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Of
PLANE ANALYTIC GEOMETRY
For Scientific and Technical Workers
M. A. ROSANOFF, Sc.D.,
Sometime Research Professor of Chemistry in Clark University
LONG ISLAND UNIVERSITY PRESS
300 PEARL STREET BROOKLYN 1, N. Y.
Copyright, 1944, by
H. G. & L. R. Lieber
ALL RIGHTS RESERVED
This book or any part of it
shall not be reproduced in any
form whatsoever without the
written permission of H. G. &
R. Lieber.
THE SCIENCE PRESS PRINTING COMPANY
LAN CASTER, PENNSYLVANIA
PREFACE
This little book aims to supply a need widely felt by laboratory
workers in science and technology, by teachers of elementary
algebra, and by others, the need, namely, of a clearer grasp of
the Cartesian method than that usually retained from the Ortho-
dox college course in Analytic Geometry. In the equipment of
the scientific or technical worker who is confronted with problems
of curve fitting, nothing is more useful than a fluent knowledge
of equations of the first and second degrees (which form the main
content of ordinary Analytic Geometry). Again, linear and
quadratic equations in two unknowns, and their graphs, consti-
tute an important part of the high-school course in Algebra, and
the teacher of the subject is not adequately equipped if he does
not possess a thorough knowledge of such equations. -
With my prospective semi-mathematical reader in mind, I have
put more emphasis upon the algebraic than the geometric aspects
of the subject, more emphasis upon the equations and their pecu-
liarities than upon the geometric properties of the curves repre-
sented by them. And with a view to the use that the scientific
or technical worker is likely to make of the method, I have treated
straight lines and the parabola more fully than the circle, the
ellipse, and the hyperbola.
The interpretation of numerical equations of the second degree
is greatly simplified by my teaching how, while leaving them
essentially unchanged, to put them in an explicit form exposing
to view both the conic’s nature and position, which are implicit
in any given equation. Informed readers will note also that I
have replaced the ‘‘discriminant” of the general equation of the
second degree by a simpler expression and that I have stressed
throughout the significance of the constant term. Appended,
finally, is a brief chapter on the more characteristic invariants of
equations of the second degree, approached in a simple and
somewhat novel way.
Notwithstanding these and other new departures, I am in-
debted to a number of standard works on analytic geometry,
including of course Charles Smith’s Comic Sections.
The manuscript was examined by Dr. T. H. Dunkelberger,
Professor of Chemistry in Alfred University, and by Dr. George
111
iV Preface
E. Davis, Professor of Physics in the University of Oklahoma.
Both encourage me in the belief that the little book should be a
useful aid in capturing, or recapturing, a sufficient minimum of
the algebra (Descartes would have said, the arithmetic) of the
conic sections. The figures were drawn by Edward M. Kihn,
Ph.D. (London), an expert mechanical draftsman. My thanks
are due also to Mr. E. H. McClelland, Head of the Technology
Department in the Carnegie Library of Pittsburgh, for his gem-
erous and intelligent cooperation.
The student is urged not to read this booklet without the use
of a pencil and a ream of paper. I have made every effort to
write it in simple form. But we are dealing here essentially with
a mathematical technique, and no technique can be acquired by
mere reading. As an inducement to my busy readers to spend
an unavoidable minimum of time on numerical exercises, I have
outlined the solutions of all the problems in the book. But the
reader must not neglect to work them all out in detail; else—as so
often happens—his knowledge of this fundamental subject will be
tenuous to begin with, and in time will dissolve to practically
nothing.
M. A. ROSANOFF
1944.
CONTENTS
PAGE .
Introduction . vii
CHAPTER I. THE STRAIGHT LINE
§1. Equation of the Straight Line 1
§2. A Line Passing through a Given Point 3
§3. A Line Passing through Two Given Points 3
§4. Solution of Simultaneous Linear Equations 4
§5. Parallel and Perpendicular Lines 4.
§6. Calculation of DISTANCES:
(1) Distance between Two Points (and equations of the Circle,
Bllipse, and Hyperbola)
(2) Distance between Two Parallel Lines
(3) Distance between a Point and a Straight Line ....................................
§7. Addition of Linear Equations
§8. Multiplication of Linear Equations
(1) Two Parallel or Coincident Lines
(2) Two Intersecting Lines (and the Hyperbola) .......................................
Numerical Examples
CHAPTER II. THE PARABOLA
$9. The Equation y = ca.”
§10. Other Forms of Parabolic Equations:
(1) The Equation y = a + ba + ca”
(2) The Equation a = cy”
(3) Parabola Axis at an Angle of 45°
(4) General (Explicit.) Equation of the Parabola ..........................................
Numerical Examples
§11. An Important Property of the Parabola
§12. A Theorem Relating to the Linear Part of All Parabola Equations ......
11
12
13
15
16
20
28
30
CHAPTER III. INTERPRETATION OF ANY EQUATION OF
THE SECOND DEGREE
(1) Two Parallel or Coincident Lines
(2) A Parabola
(3) Two Intersecting Lines
(4) A Hyperbola
(5) Identification of an Ellipse or Circle Equation ....................................
(6) Explicit Form of an Ellipse or Circle or Hyperbola. Equation
Summary of Identification of Comic Equations
Numerical Examples
35
35
37
39
44
49
52
55
55
55
56
56
56
58
59
V
Vi Contemis
CHAPTER IV. INVARLANTS OF EQUATIONS OF THE SECOND
DEGREE (witH NUMERICAL ExAMPLEs)
(1) Equations of the Ellipse (or Circle)
(2) Equations of the Hyperbola
(3) Equations of Intersecting Lines
(4) Equations of the Parabola
(5) Equations of Parallel Lines
Summary
69
72
73
74
74
76
INTRODUCTION
Analytic Geometry deals with the representation of geometric
figures by algebraic equations and vice versa. This is achieved
by introducing into geometry the concept of position, and by de-
fining the position of any point in the plane by its distances from
an arbitrary fixed pair of perpendicular lines called the Coordi-
nate Axes. One of these axes is generally placed in a horizontal
position and is called the X-axis; the second, or vertical, axis is
called the Y-axis. The point of intersection of the coordinate
axes is obviously at a distance zero from either of them and is
denoted by the symbol (0, 0). Horizontal distances from the
Y-axis are denoted by the general symbol ac, vertical distances
from the X-axis by the general symbol y; and thus the symbol
(ac, y) denotes the position of any point in the plane.
Once the position of any point is expressed by reference to the
fixed coordinate axes, everything else follows easily. For ex-
ample, let a given straight line be located parallel to the X-axis,
at a perpendicular distance a above it; then obviously the posi-
tions of all points on the line are expressed by the equation y = a,
or y – a = 0. And conversely, this equation will represent only
the straight line so located and no other. Similarly, a straight
line in every other position, as well as every geometric curve, can
be expressed without ambiguity by an equation which will repre-
sent it exhaustively both as to its geometric properties and as to
the position of all its points; and thereby is established what is
known as the “one to one correspondence” between geometric
figures and their algebraic equations.
On account of the conciseness and flexibility of the algebraic
method, its infusion into geometry resulted in great simplification
of geometric investigations. Thus, it may safely be guessed that
Kepler, who labored twenty years to discover the laws of plane-
tary motion, would have done so within a single year, had the
vii
viii Introduction
method of ‘‘algebraic,’’ ‘‘coordinate,” or “analytic,” geometry
been known when he entered upon the task in the year 1600.
Analytic Geometry was invented by the French philosopher and
mathematician René Descartes (Latinized, Rematus Cartesius) and
published by him in 1637. It paved the way to the invention of
the calculus by Newton and Leibnitz and thus became the founda-
tion of all modern mathematics and of all modern quantitative
science.
CHAPTER I
THE STRAIGHT LINE
§1. EQUATION OF THE STRAIGHT LINE
If two quantities y and a vary together so that their ratio re-
mains constant, we say that y is proportional to ac, and we express
their relationship by the equation
g = bac, (1)
in which b is the ‘‘proportionality constant.” Geometrically this
equation is represented by a straight line; and since y is neces-
sarily zero when a = 0, the line must pass through the coordinate
Y

FIG. I.
origin. By elementary trigonometry, the ratio y/a, (- b) is the
tangent of the angle which the line makes with the X-axis, and
this tangent is used as a measure of the “slope” of the line.
A more general expression of the linear relationship between
!y and a is furnished by the equation
g = a + bac, (2)
and Fig. 1 is a geometric picture of such an equation.
I
2 Plane Analytic Geometry
The proportionality principle persists in this case also, but in-
stead of y itself being proportional to ac, it is y diminished by the
constant quantity a (which denotes the point at which our line
cuts the Y-axis) : y – a = bac. Here again b appears as the slope
of the line, or the tangent of the angle which it makes with the
X-axis.
A familiar example of the linear equation is supplied by the
relationship between the Centigrade and the Fahrenheit scales.
The temperature interval between freezing and boiling water is
measured by 180 degrees Fahrenheit, by 100 degrees Centigrade,
and for any other interval the ratio is the same : 180 to 100, or
1.8 to 1; which shows the proportionality of the two scales. Yet
we cannot transform Centigrade into Fahrenheit readings by
simply multiplying by the proportionality coefficient 1.8, because
in freezing water a Fahrenheit thermometer reads 32°, a Centi-
grade thermometer 0°. If we denote Fahrenheit readings by y,
Centigrade readings by ac, we must write:
Qy = 32 + 1.82.
This is the equation of a straight line; the proportionality coeffi-
cient 1.8 represents its slope; the constant term shows that the
line cuts the Y-axis at the point 32 above the coordinate origin.
In correspondence with the facts, the equation gives: for the freez-
ing-point a = 0(*C.), y = 32(*F.); for the boiling-point a = 100,
!y = 212; and so on for all other points. Thus mathematical equa-
tions serve at least as the shorthand of science. But the student
should strive to use them, not mechanically, but as far as possible
with concrete understanding of what they tell.
Equation (2) is called the slope equation, and for representing
scientific data it is generally preferred to other equations of the
straight line.
For a line parallel to the X-axis the slope is b = 0, and the equa-
tion of the line becomes y = a. The equation of the X-axis itself
is y = 0. However, the slope equation is not usable with lines
parallel to the Y-axis: such lines make with the X-axis an angle
of 90° and their slope b (i.e., tan 90°) is infinite; also, such lines
meet the Y-axis at infinity and therefore the constant term a is
likewise infinite. Our equation thus assumes the unusable form :
g = co -- Co.ac. In such cases it is best to reverse the variables and
For Scientific and Technical Workers 3
use the form : a = a + bºy, in which a’ is the point where the line
cuts the X-axis and bº is its slope with respect to the Y-axis. In
fact, since the line is parallel to the Y-axis, its slope bº = 0 and the
equation becomes simply a = a'. The equation of the Y-axis itself
is a = 0.
§2. A LINE PASSING THROUGH A GIVEN POINT
Since the general linear equation has two constants, i.e., two
degrees of freedom, two separate conditions are required to fix the
position of any particular line: for instance, one point on the line
and its slope; or two points through which the line must pass; or
the like. If only one point is given, the line is free to spin round
it as a pivot, and that freedom must be eliminated before the posi-
tion of the line is completely fixed. To obtain the equation of
such a half-free line, let the coordinates of the given point be aca,
91. The equation of the straight line
g = a + ba:
will be satisfied by this pair of coordinates, since the point they
represent is itself on the line. So we have :
91 = a + ba 1.
This gives the value of a in terms of the given aci and y1 : viz.,
a = y1 – baci. This, obviously, still depends on the undetermined
value of b ; corresponding to the geometrical fact that the point
at which the Y-axis will be cut by a line passing through a given
point will depend on the line’s slope. Substituting this half-
determined value of a in our general equation, we get:
g = (y1 — bacı) + bac; (3)
and this is the required equation of any line passing through the
point (21, y1).
§3. A LINE PASSING THROUGH TWO GIVEN POINTS
If the two given points of the line are (acı, y1) and (22, y2), we
have
$/1 = Q, -- ba.1
!/2 = 0 + bac2.
Solving these two as simultaneous equations in a and b, we find:
_301/2 - 224/1. b = 91-92,
= —; v = —
Q, 5
3C1 - 302 2C1 - 0:2
4 Plame Amalytic Geometry
and substituting these values in the general equation y = a + bac,
we get the required equation of a line passing through two given
points, viz.:
- 2013/2 - £2?/1 +91 – ?/2 .30 (4)
3C1 -*. 2C2 £1 º 302
Q/
For a simple example, let acı = 0, yi = 0; a 2 = 1, y2 = 1. In that
case our equation becomes y = a ; we have the line passing through
the origin and making an angle of 45° with the X-axis (since the
slope b = 1 = tan 45°).
§4. SOLUTION OF SIMULTANEOUS LINEAR
EQUATIONS
Given two lines
!/ = 0,1 + b1a.
g = d2+ boa,
every point on each line will represent a pair of corresponding
values of 2 and y. At the point of intersection of the limes one
pair of values of a and y will satisfy both equations, and “solv-
ing” the equations means finding that pair of values, i.e., the
point of intersection. If a 3, yi denote the coordinates of the point
of intersection,” our two given lines are thus found to intersect at
0.1 - 0.2 . alb2 - d.2b1
E.T., ; 94 = Ti,
If the given lines happened to be parallel, their slopes would
of course be equal: ba = bi, or be — bi = 0, and in our “solution.’’
both a and yi would be infinite, corresponding to the conception
that parallel lines intersect at infinity. In elementary algebra
courses, certain pairs of linear equations are described as “incon-
sistent” and “unsolvable,”—which may be mystifying to the
young pupils. In the ordinary sense of the word, there is really
nothing “inconsistent” about those equations; they merely repre-
sent parallel lines and for this reason have no finite values of a
and y in common. &-
3C i =
§5. PARALLEL AND PERPENDICULAR LINES
We have pointed out the obvious fact that lines having the same
slope are parallel. Given a pair of linear equations, the student
* The subscript i is usually reserved for a different purpose. In this book
it is used merely as a reminder that (a, i, yi) is a point of intersection.
For Scientific and Technical Workers 5
will readily determine whether or not they represent parallel lines
by putting them in the form adopted above, in which y always has
the same coefficient 1; then, if the coefficients of a are equal, as in
the pair:
!y = a1 + ba:
y = a + bac,
the two lines are parallel.
We proceed, with the aid of the accompanying figure, to inves-
tigate the condition of perpendicularity. In the figure, the
auxiliary line BC is made perpendicular to the X-axis.
Y
AT
FIG. 2
Let the line y = a1 + bia: form with the X-axis the angle 0; there-
fore b1 = tan 0. The perpendicular line y = ax + baa forms with
the X-axis the angle 0 +90°, and therefore its slope ba = tan
BC 1. 1.
O = — = — —- = — — . d ffi-
(6+90°) == CA tan 0 b1 Thus, the necessary and su
cient condition of perpendicularity is expressed by the equations:
v = a1 + ba:
=a,-4,
{ſ = 0:2 b “
Or in words: a line will be perpendicular to a given line if its
slope is the reciprocal of the slope of the given line and has the

6 Plane Amalytic Geometry
opposite sign, so that the product of the two slopes is — 1. For
example, given two lines: 32 + 2y = 1 and 23 – 3y = 5; are they
perpendicular 3 The first equation changes to v-,-,-, the
second to y = -*** The answer is, the lines are perpendicular,
because the product of their slopes (-; and + # is — 1.
§6. CALCULATION OF DISTANCES
All algebraic representation of geometric figures is based on
DISTANCES, and consequently the calculation of distances is
as fundamental to analytic geometry as the four basic operations
are to arithmetic. The required distances are: between two
points, between two parallel lines, and between a point and a line.
The reader must not fail to acquire a fluent and permanent knowl-
edge of the simple methods used, by studying the present Section
with especially great care.
(1) Distance between Two Points.
Y
(wi, !/1)
(a 2, y2) - (r1, y2)
FIG. 3
Given two points, (aca, vi) and (aca, y2), let us denote the distance
between them by the letter r. Fig. 3 shows r to be the hypotenuse
of a right triangle whose legs are a 1 – a 2 and y1 – ya. Therefore,
r” = (a, – aca)* + (y1 — ya)*, (5)
For Scientific and Technical Workers 7
and the required distance r equals the square root of the right
member of this expression. Some examples may serve to illus-
trate the wide usefulness of this simple proposition.
First, let us employ it to find the equation of the circle. The
circle is defined by any point (a, y) on its circumference being
at the same distance r from its center. Let any fixed point
(h, k) be the center. If in our expression (5) we substitute ac, y
for a 1, y1, and the fixed point (h, k) for a 2, y2, we immediately get
for the defining equation of the circle:
(a — h)* + (y – k)* = rº,
Or
ac” + y” – 2ha – 2ky + h^+ k” — r" = 0.
If we write this equation in the more generalized (or rather,
slightly less explicit) form:
ac” + y” + Da; -- Ey + F = 0,
it is easy to see, by comparison with the preceding form (with
which the last is essentially identical), that the center is at the
point (h = -#p, k = – #P) ; also, since h’ + k” — r" = F, the square
of the radius is rº = h^+ k” — F= #(D. + E°–4F).* If the center
happens to be at the coordinate origin, then — #p = h = 0;
-: E = k = 0; F = — r*; and our general equation of the circle as-
sumes the familiar special form :
ac” + y^ = r^.
Given, for instance, the circle ac” + y^ = 2a:. Find its radius and
the location of its center. We write : ac” + y” – 2a: = 0. Hence
D = -2; E = 0; F = 0. The radius r-MIFTEFIF-1. The
center is at (-; p. -#E ), therefore at (1,0).
As a second application of the distance expression (5), we will
use it to find the equation of the ellipse; but for simplicity’s sake
we will assume the center of the ellipse to be at the origin, and its
* Before applying these formulae for the center and radius, the coefficients
of ac” and y” must be reduced to 1.
Plane Analytic Geometry

s
*
§
g
For Scientific and Technical Workers 9
major axis to lie on the X-axis of the coordinates. Fig. 4 shows
an ellipse so located.
The circle is defined by a single distance—that between any of
its points and the center—which is its radius and which remains
of constant length. The ellipse is defined by the sum of two dis-
tances, viz., the distances between any one of its points and its
two foci F and F'. Those two distances are called the “focal
radii,” and for all points of the ellipse the sum of the two remains
constant and (nota bene :) equal to the major axis. For instance,
by the definition of the ellipse, BF" + BF is the same as PF" + PF,
and either sum is equal to A^A. Let c denote the distance of
either focus, say F, from the center O ; let OA, which is half the
major axis, be denoted by a ; and let OB, which is half the minor
axis, be denoted by b. Since BF = BF, and since their sum equals
the major axis 2a (because B is a point on the ellipse), it follows
that BF = a. The right triangle BOF thus gives: c” = a” — b%. We
further note that the foci are respectively at the points (c, 0) and
(— c, 0). And now we can formulate the law of the ellipse by
employing our distance expression (5).
Let (ac, y) be any point P on the ellipse. Its distance from the
focus (c, 0) is V(x - c.)2 + (y – O)2. Its distance from the focus
(- c, 0) is V(x +c)? - (y – 0)*. And the sum of the two dis-
tances equals 20 :
V(a – c.)* + y” + V (a + c)* + y^ = 2a,
Or
Var* -- y? H. cº - 2ca; - Vicº Hºy” cº-2cc = 2a.
To simplify the algebraic work, we put ac” + y” + c2 = m. Then
Vm – 26a. -- \/m + 2Ca = 2a.
Squaring and dividing by 2:
m-2a4 =–Vng Tacº.
On squaring a second time, the term 'mº cancels out; and, dividing
by 4, we obtain:
a 4 – ma? -- cºacº = 0.
By substitution of a” — b% for c”, and of a 4 + y” + a” — b% for m, we
find :
b222 + a”y” - a*b*,
OI’
acº y?
ſatiº =
which is the familiar equation of the ellipse.
1,
10 Plame Analytic Geometry
The reader will find it instructive to undertake still a third
exercise in the use of the distance expression (5): namely, to
derive the equation of the hyperbola associated with the ellipse
in Fig. 4. In general, the ellipse and the hyperbola are close
mathematical relatives. The point O is the center of both ;
A'A = 2a is their common major axis; BB = 2b is their common
minor axis (though in the case of the hyperbola it does not meet
the curve in any real points). Only the distance, which we will
call cº, between the center and each of the foci (Fi, and FY) of
the hyperbola is greater than a ; namely, c’ = OF = OC, so that
c’” = a” + b%, whereas in the case of the ellipse we found cº = a” — bº.
The two foci are thus respectively at the points (– c’, 0) and
(c’, 0).
The hyperbola is defined by the difference of two distances:
viz., the distance of any of its points (a, y) from the focus
(– c', 0) and the distance of the point from the focus (c’, 0), this
time the difference being equal to 2a, the common major axis of
the hyperbola and the ellipse. With these leads the reader will
have no difficulty in obtaining the equation of our hyperbola:
ac2 y”
º, Tū - +.
The hyperbola always remains within the vertical angles formed
by the intersecting lines OC and O'C', called its “asymptotes,’’
which the two branches of the curve approach nearer and nearer
but touch only at infinity. In fact, an asymptote is defined in
general as a tangent to a curve touching it at infinity.
In passing, a few more words about the ellipse and hyperbola.
As the foci of an ellipse are moved nearer to the center, the ellipse
widens, its minor axis becomes longer. If foci and center coin-
cide, the minor axis becomes equal to the major, and the ellipse
becomes a circle. The circle is thus only a special case of the
ellipse: it is an ellipse with equal axes, and therefore might be
termed an “equilateral ellipse.” Algebraically we have then :
b = a = r, the radius of the circle. The associated hyperbola
changes in correspondence with this change in the ellipse. The
asymptotes, between which it is confined, now form an angle of
90°, and the hyperbola is said to be “rectangular”; or since now
b = a, the hyperbola is said to be “equilateral.” Its equation
becomes ac” – y” = r^, in which r is the radius of the associated cir-
For Scientific and Technical Workers 11
cle. In other words, before the change the curve might be
regarded as related to an ellipse, after the change it might be
regarded as related to a circle. These suggestions may help
clarify in the reader’s mind the character of the equilateral
hyperbola. The distance of the foci of such a hyperbola from
the center is + OC = + \/r2 Tr2 = + rV2, where r, again, is the
radius of the associated circle. Fig. 5 shows a circle and the

N
F,’ 7. (! | F',
AT
N
N.
FIG. 5
equilateral hyperbola corresponding to it. Note the right angles
formed by the asymptotes. But we must return to our imme-
diate topic of linear distances.
(2) Distance between Two Parallel Lines.
The vertical distance between the two parallel lines in Fig. 6 is
obviously a 1 – ae. The perpendicular distance r, which is usually
required, is easily determined from the vertical distance and the
common slope b of the parallel lines. In our figure, the two
12 Plame Analytic Geometry
angles marked 6 are equal because made by mutually perpendicu-
lar lines. Hence in the right triangle an Aa, l/r (= tan 0) = b ;
or l = rb, and lº = rºb”; also lº -- rº = (a, - az)*. Therefore rºb” +
* = (a+ — a 2)*, and
01 - Q2
r-vºi, (6)
which is the required perpendicular distance: it is, namely, the
vertical distance divided by Vb% + 1, where b is the slope of the
lines. If the two lines happened to be parallel to the X-axis,
FIG. 6
their slope b would be zero, and the perpendicular distance r
would equal the vertical distance, i.e., the difference a1 ~ a2 of the
constant terms of their equations.
(3) Distance between a Point and a Straight Line.
Let the given point be (act, y1) and let the given line be y = a1
+ bac. A line through (a 1, y1) parallel to this given line (same
slope b) is expressed by the equation y = (y1 – bacı) + ba (see §2).
In this equation the constant term is (y1 – bacı), and therefore the
vertical distance between the two parallel lines is al – (y1 – bacı),
i.e., the difference between the constant terms of their equations.

For Scientific and Technical Workers 13
Hence, according to the preceding section, the perpendicular dis-
tance r between the lines is
a1 – (y1 – bach) (7)
Vb2 + 1
and of course this is also the required perpendicular distance
between the given point and the given line.
The rule generally given for finding the perpendicular dis-
tance from a line to a point is as follows: Write the equation of
the given line in the form, ba – y + ai = 0; in the left member sub-
stitute for a and y the coordinates a 1, y1 of the given point
(whereupon the left member no longer equals zero, since ach, yi is
not on the line); and divide the resulting eacpression by the square
Yoot of the sum of the squares of the coefficients of a; and y. This
rule may be puzzling, until the student realizes that substitution
of a 1, y1 in the linear expression yields the vertical distance be-
tween the point and the line,” and that the subsequent division
transforms the vertical into the required perpendicular distance.
§7. ADDITION OF LINEAR EQUATIONS
Let the equations of two given lines be:
v = a1 + bia:
9 = a 2 + baſe.
What would be the result of adding the two equations? Obvi-
Ously, the resulting equation, viz.,
a1 + as b1 + bo
v=-5 3-3,
would again represent a straight line. This new line would have
a different slope from either of the given lines, and it would cut
the Y-axis at a different point. But it would pass through the
point of intersection of the two given lines. For the point of
intersection (aci, yi), which will satisfy both (y – a – bia: = 0) and
(y – as – boa: = 0) will necessarily satisfy also their sum, (y – ai
– blac) + (y – as – beac) = 0. Similarly, the same point (aci, yi)
will also satisfy their difference.
The same thing would still hold true if before adding the two
linear expressions we should multiply one of them by an arbi-
* This holds literally only for the “slope” form of the linear equation, but
of course any linear equation can be expressed in that form.
14 Plame Analytic Geometry
trary constant k, which might be either positive or negative.
For the values (act, yi) which satisfy the original equation
9 – a 2 – boa: = 0 will also satisfy k(y – as – beac) = 0; and since they
also satisfy the original equation y – al – b1a: = 0, they must neces-
sarily satisfy the algebraic sum (y – a – blac) + k (y – as – beac) = 0.
This sum is readily transformed into the equation,
a1 + kaa b1 + kbe
==ETH-T-TT-z. (8)
This represents a different straight line for every value of k.
But all those lines will pass through the point (act, yi), the point
of intersection of the two originally given lines, and therefore
Our equation is said to represent a “pencil” of straight lines, and
their common point of intersection (act, yì) is said to be the “cen-
ter” of the pencil. It would be truer to say that our equation
represents each of a ‘‘pencil” of lines, for two or more lines
together are expressed by equations of the second or correspond-
ingly higher degrees, as will appear in the next section of this
Chapter.
While the characteristic property of the Pencil equation is
sufficiently established by the above considerations, the reader
would profit by the trouble of actually solving the Pencil equa-
tion (8) simultaneously with one of the two originally given
equations: he will find that the arbitrary constant k will cancel
out, and the point of intersection (aci, yi) will be the same as that
of the original lines (see $4, p. 4).
The Pencil equation may be utilized in the solution of any
problem involving the point of intersection of straight lines. By
way of a simple illustration, we apply it to the following problem:
Show that the line y = 32/7 – 3a, passes through the point of inter-
section of the two lines, y = 2 + 3a, and y = 5–4a. If our third
line belongs to the same pencil as these two, its slope (–3) must
equal the coefficient of a in their Pencil equation (8), in which
bi = 3, b2 = — 4. Substituting these values in our general Pencil
equation, we have:
_3–4k
~7, IT,
whence k = 6. Next, substituting k = 6, and a1 = 2, a2 = 5 from the
two given lines in the constant term of the general Pencil equa-
tion (8), we get (2 + 30)/7 = 32/7, which shows that the line
– 3
For Scientific and Technical Workers 15
Aſ = 32/7 – 3a, belongs to the same pencil as the two given lines
and therefore does pass through the point of their intersection.
It may be observed that Pencil equations are not indispensable
in problems involving intersection points, that the general solu-
tion of simultaneous equations given in $4 would serve equally
well. The subject of line pencils is introduced here mainly to
show what happens when two linear equations are added or
subtracted, as is frequently done in solving simultaneous equa-
tions in elementary algebra.
§8. MULTIPLICATION OF LINEAR EQUATIONS
The product of two linear expressions, equated to zero,
(y – a 1 — blac) (y – as – bga) = 0,
is obviously an equation of the second degree. All values of a
and y for which the factor (y – a – blac) equals zero will give
points on a straight line; and all those values will satisfy our
quadratic equation as a whole, since the product is zero if one
of the factors is Zero. Similarly, all values of a; and y making the
factor (y – a 2 – boa) Zero will give points on a second straight line,
and they again will satisfy our quadratic expression, i.e., make it
equal to zero. From this it follows that any equation of the sec-
ond degree which can be resolved into two linear factors repre-
sents a pair of straight lines.
Further on we will find that any equation of the second degree
which cannot be so resolved necessarily represents one of the conic
curves: namely, a parabola, an ellipse (or circle), or a hyperbola.
It will thus be seen that every equation of the second degree rep-
resents either a comic curve or a pair of straight lines.* We pro-
ceed to investigate how to tell whether a given second degree
equation represents straight lines, and what can be learned about
those lines from the constants of the equation. Gradually we will
also show how to determine the nature and position of any conic
curve from an equation representing it.
* In addition, certain equations of the second degree represent single
points. Thus, a” + y^ = 0 is said to represent the single point (0, 0). But
this may be viewed as the circle & 4 y” = r" in case its radius r is infinitely
Small (approaching zero) and therefore all its points coincide with the center
(0, 0). Such a circle is called a “point-circle.” Similarly, if the axes of
an ellipse approach zero, we get a ‘‘point-ellipse.”
16 Plane Analytic Geometry
In our investigation we will employ the general equation of
the second degree, which represents all conic curves and all pairs
of straight lines: namely,
Aacº -- Bacy + Cy^+ Da: -- Ey + F = 0. .............................. (9)
We hasten to emphasize that while this equation contains six con-
stants, only five of them are independent; for if we divide the six
constants by any one of them, that constant assumes for all cases
the fixed value 1, yet in every case the geometrical entity corre-
sponding to the equation will remain unchanged. This means
that a conic can have at most five degrees of freedom; that in no
case, for instance, can more than five points (but not any three
of them on a straight line !) be required to determine the nature
and position of a comic curve.*
(1) Two Parallel or Two Coincident Lines.
For comparison with the general equation (9) we will build up
a quadratic representing a pair of parallel lines, by multiplying
two linear expressions. For the latter we will employ the slope
equation y = a + ba in somewhat modified form : namely,
tºms S 7%,
g = — m * m .90.
Two parallel lines will have, say, - s1/m, - S2/m, and their common
coefficient of ac, namely — m/m : altogether three independent con-
stants. By slight transformation and multiplication:
(ma + my + 81) (ma + my + 82) = 0,
whence:
m*a* + 2mmary + mºy” + m (S1 + sa) a + n (s1 + 82) y + S182 = 0. ...... (10)
Comparing this with the general equation (9), we find: A = m^;
B = 2mm; C = m”; D = m (81 + 82); E = n (s1 + 82); F = 81s2.i
* Frequently fewer points are sufficient. For instance, three points suffice
to determine a circle. That is because when we decide that a circle shall
pass through the three points, them in the general equation (9) the coefficient
B becomes zero, and A becomes equal to C; whereby two of the five indepen-
dent constants are eliminated and there remain only three. See again the
general equation of the circle in §6(1) of this Chapter (p. 7).
f Had we used the unmodified linear form y – a – ba: = 0, we would have
obtained for the coefficient C (= n°) the fixed value 1, and our parallel-line
quadratic would have been unusable in those cases in which the yº-term is
absent (coefficient of y” zero).
For Scientific and Technical Workers 17
Two relations will reduce the five independent constants of
equation (9) to the three that are necessary and sufficient to
determine two parallel lines.
The first of these relations results from the fact that in the
case of parallel lines the terms of the second degree, namely
m*a* + 2mmary -- m^y” in equation (10), necessarily form a perfect
square, since they are the product of the multiplication of the
binomial ma; + my in one linear factor by the very same binomial
in the second factor. And therefore, by a rule of elementary
algebra, the square of the coefficient of the acy-term, that is,
4mºn*, must necessarily equal 4 times the product of the coeffi-
cients of aº and y”. In the notation of the general equation (9):
B* = 4AC ; and this is the first characteristic relation to be found
in every equation of parallel lines.
The necessary second relation is furnished by the fact that m
times the coefficient of y equals n times the coefficient of a ; that is,
nE = m.D, or mR) – m D = 0. However, it will be found best to use
this expression in the form of its square: namely, (mL – m D)* =
m*E* + n°D°–2mm DE = 0, or BDE = AE* + CD*; and this is the
second characteristic relation to be found in all equations of
parallel lines.
To recapitulate: the two relations necessary and sufficient for
an equation of the second degree to represent a pair of parallel
lines are: B4 = 4AC and BDE = AE * + C D*. We shall find later
that the first of these relations holds true also in the equations of
all parabolas (pointing to a mathematical relationship between
the Parabola and parallel lines). The second of our relations
holds true also in all equations of non-parallel lines intersecting
in the coordinate origin (and in certain other cases). But both
our relations hold true exclusively in the case of parallel lines,
and therefore the two together can always be relied upon to show
whether or not a given equation represents a pair of parallel lines.
It will be noted that the constant F is absent from either of the
two relations; which means that the comstant term of an equation
plays no part in determining whether the equation will stand for
parallel limes. The constant F, however, does come into play if,
having discovered that a given equation represents parallel lines,
we further apply to it a principle relating to all real numbers:
namely, two numbers will not be real if the square of their sum is
18 Plame Analytic Geometry
less than four times their product. Thus, in our equation (10),
the constants s, and so (and therefore the lines themselves)
will not be real if (s1 + sa)* is less than 4S1s2. This is because
(s. --S2)*–4S182 = (s1 — sa)*, and if this square were a negative
quantity, S1 and so would be imaginary. Returning to the nota-
tion of the general equation (9): the two parallel lines will be
imaginary if 4F S. D*/A (or 4F SEA/C). If 4F ~ D*/A (or
4F 3 E*/C), then s, and so are unequal and we have two separate
real lines. If 4F = D*/A = E4/C, then si = sa, and our lines are
real but coincident: practically speaking, we have them a single
real line. -
We thus learn that, while ordinarily a straight line is repre-
sented by an equation of the first degree, there are cases in
which an equation of the second degree may represent a single
line. We repeat: that will be the case whenever Bº = 4AC ;
BDE = AE2 + C D*; and 4F = D*/A = E°/C. This last relation
is obviously independent of the other two, and the three together
reduce the five independent constants of the general equation (9)
to two; which is in proper accord with the fact that two constants
are necessary and sufficient to determine a single straight line.
The constants of the two linear equations combined in our
quadratic are — sa/m, - S2/m, and the common slope — m/m. We
will express these in terms of the constants of equation (9). For
the common slope we get directly : — m/m = – D/E = – 2A/B. For
the two constant terms we have : -
$1, $2 I' S1 82 F.
m " n " O and m m C'
which give :
s, - E + \/E2–4CF s, - E - VE2 – 4CF
* — - and — if = e
%, 2C 70, 2C
Thus we can readily obtain the linear equations of the two parallel
lines from the constants of the quadratic equation representing
them. The case of lines parallel to the Y-axis (C=O) will be dis.
cussed separately.
Let us consider a numerical example: the equation 162°– 8acy
+ y” + 24a – 6y +8 = 0 as a special case of the general equation
(9). Here: A = 16; B = -8; C = 1; D = 24; E =–6; F = 8. We
find: B4 (= 64) = 4AC (=64); BDE (= 1152) = AE*(= 576) + CD2
(= 576); and 4F (= 32) < D*/A (= 36), or 4F (= 32) < E*/C
For Scientific and Technical Workers 19
(= 36). We conclude that the given equation represents two dis-
tinct parallel real lines. Their common slope is — D/E = (–24/–6)
=–2A/B (= – 32/–8)= 4. The constant terms of their linear equa-
tions, according to the above formulae, are 4 and 2. The two
equations are therefore : y = 4 + 4a, and y = 2 + 42.
It might further be asked, what change would cause the given
equation to represent a single line? The answer is, the equation
would have to show : 4F = D*/A = E*/C = 36, or F = 9. The equa-
tion of the single line would therefore be 16a:” – 8ay + y” + 243:
– 69 + 9 = 0. Applying our formulae for the constants of the linear
equations, we would find for the slope still – D/E = -2A/B = 4,
but the two constant terms would now be identical and equal to 3:
we would have the single linear equation, y = 3 + 4a. This single
line obviously runs midway between the two lines obtained when
F' equaled 8. If F were greater than the critical value 9, we
would get imaginary lines. But, generally, for every value of
F less than 9 we will get a pair of lines parallel to y = 3 + 4a and
equidistant from it on both sides, the spread between the lines
becoming greater and greater as the difference between 9 and the
actual value of F is increased. -
Thus the second degree equation of a single line correlates it
with an infinite number of pairs of lines parallel to it and sym-
metrically located to either side of it. And obviously, the equa-
tion of any pair of parallel lines can be represented by the quad-
ºratic equation of the single line running midway between them,
with only the comstant term changed. The constant term for the
midway line may be denoted by the symbol Fo, as a reminder that
theoretically we are dealing with a pair of parallel lines between
which the distance is zero. The constant term for any pair of
distinct parallel lines related to this midway line may be denoted
by the more general symbol F. The “spread” between the two
distinct lines may be expressed in terms of the vertical distance
between them,” with the aid of the s/n formulae given above:
s, , sa VE2 – 4CF 2 E2 2 VF.I.F
* = −75. A |: – F = −75. VF'o - F,
70, 70, C Vä - Nia-F-Rya Y”
since, as previously pointed out, for a pair of coincident lines
E*/C = 4F or, using the symbol Fo, E*/(40) = Fo.
* See §6(2), p. 11.
20 Plame Amalytic Geometry
Again according to $6 (2), and since — m/m is the slope of the
lines, the perpendicular distance between them is obtained by
dividing the vertical distance by Vmº/nº-1, or VA/C+1. The
perpendicular distance is thus: 2VF, F/(VC. VA/C+1), or
2\/F, - F
VA + C
In the light of the preceding, let us briefly examine the quad-
ratics of elementary algebra, which we will write in the form
ca:* + ba + a = 0. In the notation of our general equation (9), this
becomes: Aacº -- Da; -- F = 0. Here the coefficients B, C, and E of
equation (9) are zero. Hence, B*(= 0) = 4AC (= 0); BDE (= 0)
= AE* + C D*(= 0). We conclude that every elementary quad-
ratic in one variable represents two parallel lines. The lines will
be imaginary if 4F - D*/A; they will be real but coincident if
4F = D*/A; they will be real and distinct if 4F 3 D*/A. The
Quadratic obviously resolves itself into two linear equations of
the form a = a1 and a = a-', and so the lines are necessarily verti-
cal, i.e., parallel to the Y-aacis (see end of $1). “Solving” the
quadratic means finding the distances of the two lines from the
Y-axis. All this should be clearly understood by teachers of
elementary algebra.
(2) Intersecting Lines.
We have referred to a mathematical relationship existing be-
tween parallel lines and the Parabola. There exists an even more
intimate relationship between non-parallel lines and the Hyper-
bola; and as a knowledge of this relationship is helpful in deter-
mining whether a given equation of the second degree represents
a pair of intersecting lines, we will use it as a starting point in
our discussion of such lines.
The equation of the hyperbola assumes a simple form if its
major axis lies on the coordinate X-axis and its center is at the
origin. The equation is then :
b°w” – a”y2 = a”b°.
In §6 (p. 10) it was indicated how this equation can be shown to
express the defining law of the hyperbola. In §6 also the two
lines OC and O'C', intersecting at the origin, were described as
For Scientific and Technical Workers 21
asymptotes, each meeting the hyperbola at plus and minus
infinity.
Let us briefly verify this statement. Using the same notation
as with parallel lines above, we write the equation of a line pass-
ing through the origin in the form : y = — #a. To find where this
line will meet the curve, we substitute this value of y in the equa-
tion of the hyperbola. We get:
Q,
2A-2 — a 2/r2 – a 2 h 2
b2a: ºr a rea b°,
whence, by simple transformation :
b? b
ac? - and QC = −H —=.
b2 m.” b2 m2
2Tm2. a 2 m2.
Now, the slope — m/m of the line OC is obviously CA/OA = b/a
(see Fig. 4 in §6, p. 8); therefore b%/a” = m”/m”, the above
denominator is zero, and a = + Co. That is, the line OC will meet
the hyperbola only at plus and minus infinity and is therefore,
by definition, an asymptote. The same holds true of the line O'C',
whose slope — m/m is — b/a, so that again b”/a” = m”/m” and
again a = + Co. Between plus and minus infinity, therefore, the
hyperbola will be confined within two vertical angles formed by
its asymptotes.
The equations of the two asymptotes can be written: bac – ay = 0
and ba: + ay = 0. Multiplication of these linear expressions gives:
b222 * a”y” - 0,
an equation of the second degree representing both asymptotes.
Comparing the equations of the hyperbola and the asymptotes,
we see that they differ only in their constant terms and are other-
wise identical. If the curve together with its asymptotes were
moved, say, h units to the right and k units upward, the slopes
of the curve’s axes and asymptotes remaining unchanged, then
the ac's and the y's of all the points would be increased respec-
tively by h and by k. But if we restore the original magnitudes
by subtracting h and k from the increased ac's and y's, we will
obtain equations essentially similar to the original ones, only
corrected for the new position of curve and asymptotes. For the
hyperbola we will thus have :
b°(ac – h)” – a” (y – k)* = a”b°;
Plame Analytic Geometry

§
#.
º
For Scientific and Technical Workers 23
for its asymptotes:
b°(a – h)” – a” (y – k)* = 0.
Expanding these equations, we get: for the hyperbola, bºar” –
a”y” – 2b^ha, +2aºky + bºh” – a”k” – a”b° = 0; for the asymptotes,
b°ac” – a”y” – 2b^ha, +2aºky + b%h” – a”k” = 0. Obviously, the two
equations again differ only in their constant terms. We would
obtain a similar result if we changed also the slopes of the hyper-
bola's axes and asymptotes, because again the ac's and y's would
be similarly changed for the curve and its asymptotes. This rela-
tionship between the equations of a hyperbola and its asymptotes
will presently appear as a criterion for distinguishing whether a
given quadratic equation represents intersecting straight lines
or a hyperbola.
In passing, we will briefly introduce another mathematical
relative of our hyperbola and asymptotes. If, as in $6, we should
place two hyperbolic foci at equal distances cº from the center,
but this time on the Y-axis, so that the two foci would be (0, c’)
and (0, - cºy; and if, not 2a as in §6, but 2b (the vertical axis of
the associated ellipse) were adopted as the constant difference
of the distances between any point (ac, y) of the hyperbola and
the new foci, we would get as the equation of the new hyperbola :
b°ac” – a”y* = — a”b°. Fig. 7 is a duplicate of Fig. 4 in §6, but in
addition it pictures this new hyperbola, touching the ellipse at
the upper and lower extremities of its minor axis. The new
hyperbola is said to be the “conjugate’’ of the original hy-
perbola (and vice versa!), and they have in common the same
pair of asymptotes. It is important to note that the constant
term in the equation of the asymptotes (which in our case hap-
pens to be zero) is the arithmetical mean of the constant terms
in the equations of the two conjugate hyperbolas. This holds true
for all conjugate hyperbolas and their asymptotes.
If we should keep the ratio b/a constant, but increase or de-
crease both b and a in the same proportion, our asymptotes (be-
cause of the unchanged slopes) would remain unchanged, but the
value of the term a”b% would change, and that would affect both
hyperbolas and also, of course, the associated ellipse. Starting
with the original hyperbola, let us gradually diminish a and b.
At each step the hyperbola will have undergone some change
and will have moved nearer to the asymptotes. By the time
24 Plame Analytic Geometry
a and b have reached values, say, like 1/10-millimeter, the naked
eye will scarcely distinguish the hyperbola from the asymptotes.
When a and b have reached zero, the hyperbolic foci will have
met at the center, and the hyperbola will have completely merged
with the asymptotes: it will have become a pair of intersecting
straight lines.* Some writers speak of pairs of lines as “de-
generate’’ comics, the contemptuous term vaguely suggesting
human degenerates, or criminals. Perhaps, if a pair of inter-
secting lines could raise a voice in self-defense, they might claim
that they are a sort of ideal rectilinear hyperbola, that they are
quite normal and straight, and it is the conic curves that are
degenerate and crooked. At any rate, for every pair of inter-
secting lines, viewed as potential asymptotes, there is an infinity
of hyperbolas and conjugate hyperbolas, the lines and all the
hyperbolas having the selfsame equation except for its constant
term.
For example, ac” – 5acy + 4y” + 32 – 4 = 0 represents two straight
lines. The similar equation, ac” – 5acy + 4y” + 3a – 3 = 0, is then a
hyperbola; and the equation of its conjugate hyperbola is
a:” – 5acy + 4y” + 3a – 5 = 0, because the constant term – 4 of the
bilinear equation is the arithmetical mean of — 3 and – 5. The
equation, acº — 5acy + 4y” + 3a – 0 (the constant term = 0) is an-
other hyperbola of the same asymptotes; its conjugate hyperbola
is then ac” – 5ay + 4y” + 3a – 8 = 0, because – 4 is the arithmetical
mean of 0 and – 8; and so forth ad infinitum.
We proceed to synthesize for investigation a general quadratic
equation of two intersecting lines by multiplication of their
linear expressions. The latter we will employ in a form showing
that the lines pass through the point of intersection (act, yi), and
we will denote their different slopes by — m/m, and — ma/m2 :
7%. 7%)
(y – yi) =-º (2–2) and (y – yº) =–4: (a – wi).
li 71.2
After slight transformation we have:
[m] (a – a .) + m1(y – yi)] . [me (a – wi) + m2 (y – ya)] = 0,
* At the same time the associated ellipse will have shrunk to a single point;
its equation will have become bºx” + a”y*= 0, which will be satisfied only by
the point (0, 0).
f It was obtained by multiplying together a – 4y + 4 = 0 and a — y – 1 = 0.
For Scientific and Technical Workers 25
and this leads to the following quadratic equation, which neces-
sarily represents two non-parallel lines:
m1 m2a:” + (minº -- mana) a y + m1 may”
— [2m.1m2a: + (mima + m,n) yija – [2n 1m29; + (mima + mani)a's]y
+ m1 m2a : + (mine + m2n+) wivi + m17.29% = 0.
Comparing this with the general equation (9) of the second de-
gree, we obtain the following identities between their correspond-
ing constants:
A = m1 m2; D = -2m.1m2a: ; – (mama + m2n+) yi;
B = m172 + m2n+; E = – (mine + m2nd) act – 2ninaya;
C = m1772; F = m1 mga tº + (mima + m2n+) wiyi + m1 may it.
We easily get:
D = — 2Aact * By;
E = — Bacº — 20 y;.
And solving these as simultaneous equations in 2, and yi, we find:
QC; = 2CD – BE and u , = 2AE – BD
* T B2 – 4AC !/? - TEITACſ.
by which we are enabled to determine the point of intersection
(aft, yi) directly from the coefficients of a given equation repre-
senting non-parallel lines.
Examining the coefficients A, B, C of the three quadratic terms,
we discover next that in the case of non-parallel lines Bº must
necessarily be greater than 4AC. Indeed, B* – 4AC = m1°m2°
+ m2*n-1” – 2m.1m2 minº = (mºna – mani)*. If Bº were less than 4AC,
this square would be a negative quantity: the binomial which is
its square root, and therefore the slopes of the lines, and there-
fore the lines themselves, would be imaginary. If Bº were equal
to 44C, the slopes of our lines, – mi/ni and — m2/m2, would be
equal, and the lines would be parallel. For intersecting real
lines, therefore, we can only have Bº X- 4AC.
This, however, must also hold true in the equation of a hy-
perbola, since any pair of intersecting lines may be the asymp-
totes of hyperbolas, and since the equations of hyperbolas and of
their asymptotes are identical in all their variable terms. For a
characteristic permitting to tell whether a given equation repre-
sents a hyperbola or intersecting lines we must turn to the
COnStant term.
In our equation of non-parallel lines, the three variable quad-
26 Plame Analytic Geometry
ratic terms given first and the three constant terms at the end are
similar in form and would become identical if in the first three
terms we should substitute the fixed aci and yi for the variable
a; and y. Since the coordinates of the point of intersection satisfy
the equation, we can make this substitution throughout the equa-
tion. In the notation of the general equation (9), except using
the symbol F., instead of F (to signify that we are now dealing
with straight lines), we have : Fs + Da: ; + Ey + F s = 0, or F's =
— # (Daci + Eys). And substituting here for act and y; their values
given above in terms of the coefficients of the general equation,
we get:
F's = BDE – AE? — CD2
& – B* – 4AC
This holds for intersecting lines only, and not for any hyperbola;
because only in the case of intersecting lines do the coordinates
a'i, y, satisfy the equation and can be substituted for ac, y.
For a numerical example, let us examine the equation intro-
duced once before: viz., ac” – 5ary + 4y” + 3a – 4 = 0. Here A = 1;
B = — 5; C = 4; D = 3; E = 0; F = – 4. We note that B” (= 25)
> 4AC (= 16); hence the equation represents either a hyper-
bola or non-parallel lines. Applying the Fs-formula, we find:
F's =–36/(25–16) = — 4, with which the constant term F of the
given equation is identical; hence the given equation represents
a pair of lines. If the constant term of the given equation were
anything else than — 4, the conclusion would be that the equation
represents a hyperbola.
We have now learned how to ascertain whether a given equa-
tion represents non-parallel lines; also, how to determine (aft, yi),
the point of intersection of the lines, from the coefficients of the
given equation. If in addition we deduce from the given quad-
ratic the slopes (— m/mi and – ma/m2) of the two lines, the two
linear equations will become known, and our interpretation of the
Quadratic will be complete for all ordinary purposes. From the
values of A, B, and C given above, we have:
"GT m, "7. and a m, m,
Solving these as simultaneous equations in ma/ml and m2/m2, we
get:
* = B-VB-4AC and ma_-B+VB-4AC
711 2C 71.2 2C e
For Scientific and Technical Workers 27
A minor complication will arise in those cases where one of the
limes is parallel to the Y-aacis. The equation of such a line is of
the form a = a' (see $1), and consequently the y”-term will be
absent in the quadratic equation: we will have C = 0. In such
cases the first of our slope-formulae will clearly indicate, – ma/mi
= Co for the vertical line. But for the second line we would get:
– m2/m2 = 0/0, an indeterminate quantity. We can get around
this complication as follows. Since C = m1772 = 0, either ml or nz
must be zero; say mi = 0. Then B = m172 + m2n 1 = m172. And
since A = m17m2, we have : A/B = m2/m2, and the slope – m2/m2 =
– A/B, while — m/mi = – Co (for the line parallel to the Y-axis).
Before concluding our review of intersecting lines, we will
mention some further characteristics of their equations.
First, since A = m17m2 and C = m1772, their ratio A/C equals the
product of the slopes of the two lines, (– mi/mi) × (– m2/m2); and
since if this product equals – 1, the lines are perpendicular ($5),
we infer: that if in a given quadratic equation the ratio of the
coefficients of acº and y” is – 1, that is, if these coefficients are
numerically equal but opposite in sign, the two lines represented
by the equation are perpendicular to each other. Incidentally,
if the given equation represents a hyperbola, then C = — A will
signify that the hyperbola is rectangular (since its asymptotes
are perpendicular).
We next examine the significance of the absence of certain
terms in an equation of intersecting lines. We have seen that if
one of the lines is parallel to the Y-axis, its linear equation, a = a',
has no y-term, and consequently a quadratic equation of which
it is a component will have no y”-term (in such a quadratic, C = 0).
Similarly, absence of the acº-term in a given quadratic shows that
one of the lines is parallel to the X-axis (we have then, A = 0).
If both A = 0 and C = 0, the only term of the second degree is Bacy,
and then one of the lines is parallel to the X-axis, the other to the
Y-axis.
Now as to the absence of the acy-term, that is B = 0. Both B
and A or both B and C cannot be absent, for in such a case
B” (= 0) would equal 4AC (= 0), and the lines would be parallel,
not intersecting. But what is the meaning of the absence of the
acy-term while both Aacº and Cy” are present? The answer is
readily found in the multiplication of the two linear components
28 Plane Amalytic Geometry
of the quadratic equation: the acy-term will cancel out whenever
the slopes of the two lines are numerically equal but opposite in
sign, their algebraic sum therefore being Zero. Thus, the asymp-
totes shown in Figs. 4 and 7 have the slopes b/a and — b/a, and
their combined equation, bºat” – a”y* = 0, has no acy-term. The
angles formed by these asymptotes are bisected by the coordinate
axes. In correspondence with this, the axes of the conjugate
hyperbolas in the same figures lie on the coordinate axes, and the
equations of the two hyperbolas have no acy-terms. For the same
reason the acy-term will be absent whenever the axes of a hyper-
bola, which are also bisectors of the angles formed by the asymp-
totes, will be parallel to the coordinate axes.
NUMERICAL EXAMPLE'S
Let us apply the foregoing propositions and formulae to a few
numerical examples. First, we return once more to the equa-
tion, aº – 5acy + 4y” + 3a – 4 = 0, which was previously shown to
represent two lines. Again: A = 1; B = — 5; C = 4; D = 3; E = 0.
Hence:
a, 2CD + BE 24–0, 8, , .24E-BD 0+ 15 5,
** EIHTC = 25II6 = 5 ; 94 = ETIC = -0T =5;
_mi_5–V9–1. _*-5+V9-1
m1 8 4 * 71.2 8 tº
The two lines are therefore:
3 3’
or y = 1 + æ/4 and y = –1 + ac; or a – 4y + 4 = 0 and ac – y – 1 = 0
(see p. 24). -
As a 2nd example we take: y” — ac” – y + 4 = 0. Here, A = -1;
B = 0; C = 1; D = 0; E = − 1. We find B2 (= 0) > 4AC (= —4),
hence we have either a hyperbola or intersecting lines. For
straight lines we must have Fs = 1/4, which we find to be the
constant term of the given equation; hence the equation repre-
sents a pair of lines. For the point of intersection we get: a, i = 0;
Ayi = 2/4 = 1/2. Since C (= 1)=– A (= 1), we conclude that the
lines are perpendicular. In fact, we find — m/mi =–2/2=-1;
— m2/m2 = 2/2 = 1, and obviously the product of the two slopes
equals — 1. The two linear equations are thus: y – 1/2 = — a + 0
and y – 1/2 = a – 0; or y = 1/2 – a and y = 1/2 + æ.
For Scientific and Technical Workers 29
For a 3rd example we choose: y” — acy — 5a, + 5 y = 0, with the
a 4-term missing. We have: A = 0; B = –1; C = 1; D = – 5; E = 5;
F = 0. Here, B* (= 1) > 4AC (= 0); if the equation represents
lines, Fs must be zero, which is indeed the value of F in the equa-
tion. The point of intersection is act = — 5, yi = – 5. The absence
of acº indicates that the slope of one of the lines is zero; we find:
— m/1/n 1 = (1–VT)/2=0; – m2/m2 = (1+ VT)/2= 1. The two lines
are therefore: y + 5 = 0 and y + 5 = a +5; or y = – 5 and y = a-.
As a 4th example we consider an equation with both the ac” and
the y” terms absent: ary–2a – y +2=0. A = 0; B = 1; C = 0; D =–2;
E = –1; F = 2. Bº (= 1) > 4AC (= 0); Fa – 2/1 = 2, which is the
value of F in the equation. At the intersection, act = 1/1 = 1;
Ayi = 2/1 = 2. The slopes we expect to be 0 and Co; we find:
— m/1/n 1 = CO ; for the second slope we get by the – A/B formula:
— mana = – A/B = 0. The two lines are therefore: y –2 = 0 and
a — 1 = 0. The reader is invited to plot the lines of the above four
examples.
As a 5th example we will consider the equation representing
Boyle's law of gases. The usual statement of the law is: pºv = con-
stant. For uniformity of notation, we will write acy for pv.
Further, if the amount of a gas employed is its molecular quan-
tity in grams (e.g., 32 grams of oxygen), if the pressure is
measured in atmospheres and the volume in liters, then at 0°C.
the constant of Boyle’s law is 22.42. We thus have the equation:
acy – 22.42 = 0. Here A = 0; B = 1; C = 0; D = 0; E = 0; F = – 22.42.
To begin with, Bº (=1) > 4AC (= 0); hence we have either a
hyperbola or intersecting lines. In this case, the Fs-formula for
straight lines requires F, - 0, whereas the equation gives F =
– 22.42; hence the equation represents a hyperbola, whose asymp-
totes are acy = 0. The point of intersection is act = 0, yi = 0. Since
one of the two lines is parallel to the Y-axis (absence of y”-term 1),
we use the – A/B formula, and we get: – miſm.1 = – Co ; – m2/m2 =
– A/B = 0. Hence the two lines are y = 0 (the X-axis) and a = 0
(the Y-axis). The coordinate axes are thus the asymptotes of our
hyperbola, and since the axes are perpendicular, the hyperbola is
rectangular. We conclude that if and when a gas obeys Boyle's
law, its behavior is geometrically represented by a rectangular
hyperbola.
CHAPTER II
THE PARABOLA
$9. ITS SIMPLEST CARTESIAN EQUATION
In §1 it was pointed out that the straight line represents essen-
tially the proportionality of two varying quantities a, and y, as
expressed by the equation y = bac, or by the more general equation
v = a + bac. The shape of a straight line is of course always the
Same, and so the constants of its equation have nothing to do with
its shape; they serve to determine only its position with respect
to the coordinate axes.
The parabola too represents a proportionality, but the propor-
tionality is between one variable and the square of a second, as
expressed by the equation: *
g = ca:”.
The line y = bac passes through the coordinate origin, since y is
necessarily Zero when a = 0. Likewise, the parabola y = ca:* passes
through the origin; at which only one value of a (viz., a = 0) cor-
responds to the ordinate y = 0. From there on, if the proportion-
ality constant c is a positive quantity, y will increase upward;
for every positive value of y there will be two equal values of ac-
One positive, the other negative, and the distance between the
two will grow indefinitely with increasing y. In other words, the
parabola starts at a single point (its “vertex’’) lying at the
origin, then opens upward and grows wider and wider without
end. All this is indicated in Fig. 8.
If the coefficient c is negative, the parabola y = ca:* opens down-
ward, as shown in Fig. 9. The line through the vertex, by which
the parabola is symmetrically divided, is called its ‘‘axis,” or its
‘‘axis of symmetry.” In Figs. 8 and 9 the axes of the parabolas
coincide with the coordinate Y-axis.
The simplest example of a relationship represented by a pa-
rabola is furnished by the natural numbers, 0, 1, 2, 3, 4, etc., and
their squares, 0, 1, 4, 9, 16, etc. If a denotes the numbers and y
their squares, then y = azº. In this case the coefficient c of the
equation y = ca:” is unity. However, it should be noted that while
the natural numbers and their squares grow by finite increments,
30
For Scientific and Technical Workers 31
y
O
FIG. 8
the equation y = a + and the parabolic curve corresponding to it
are continuous, and extend to all the infinity of quantities be-
tween any two consecutive numbers as well as to the numbers
themselves.
Y
FIG. 9
32 Plame Amalytic Geometry
Another familiar example of a parabolic relationship is pre-
sented by the law of falling bodies. The distance s traversed by
a body falling in vacuo from a position of rest is proportional to
the square of the time t, reckoned from the beginning of the fall.
This is expressed by the familiar parabolic formula s -** in
which, if s is measured in centimeters and t in seconds, the con-
stant g, called ‘‘the gravity acceleration constant,” has the
numerical value 980.6 (at sea level and 45° Latitude).
The constant c of the equation y = ca:*, or rather its reciprocal
1/c, has an important geometric significance. The numerical
value of the reciprocal 1/c represents, namely, the length of the
chord AB (Fig. 10), which cuts the parabola's axis at the point
F, called the Focus. The chord AB itself is called the parabola's
y /
ſ P(x, y)
A. - e
X
O
D
E
FIG. 10
- º º º 1
Latus Rectum. The abscissa of the point B is obviously *=3;
(i.e., 1/2 of 1/c); what is its ordinate? Since the equation of
our parabola is y = ca”, and since B is on the parabola, we have
2
for its ordinate : y = c (...) F i. ; that is, the distance of the point

For Scientific and Technical Workers 33
B, and therefore also of the latus rectum, from the X-axis equals
one-fourth the length of the latus rectum. And since the focus F
is at the intersection of the parabola's axis and latus rectum, its
coordinates are obviously (0. #).
The reader will also note in Fig. 10 the auxiliary line CD,
parallel to the X-axis and located at the same distance, #. below
the X-axis as the latus rectum is above it. This auxiliary line is
called the parabola's Directria. The parabola has an interesting
property which is often used as its defining law: namely, the
distance PE of any of its points (ac, y) from the directrix
is the same at its distance PF from the focus. PE is obviously
vº; and according to $6(1), p. 6, the distance PF is

2
N (ac – 0)” + (v-#) º The reader is urged to satisfy himself
that if these two distance expressions are equated, then squaring
and simplification will lead to the equation of the parabola y = ca”.
As the diameter of a circle determines its size, so the latus
rectum of a parabola determines its size, therefore its shape.*
If we write our equation in the form *=}, and if the latus
rectum 1/e is reduced to zero (c = oo), then aft a 0 and a = 0:
the parabola has become infinitely narrow and has shrunk to a
straight line (viz., its own axis, coincident with the coordinate
Y-axis). On the other hand, if c = 0 (i.e., the latus rectum
1/c = Co.), we get y = ca.” = 0: now the parabola has become infi-
nitely wide and has spread out into another straight line, coinci-
dent with the coordinate X-axis. And since parabolas can only
differ in their width, all possible parabolas are included between
the two extremes of zero and infinite width, and all are repre-
* Properly speaking, all parabolas have the same shape, just as all circles
are really one and the same: any “narrow’’ parabola can be transformed
into any ‘‘wide’’ one merely by enlargement. This is generally expressed by
saying that all parabolas are similar. The same is true of all equilateral
hyperbolas, because they are essentially determined by a single quantity—
the magnitude of their equal axes (just as a circle is determined by its radius
and a parabola by its latus rectum).
34 Plame Amalytic Geometry
sented by the simple equation, y = ca.”. If the equation of a
parabola appears in a more complex form, this is not because the
parabola itself is any different, but only because it is differently
located with respect to the coordinate axes. By suitably chang-
ing the coordinates the equation of any parabola can be trans-
formed into y = ca:*. However, such transformation is very sel-
dom, desirable.
§10. OTHER FORMS OF THE PARABOLIC EQUATION
In the preceding section, the parabola was presented as the locus
of a point whose ordinate is proportional to the square of its ab-
Scissa : y = ca.”. Of course, the ordinate of a point is its distance
from the X-axis, the abscissa is its distance from the Y-axis. Our
description might therefore be worded as follows: the parabola is
the locus of a point whose distance from the X-axis is propor-
tional to the square of its distance from the Y-axis. As thus de-
scribed, the parabola is free to assume every possible width, or
size, but its position with respect to the coordinate axes is fixed:
namely, its axis of symmetry is made coincident with the coordi-
nate Y-axis and its vertex is fixed at the coordinate origin. A
slight change in our wording will set the parabola free also to
occupy every possible position.
Leaving the coordinate axes undisturbed, we will refer the
parabola to any arbitrary pair of lines perpendicular to each
other, and we will define it as the locus of a point whose distance
from one of the two lines is proportional to the square of its dis-
tance from the second line. This second line will then be the
parabola’s axis of symmetry, and the vertex will lie at the inter-
section of the two perpendicular lines. The proportionality con-
stant c will be the same as before and will have the same geometric
significance: i.e., irrespective of the sign of c, its reciprocal 1/c
will numerically represent the length of the latus rectum and will
be free to assume any fixed value between 0 and co ; by which
parabolas of every possible width will be included under the new
definition.* .
* Before proceeding further, the student is advised to read again $6(3),
dealing with the distance of a point from a straight line, as that distance
will repeatedly come up in the following paragraphs.
For Scientific and Technical Workers 35
(1) The Equation y = a + ba + ca:*.
This equation has very special importance in the present work,
since of all equations of the parabola it is the most frequently
used in describing or summarizing scientific data. It can be ob-
tained by referring the parabola to any two perpendicular lines
parallel, respectively, to the coordinate axes. The equations of
two such lines can be written: y = k and a = h; or y – k = 0 and
a — h = 0. Their point of intersection is (h, k). According to
$6(3), the distance of a point (ac, y) from the first of these lines
is y – k; the square of its distance from the second line is (a – h)*.
The parabola is therefore, y – k = c(a – h)*, or y = k + ch” – 2Cha, +
ca”. If we write a for k + ch” and b for – 26h, the equation as-
Sumes the form : y = a + ba + ca:*.
The length of the latus rectum is again 1/c, taken with the
positive sign (whether c is positive or negative). The point of
intersection (h, k) of the reference lines, which is also the vertex
of the parabola, can be expressed in terms of the constants a, b, c :
h = — b/(2c); k = a – bº/(4c). The vertex is thus at the point,
[– b/(2¢), a - bº/(4c)]. The focus lies at a distance 1/(4C)
directly above the vertex and is therefore at the point [– b/(2C),
b 4dc — b% + 1
a — b%/(4c) + 1/(4c)], or (-; —I-
is the vertical line a = — b/(2c).
) The parabola’s aa is
Fig. 11 shows such a parabola. The proportionality constant
c is assumed to be positive, and accordingly the parabola is shown
opening upward. In Fig. 12 the very same parabola, having the
same width (i.e., the same constant c), the same vertex, and the
same vertical axis, is shown opening downward, c being assumed
to be negative.
(2) The Equation a = cy”.
A parabola corresponding to this equation is shown in Fig. 13.
The parabola's axis of symmetry now lies along the coordinate
X-axis, and for every positive value of a there are two values
of y, one positive, the other negative, but numerically equal.
The vertex is again at the coordinate origin. The coefficient c is
assumed to be positive, hence the parabola opens toward the right.
If c were negative, the parabola–its vertex still at the origin—
36 Plane Analytic Geometry
Y
NA F |B
N L^
k
O
FIG. 11


h
FIG. 12
For Scientific and Technical Workers 37
would open toward the left. The latus rectum AB (=1/c) being
the same as in the preceding figures, the parabola itself is the same
as before, only differently placed with respect to the coordinate
system; whence the difference in the form of its equation.

Y
FIG. 13
(3) Parabola Axis at an Angle of 45°.
Before generalizing the equation of the parabola with respect
to both width and position, we will examine the equation of our
parabola in one more special position. Let the latus rectum again
be 1/c.; let the vertex be at the origin; and let the axis of sym-
metry form an angle of 45° with the coordinate X-axis.
For the perpendicular reference lines of the parabola we
choose: its own axis of symmetry and the line through the origin
perpendicular to it. The equations of the two lines are: y = a, and
38 Plane Analytic Geometry
y = — ac, or a – y = 0 and a + y = 0. The perpendicular distance of a
point (a, y) on the parabola from the line a + y = 0 is (a + y)/\/2.
The square of the distance of the same point from the line
a – y = 0 is (a — y)*/2. The equation of the parabola in the
specified position is therefore:
*** - c. (*= 9):
V2. " 2 ”
which readily transforms into :
V2 V2
ac” – 2a:y + y” — — tº ac – – t . y = 0.
C C
Fig. 14 is a graph of this equation.
Y
Yº
EIG. 14
The reader will note that as soon as the parabola’s aa is of sym-
metry was turned from the vertical or the horizontal position, an
ay-term appeared in the equation. (Compare with the last para-
graph of $8(2), p. 28.)

For Scientific and Technical Workers 39
(4) General Equation of the Parabola.
We will now build up an equation of a parabola that will be
free to assume any possible shape and any possible position in
the plane. From that equation we will learn how to tell whether
any given numerical equation of the second degree represents a
parabola; and if so, how to calculate the width and position of
the parabola from the given numerical coefficients.
As before, we will take for our two reference lines the parab-
ola’s own axis of symmetry and the perpendicular to it through
the vertex. Let the vertex be (aw, yo), which may mean any
point. To impart to the parabola complete freedom of location,
we adopt for the equation of its axis the general equation of the
straight line, which we write in the form: mæ – my + s = 0. Since
the axis passes through the vertex, we have also: maco — myo + s = 0.
And if we eliminate s between the two equations by subtracting
the second from the first, we obtain for the equation of the axis:
ma. — my – (mayo – myo) = 0; which denotes the axis as any line
passing through any point (aco, yo). The perpendicular distance
from the axis of a point (ac, y) lying on the parabola is given by
the left member of the axis equation, divided by V m” + m”.* The
square of that distance is therefore :
[ma' – my – (maco — mya)]*
- m” + m,” g
The equation of the line perpendicular to the axis is na; + my
+ k = 0; and as this line too passes through the vertex, its equation
may be written: ma; + my – (maco + myo) = 0. The distance of our
parabola point (ac, y) from this line is
na, + my – (ma, + myo)
Vm2 + m2 º
For all points on the parabola this distance will be proportional
to the square of the distance of the point from the axis. We
again denote by the letter c the proportionality constant, which
may be either positive or negative (though its reciprocal 1/c,
representing the length of the latus rectum, will always be taken
as a positive quantity). We thus get the general equation of the
parabola in the following form :
na: + my – (maco + myo) [ma' — my – (mayo – mye)]*
= C : ... (11)
Vm2 + m2 m” + m,”
* $6(3), p. 13.
40 Plame Analytic Geometry
To decipher any given parabola equation, it will suffice to put
it in the form of (11): for then the bracketed expression in the
right member will yield the equation of the axis, the numerator
of the left member will yield the equation of the perpendicular
at the vertex, and the width and direction of opening of the
parabola will be indicated by the magnitude and sign of the pro-
portionality coefficient C ; and thus the shape and position of the
parabola will be fully determined. To make this possible we pro-
ceed to develop formulae for calculating the constants m, n, c, aco,
and ye from the coefficients of any given equation.
By obvious algebraic transformation, equation (11) turns into :
n\/m” + º
— | * 0.
C
T
m*ac” – 2mmacy + m?y” — |ºm (maco — mye) +
2 a 2
+ |2. (mº-ny) – "Y" tº:
C
| y + (mayo – myo)*
+ (nºw + myo)Ym’: nº 0
C
Comparing this general equation of the parabola with equation
(9), namely,
Aacº -- Bacy + Cy^+ Dae H-Ey + F = 0,
which represents all conics, we get the following six identities:
A = m^; D--2monº-ny.)-nVº
B = -2mm; E = 2n (mayo – myo) —mºvº,
(ma, + my,) Vmºnº
C
C = n°; F = (maco — myo)* +
The first thing we note is that in the equation of a parabola the
Square of the coefficient of the acy-term, that is (–2mm)*, equals
four times the product of the coefficients of ac” and y”, that is
4mºm”; or, in the notation of equation (9): B4 = 4AC. This is the
same as we found in the case of parallel lines.* And as in the
case of parallel lines, so here also this fundamental relation results
from the terms of the second degree being the square of a
binomial: in this case, the (ma — my)* of equation (11). A little
later it will be shown that this relation alone, if it subsists in a
* $8(1).
For Scientific and Technical Workers 41
given equation, is sufficient proof that the equation represents a
parabola, provided the relation BDE = AE* + C D*, which is
characteristic of parallel lines, does not hold in the given equation.
It should be noted that since B^+4AC, the product BDE cannot
be greater than the sum AE* + C D*, if the quantities involved be
real.” For AE*4. CD2 – BDE = AE* CD2–2DEVAC = (EVA-
DVC) *, and if this square were a negative quantity, the expres-
sion would be imaginary. Therefore, the coefficients being real,
all parabola equations are necessarily and sufficiently charac-
terized by the two eacpressions: Bº – 4AC = 0 and BDE 3 A.E.”
+ C D*. The second of these expressions, however, must not be
mistaken for an independent relationship among the coefficients;
it should rather be viewed as the absence of a relationship, since it
leaves BDE – AE* – C D* free to assume an infinity of values.
The two expressions characteristic of the equation of a parabola
do not contain the term F ; and any change of F in a parabola
equation has no effect on the parabola as such (or even on its
dimensions), and merely changes its position.f
And now as to the constants m, m, c, aco, and ye. The first and
third of our six identities give directly: m = VA and n = V/C.
The identities for D and E can be solved as simultaneous equa-
tions in 1/c and (maco — myo). Having these two values, we can
next obtain the value of (maco -i- myo) from the identity for F.
And finally, having expressions for (mayo – mye) and (maco -- myo),
we will solve them as a pair of simultaneous equations and thus
obtain general formulae for a, and ye.
The identities for D and E give, in the first place:
only) VºIº TT (ATC) VATC’
In our notation, B = -2mm, whence m = — B/2n = – B/(2 VC).
Substituting in the preceding expression VC for n and – B/
(2VC) for m, we get
* A and C are supposed to be, as usual, positive.
f We shall see later that, unlike parabola equations, which always stand
for real curves, the equation of any ellipse may be transformed into that of
an imaginary ellipse merely by increasing the constant term beyond a certain
value. In the equation of a parabola, increase of the constant term would
merely move the vertex along the axis of symmetry, in the direction in which
the parabola opens.
1 'm D + m E ſm D + m E
C
42 Plane Analytic Geometry
Formula I:
c - 204 + C) V(A + C) C
BE – 2CD
which will always give the proportionality coefficient c with its
correct sign if V(A + C) C is always taken with the positive sign.
The sign of c is important since, when the axis of symmetry has
been determined, it permits of telling in which of the two opposite
directions of the axis the parabola will open. Fig. 15 indicates
the positive and negative directions for different positions of the
axis.” The directions, corresponding to the sign of c, are dic-
tated by the positive or negative values of the distances repre-
sented by the linear member of equation (11), if we always write,
+

A X
º
FIG. 15
the linear member beginning with the ac-term positive. The stu-
dent should note that Fig. 15 is intended as a practical guide in
graphing parabolas in harmony with our formulae and equations,
and not as an attempt to reverse any accepted convention regard-
ing positive and negative directions generally.
* Formula I readily gives:
1 (BE-2CD)* B*E* +4C-Dº-4BCDE
(a) = IGCHIa) - 4C (A + C)*
_4ACE* +4C*D*-4BCDE AE*4. CD3 – BDE .
4C (A + C)* * (A + C)* '
Or
l 1.
C V (A + C)*
This formula for the latus rectum brings out its proportionality to the square
root of the difference AE* + CD*— BDE. If this difference is reduced to zero,
the latus rectum too becomes zero and the parabola narrows down into a
For Scientific and Technical Workers 43
For the constant (maco — myo) the solution of the simultaneous
identities for D and E gives:
(mayo – myo) = *E — m/D - *E – mp.
2(m2 + m2) T2(A + C)
Again substituting – B/ (2VC) for m and VC for n, we get:
BD + 2CE
4(A + C)VC
The identity for the constant term F gives:
[F – (mayo - nye)*] c_[F– (maw-myo)*] c.
Vm2 + m2 *ms VA. H. C. 5
and if in this we substitute the value of (maco — myo) from (13)
and the value of c from Formula I, we get :
16CF (A + C)” – (BD + 2CE)*
8(A + C) (BE – 20 D)\/C
And now, by solving (13) and (14) as simultaneous equations
in we and y, and once more writing VC for n and – B/(2N/C)
for m, we get, for the coordinates of the vertex,
(maco — myo) =
(ma', + ºnyo) -
(ma's 4 mya) =
Formulae II:
QCv =
160°F (A + C)” – (BD+2CE) [B(BE – 20 D) + C (BD+ 2CE)].
8C (A + C)*(BE – 2CD)
!/v =
(BD + 2CE) [B (BD + 2CE) – 4C(BE – 20 D)] – 16 BCF (A + C)?
16C (A + C)*(BE – 20 D) te
Equation (11), in which the values of the constants m, n, c,
ace, yo are to be substituted, corresponds to negative coefficients
of the acy-term in given equations, since we assume m and n to
be positive and since in our notation B = -2mm. In all cases
where B is positive, some of the signs will be reversed in equation
(11) and it will have to be used in the form of equation (117):
na. – my – (mayo – myo) C [ma' + my – (ma, + myo)]* (11')
Vm2 + m2 m” + m”
The formulae for m, n, c, aco, yo will remain unchanged.
straight line: namely, as may be seen from equation (11), into the parabola's
axis of symmetry. However, this instructive formula should not be used in
place of Formula I, since in squaring the latter, as we just did, the proper
sign of c, and therefore the direction in which the parabola opens, is lost
from view.
44 Plame Amalytic Geometry
These formulae furnish a pre-solution of all parabola equations
eaccept those in which C = 0 and therefore (since B4 = 4AC) also
B = 0; in other words, eaccept equations in which the y”-term and
the acy-term are absent. In those cases our formulae will yield
the indeterminate quantity 0/0. But in any such case the given
equation can obviously be put in the form, y = a + ba + ca:*, or
ca:* + ba – y + a = 0, and for the constants of equation (11) we
will have : m = Vc, m = 0; c, the proportionality constant of (11),
will be simply the coefficient of ac” in the given equation; and
finally, as shown in sub-section (1) of the present $10, a = — b/2c
and y = (a – bº/4C). Thus the “exceptional’’ case where C = 0
and B = 0 will give us no trouble.
The student should not be intimidated by the somewhat lengthy
appearance of our formulae I and II. In actual practice each of
them requires but a few minutes of simple arithmetical work, and
(nota bene:) the work will be further simplified if the computer
begins by evaluating the expressions (BD + 2CE) and (BE
– 20 D), which occur in the formulae several times; and of course
if he keeps his eyes open for possible cancellations before begin-
ning to multiply and divide.
We are now ready to introduce a few numerical examples, and
the student will be well repaid if he takes the trouble to work
them all. To visualize his results, he should in each case make
an approximate picture of them by drawing a parabola-like curve
through the vertex and the ends of the latus rectum (as we will
do in Example 1). In $9 it was shown that in the case of the
simple equation y = ca.” the distance from the vertea to the latus
rectum equals one-fourth the length of the latus rectum; but that
distance must remain the same for any other form of equation,
since, as already explained, the parabola itself, with its focus and
its latus rectum, remains unchanged when it is moved to a dif-
ferent position and its equation is changed accordingly.
NUMERICAL, EXAMPLES
Eacample 1. Given, ac” – 2acy + y” – 8a; +16 = 0. Here: A = 1,
B = -2, C = 1, D = – 8, E = 0, F = 16. We find: Bº (= 4) = 4AC
(= 4 × 1 × 1) and BDE (= 0) < AE* + CD*(= 64); hence we have
a parabola. And first, (BD + 2CE) = 16; (BE – 20 D) = 16;
m = VA = 1; n = VC = 1. By Formula I; c = 4\/2/16 = V2/4.
For Scientific and Technical Workers 45
By Formulae II: aco - 5/2; yo = 1/2. Since B is negative, we fol-
low equation (11), and find : (maco 4 mya) = 3, (ma, - myo) = 2,
and therefore the given equation assumes the explicit form :
a + y – 3 V2 (a – y–2)*
V2 TT 2
The axis is thus a: — y –2 = 0; the perpendicular through the ver-
tex is a + y – 3 = 0; the latus rectum is 4/ V2, and its distance
(toward the right) from the vertex is 1/ V2. Fig. 16 is an
approximate picture of this parabola.
–2 y
FIG. 16
Eacample 2. Given, ac” + 2acy + y” — a + y = 0. Here: A = 1,
B = 2, C = 1, D = –1, E = 1, F = 0. Bº (= 4) = 4AC (= 4) and BDE
(=–2) < AE* + C D*(= 2); hence we have a parabola. (BD
+2CE) = 0; (BE-2CD) = 4; m = VA = 1; n = VC = 1. By For-
mula I; c = 4\/2/4 = V2. By Formulae II: a = 0; y = 0. Since
B is positive, we follow equation (117): (naco — myo) = 0; (maco
+ myº) = 0; and the given equation becomes:
ac – y (a + y)*
V2 V2 2
Hence the axis is a + y = 0; the perpendicular through the vertex
is a − y = 0; the latus rectum is 1/ V2, and its distance from the
vertex is 1/ (4V2). Using these data, the student will roughly

46 Plane Analytic Geometry
sketch the curve and note that, with the vertex at the coordinate
origin, the parabola opens downward toward the right (coefficient
c positive!) and lies mostly in the fourth quadrant. Compare
it with the parabola discussed in subsection (3) of the present
§10.
Eacample 3. Given, ac” +2acy + y” — 2a – 2y+ 1 = 0. Here: A = 1,
B = 2, C = 1, D = -2, E = – 2, F = 1. We find: Bº (= 4) = 4AC (= 4);
BDE(= 8) = AE* + CD*(= 8). Hence we have a pair of parallel
lines. Using the formulae of $8(1), show that the two lines are
coincident and that the simple equation of the really single line
is a + y –1 = 0. Show that the square of the left member of this
simple equation is identical with the left member of the given
quadratic. What single change in the given equation would
separate its coincident lines? (Answer, any diminution of its
constant term; an increase would make the lines imaginary.)
Eacample 4. Given, Vºl. Vy= 1. Squaring and rearranging
terms: 2Vºy–– (a + y – 1). Squaring again and collecting
terms, we get: ac” – 2acy + y” — 2a – 2y+ 1 = 0 (which differs from
the equation of Example 3 only by the sign of B). Here: A = 1,
B = – 2, C = 1, D = – 2, E = – 2, F = 1. Bº = 4AC and BDE 3 AE”
+ CD*; hence the given equation represents a parabola. Show
that it can be put in the following form, corresponding to equa-
tion (11), since B 30:
* + y = 1/2 - 1 (2-9)*
V2 V2 2
Write the equations of the axis and the perpendicular through
the vertex, and make a rough graph of the parabola. Show that
Vic- Vy= 1 represents the same parabola. -
Eacample 5. Given, 3y” +2a: 4- 1 = 0. Here: A = 0, B = 0, C = 3,
D = 2, E = 0, F = 1; m = VA = 1, n = VC = V3. We find:
B” (= 0) = 4AC (= 0) and BDE(= 0) < AE* + C D*(= 12); hence
we have a parabola. Show that its vertex is at the point
(–1/2, 0) and that, in conformity with equation (11), the given
equation can be written:
V3: a +V3/2 – 3 (–V3 y)*
V: T 2 3
Where is the parabola’s axis.” What is the equation of the per-
For Scientific and Technical Workers 47
pendicular through the vertex? Note that (since there is no
ac”-term) if the given equation is written, a = – 1/2 – (3/2) y”, the
coefficient of y” is c, the reciprocal of the latus rectum. Make a
rough plot of the parabola.
Eacample 6. Given, 4a:* + 4acy + y” — 5ac = 0. Here: A = 4, B = 4,
C = 1, D =–5, E = 0, F = 0; m = \/A = 2, n = VC = 1. Bº (=16)
=4AC (= 16) and BDE(= 0) < AE* + CD” (= 25), hence we have
a parabola. BD + 2CE = – 20; BE – 2CD = 10. By Formula I:
c = \/5. By Formulae II: aco = 1/5; yo = 3/5. Since B > 0, we
conform to equation (117):
- a — 29 + 1 (2a, + y – 1)*
V5 V5 wº
What are the equations of the axis and the perpendicular through
the vertex? Draw the latus rectum and a curve through its ends
and the vertex.
Eacample 7. Given, 4ac” – 12ay +99°–3a – 29 + 4 = 0. Here:
A=4, B =–12, C =9, D =–3, E=-2, F =4; m =VA =2, n = VC =3.
B” (= 144) = 4AC (= 144) and BDE (= —72) < AE* + C D* (=97),
hence we have a parabola. BD + 2CE = 0; BE – 20 D = 78. By
Formula I; c = V13. By Formulae II: c. – 12/13; y = 8/13.
Since B 30, we conform to equation (11):
3a; + 2y — 4 H; (22 – 39)*
V13 T V13 13
Hence, the parabola’s axis is 2a – 3y = 0; the perpendicular at
the vertex is 3a; + 2y – 4 = 0. What is the latus rectum ? Draw
a parabolic arc through the vertex and the ends of the latus
rectum.
Eacample 8. Given, ac” — 2\/3. acy + 3y” — 6V3 a - 6y = 0.
Here: A = 1, B =-2\/3, C =3, D=–6V3, E =–6, F = 0; m = 1,
n = V3. Bº (= 12) = 4AC(= 12) and BDE(= – 216) < AE* +
CD*(= 360), hence we have a parabola. BD + 2CE = 0 and BE –
2CD =48V3. By Formula I: c = 1/3. From Formulae II it is
obvious that since in this case both F = 0 and BD + 2CE = 0, the
vertex (a.e., yo) must be at the coordinate origin. Finally, since
B - 0, we can write the given equation in conformity with (11):
V3: a 4 y_1 . (2–V3 y)*
VTT 3 4 ſº
48 Plame Analytic Geometry
What are the equations of the axis and the perpendicular through
the vertex? What is the distance of the focus from the origin Ż
Sketch the parabola.
Eacample 9. Given, y” – 2a – 8 y + 10 = 0. Here: A = 0, B = 0,
C = 1, D = – 2, E = – 8, F = 10; m = 0, n = 1. B*(= 0) = 4AC (= 0)
and BDE (= 0) < AE* + C D*(= 4), hence we have a parabola.
BD + 2CE = –16; BE – 20 D = 4. By Formula I; c = 1/2. By
Formulae II: aco = — 3, ye = 4. Since B = 0, we conform to equa-
tion (11): º
{e & º 0) 4? +l- *ze ** * * ! * -:
What is the equation of the axis? Of the
—a. V/-R-IL V-1. U.t-i \->
vertex? Sketch the parabola.
Eacample 10. Given, 12y” + 82 – 12y+ 7 = 0. Here: A = 0,
B = 0, C = 12, D = 8, E = –12, F = 7; m = 0, n = VC = 2\/3.
B” (= 0) = 4AC (= 0) and BDE(= 0) < AE* + CD” (= 768), hence
we have a parabola. BD + 2CE = – 288; BE – 2CD = –192. By
Formula I; c = — 3/2. By Formulae II: aco = – 1/2; yo = 1/2.
Since B = 0, we conform to equation (11):
2V3 ºr V3__3 (-2\/3 yr V3):
2V3 2 12 *
Obviously, the axis is 29 — 1 = 0, the perpendicular at the vertex
is 2a: -- 1 = 0. The axis of symmetry is thus horizontal, and
since c (= – 3/2) is negative, the parabola opens toward the left.
Sketch it.
Eacample 11. Given, ac” – 2a: + 4y — 3 = 0. Here: A = 1, B = 0,
C = 0, D = – 2, E = 4, F = – 3; m = 1, ſm ºr 0. B” (= 0) = 4AC (= 0)
and BDE(= 0) < AE* + CD*(= 16); hence, we have a parabola.
But since C = 0, our Formulae I and II would yield 0/0. There-
fore we must put the equation in the form of all vertical parab-
olas, namely, y = a + ba + ca:*, and use the formulae of $10(1).
We get: y = 3/4 + 1/2 - a - 1/4 ac”. The proportionality constant
c is now given directly by the coefficient of ac”: namely, c = −1/4
(whence the length of the latus rectum is 4). Further, a = 3/4,
b = 1/2. Hence, a , = — b/(2C) = 1; yo = a – bº/(4C) = 1. Conform-
ing to equation (11), we get:
y - 1 1 (a – 1)*
* • – 8
VI T 4 1
For Scientific and Techimcal Workers 49
The axis of symmetry is thus a: –1 = 0; the perpendicular at the
vertex is y – 1 = 0; the curve opens downward. Sketch the
parabola.
Eacample 12. Given, 9a:” – 6acy + y” + 4a – 3y – 1 = 0. Here:
A = 9, B = – 6, C = 1, D = 4, E = −3, F = –1; m = 3, n = 1. Bº (= 36)
=4AC (= 36) and BDE(=72) < AE*4 CD*(=97), hence we have
a parabola. BD + 2CE = – 30; BE – 2CD = 10. By Formula I:
c=2\/10. By Formulae II: æ, =–43/80; y =-69/80. Since
B 3.0, we conform to equation (11): -
a +39 + 25/8 = 2\/10. (32-y: 3/4).
V10 10
The axis is thus 12a – 4y + 3 = 0, the perpendicular through the
vertex is 8a, + 24y + 25 = 0. Sketch the parabola.
Some further parabola and parallel line problems will be found
among the Additional Numerical Examples following Chapter
III.
$11. AN IMPORTANT PROPERTY OF THE PARABOLA
In the preceding sections we defined the parabola with refer-
ence to two perpendicular lines: namely, as the locus of a point
the square of whose distance from one of the lines (the parabola's
axis of symmetry) is proportional to its distance from the second
line (the tangent to the parabola at its vertex). This, we have
Seen, leads to an equation, of which one part forms a perfect
Square and the other part is linear. But the same type of an
equation would also result if the two intersecting reference lines
were not perpendicular, yet the square of the distance from one
of the lines were proportional to the distance from the second
line. The question arises, what species of curve would result in
such a case?
The answer is, we would still have a parabola. For, as we shall
presently see, it is a characteristic of this curve that what is true
of the distances of its points from the tangent at the vertex and
from the axis of symmetry is equally true of their distances from
any tangent and from the line through its point of tangency
parallel to the axis of symmetry: that is to say, the distance from
any tangent is proportional to the square of the distance from the
line through the point of tangency parallel to the axis of sym-
metry; though this line and the tangent are never perpendicular
except at the vertex.
50 Plame Analytic Geometry
If we prove this proposition, we shall be able to draw the fol-
lowing instructive conclusions: (1) That every parabola can be
represented by an infinite number of equations, corresponding to
the infinite number of tangents that it can have; (2) conversely,
that any pair of intersecting lines, no matter what the angle
between them, can serve as the reference lines of a parabola; and,
above all, (3) that a comic in whose equation the terms of the
second degree form a perfect square (that is, B* = 4AC), if at all
a curve, is necessarily a parabola.”
The last of these conclusions has a particularly direct bearing
on our problem of deciphering equations of the second degree,
and hence the importance to us of proving the proposition which
leads to it.
For the sake of algebraic simplicity, we will employ the parab-
ola equation in the form y = ca”. To find the points where a line
g = a + ba will intersect the parabola we write, y = ca:* = a + bac,
whence ca:* – ba: — a = 0. Solving this by the ordinary quadratic
formula, we get:
- *-*. ++. V b% + 4ac,
2C T 26
If the line y = a + ba is a tangent, it will meet the parabola in only
one point, and therefore the expression under the square root
must be zero; whence a = — b%/(4C). Substituting this in y = a + bac,
we obtain the general equation of a tangent to our parabola :
g = — bº/(4c) + bac. Further, since the expression under the square
root is zero, the abscissa of the point of tangency is obviously
a = b/(2C), and the equation of the vertical line through this
point is a - b/(2c) = 0. The square of the distance from this line
2
of a point (ac, y) on the parabola is ( *-*) . The distance of
Ay — bac + b%/(4C)
VS2 TT
two quantities were proportional to each other, the proportion-
ality coefficient being c/V b° 4-1, we would get:
y – ba: + bº/(4c) C * I (C b \*
Vbº + 1 V b% + 1 2c /
* If the angle between the reference lines is reduced to zero, then BDE =
AE* + CD* and the parabola turns into a pair of parallel straight lines.
the same point from the tangent is Now, if these
For Scientific and Technical Workers 51
If the tangent should happen to be perpendicular to the axis of
symmetry, its slope b would be Zero, and the proportionality
coefficient would be simply c. The equation as a whole would
become y = ca:*. But it is easy to see that the equation changes
to this simple form not only if b = 0, but no matter what the value
of b. Which proves that our parabola y = ca.” can be represented
by any equation showing the distance of its points from one of
its tangents to be proportional to the square of the distance of the
points from the line through the point of tangency parallel to the
axis of symmetry. In other words, we have a parabola whether

Y - P
, j), 8
`s ~
nº
FIG. 17
Q
the intersecting reference lines are perpendicular to each other or
ſnoi.
Fig. 17 shows a pair of reference lines PQ and RS chosen at
random for the parabola y = ca:”, with c = 1/4 (whence the latus
rectum equals 4). PQ is vertical (that is, parallel to the axis of
Symmetry), and since we have made it pass through the right end
of the latus rectum, its equation is a -2 = 0. The square of the
distance of a point (a, y) on the parabola from this line is (a, -2)*.
On the other hand, we have seen that the equation of any tangent
to the parabola y = ca” is y = — b% (4C) + bac; or, since in our case
52 Plame Amalytic Geometry
C = 1/4, y = – b% + bac. Further, since the tangent RS passes
through the end of the latus rectum, that is, through the point
a = 2, y = 1, we have: 1 = — b% + 2b. Hence b = 1 and the tangent
RS makes an angle of 45° with the X-axis; and obviously also an
angle of 45° with the second reference line PQ. Since b = 1 and
C = 1/4, the equation of the tangent is y = − 1 + æ, or y – a + 1 = 0.
The distance of our point (a, y) from this tangentis (y – a + 1) /V2,
and by the above proposition this distance should be proportional
to the square of the distance of (a, y) from PQ, that is, to (a – 2)”.
The following table shows this proportionality for several points
of the parabola y = wº/4, and thus furnishes a simple numerical
test of our important proposition; the ratio of (y – 2 + 1) /w/2 to
** v-u- \ 9
(a – 2)” remains constant: namely, 1/ (4\/2).
3C Qſ (y – c + 1)/V2 (a – 2)”
0 0 1/V2 4
4 4 1/V2 4
6 9 4/\/2 16
8 16 9/V2 36
10 25 16/V2 64
– 2 1 4/V2 16
— 4 4 9/V2 36
– 6 9 16/V2 64
– 8 16 25/V2 100
– 10 25 36/V2 144
§12. A THEOREM RELATING TO THE LINEAR PART
OF ALL PARABOLA EQUATIONS
In §§10 and 11 we furnished principles and formulae sufficient
for identifying a given equation as that of a parabola and for
readily deciphering such an equation. In the present section we
will discuss an aspect of the subject that will yield some further
insight into equations of the parabola.
The general equation (9), which stands for all conics, may be
written in the form,
Aacº -i- Bacy + Cy^ = — (Dac H. Ey + F).
For Scientific and Technical Workers 53
If this equation as a whole represents a parabola or parallel lines,
then the left member is a perfect square, and either member may
be equal to Zero or to any positive quantity. The generality of
our discussion will not be affected if we concentrate on the case
when either member is zero:
Arº-Bry: Cyº-0;
Dac + Ey + F = 0.
The first of these partial equations represents a pair of coincident
lines passing through the origin and having the slope — VA/ VC.
The second equation represents a single line whose slope is — D/E.
If in a given equation VA/ VC = D /E, that is, if the lines rep-
resented by the two partial equations have the same slope, then
EVA – DVC = 0, by squaring which we get BDE = AE* + C D*
and, as we have seen, the equation as a whole represents parallel
or coincident lines.
If VA/VC and D/E are unequal, that is, if the lines repre-
sented by the two partial equations are non-parallel, then E VA –
DVC is either greater or less than zero, BDE 3 AE* + C D* and,
as we have seen, the given equation as a whole represents a
parabola.
In the latter case the coincident lines Aacº + Bacy + Cy^ = 0 are
parallel to the parabola’s axis of symmetry and hence cut the
curve in a single point. But since the coordinates of that point
of intersection satisfy both Aacº + Bacy + Cy^ = 0 and the total
equation of the parabola, they must also satisfy the partial equa-
tion Dac -- Ey + F = 0. Moreover, the value of y derived from
this latter partial equation, namely, y = — (Da; -- F)/E, if substi-
tuted in Aacº + Bacy + Cy^ = 0, will lead to a single value of ac, be-
cause Aac” + Bacy + Cy^ is a perfect square. Similarly, the value
of a derived from Dac -- Ey + F = 0 will lead to a single value of y.
We conclude that the point of intersection of the parabola and
the coincident lines Aacº -i- Bacy + Cy^ = 0 is also a point, and the
only point, at which the line Da; -- Ey + F = 0 meets the parabola.
We thus arrive at the interesting theorem: that in any equation
of a parabola the linear part, equated to 26 ro, represents a tangent
to the parabola.
54 Plame Analytic Geometry
By way of a numerical illustration, we examine at random
the equation in the 7th of the Examples at the end of $10. The
equation is 4a:” – 12ay +99°–3a – 2y+ 4 = 0. According to our
theorem, 3a; + 2y — 4 = 0 should be a tangent to the parabola. We
have: y = 2 – 3a/2. Substituting this in the given equation, we
get: 169a,” – 312a: + 144 = 0; whence,
a =}+ Viº,
13 169 °
that is, a has the single value 12/13. Similarly, y is found to
have the single value 8/13. The line 3a 4-2y –4 = 0 is therefore
a tangent to the parabola. In this case the point of tangency
happens to be the parabola's vertex. The reader is advised to
test the theorem on a few other numerical equations.
CHAPTER III
INTERPRETATION OF ANY EQUATION OF THE
SECOND DEGREE
Every possible equation of the second degree is included in the
general pattern,
Aacº -- Bacy + Cy^+ Dac -- Ey + F = 0,
and therefore this general equation may represent any of the
following geometric entities: (1) a pair of parallel straight lines
or a parabola; (2) a pair of intersecting straight lines or a hyper-
bola; and (3) an ellipse or a circle. These are collectively re-
ferred to as the conics or the comic sections. We have seen that
the nature and location of any particular conic are implicit in
the constants of its equation, and we will now summarize the
characteristics by which any given equation of the second degree
may be deciphered and the conic represented by it identified.”
For this purpose we recommend the use of two KEY EXPRES-
SIONS, namely, in the notation of the general equation: Bº – 4AC
and BDE – AE? — CD2.
(1) If in a given equation both Bº – 4AC = 0 and BDE –
AE* – CD* = 0, the equation represents a pair of parallel line.
If in addition the constant term F = Fo = D*/4A = E4/4C, then
the two lines are coincident; that is, really a single straight line
is represented by the equation. Formulae for finding the slope
and positions of the lines, whether they are coincident or distinct,
and for finding the distance between the lines if distinct, are
given in $8 (1), pp. 18–20.
(2) If in an equation B” – 4AC = 0 but BDE 3 AE* + CD*,
the equation represents a parabola. In $10 (4), pp. 42f, are given
formulae and instructions for finding: the latus rectum, the di-
rection in which the parabola opens, the coordinates of the vertex,
and the equation of the axis of symmetry; by which the parabola
of the given equation is fully described. If the y”-term is absent
in the given equation, then the simple formulae of $10(1), p. 35,
should be used instead.
(3) If in a given equation B” X- 4AC, the equation repre-
sents a pair of intersecting lines if its constant term F equals the
* As heretofore, the coordinate axes are assumed perpendicular.
55
56 Plame Amalytic Geometry
ratio of our two Key Expressions. F is then denoted by the
symbol F., (to indicate that it pertains to straight lines), and its
value is given by the equality
BDE – AE2 — CD2
B” – 4AC
Formulae for finding the point of intersection (a, i, yi) of the lines
and for finding their separate linear equations are given in $8 (2),
pp. 25–26.
(4) If in a given equation B” X- 4AC, but the constant term
F is either greater or less than Fs as defined above, the equation
represents a hyperbola. If in place of the given value of F we
introduce the value of Fs, the equation is changed into that of the
hyperbola’s asymptotes (whose point of intersection ach, yi is the
center of the given hyperbola). Further, if F′ denotes the con-
stant term for the conjugate hyperbola, then F's = (F -- F^)/2, or
F’ = 2F's – F, and by introducing this value of F′ in place of the
given F we obtain the equation of the conjugate hyperbola.
(5) If in a given equation B” < 4AC, the equation repre-
sents either an ellipse or a circle. If Bº 34AC and in addition
to that B = 0 and A = C, the equation represents a circle. The
necessary condition B = 0 (that is, absence of the acy-term) corre-
sponds to the perfect symmetry of the circle, in consequence of
which, and unlike all other conics, the circle cannot assume an
‘‘inclined’’ position with respect to the coordinate axes.* If in
a given equation B” < 4AC but either one or both of the addi-
tional conditions of a circle are not fulfilled, the equation repre-
sents an ellipse.
(6) If a given equation represents either an ellipse or a
hyperbola, complete transparency may be imparted to it with the
aid of a single model equation analogous to equation (11) or (11)
for the Parabola. The model equation is nothing but a generali-
zation of the familiar ac”/a” + y”/b° = 1 and acº/a” — y”/b° = 1. In
these latter, ac” is the square of the distance of a point (ac, y) from
the minor axis (that is, from a = 0), and it is divided by a”, which
is the square of half the major axis; y” is the square of the dis-
tance of the point from the major axis (that is, from y = 0), and
F's =
* Compare this with the last statement of $10(3), p. 38, concerning the
acy-term in equations of the Parabola; also with the last paragraph of $8(2),
p. 28.
For Scientific and Technical Workers 57
it is divided by b%, which is the square of half the minor axis. To
generalize these familiar equations, we replace the distances from
the particular lines a = 0 and y = 0 by distances from any two per-
pendicular lines intersecting at the curve’s center (act, yi). But
there is no need of two separate equations, respectively with plus
and minus connecting the two square expressions, and the follow-
ing single equation will be seen to represent hyperbolas as well
as ellipses:
(9-9); 'ºrº)lf, 10/-tº-ººrºol: 1. (15)
(ta” + 1) a” (t,” + 1) b%
The constants ai, y, may be found by the same formulae as given
is $8(2), p. 25, for finding the intersection of two straight lines.
The slopes to, it of the curve’s axes are given by:
C – A + V (C–A)2 + B2 C – A – V(C–A)2 + Bº
* B and to = P 2
the product totb being obviously — 1 (perpendicularity of the
axes!).” The squares of the semi-axes are given by:
ta
2(F – F.)
* --→ ~ *=s 2 2
=###(C + 4 + V(0 A)2 + B2)
and -
_2(F–Fs) 2 2 Y
bº-º-º:#(or A-V(C-4) + B*),
where Fs has the same meaning as before, the same as in sub-
section (3) of this Summary.
If the given equation represents an ellipse, both a” and b* will
come out positive. If the given equation represents a hyperbola,
then either aº or b% will come out negative, wherefore some au-
thors, with doubtful accuracy, refer to one of the hyperbola's
axes as ‘‘imaginary”; practically, a negative a” or b% merely in-
dicates that the sign connecting the two real quadratic expres-
sions in (15) is minus. If both a” and b* come out negative
* The significance of the ta and to slope formulae is not limited to the
ellipse and the hyperbola: for intersecting straight lines (asymptotes), they
give the slopes of the bisectors of the angles formed by the lines; for parallel
lines, to gives their slope; for the parabola, to gives the slope of the axis of
symmetry, to gives the slope of the tangent at the vertex. The formulae for
ta and to show that whenever C = A, while an aſy-term is present, the angle
formed with the coordinate axes is 45°. These propositions are worth
remembering.
58 Plame Analytic Geometry
we will have an imaginary ellipse (or if a” = b%, an imagimary
circle), since its equation leads to the sum of two quadratic ex-
pressions appearing equal to a negative quantity. If F should
happen to equal Fs, both a” and b% would obviously be zero; then,
if Bº 34AC we would have a point-ellipse (that is, an ellipse
shrunk to a single point, namely the center); and of course if
B* > 4AC we would have a pair of intersecting lines (asymp-
totes). Upon substitution of the values of the constants, the
bracketed expressions in equation (15), each equated to zero, will
give the equations of the curve’s aaces.
If in a given equation B = 0 (no acy-term), the formulae to be
wsed are: a* = — (F–Fs)/A and b% = - (F–Fs)/C. In all such
cases the explicit form of the equation will be simply
(a – wa)*/a” + (y – yi)*/b% = 1.
By following the directions in the above summary, as illustrated
by the numerical exercises furnished in this booklet, the reader
will have no difficulty in obtaining insight into any equation of
the second degree.
The following Table reproduces in condensed form the instruc-
tions for identifying any conic by its equation. The generalized
equation being
Aacº 4- Bacy + Cy^+ Da: 4: Ey + F = 0,
the symbols Fo and F's signify as before :
2 Hº 2 BDE – AE2 — CD2
Fo = − = Hº ; F ==Hiia
The sign A means greater or less.
F = Fo: coincident
f < Fo: distinct
F' > Fo: imaginary
B% = 4AC (BDE = AE* + C D*: parallel
Parallel lines lines
or parabola (BDE 3 AE* + CD*: parabola
B* > 4AC
Intersecting lines
or hyperbola
F = Fs: intersecting lines (asymptotes)
F #4 F, : hyperbola
B* < 4AC [B #4 0 and/or A A C : F' < F's : real curve
Ellipse or ellipse B' = Fs: one-point curve
circle = 0 and A = C : circle | F - F. : imaginary curve
For Scientific and Technical Workers 59
ADDITIONAL NUMERICAL, EXAMPLES
1. Show that 14a:” – 11acy — 15y” + 1 = 0 represents a hyperbola.
What is the quadratic equation of its asymptotes? What are the
coordinates of its center? What are the separate linear equations
of its asymptotes? What is the equation of the conjugate
hyperbola Ž
2. Show that 4a:* + 12acy +99°–20a: – 30y + 25 = 0 represents a
pair of coincident lines. What is their slope (=– D/E =–2A/B) ;
What is the simple linear equation? What would happen if the
constant term 25 were increased to 25.01% (To answer this, sub-
stitute 25.01 for 25 in the given equation, assume that y is con-
stant, and solve the equation as a quadratic in the single variable
a;). What would happen if the constant 25 were reduced to zero?
In the latter case what would be the perpendicular distance be-
tween the lines? What would be their separate linear equations?
3. Starting with two random linear equations representing
non-parallel lines, say, 3a; +4y = 0 and 2a: + y = 0, square the left
member of the second : (2a, + y)* = 4a:* + 4acy + y^ = 0. Add this
squared and the first of the given equations. Show that the re-
Sulting equation, 4a:* +4ay + y^+ 3a; +4y = 0, represents a parab-
Ola and that 32 + 4y = 0 is now a tangent to the parabola. Show
that the parabola opens in a positive direction, since the pro-
portionality coefficient c, by Formula I of $10(4), is +V5.
Show that the vertex is at the point (– 3/5, 1/5). Make a rough
sketch of the curve. Show that if its axis of symmetry and the
tangent (that is, the perpendicular) at the vertex are chosen as
reference lines, the equation, in conformity with equation (11%),
becomes:
a – 2 y + 1 (2a, + y + 1)*
\/5 T V5 5 º
Note that the axis of symmetry, 2a: + y + 1 = 0, is parallel to the
line 2a: -- y = 0, which we chose to square. Note that the parabola
passes through the point of intersection of 34: 4-4y = 0 and
2a: + y = 0. Why?
The student’s attention is called to the important difference
between the linear and the quadratic equations of one and the
same straight line. In §7 we learned that the addition of two
60 Plame Amalytic Geometry
linear equations of non-parallel lines results in a third line (pass-
ing through the intersection of the first two). In the preceding
exercise, addition of the linear equation of one straight line and
the quadratic equation of a second straight line resulted in a
parabola (passing through the intersection of the two lines);
which shows that the quadratic equation of a straight line pre-
sents very different geometric possibilities than the simple linear
equation of the same line.
4. Starting with two random parallel lines, namely, a -2y+ 1 = 0
and 2a – 4y + 1 = 0, we square the left member of the first equa-
tion: (a – 2y+ 1)* = a," – 4ay + 4y” + 23 – 4y + 1 = 0. Add this
squared and the second of the given equations. We get: 3: -43 y +
4y” +44 – 8y + 2 = 0, where A = 1, B = — 4, C = 4, D = 4, E = – 8,
F = 2. Show that this resulting equation represents two dis-
tinct parallel lines: distinct because F |=2} < Foſ= D*/(4A) =
E*/(40) = 4], Fo being the greatest value for real lines. Using for-
mulae given in $801), show that the perpendicular distance be-
tween the two lines (= 2VF-F/VA - C) is 2\/2/5; that their
common slope (= — D/E = – 2A/B) is 1/2; and that their separate
linear equations are a -29 + 2 + V2 = 0 and a -2ſ +2– V2 = 0.
Note that the slope of this pair of lines is the same as that of the
two original lines.
5. We have pointed out that only five of the six constants of the
general equation are mutually independent. The equation may
therefore be of the form, a 4 + Bacy + Cy^+ Dae + Ey + F = 0. If
the equation represents a parabola, the relation B” = 4AC then
becomes Bº-4C, or B =2\/C, and the equation can be written,
wº-2VC wy Cy” + Da 4 Ey + F = 0. It might be thought that
four simultaneous equations based, say, on four given points of
the parabola would suffice to fix the values of the constants C, D,
E, and F, and thus determine the equation of the curve. In re-
ality, when D, E, and F have been eliminated, we get an equation
containing both C and VC, which is a quadratic in VC and will
yield two different values of C, in consequence of which we will
get two different values also for each of the constants D, E, and F.
In brief, we will obtain two different equations; showing that two
different parabolas can pass through four given points. The
For Scientific and Technical Workers 61
following numerical exercise will illustrate this noteworthy propo-
sition. The reader will work out the details.
Given four points: (2, 2), (10, 2), (4, 0), (5/2, 1/2). Hence
four simultaneous equations: 4 + 8 VC F40 + 2D + 2E + F = 0;
100 + 40VC-1 40 + 10D + 2E + F = 0; 16 + 4D + F = 0; and
25+10VC+C+10D +2E +4F =0. Upon eliminating D, E, and
F, we get: C – 6VC – 7 = 0, whence VC = 3 + 4. Using the sub-
scripts 1 and 2 to denote corresponding values, we get: (VC) 1 = 7,
(VC), E-1; C, =49, C, -1; D, E-40, D,--8; E, E–160, E, E0:
F1 = 144, F2 = 16. The two parabolas passing through the four
given points are therefore: ac” +14ay + 499°–40a: – 160y + 144 = 0
and ac” – 2acy + y” – 8a; +16 = 0. The second of these equations was
deciphered in the 1st of the Numerical Examples following $10.
6. Given five points: (0, 0), (0, 3), (2, 3), (2,5), (4, 5). Show
that the only conic passing through them all is 5ac” – 10ay + 4y”
+20a: – 12y = 0. Show that this is a hyperbola, whose center is
at the point (2, 4). Show that the combined equation of its
asymptotes is 5ac” – 10ay + 4y” + 20a: – 12y + 4 = 0, their separate
linear equations (5 –V5)2-4y +6+2V5=0 and (5 + V5)2-4y
+ 6–2V5=0. Write the equation of the conjugate hyperbola
(F’ = 2F, - F = 8 – 0 = 8).
7. Given, 52°4-2ay + 5 y” – 12a – 12y = 0. Here: A = 5, B = 2,
C = 5, D =–12, E =–12, F = 0. The equation represents an ellipse:
because Bº (=4) < 4AC (=100) and an acy-term is present. Fa = 12;
and since F 3 F, the ellipse is real. The center is at the point
a's = 1, yi = 1. The slopes of the axes are: to = 1, to = –1 (that is,
the inclination is 45°, which is in accord with C = A). The
squares of the semi-axes are: a* = 3, b% = 2. Show that the given
equation therefore assumes the explicit form :
(a — y)* (a + y – 2)*
=#–4 =#––1,
which shows the equation of the major axis to be y = 2 - a, the
equation of the minor axis y = az. Draw the axes ( 2V3 and 2V2) 2
then roughly sketch the curve. Show that since Fs = 12, the equa-
tion 5ac” +2acy + 5 y” – 12a – 12y+ 12 = 0 represents a point-ellipse
and is satisfied solely by the coordinates of the center (1, 1).
Show that the same point-ellipse is obtained by equating the left
62 Plame Amalytic Geometry
member of our explicit equation to zero. For any constant term
greater than 12 (= Fe) the ellipse would be imaginary. In this
and the following eacercises, show that performance of the opera-
tions imdicated in the eacplicit equation leads back to the given
equation; hence, that our eacplicit equation and the given equation
are really one and the same.
8. Given, 4a:* + 99% + 82 + 36 y + 4 = 0. Here: A = 4, B = 0, C = 9,
D = 8, E = 36, F = 4. Since Bº (= 0) < 4AC (= 144) and A and C
are unequal, we have an ellipse; and since B = 0, its axes are
parallel to the coordinate axes. We find: act = -1, yi = -2; Fe = 40;
a*=— (F–Fs)/A =9; b% =— (F–Fs)/C = 4. Therefore the explicit
foLIII of the given equation is
(a + 1)* (y | 2)*_
9 + 4 = 1,
showing the equation of the major axis to be y =–2; that of the
minor axis, a = − 1. Draw the axes (2\/9 = 6 and 2V4 =4), then
roughly sketch the ellipse. Show that if F = 4 is replaced by
F's = 40, the equation will be satisfied only by the point (– 1, -2);
and if F′ × 40 the ellipse is imaginary (for instance, substitute
F = 49, then try to find the points of intersection of the minor axis
a = –1 with the ellipse).
9. Given, 2a:” – 3ay – 2y” + 3a – y +8 = 0. Here: A = 2, B = – 3,
C = – 2, D = 3, E =–1, F = 8. We find: Bº (=9) > 4AC (= – 16)
and F(= 8) > F. (= 1); whence the equation represents a hy-
perbola; and since C = – A, the hyperbola is rectangular (equi-
lateral). The center is at the point acs = – 3/5, yi = 1/5. The
slopes of the axes are: to = -1/3, to = 3. The squares of the semi-
axes are: a* = 14/5, b% = – 14/5. Show that the given equation
therefore assumes the explicit form:
(~ +39)* (32-y:#2)* = 1
10 : 14/5 10 : 14/5 5
which shows the equation of the transverse axis (the one that
intersects with the hyperbola) to be y = 2+3a: ; the equation of the
second axis to be a + 3y = 0. The combined equation of the asymp-
totes is 22°–3acy – 29* + 33 – y + 1 = 0 (since Fe = 1). Show that
this equation of the asymptotes is also obtained by equating the
left member of our explicit equation to zero (instead of 1). The
For Scientific and Technical Workers 63
equation of the conjugate hyperbola is 2a:” – 3acy — 29* + 3a – y – 6
= 0 (since Fs = 1 is the arithmetical mean of 8 and – 6).
10. Given, 9a:*— 4y” – 183: 4-89 – 31 = 0. Here: A = 9, B = 0,
C = — 4, D = — 18, E = 8, F = – 31. We find: Bº (= 0) > 4AC
(=–144) and F's = 5(> – 31); hence the equation represents a
hyperbola. And since there is no acy-term, the hyperbola's axes
are parallel to the coordinate axes. The center is at the point
acs - 1, yi = 1. The squares of the semi-axes are: a* = — (F–Fs)/
A = 4; b% = — (F–Fs)/C = – 9 (which makes the lengths of the
axes 2\/4=4 and 2V9=6). The explicit form of the given
equation is therefore
(a -1)* (y – 1)*_
+-------1,
the axis of length 2\/4 = 4 being on the horizontal line y = 1, the
axis of length 2V9 = 6 on the vertical line a = 1. What is the
combined equation of the asymptotes? What is the equation of
the conjugate hyperbola 2 Show that the line a = 1 does not meet
the given hyperbola. Draw the axes and the asymptotes (diag-
onals of the rectangle determined by the axes), then roughly
sketch the hyperbola. -
11. Given, 92% + 9y” — 360, H-69 + 10 = 0. Here: A = 9, B = 0,
C = 9, D = — 36, E = 6, F = 10. Since B*(= 0) < 4AC (= 324),
while A = C and there is no acy-term, the equation represents a
circle. We find: Fa – 37; a” = — (F–Fs)/A = 3; b% = — (F–Fs)/
C = 3; and since F 3Fs, the circle is real and more than a single
point. The center is at a:; = 2, yi = -1/3. The explicit equation
is therefore
(a – 2)” (y + 1/3)*
=s=–4–3–––1,
V3 being the radius. Since F, - 37, the equation 9a.” -- 99%–36a,
+ 6y + 37 = 0 would represent a point-circle and would be satisfied
solely by the point (2, — 1/3). Show that the same point-circle
is obtained by equating the left member of our explicit equation
to zero. If in the given equation F were greater than 37, we
would have an imaginary circle.
12. School books of elementary algebra frequently include as
simultaneous linear equations problems of which the following
64 Plane Amalytic Geometry
is typical: 6/a: + 12/y = -1; 8/a: – 9/y = 7. The pupil is directed
to eliminate y without clearing of fractions. The equations are
solved as if they were: 6v -- 12V = –1 and 8w – 9 w = 7, giving
w = 1/2, v = -1/3, and therefore a =1/u = 2 and y = 1/v = – 3.
This is a rather misleading practice, for the two given equations
are really of the second degree and represent, not straight lines,
- Y
FIG. 18
but rectangular hyperbolas. Cleared of fractions, they become:
acy + 12a: + 6y = 0 and 7acy + 9a, -89 = 0. (1) In the first of these :
A = 0, B = 1, C = 0, D = 12, E = 6, F = 0; thus B2 (= 1) > 4AC (= 0)
and F. (= 72) > F(= 0). We therefore have a hyperbola whose
asymptotes are together represented by acy + 12a: + 6y + 72 = 0 and
separately by a + 6 = 0 and y + 12 = 0, so that the center of the
hyperbola is at (– 6, -12). (2) In the second equation: A = 0,
B = 7, C = 0, D = 9, E = – 8, F = 0; thus Bº (=49) > 4AC (= 0) and

For Scientific and Technical Workers 65
F's (=–72/7) < F(= 0). So we have a hyperbola whose asymp-
totes are 7ay + 92 – 8y – 72/7 = 0 or, separately, 7a: – 8 = 0 and
7y + 9 = 0; therefore the center of the hyperbola is at (8/7, -9/7).
Treated as they should be, as simultaneous quadratics in a and y,
the given equations yield two solutions: a = 2, y = – 3; and a = 0,
g = 0 (showing that one of the two points of intersection is at
the coordinate origin). Fig. 18 shows both branches of the
hyperbola (2) and the points of their intersection with one of
the branches of hyperbola (1). The second branch and the
asymptotes of (1) are out of the picture. The center of hyper-
bola (2) is shown as the intersection of its asymptotes.
13. Given, a.” – 5ay + y” + 82 – 20 y + 15 = 0. Here: A = 1; B = — 5;
C = 1; D = 8; E =–20; F = 15. We find: Bº (= 25) > 4AC (= 4)
and F(= 15) < F. (= 16); hence we have a hyperbola. The cen-
ter is at act = -4, yi = 0. The slopes of the axes are: to - – 1, to - 1
(which is in accord with the fact that C = A while an æy-term is
present). The squares of the semi-axes are: a* = – 2/3, b% = + 2/7.
Show that the given equation assumes the explicit form:
(w-y-4)*_ (ºr 944): 1
2, 2/7 2, 2/3 “
Since the smaller semi-axis square 2/7 is positive, it is the shorter
axis, lying on the line a + y + 4 = 0, that intersects with the hyper-
bola. Compare in Fig. 7 (p. 22) the hyperbola which touches
the accompanying ellipse at the ends of its minor axis. Show
that our explicit equation resumes the given form if the opera-
tions indicated in it are carried out. Show that if in the explicit
equation the left member is equated to 0 instead of to 1, we get
the combined equation of the asymptotes.*
14. Let us build up the equation of a circle of radius r on the
basis of the distances of its points from a pair of diameters lying
* The student’s attention is called to the fact that to determine the posi-
tion of an ellipse it is sufficient to know the equations of the major and minor
axes; while to determine the position of a hyperbola it is further necessary
to note which of the axes intersects with the curve, that is, the square of
which semi-axis comes out with the positive sign. This, obviously, is because
for every pair of axes there are two conjugate hyperbolas and we must dis-
tinguish which of them is represented by the given equation. If the major
semi-axis square is positive, the hyperbola will lie within the acute vertical
angles of the asymptotes; if the major semi-axis square is negative, the
hyperbola will lie within the obtuse angles.
66 Plame Amalytic Geometry
on two random perpendicular lines, a - y + 2 = 0 and a + y – 3 = 0.
We get:
(ac – y + 2)* , (a + y – 3)*
2 . r2 + 2 : 7-2 = 1
Show that this equation reduces to 2a:* + 2y” – 2a – 10y – 2r”
+ 13 = 0. By way of a check: we have A = 2, B = 0, C = 2, D = -2,
E = –10, F =–2r” + 13; since B4-4AC (=–16) < 0 and since
B = 0 and A = C, the equation indeed represents a circle. We
find: a = 1/2, yi = 5/2, Fs = 13 and F – F. - – 2r”. Hence:
a” = — (F–Fs)/A = r^; b% = — (F–Fs)/C = r^. Applying the
explicit formula (ac – act)*/a” + (y – yi)*/b° = 1 (p. 58), we get
the equation of our circle referred to diameters parallel to ille
coordinate axes:
2 2
(a: ºr, tº = 1.
This equation too reduces to the form, 2a:* + 2y” — 2a – 10y – 2r”
+ 13 = 0; showing, as might be expected from the perfect sym-
metry of the circle, that any pair of perpendicular diameters, no
matter what their slopes, may be used as reference lines for the
algebraic description of a circle. Verify that, since Fs = 13, 2a:*
+ 2y” – 2a – 10y + 13 = 0 is a point-circle and is satisfied only by
the point (1/2, 5/2).
15. We will express in our explicit form the equation acy
–22.42 = 0, representing Boyle’s law of gases at 0° C. (see the 5th
example on p. 29). In this equation, A = 0, B = 1, C = 0, D = 0,
E = 0, F = — 22.42. Hence F, - 0 and F – F, -– 22.42; act = 0,
Ayi = 0; ta = 1, to = -1; a” = — 44.84, b% = + 44.84. Therefore the
explicit equation is
– 9:2. (9:2-1
2 : 44.84 2 44.84 ’
OI’
(*#9) - (*-9): 1
2 : 44.84 2 44.84
Evidently, the curve is an equilateral hyperbola; and since the
positive semi-axis square (+44.84) corresponds to the line
ac – y = 0, it is this line which intersects with the curve. Show
that our explicit equation reduces to the given form acy = 22.42.
For Scientific and Technical Workers 67
16. As a special test of the general theorem relating to the
linear part of parabola equations (Chapter II, §12, p. 53), show
that if a parabola's axis of symmetry is vertical, that is, if the
equation of a parabola is of the form y = a + ba + ca:”, its linear
part y = a + ba represents a tangent to the parabola and that the
point of tangency is (0, a ): in other words, that in such parab-
olas the point of tangency is always at the intersection of the
parabola with the coordinate Y-axis. By way of a graphic test,
plot the parabola y = 3 + a 4- }* and the line y = 34-2, to demon-
strate that the line is the tangent to the parabola at the point
(0, 3).
17. In curve-fitting, namely, in expressing a set of scientific
or statistical data in the form of a parabolic function,--the para-
bola equation most generally employed is y = a + ba + ca:”; which,
as we have seen, can only represent a parabola opening verti-
cally, upward or downward. Less widely known is the somewhat
more flexible equation y = S + rV24. pac, whose parabola may open
in any direction to the right of the vertical, upward or down-
ward, though not in either vertical direction itself. In some
cases this will yield a “closer fit” between the curve and the
observations than that yielded by the “vertical” parabola. With-
out attempting here to illustrate the curve-fitting power of the
equation y = S + rV24-pa, we will give three numerical examples,
to indicate some of the positions that the parabola represented by
it may assume.
(1) y = 1 + 2V.g. a. We write, 2Vºc = gy – a – 1. On squar-
ing and rearranging terms: a*–2ay + y”–2a – 2y+ 1 = 0, which
is identical with the equation of Example 4 on p. 46. The axis of
symmetry makes an angle of 45° with the X-axis (A = C), and
since in the explicit form of the equation the proportionality con-
stant is positive (namely, c = + 1/ V2), the parabola opens upward
toward the right.
(2) y = 1+2\/2-2. We get: acº +2ay + y” – 6a, – 29 + 1 = 0.
Show that the axis of symmetry is on the line a + y –2 = 0, and
since again c = +1/ V2, the parabola opens downward toward the
right (see Fig. 15, p. 42).
(3) Show that if in the equation y = S + rVa 4. pa, we set p = 0,
68 Plame Analytic Geometry
the axis of symmetry is parallel to the X-axis, and that again the
parabola opens in the positive direction (toward the right). For
instance, if y = 1 + 2\/z, we get a = 1/4- (1/2) y + (1/4) y”; or, as
this may be written explicitly: a, -} (y – 1)*; which shows the axis
of symmetry to lie on the horizontal line y – 1 = 0, the proportion-
ality constant to be + 1/4, and the vertex to be at the point (0, 1).
For curve-fitting purposes the equation will of course be used
in its general form, and without squaring it. On account of the
term rVg, it will not be applied to observations including mega-
tive values of a; ; any real parabola represented by the equation
will lie entirely to the right of the Y-axis, tottching it at the point
(0, s). These limitations of position, added to minor arithmeti-
cal complications, will ordinarily lead the computer still to give
preference to the form y = a + ba -- ca”.
CHAPTER IV
INVARIANTS
We have repeatedly emphasized that the equation of a geo-
metric figure expresses both its nature and its position with re-
spect to the coordinate axes. For each of the infinity of positions
that one and the same curve may occupy, it will have a different
equation. Yet all these equations must obviously have something
in common which corresponds to the sameness of the curve, and
which will remain invariant while the equations otherwise vary
from position to position. We will briefly investigate here the
more characteristic ‘‘Invariants’’ of equations of the second de-
gree corresponding to the several conics.
(1) Equations of the Ellipse and the Circle
The size and shape of an ellipse are defined by two quantities,
namely, the lengths of its major and minor axes. Hence, so long
as the ellipse itself remains the same, the right members of our
formulae [Chapter III (6)] for evaluating the squares a”
and b% of the semi-axes will remain invariant through all changes
of the curve’s position. For all positions we have:
*-###(c., A. VCETFTE),
T B2 – 4AC
2(F – F.
*-*#(c., 4-VEETFE).
Plainly, the product and the sum of these invariant quantities
must likewise remain invariant. By multiplication:
4(F–Fs)? 4(F–Fs)*
(B2 - 4AC)* B” – 4AC
Dropping the minus sign and the numerical coefficient, we may
adopt, as a 1st Invariant of the ellipse equation, the expression
(F – F.)?
B2 TAAG
Further, simple algebra gives for the quantity — a”b°/(a” + b%)
the expression
a?b? = . (4AC – Bº) = —
II.
F – F,
A + C '
I2
69
70 Plame Analytic Geometry
and this we may adopt as a 2nd Invariant of the ellipse equation.
Finally, by dividing the square of the 2nd Invariant by the 1st, we
get the expression
B? – 4AC
(ATC)7. -
which we may adopt as a 3rd Invariant of the ellipse equation.
For convenience we have denoted our first Invariant by the
symbol II, the second by I2, the third by Is.
By way of a numerical illustration, we will place in four dif-
ferent positions the ellipse of major axis 2a = 6, minor axis 2b = 4,
wherefore a* = 9, b% = 4, and the student is urged to roughly draw
the curve in its different positions. This ellipse was examined in
example 8 on p. 62, where it was placed with the center at (-1, -2),
with the major axis on y + 2 = 0, the minor axis on a + 1 = 0. The
equation was then 4ac” + 9 y” + 8a; + 369 + 4 = 0. Here I's = 40,
F–Fs = — 36; Bº – 4AC = – 144; A + C = 13; therefore Ii = -9,
I2 = — 36/13, Is = – 144/169.
Next we place the same ellipse with its center at the origin (0,
0), with the major axis on a – y = 0, the minor axis on a + y = 0.
Our explicit equation is them
Is
(a + y)* (ac – y)*
#–4 =#4––1,
and this reduces to the form, 13ac” – 10ay + 13y” – 72 = 0. Here
F's = 0; F – F. - – 72; Bº – 4AC = — 576; A + C = 26; therefore
again: Ii = – 9, I2 = — 36/13, Is = – 144/169.
We next place the same ellipse with its major axis on the line
3a; +4y + 1 = 0, its minor axis on 4a – 3y — 1 = 0, and its center at
the point of intersection of the two lines. The explicit form of
its equation will now be
(4a – 3y – 1)* + (3a; -- 4y + 1)* = 1
25 - 9 25 - 4 5
and this reduces to 145a:* + 120ay + 180y” + 22a 4- 969 – 887 = 0.
In this new equation of our ellipse, Fa – 13, F – Fe = – 900; Bº
– 4AC = – 90000; A + C = 325; therefore again Ii = -9, I2 = — 36/
13, Is = – 144/169.
Finally, we will place the center of our ellipse at the origin,
its major axis on the line y = 0 (the X-axis), the minor axis on the
line a = 0 (the Y-axis). Its equation will then appear in the
familiar form,
For Scientific and Technical Workers 71
*** #y-1-0.
Here F. - 0; F – F, -–1; Bº – 4AC = – 1/9; A + C = 13/36; and
once again Ii = – 9, I2 = — 36/13, Is = – 144/169.
It is seen that for the four different positions of our ellipse the
quantities II, I2, and Is remain invariant, and the same would hold
strictly true for all other equations of the ellipse of semi-axes
squares 9 and 4.
The student may properly inquire as to the use, or practical
significance, of the Invariants. The answer is, that while the
expression B” – 4AC < 0 characterizes an ellipse qualitatively,
the Invariants determine the magnitude of its axes and thus de-
scribe both its size and shape quantitatively. The following
formulae may readily be derived:
2I , 2I mºmºmº-mºm-º:
*-*(1, VIII). b% = #(i-VT).
The Invariant I s, by itself, determines the ratio a”/b°:
a’ - 1 + \/ Is + 1
b° 1–VIII'
Thus Is determines, not the size, but only the shape of an ellipse
and therefore embraces all “similar” ellipses of a given family,
including the point-ellipse and, theoretically speaking, even all the
imaginary ellipses of the family. Since Ia is derived from I1 and
I2, it will be seen that the ellipse has only two independent Invari-
ants, corresponding to the fact that its size and shape are fully
determined by the lengths of its two axes.
As to the quantities A + C and Bº – 4AC, the above discussion
shows the former to vary proportionally with F – Fe, the latter
proportionally with (F–Fs)*, and both therefore to be likely
to vary with the position of the ellipse.*
* It can readily be shown that A + C and B4-4AC (as well as certain re-
lated expressions) WOULD remain invariant if the coefficients of equations
corresponding to different positions of a conic were not differently affected
by the usual clearing of fractions and other incidental simplifying operations.
Thus above, in the second of the positions chosen for the ellipse, clearing of
fractions resulted in multiplication of A + C by 2 (= sum of the squares of
the coefficients of a; and y in the equations of the elipse's axes); and, Cor-
respondingly, it resulted in multiplication of Bº – 4AC by 4 (= 2*). Simi-
larly, in the third position of our ellipse, clearing of fractions multiplied
A + C by 25 (= 3°4-4*) and Bº – 4AC by 625 (= 25°).
72 Plane Analytic Geometry
Obviously, the expressions II, I2, Is must remain invariant also
for equations of the Circle. But here, since B = 0 and C = A,
there will be a simplification. We will have : I. = — (F–Fs)*/4A*
and I2 = (F–Fs)/2A, so that I = — I.” (really a single Invariant,
because a” = b% = r^), while for all circles Is = I2°/Ii = − 1 (because
all circles are “similar”).
(2) Equations of the Hyperbola
Since our formulae for the semi-axes squares a” and b* are the
same for the hyperbola and the ellipse, the Invariants II, I2, Is
must likewise be the same for the two comics. However, their
numerical values in the two cases will show the following differ-
ences in sign: for all real ellipses the three ſnvariants wiłł have
negative values (because Bº – 4AC < 0 and F – F, 30); for all
pairs of conjugate hyperbolas, I1 and Is will have positive values
(because Bº – 4AC > 0); for one of a pair of conjugate hyper-
bolas (namely, the one in which the major axis meets the curve),
Ia will have a positive value, for the second a negative value. Ia
will remain the same, not only for all “similar” hyperbolas, i.e.,
all hyperbolas confined within a pair of vertical angles formed
by the asymptotes (because there the ratio a”/b° remains con-
stant), but also for all their conjugate hyperbolas and for the
asymptotes themselves (see the next subsection). The formulae
for calculating a” and b% from the Invariants are the same as for
the ellipse, and will always yield these quantities with their cor-
rect signs. [However, for the purpose of putting an equation in
its explicit form, a” and b* should be calculated, not from the In-
variants, but by the formulae given on pp. 57–58, in which the
position of the conic may occasionally make a* < b” numerically;
whereas the Invariants would always yield numerically a” - b%.]
For an illustration we choose a hyperbola of the same axes as
in the ellipse of the preceding subsection, viz., 2a = 6, 2b =4, and
we will have the major axis meet the hyperbola; wherefore a” = 9,
* = — 4. We will place this hyperbola, together with its asymp-
totes, in four different positions corresponding to those chosen for
our ellipse. To begin with, we place the center of the hyperbola
at (–1, -2), the major axis on the line y + 2 = 0, the minor axis on
a 4- 1 = 0. Then
(a + 1)*_(y | 2)” –
9 4 T
1,
For Scientific and Technical Workers 73
which reduces to 4a:” – 99%+8a; - 36 y – 68 = 0. Here F, - – 32,
F – F. - — 36; Bº – 4AC = + 144; A + C = – 5; therefore Ii = 9,
I2 = 36/5, Is = 144/25. For the conjugate hyperbola we have
F" = 2F's – F = – 64 + 68 = + 4, and therefore the equation is
4a:” – 9 y” + 82 – 36 y + 4 = 0. Now F – F. - + 36, B* – 4AC and
A + C are the same as for the first hyperbola, and therefore, I1 = 9,
I2 = — 36/5, IA = 144/25.
Next we place our first hyperbola with its center at (0, 0),
with the major axis on a – y = 0, the minor axis on a + y = 0. This
gives the equation 5a:” – 26acy + 5 y” + 72 = 0. Here Fs = 0, F – Fe
= 72; Bº – 4AC = 576; A + C = 10. Hence again Ii = 9, I2 = 36/5,
Is = 144/25. For the conjugate hyperbola we now have 5a:” – 26ay
+ 5y” – 72 = 0; whence F – F. - – 72, and again Ii = 9, I2 = — 36/5,
Ia = 144/25.
Next, as in the case of the ellipse, we place our first hyperbola
with the major axis on the line 3a 4-49 + 1 = 0, the minor axis on
4a – 3y – 1 = 0, and the center at the intersection of the two lines.
This leads to the equation, 17a:* + 312xy + 108y” + 86a +48y +905
= 0. We find: F. - 5; F – F. - 900; B2 – 4AC = 90000; A + C
= 125; therefore in this new position again Ii = 9, I2 = 36/5, IA
= 144/25. The equation of the conjugate hyperbola will now be
17a:* + 312acy + 108y” + 86a, + 48y — 895 = 0, and we will find :
F – I's = – 900, and again Ii = 9, I2 = — 36/5, Is = 144/25.
Finally, as in the case of the ellipse, we will place the major
axis of the hyperbola on the coodinate X-axis, its minor axis on
the coordinate Y-axis, and its center at the origin. The equation
of our first hyperbola will now be
- 1
1
– or” – –au” – 1 =
g” 49 1 = 0.
Here F. - 0; F – F. - – 1; Bº – 4AC = 1/9; A + C = – 5/36; there-
fore again Ii = 9, I2 = 36/5, Is = 144/25. For the conjugate hyper-
bola, F = 1; F – F. - 1; and again II – 9, I2 = — 36/5, Is = 144/25.
Needless to say, the two independent Invariants II and I2, as
well as the derived Invariant Is, will have the same values for
all other positions of our hyperbola (a” = 9, b% = -4), and likewise
constant values for all other positions of its conjugate hyperbola
(a” = – 9, b% = 4).
(3) Equations of Intersecting Lines
The single characteristic by which a given pair of intersecting
74 Plame Analytic Geometry
lines may differ from other pairs is the angle formed by their
intersection. By elementary trigonometry, if mi and me are the
slopes of the two lines, and if 0 denotes the angle formed by
them, then tan 0 = (m. – m2)/(1 + m1m2). In Chapter I (p. 26)
the slopes of intersecting lines, expressed in terms of coefficients
of the general equation, were found to be
– B + \/B2 - 4AC – B – VB°–4AC
and g
2C 2C
And if these expressions are substituted for mi and me in the trig-
onometric formula, we readily find: tan 0 = \/B°–4AC/(A + C).
Squaring this expression, we get
B” – 4AO
(A + C).”
We thus arrive at the same Invariant Is which we obtained for
all similar ellipses and all similar hyperbolas; and obviously the
value of this Invariant will be positive and remain the same for
all positions of the intersecting lines—so long as their angle 6
remains the same. Thus, for the asymptotes in the four different
positions chosen in the preceding subsection, IA = + 144/25.
tan” (9 = = I2.
(4) Equations of the Parabola
In Chapter II it was pointed out that all parabolas are “simi-
lar,’’ i.e., have the same shape and, like circles, can differ only
in size; and that the size of a parabola is determined solely by
its latus rectum. The parabola will consequently have only one
independent Invariant; and for that single Invariant, which may
be denoted by the symbol I4, we will adopt the square (1/c)*
of the latus rectum expressed in terms of the coefficients of the
general equation (see footnote p. 42):
(. *_AE*H CD*– BDE = I
) - (A + C)* - L 43
which will be positive for all positions of the parabola, irre-
spective of the signs of c in the different positions.
C
(5) Equations of Parallel Lines
We turn to the one remaining conic—a pair of parallel lines.
The only characteristic by which a given pair of such limes is
distinguished from any different pair is their perpendicular dis-
For Scientific and Technical Workers 75
tance from each other, for otherwise of course all straight lines
are the same. Hence the sole Invariant of a quadratic equation
of parallel lines, independently of their joint position in the plane,
must express their distance apart in terms of the constants of
the general equation. Such an expression we obtained in Chapter
I (p. 20); and here we will adopt it as a fifth Invariant of equa-
tions of the second degree and will denote it by the symbol Is :
Is = 2\/Fo – F
" VATC
where Fo (= D*/4A = E4/4C) stands again for the value of the
constant term F when the distance between the lines is zero
(coincidence of the lines).
For illustration we will place in different positions the pair of
vertical lines represented by the simple quadratic ac” – 2a – 3 = 0.
Here A = 1, B = 0, C = 0, D = – 2, E = 0, F = – 3; Fo (= D*/4A) = 1.
Therefore A + C = 1 and Is = 4. [The separate linear equations
are a = –1 and a = 3, and obviously the distance between the
lines is 4].
Next we will place the same pair of lines in a direction parallel
to that of a = y = 0; and in order that the new equation (as well
as those for subsequent positions) shall be analogous to the above
quadratic, we will express the variables in the form of perpen-
dicular distances: a/V1, (a — y)/ V2, etc. We will now have,
(*!)-2(*#)-3-0
V2 V2 2
which reduces to acº–2acy + yº–V8 w + V8 y – 6 = 0. Here A = 1,
C = 1, F = – 6, Fo = 2. Therefore A + C = 2 and again Is = 4.
We next place our limes parallel to 3a, + 4y = 0, as follows:
2
(**) –2 (**)-8-0
5 5
which gives, 9a:* + 24acy + 169°–30a: — 40y – 75 = 0. Here A = 9,
C = 16, F = -75, Fo = 25; therefore A + C = 25 and again Is = 4.
Finally, for ac (= w/V1) in the original quadratic, we will
substitute (2a – 31ſ +5)/\/4 + 9, and get:
(**** |-2(***) – 3 = 0
\/13 \/13 5
whence 44° – 12xy + 9 y” + (20–4V13)a - (30 – 6V13) y – 14
76 Plane Analytic Geometry
– 10V13 = 0. Here A = 4, C = 9, F =–14–10V13, and F, -38
– 10\/13. Therefore A + C = 13 and once again Is = 4.
Summary
We have seen that, irrespective of position, the Invariants
reflect the intrinsic characteristics of the conics both qualitatively
and quantitatively. We have further seen that although the ex-
pressions Bº – 4AC and A + C are themselves subject to variation
with the position of the comics, they nevertheless constitute essen-
tial elements in the true Invariants of their equations.” In addi-
tion, it may be noted that since the same change in relative posi-
tion may be effected by moving either a given conic or the
coordinate axes themselves, it follows that the Invariants will
remain unaltered through any translation or rotation of the axes
in their plane—a process occasionally called for in scientific work.
* Readers familiar with the arithmetical details of the topic will observe
that our several Invariants have the form of ratios, -each being a ratio of
homogeneous expressions of the same degree in coefficients of the general
equation. Otherwise no actual invariance might be expected to persist
through the operations involved.
Iliſii
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