therefore the affymptote coincides with the axis
TP.
Ex. 2.
Let aayy2abbx + bbxx.
2abbx+2bbxx, and x =
yx
12abbx+bb xx
aa
In fluxions 2aayý
aayy
there-
abb + bbx
aayy
abb bbx
and
,
y =
Make x infinite, then PT =
2abbx+bbxx
fore PT =
J2abbx
aayy
abb + bbx
aayy
bbxx
bbx
bbx
bbx
X
= infinity, and PM = y =
Jb b x x
bx
in-
finity. Therefore PT: PM :: * :
aa
bx
::a: b. Al-
a:b.
a
a
fo
106
B. It
CURVE LINES.
Fig.
aayy
fo ATTP - AP =
98.
abb + bbx
2abbx + bbxx abbx
bbxx
abbx
abb + bbx
abb + bbx
abbx
(when x is infinite)
bbx
a. Therefore the point
T is determined, and the pofition of the affymp-
tote TM.
Alfo fince y may be
bbxx
aa
bx
by, this
a
gives another affymptote on the other fide of the
axis AP, and making the fame angle.
axy
Ex. 3.
Let the equation of the curve be, y3 aay +
-3o. Extracting the root by infinite fe-
ries, then yx — — a + &c. The equation
3
aa
3x
of the curve put into fluxions, gives 3y'y
+axy + ayx 3xxo, whence x =
aa
ax
ЗУУ
ay3xxy, and PT –
aay
yx ax + 3yyaa
Y
y's
3xx ay
ax + 3yy
aa
and AT =
У
*
3xx ay
2axy + 3y³
aay
3223
3xx
ay
2axy aay
3x3 + 3aay
3xx
ay
zaay axy
3xx ay
axx
3xx
(expunging y)
(expunging y³)
3
a. But PT-
zaxy + 3x3
ax + 2aa × x
3xx ax &c.
axy + 3y³ = aay
A
||
3xx ay
(expunging y³)
axy-aay + 3x²+3aayзaxy
3XX ay
2aay
Sect. III.
CURVE LINES.
107
2ax zaa Xy Fig.
= 98.
2aay - 2axy + 3x³
323
3xx
ay
3xx
аху
(expunging y)
3x3
2ax² &c.
323
32
ax &c.
I
3
PT: PM: :x:y or x-
PT PM. And therefore
T falls between A and P;
322
=x, whence
*,
a::x:x; therefore
fince AT is negative,
and the affymptote
therefore lies between the curve and abſciſſa.
Ex. 4.
Let the equation of the curve be y³ + axy
I
3o, whofe root is y = x — — a +
उ
03
81x²
&c.
then the equation in fluxions is 3y2y + axỳ + ayx
3x²o, whence x =
3yy + ax
3xx―ay, and PT
yx
3y² + axy
y 3xx ay
= (expunging y³)
3x3
3axy+axy
3xx
ay
(expunging y)
I
3x3
2ax X x
a
3
3203
2ax² &c.
3x3
3xx
a x x
a
3xx ax &c.
3
3xx
=X.
Whence PT : PM : : ≈ : y or x
&c. ::x:x. Whence again PT
33 a3
उ
PM.
Alfo
3223
2axy
axy
TP X =
X
зах
= (ex-
ay
3**
ay
ȧx X x
I
a
punging)
3
3xx
I
ах х
a
3
I
axx
༣་ aax & c
axx
3xx
ax + + aa &c.
a,
3xx
3
=-a; whence
the fame conclufion will follow as in the laft.
Ex. 5.
axy
Suppoſe the equation of the curve is y
x3=0; this in fluxions is 3y-axỳ — ayx-3xx
108
B. II.
LINES.
CURVE
3xx + ayy, and PT –
Fig.
3yy ax
98.
= ö; and x
=
3y3
axy
3xx + ay
=(expunging j³)
3223
3x² + 2axy
3xx + ay
i
yx
y
3x³ +3axy-axy
3xx + ay
n
And the root of the equation is
y = x + 3 a &c.
I
Therefore PT =
3x3 + 2ax × x + a
I
3xx + a x x + a
3
3x³ +2ax² &ċ.
3x² + ax &c.
=(when
x is infinite)
3x3
3x
I
*.
Therefore PT: PM: x:
y or x + 3 a::x:x, and PT = PM.
उ
99.
11
Alfo PT
3x3 + 2axy
axy
X
3xx + ay
3xx + ay
ax x x +
I
a
ax² &c.
3xx + a × x + — a
3x² &c.
= ÷ a = AT.
I
3
3
Therefore fince AT is affirmative, the curve is
concave to the abfciffa, and the curve lies between
the abfciffa and the affymptote.
PRO P. XI.
Let AMN be any curve, CN its affymptote; then
if the ordinates be parallel to the affymptote; they
are depreffed a dimenfion lower in the equation of
the curve.
Let AP = x, ordinate PM y, and let the
y, and let the equa
tion of the curve be y²+qx" -1 &c. = "x", and
C
the root y=rx + B + &c. Take s infinitely
X
near M, and draw so parallel to AP, and DM
parallel to CN; then by fimilar triangles, Mo (y)
yx
: so (x) : : MP (†) : PT =
the fubtangent. But
y
Sect. III.
109
CURVE
LINES.
CX
yx
&c. and therefore PT =
Cx
XX
rx
XX
r
al
r
y
C
xx
;
and when x is infinite, the ſubtangent
= CQ, CN the affymptote being a tangent
at an infinite diftance. Draw AB parallel to PM,
and let CA t, CB = d, AB = b, CD = v₂
DM 2. Then by fimilar triangles, we have CQ
(2): QN
=
v,
: QN (y) : : CA (†): AB (b), whence ty =
and trb. Alfo AB (b): BC (d): PM
by,
r
bz
(y): DM (≈), whence y =
d
Alſo AB (b) :
ty
Then AP = PC
Fig.
99.
AC (t): PM (y): PD =
tz
d
ty
AC; that is, x = 2/+v―t=(expunging y)
+vt. Then fubftituting theſe quantities
inſtead of y and x, in the equation of the curve,
72
X
bn zn
tz
dn
+ax
+ v t
n-1 &c. = r" ×
22
12
tz
d
zn in zn
dn
72
&c. But brt, and b"
I
; therefore 2" the higheſt power of the or-
dinate vanishes; therefore when the ordinate DM is
parallel to the affymptote, it lofes one dimenſion
in the equation of the curve,
C.c
PROP.
110
CURVE LINE S. B. II.
Fig.
PROP.
XII.
100. In any curve referred to an axis, if the fecond
fluxion of the ordinate PM be negative; the curve is
concave towards the axis AP. But if the Second
fluxion of the ordinate be affirmative, the curve is
convex towards the axis; fuppofing the fluxion of the
qxis given.
Draw the ordinates pm, qn parallel to PM, at
equal but infinitely fmall diſtances. Produce the
infinitely ſmall portion of the curve Mm to s. And
draw Mo, mr parallel to the axis AP. Then the
fluxion of the axis is Mo mr, and confequently
mo sr, the fluxion of the ordinate. Now rn-
mo is the fecond fluxion of the ordinate; that is,
rnrs is the fecond fluxion, and this will be ne-
gative when rn is lefs than rs, in which cafe Mmn
is concave towards AP. But if rn rs is affir-
mative, then rn is greater than rs, and then ʼn falls
above s, and therefore the curve is convex towards
AP.
Cor. From hence may be known whether a curve,
referred to an axis, is concave or convex towards the
axis in any given point. Put the equation of the
curve into fluxions, and find the value of the fecond
fluxion of the ordinate, making the fluxion of the
axis conftant. Then if the value of that fecond flux-
ion be negative, the curve is concave; if affirmative,
'tis convex to the axis.
Ex. I.
Let the equation of the curve be aayy = 2bbrx
bbxx, this put into fluxions, and divided by
2, gives aayy = bbrx+bbxx, and put in fluxions
again, aayy + aayÿj = bbxx, and aayÿj = bbxx
aayy,
Sect. III.
III
CURVE
LINE S.
bbx
64x2
aajy, but y =
xr+x, and j² =
Fig.
X
aay
a+yy
100.
64x2
2
r+x, whence aayÿ = bbxx
2
xr + x =
aayy
2
bbaayy b4 X rŵ x² = (expunging yy)
b4rrxx
aayy
; therefore ÿ being negative, the curve is
aayy
concave to the axis.
Ex. 2.
Let the equation of the curve be aay + xxy =
axx; then y =
axx
aa + xx
2axx
› and y =
aa + xx
aa + xx
and ÿ =
2ax³x 2a³xx + 2ax³ x 2ax³ x
2.
2a3xx
aa + xx
2a3xx X aa + xx
aa + xx
2
2a³xx X 4xx
aa + xx
3
2a3xx
aa + xx
8a³xxxx
295x2
6a3x²x²
3
aa + xx
3
aa + xx
2032
× aa
aa + xx³
3
3xx. Therefore if 3xx be
greater than aa, the curve is concave towards the
axis; but if aa be greater than 3xx, it is convex.
Ex. 3.
=
If the equation of the curve be aax- a³ — y — b³.
3ỳ × y — b², and again 3ÿ × y — b
3ỷ × y
Then aax
+ 3ỷ × 2ỷ × y —bo,
and ÿ x y
b + 2ÿÿ
2yy
aax
= o, and j =
y-
b
But y =
ЗХУ
b
294xx
therefore ÿ
39
; confequently if y
9xy-
b
be greater than b, the curve is concave towards the
Cc 2
axis;
1 12
LINE S. B. II.
CURVE
Fig. axis; but if b be greater than y, the curve is con-
100. vex towards the axis.
91.
PRO P. XIII.
In a curve where the ordinates meet in a fixed
point, let the ordinate CM or Cm = y, perp. Mo =
x, Mm ; then fuppofing x to be conftant, if y
-żż be negative, the curve is concave towards the
center; but if affirmative, 'tis convex.
For by Prop III. Fig. 91. sn =ÿ
ZZ
y'
which
is affirmative if n lie above s, but negative when
'tis below s. And therefore yÿ żż will be af
firmative in the former cafe; that is, when the
curve is convex; and negative in the latter, or
when the curve is concave.
Cor. Hence if a curve is referred to a fixed point,
to know whether it is concave or convex; put the
equation of the curve into fluxions, and making x
conftant, find the value of yÿj
ZZ; then if it be
negative the curve is concave towards the center; if
affirmative, 'tis convex.
Ex. I.
Let the equation of the curve be yx =
Z√ ay
2.
Then x²=
2× ay az
a
2
xx²+ y², whence
4yy
4y
4y
41
X
j² =
a
zỷ
X
Alfo =
a
a
дyyxx
ay
4уxx
43a, and ÿ
× ¹³, and y = √√ 49
2yx
ż
Vay
; thereforeÿ-żż
and żż =
zz
гуйх
Sect. III. CURVE LINES.
113
2.
4YXX
2y
X
a
2yx²
√ 49
4уxx
a
a
2yxx
a
a
a
4уx²
Fig.
a
91.
which being negative
the curve is concave towards the center.
Ex. 2.
=
Let ABM be a fpiral, FBR the generating cir-
cle, radius CRr, arch RD v, CM = y, and
let the nature of the curve be denoted by the equa-
tion vyaa. By the ſimilar ſectors DCd, MC,
r: vy, whence yur. And putting the
equation of the curve into fluxions, we have vỷ +
rx, then y =-
уч
yʊ = o, and vỷ
2
and yÿ =
rx
aa
X
aa
Alfo żż = x² + j;²
= - yv =
rx
ryx
› and ÿ =
ryx
V
aa
aa
rryxx
a4
x² +
a4
a+
Therefore yÿ
ZZ
żż =
rryyxx
at + r²y²
a4
²x²; which
a4
rryyx²
04
a+x² + rryyx²
95.
being negative the curve is concave to the center.
Ex. 3.
But if the equation of the curve be vy3a4,
and r = a, then 3vy'ỳ +y'v=0; that is (becauſe
3a4j
yv = ax)
y
+ ay³xo, whence y=
y³ x
and ÿ =
32xy
325
a3
-, and yÿ
313x
y3x
заз
3a°
16 x12
C c 3
yo xi z
Alfo = x² + y²
**
yas Therefore yÿ — żż —
345
аз
9ax²
9a
90° +
gas +ys
a3
X
90s
336
114
B. II-
CURVE
LINES.
Fig. 3y6 — 9a6 — yo
2y6
даб
x²
95.
9a6
x. Therefore if
9a6
ΙΟΙ.
2y be leffer than 9a6, the curve is concave; but
if greater, convex towards the center.
PROP.
XIV.
=
Let AMC be a curve, AP = x, PM y, AM
=z, Mm = 2, Mn = x, mny, and let y be
given; and if a and b be given quantities, and ax
bz be the nature of the curve, then will aż bx be
a maximum or minimum.
2.
2
For taking the fluxion of aż -bx, we have až
bxo, and až = bx. But ²² + y²;
therefore 222 2xx, and żż =
xx; whence by
XX
axx
dividing, z=
and až =
,
; therefore
Z
Z
axx
Therefore on the con-
Z
bx, and ax = bz.
trary when axbż, then aż - bx will be a maxi-
mum or minimum.
Cor. If ax = bz be the nature of a curve, then
the Fl: aż - Fl: bx will be a maximum or minimum.
For aż bx is a maximum or minimum, and
therefore the fum of all the aż the fum of all
the bx will be a maximum or minimum; that is,
fluent of aż fluent of bx will be a maximum
or minimum.
PROP.
P
Fig. 89.
M
72
A
17
72
go.
D
92.
93.
n
P Q R S T
96.
KAD
m
1
9.9.
TA P
102.
m
མ༡༦
A
B C D
!
R
m
94.
M
10
T
A
P
100.
m
91.
m
10
M
M
B
m
10
n
95.
R
C
97.
K
P
IM
98.
M
B
ת
M
m
M
In
101.
A
P
Pp
D
m
PIVIII the End
UN
OF
MICH
Sect. III.
115
CURVE LINE S.
PROP XV.
Fig.
If Bm, Dr, be two ordinates of a curve, infinite- 102.
ly near; and let the ordinate Cn be drawn, fo that
the differences of the ordinates on, tr, may be equal.
=
Then put mo=x, os or nt = v, on or tr = ÿ;
then if ax + bv = d, a given quantity; and p x
x² + j²² + q × v² + j² = a maximum or mini-
mum; then pbx× ï² + j² = qav × v² + j²
will denote the nature of the curve.
-HI
-I
-22 I
For putting the two given equations
into fluxi-
ax
ons, ax + bvo, and v = -
b
Alſo 2pnxx
-22 I
•2 I
x x² + y²
+ 2qnvv × vv +ÿÿ
=o, and
·NI 2qnaïv
-1- I
2pnxx × x² + jjz
b
x v² + j²
•12 I
o, and tranfpofing and dividing bpx × xx + ÿÿ
= aqv x vv + ÿÿ
7 I
P
Cor. Hence
2122
× mo × mn
a
21-2
Ny
0/0
× nt X
b
For xx + ÿÿ = mn², and ¿¹ + ÿ² = nr,
&² jnr, whence
pbx × mn²n—2
b
qv
X nr
212-2
11
2122
px
= gav X nr 2n−2, and × mn
a
212--20
Many more properties of curves might be ob-
tained by the method of fluxions; but I forbear
purſuing them any further; fince what have been
here delivered, are fufficient for an introduction.
FINI S.
THE
CONTENT S.
BOOK I.
Page
1
Sect. I.
Sect. II.
T
THE Conchoid,
The Ciffoid,
Sect. III:
The Cycloid,
I
5
II
Sect. IV.
The Quadratrix,
16
Sect. V.
The Logarithmic Curve,
19
25
31
38
Sect. VI. The Spiral of Archimedes,
Sect. VII. The Logarithmic Spiral,
Sect. VIII. The Hyperbolic Spiral,
BOOK
II.
Sect. I. The generation of curves from one ano-
ther by transformation; by which their
lengths, areas, furfaces or folidities are
determined,
49
Sect. II. The general properties of curves, the de-
termination of their axes, ordinates, and
affymptotes,
65
Sect. III. Several propofitions drawn from fluxions,
very neceffary for inveftigating the na-
ture of curves.
94
ERRA T A.
For the A R.
INFINITES.
b fignifies reckon from the bottom.
page line [read
33bs = n + 1. a
713 all the xes
m
12 11 C - A+ 26 + C-
44 ob FINIS. Dele THE.
CONIC
SECTIONS.
Where b follows, reckon from the bottom.
[page] line read
18
290 +O
34-
56
60
131
66 XIII. Cor. 1.)
56 OD × DN : :
7 lines el,
2 ib. Prop. III.)
4 multiplying by 4.
CURVE LINES.
89
7 whence a
99
2 |= jà ×
y* × 2
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