C_Plumpton_and_W_A_Tomkys_Auth_Sixth_Form_Pure_Mathematics_Volume (2)

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Some other Pergamon Press titles of interest 
C. PLUMPTON & W. A. TOMKYS 
Theoretical Mechanics for Sixth Forms 
2nd SI Edition Volumes 1 & 2 
D. T. E. MARJORAM 
Exercises in Modern Mathematics 
Further Exercises in Modern Mathematics 
Modern Mathematics in Secondary Schools 
D. G. H. B. LLOYD 
Modern Syllabus Algebra 
Sixth Form 
Pure Mathematics 
VOLUME ONE 
C. PIUMPTON 
Queen Mary College, London 
W. A. TOMKYS 
Belle Vue Boys Grammar School 
Bradford 
SECOND EDITION 
PERGAMON PRESS 
OXFORD NEW YORK TORONTO SYDNEY 
PARIS KRANKKURT 
U. K. Pergamon Press Ltd., Headington Hill Hall, 
Oxford OX3 OBW, England 
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Elmsford, New York, 10523, U.S.A. 
C A N A D A Pergamon of Canada Ltd., 75 The East Mall, 
Toronto, Ontario, Canada 
A U S T R A L I A Pergamon Press (Aust.) Pty. Ltd., 19a Boundary 
Street, Rushcutters Bay, N.S.W. 2011, Australia 
F R A N C E Pergamon Press SARL, 24 rue des Ecoles, 
75240 Paris, Cedex 05, France 
FEDERAL RE PUBLIC Pergamon Press GmbH, 6242 Kronberg-Taunus, 
OF G E R M A N Y Pferdstrasse 1, Federal Republic of Germany 
Copyright © 1968 Pergamon Press Ltd. 
All Rights Reserved. No part of this publication may be 
reproduced, stored in a retrieval system or transmitted in 
any form or by any means: electronic, electrostatic, 
magnetic tape, mechanical, photocopying, recording or 
otherwise, without permission in writing from the 
publishers 
First edition 1962 
Reprinted with corrections 1965 
Second edition 1968 
Reprinted 1972, 1973, 1975, 1978 
Library of Congress Catalog Card No. 67-30688 
Printed in Great Britain by A. Wheaton & Co. Ltd., Exeter 
ISBN 0 08 009374 4 Flexi-cover 
PREFACE TO THE SECOND EDITION 
THIS book is the first of a series of volumes on Pure Mathematics and 
Theoretical Mechanics for Sixth Form students whose aim is entrance 
into British and Commonwealth Universities or Technical Colleges. 
A knowledge of Pure Mathematics up to G.C.E. O-level is assumed 
and the subject is developed by a concentric treatment in which each 
new topic is used to illustrate ideas already treated. The major topics 
of Algebra, Calculus, Coordinate Geometry and Trigonometry are 
developed together. 
This volume covers most of the Pure Mathematics required for the 
single subject Mathematics at Advanced Level. Volume Two covers 
the remainder, with the exception of Pure Geometry, of the Pure Mathe­
matics required for a double subject at Advanced Level. Early and rapid 
progress in calculus is made at the beginning of this volume in order 
to facilitate the student's progress along the most satisfactory lines in 
Pure Mathematics, in Theoretical Mechanics and in Physics. 
The worked examples are an essential feature of this book and they 
are followed by routine exercises within the text of each chapter, asso­
ciated closely with the work on which they are dependent. The exercises 
at the end of each chapter collectively embody all the topics of that 
chapter and, where possible, the preceding chapters also. They are grad­
ed in difficulty and in all cases the last few of these exercises might 
well be deferred, by most students, until a second reading. Most of these 
miscellaneous exercises are taken from the examination papers of the 
University of London (L.), the Northern Universities Joint Matricula­
tion Board (N.), the Oxford and Cambridge Schools Examination 
Board (O.C.) and the Cambridge Local Examinations Syndicate (C). 
We are grateful to the authorities concerned for permission to reprint 
and use these questions. 
ix 
X PREFACE TO THE SECOND EDITION 
The sections, exercises and equations are numbered according to 
the chapters; e.g. § 5.4 is the fourth section of Chapter V; Ex. 9.5 is the 
set of exercises at the end of § 9.5; equation (3.7) is the seventh (number­
ed) equation of Chapter m . Only those equations to which subsequent 
references are made are numbered. 
In this second edition we have made some minor changes and cor­
rections. We are grateful to all those who have made suggestions for 
improvements and corrections. In particular we thank Mr. J. A. Croft 
and Mr. G. Hawkes who read the proofs. 
C. PLUMPTON 
W. A. TOMKYS 
CHAPTER I 
INTRODUCTION TO THE CALCULUS 
1.1 Coordinates and Loci 
Coordinates. The position of a point in a plane may be defined by 
reference to a pair of rectangular axes Ox, Oy with which the student 
will be familiar from drawing the graphs of functions of x. Thus in 
Fig. 1 (#1, yi) denotes the point whose x-value or x-coordinate is x\ 
and whose j-coordinate is y±. Negative jc-coordinates are by convention 
to the left of the origin and negative ^-coordinates below the origin. 
Fig. 1 shows the points A(3, 2), B(-l9 2), C(-2 , -1) and Z)(l, -2) . 
4 
•(-1,2) 
0 
C»(-2,-l) 
* *, 
y 
D.(..-21 
* (3 .2 ) 
y. 
Fig. 1. 
vi 
2 SIXTH FORM PURE MATHEMATICS 
The Equation of a Curve. The relation which is satisfied by the 
coordinates of any point on a particular curve is called the equation of 
the curve. Thus if the equation of the curve is y = f(x) [y is a function 
of x] and if (xi, y{) is a point on the curve, then y± =f(x{). The curve 
is the locus of a point which moves so that its coordinates always satisfy 
the equation. 
Example. If f(x) = ax2+bx+c9 and if the points (4, —9), ( — 1, —4), 
(0, —1) lie on the curve y =/(*), calculate the values of a, b and c. 
Since (4, —9), ( — 1, —4), and (0, —1) each lie on y = /(*), then 
- 9 = / ( 4 ) , - 4 = / ( - l ) and - l = / ( 0 ) ; 
.-. - 9 = 1 6 a + 46 + c, - 4 = a -6- fc , - l = c 
from which 
a= - 1 , b = 2, c == - 1 . 
The Gradient of a Straight Line. The gradient of the straight line 
joining the points P(a, b) and Q(p9 q) is defined as (q—b)/(p—a). It re­
presents the rate of change of y compared with x along PQ. 
Exercises 1.1 
1. Show on a diagram the positions of the following points whose coordinates 
are given: (2,4), (5, - 1 ) , ( - 3 , 2), (0,4), (5,0). 
2. In each of the following, the equation of a curve is given together with one 
coordinate of a point on the curve. Calculate, in each case, the possible values of 
the other coordinate: 
(i) 
(ii) 
(iii) 
(iv) 
(v) 
y — JC ! +5X+4 , 
x*+y* = 25, 
y = logi„ x, 
y = x3-x, 
y = 5x+4, 
(* = 3) 
(y = 3) 
0- = 2) 
(y = 0) 
(x = a^ -h). 
3. (i) If /■<*) = 3 ^ 1 > calculate/(-l); 
(ii) If f(x) = (xt+5x+l3)1", calculate/(2); 
INTRODUCTION TO THE CALCULUS 3 
(iii) If /(*) = — +4, calculate / ( - 2); 
(iv) If /(*) = log10 x, write d o w n / ( — j ; 
(v) If /(*) = sin (2x+10)°, write down /(40). 
4. Write down the gradients of the lines joining each of the following pairs of 
points: 
(i) (2,4), ( - 1 , 5 ) ; 
(ii) ( - 2 , 3 ) , (4, - 2 ) ; 
(iii) ( - 3 , - 2 ) , ( - 5 , - 3 ) ; 
(iv) (a, - * ) , ( - « , « ; 
(v) (2/, 2//), (-2/f, - 2 0 . 
b c 
5. If f(x) = a+—+—t and if f(x) = 3-5 when x = - 2 , /(*) = 7 when 
JC = — 1, and f(x) = 16 when x = 0*5, calculate the values of a, b, and c. 
/ 1 1 \ 
6. Show that the gradient of the line joining (h, k) to I - —, —r-l is equal to 
the gradient of the line joining (/i, k) to the origin and hence show that the three 
points are collinear. 
State which of the following sets of points are collinear: 
(i) (3,5), ( - 2 , 1 ) , (4,0); 
(ii) (1,3), ( - 5 , 1 ) , (7,5); 
(iii) (2m, - / ) , (0,0), (/, m); 
(iv) (2//, -1/m), (l / / ,0), (0, 1/m). 
7. Calculate the gradient of the line joining the points whose x-coordinates are 
1 and - 2 respectively on the curve y — x+4/x. Show that this chord is parallel to 
the chord joining the two points on the curve whose ^-coordinates are 4 and — \ 
respectively. 
8. The curve y = f(x) cuts the jc-axis three times at x = 2, x = 3, and x = — 2. 
If/(x) is a polynomial function of x of degree three of the form x3 + ax*+bx+c, 
find/(x). 
4 SIXTH FORM PURE MATHEMATICS 
1.2 The Idea of a Limit 
Consider the sum of the series of numbers 1 + Y + J + - | - + We 
have: 
i . X _ A i - r 2 — 2 » 
1 . JL . JL - .7.1 T" 2 T- 4 — 4 » 
1 4- - 1 -4 . i4 .JL - 15 1 + 2 T 4 T 8 — 8 > 
The addition of each number halves the gap between the sum that has 
been arrived at and 2. By taking sufficient numbers the sum can be 
made as near to 2 as we please. In these circumstances 2 is defined as the 
limit of the sum of the series as the number of terms increases indefinitely 
(or as the number of terms tends to infinity) and this statement is written, 
nlnL(14+T+-4) = 2-
A similar result to this is illustrated by a recurring decimal. Thus by 
division -|- = 0-1 and this is equivalent to the statement 
. . / 1 1 1 1 \ 1 
n
h?14To+Io-2+ios+ •• +io*) = T-
The tangent to a curve can be considered as the limiting position of a 
chord as the two points of intersection approach one another. Thus in 
Fig. 2 the tangent to the curve at P is the limiting position of the chord 
PQ as Q approaches P. 
x2—a2 Algebraic Limits. Consider lim . Direct substitution of 
x2—a2 x = a in the expression gives an indeterminate result, but x—a 
x2 — a2 lim = lim (x-\-a)= 2a. 
x-*-a ^ ^ x-*- a 
INTRODUCTION TO THE CALCULUS 
This result can be extended: 
xn-an 
lim 
x^a x-a 
if n is a positive integer 
= lim (xn-1+xn-2a+xn-*a2+. . , + a71-1) by division; 
-. xn-an n , 
.'. lim = na71"1. 
x-+a x-a 
(1.1) 
Fig. 2. 
It can be proved, but it will not be proved here, that result (1.1) is true 
also when n is fractional or negative. The result is illustrated by two 
particular cases thus: 
lim 
x^a x-a 
= l i m ^ - ^ = l i m ( V ^ ^ ) ( V ^ + V a ) 
x^a x-a x-+a (x-a)(y/x+y/a) 
lim i - \ 
f. x~x—a" lim 
_1 1_ 
x a .. 1 1 = lim = lim = — = . 
x_*a x-a x^a ax a2 
6 SIXTH FORM PURE MATHEMATICS 
1.3 The Gradient of a Curve 
If P[x9f(x)] and Q[x+h,f(x+h)] are points on the curve y =/(*) 
(Fig. 2) the average rate of change of y compared with x from P to Q is 
{f(x+h)-f(x)}lh and, in general, this quantity will have a limit as h 
tends to zero. Thus in a particular case iff(x) = x2: 
lim {f(x+h)-f(x)}/h = lim {(x+A)2-*2}//* 
= lim(2hx+h2)/h = lim(2x+/*) 
= 2x. 
This limit is defined as the gradient of the curve at P. It represents 
what is meant by the rate of change ofy compared with x at P and it is 
the gradient of the tangent to the curve at P. This idea was developed 
by Newton with special reference to the relations between distance, time, 
velocity, and acceleration of a particle moving in a straight line. Velocity 
at a certain instant is defined as the gradient of the distance-time curve 
at that instant, and the acceleration at a certain instant as the gradient 
of the velocity-time curve at that instant. 
Examples, (i) Calculate the gradient of the curve 
;;=/(*) = 3/*2 + 2 
at the point (1, 5). 
{ / ( 1 + A ) _ / ( 1 ) } / „ = {3/( l+*y+2}-(3/ l+2) = 3/(1+^-3 
_ -6h-3h2 m 
A(l+/*)2 ' 
therefore the required gradient 
= l l gl /I , L\2 = ~ 6 -
h^0 (1+A)2 
(ii) The distance s ft from a fixed point O in a straight line of a particle 
moving in the line at time / sec is given by s = 5f3+4. Calculate the 
INTRODUCTION TO THE CALCULUS 7 
velocity of the particle at time t = 2. 
Velocity - lim {*2+*)»+4}-{3(2y+4} ft 
ft-*0 A 
= l im5( 8 + 1 2 A + f + A 3 - 8 ) f t per sec 
= lim 5 I = 1 ft per sec 
fc-0 \ h J 
= lim 5 (12+6A+A2) ft per sec = 60 ft per sec. 
h—o 
Exercises 1.3 
1. Calculate the gradient of each of the following curves at the point stated: 
(i)y = 2x*+l9 (1,3); 
3 
(ii)y = —+2, (3,3); 
(iii)y = x«+2x+l , (0,1). 
2. The distance-time equation of a certain particle moving in a straight line is 
,y = t2+t+1 (with the usual notation). 
Calculate the velocity when t = 2. 
3. The velocity-time equation of a particle moving in a straight line is 
v = 3(1 + tz) (with the usual notation). 
Calculate the acceleration when / = 1. 
1.4 Differentiation 
It is convenient to refer to a small addition (increment) to the value 
of x as bx [delta x] and to the addition which results in the value of y 
as by. Here bx is a single symbol and the b is inseparable from the x. 
With this notation and with reference to Fig. 3 we have: 
by 
Average gradient from P to Q = — or by/bx. 
bx 
by 
Gradient at P = lim — or lim by/bx. 
dy 
We write this limit as — or dy/dx [dee y by dee x]. 
dx 
8 SIXTH FORM PURE MATHEMATICS 
Summary. Thus we have: 
1. The gradient of a curve at a point P(x9 y) on the curve is define-* 
as lim dy/dx. 
2. It represents the rate of change of y compared with x at P. 
3. It represents the gradient of the tangent at P. 
Fig. 3. 
The process of obtaining dy/dx in particular cases is called differen­
tiation and dy/dx is called the differential coefficient ofy with respect to x. 
[For the curve y = f(x), dy/dx is sometimes written as f'(x) and is 
called the derivative off(x).] 
Since calculus is intended primarily as an aid to calculation, it is 
desirable to discover rules which make it possible to write down the 
derivative of a function. 
If y = axn, where a and n are constants, then 
dy t. a(x+dx)n-axn t. (x+dx)n-xn -/• = lim —i ~ = a lim ^ . 
QX fo-*0 OX 4c-»o OX 
INTRODUCTION TO THE CALCULUS 9 
This limit is a special case of the limit (1.1) discussed above, in which 
(x+ dx) takes the place of x and x takes the place of a: 
.-. 4- (<*xn) = nax"-1. (1.2) 
ax 
It follows that the general rule for writing down at once the differential 
coefficient of a function of the form axn or the sum of a number of 
terms in this form is: For each term> multiply by the index of the variable 
and reduce the index by one. 
Example. If 
4 
y = 5x3+6x2+-3+2V^+3, 
then 
2. 
i.e. y = 5x3+6x2+4;c-3+2;c2 +3, 
djVdx = 15x2+12;c-12;c-4 + x~T, 
12 1 i.e. dj>/djc = 15x2+12x--T + —-. 
X *w X 
Exercises 1.4 
1. Write down the derivative of each of the following functions: 
(i) ^ + 3 ^ + 1 ; 
(ii) 3x l+2x+4; 
r-\ l 2 ( , l l ) T~? ; 
(iv) V*3; 
(v) (2x+4)(x-2) [Multiply out first]; 
, .x 3**+JC+1 p _ . # J ^ , 
(yi) j [Divide out first]. 
2. Write down the value of dy/dx at each of the points named: 
( i ) ? « x ' + 2 x + l , (1,4); 
OOy-pr. (2,D; 
(?ii)?«(x+3)(2x-l), (2,15); 
(iv)y=V*, (8,2). 
10 SIXTH FORM PURE MATHEMATICS 
3. Calculate the coordinates of the two points on the curve 
y = 2x 3 -9* l + 12x+4 
at which the gradient is zero. 
4. The distance-time equation of a particle moving in a straight line is [in the 
usual notation] 
5 = 2^ + 3 ^ . 
Calculate: (i) the velocity when / = 1; (ii) the acceleration when t = 4. 
5. Calculate the gradient of the curve y = x+l/x at the point where x = 2. 
Find the coordinates of the other point on the curve where the tangent is parallel 
to the tangent at the point (2, 2f). 
1.5 Tangents and Normals 
The equation of the line through P(xi, y±) which makes an angle 8 with 
the positive direction of the x-axis. With reference to Fig. 4, A(x, y) is 
Fig. 4. 
any point on the line, and the line through P parallel to the jc-axis meets 
the line through A parallel to the j-axis at N. Then ^.APN = 6 and 
AN 
tan0 y-yi PN ~ x - * i ' 
The equation of AP is therefore 
(y-yi) = (x-*i ) tan0. 
(1.3) 
(1.4) 
INTRODUCTION TO THE CALCULUS 11 
The equation of the line through P at right angles to AP. The equation 
obtained above is derived on the assumption that 0 is acute. The work 
of Chapter IV where angles of all magnitudes are discussed shows 
that the equation is a correct one for all values of 0. In that chapter it is 
also shown that tan (90° + 0) = - 1/tan 0. 
The line through P at right angles to AP makes an angle (90°+ 0) 
with the positive direction of the x-axis. Its equation is therefore 
(y-yd = (*-*i)tan (90°+0), (1.5) 
i.e. 0>->>i) = - - ^ - 0 ( * - * i ) - C1-5*) 
Since the gradient of a curve at a particular point is the gradient of 
its tangent at that point, the equation of the tangent to the curve y = /(x) 
at the point (a, b) on the curve is 
(y-b) =f'(a)(x-a). (1.6) 
[f'(a) means: the value off'(x) when x = a.] 
The normal to a curve at a given point is the straight line through that 
point atright angles to the tangent. The equation of the normal to 
y = f(x) at the point (a, b) on the curve is therefore 
°"*) = ~fh)(x~ay (L7> 
Example. Find the equation of the tangent and the equation of the 
normal to the curve y = x3—2x+l at the point (2, 5). Calculate the 
coordinates of the point at which the tangent meets the curve again* 
ax 
therefore the gradient at the point (2, 5) is 3(2)2—2 = 10; 
therefore the equation of the tangent at (2, 5) is 
7 - 5 = 10(x-2) 
which simplifies to 
y= 10x-15. 
12 SIXTH FORM PURE MATHEMATICS 
The equation of the normal at (2, 5) is 
which simplifies to 
\0y = - x+52 . 
The coordinates of a point of intersection of two curves must satisfy the 
equations of the curves simultaneously. 
Therefore the coordinates of the points of intersection of the curve and 
the tangent are given by the solution of 
y = JC 3 -2X+1, 
V = 10x-15. 
Eliminate y: 
x 3 - 2 x + l = 10x-15; 
.". JC3-12JC+16 = 0. 
This is a cubic equation, but one factor, at least, of the left-hand side 
is known because it is known that the tangent is drawn from a point on 
the curve where x = 2; (x—2) is therefore a factor. Completing the 
factorization 
0 C - 2 ) ( J C 2 + 2 X - 8 ) = 0. 
Thus 
(x-2)(x-2)(x+4) = 0 
and 
.*. x = 2 or —4. 
At the point on the curve where x = — 4, y = — 55. Hence the tangent 
meets the curve again at (—4, — 55).(It is a characteristic of the tangent 
to a curve that, when its point of intersection with the curve is calculated 
by this method, in general two equal roots are obtained. This is because 
the tangent can be considered as the limiting case of a chord to the 
curve; two points of intersection of the chord with the curve tend to one 
in the limit.) 
INTRODUCTION TO THE CALCULUS 13 
Exercises 1.5 
1. Find the equation of the tangent and the equation of the normal to each o/ the 
following curves at the points named: 
(i) y = 3 x 2 - 4 x - 2 , (1, - 3 ) ; 
(ii) , . = x2+~, ( -1 ,0 ) ; 
(Hi) y = +Vx9 (4,2); 
(iv) y = 3x*-x2+5x+l, (1,8). 
2. Find the equation of the tangent to the curve .y = 2x+l/jc~at the point (]> 3) 
and show that it does not meet the curve again. Find the coordinates of the point at 
which the normal at (J, 3) meets the curve again. 
3. Find the equation of the tangent to the curve y = x1 — 1/x at the point 
(h ~ J ) a°d find the coordinates of the point at which it meets the curve again. 
4. The tangent at the point P(8, 4) to the curve y = xl,z meets the x-axis at T 
and the normal at P meets the jc-axis at G. Calculate the area of the triangle PTG. 
5. Find the equation of the tangent at the point (1,0) to the curve y = x*+x2—2x 
and find also the equation of that other tangent to the curve which is parallel to the 
tangent at (1, 0). 
6. Show that the coordinates of a point of intersection of the curves 3y = 4x2 
and 24y = 24x3+ll are (—\, \). Find the equation of the tangent to each of the 
curves at this point and prove that these tangents are at right angles to one another. 
(The two curves in such a case are said to intersect at right angles or to intersect 
orthogonally.) 
7. Prove that the normals to the curve ay = x1 at the points {a, a) (2a, 4a), 
( - 3a, 9a) are concurrent and find the coordinates of their point of intersection. 
8. The tangent at the point P(a, a*) on the curve y = x* meets the x-axis at T, and 
PN, perpendicular to the jc-axis, meets it at N. Prove that OT = 3TN. 
1.6 Rates of Change 
If three variables x, y and t are such that x and y are each functions 
of t [say x = f{t), y = g(t)], then there will be a functional relationship 
between x and y [say y = cp(x)]. If now (t, x) and (t+bt, x+dx), (t, y) 
and (t+dt, y+8y), (x, y) and (x+bx, y+by) are corresponding points 
on the curves x = /(f), y = g(t), y = q>(x), then 
by _ by Ibx 
Ix~ ~bi\~bi' 
14 SIXTH FORM PURE MATHEMATICS 
It can be proved, but it will not be proved here, that 
i i m *? = / i i m *y\l( i i m * ? \ , 
*■"■■£ = £/£■ 0-8) 
d* df / df v y 
This equation can also be written in the form 
dy _ dy dx 
~di ~~ dx ~di' (1.9) 
The diflferential coefficient dyfdx is not a fraction but in this respect 
it behaves like one and in particular it can be shown that 
% - «/£• <'-10> 
These relations permit the consideration of a wider field of problems 
concerning rates of change. 
Example. Water is poured into a hemispherical bowl, of radius 10 cm, 
whose axis is vertical, at the rate of 100 cm3 per sec. The volume V cm3 
of a "cap" of height A cm of a sphere of radius R cm is given by the 
formula V = ^7th2(3R—h). Calculate the rate at which the depth of the 
water is increasing when the depth is 5 cm. 
Here one rate of change is given, the rate at which the volume of 
water is changing, i.e. dV/dt. 
A second rate of change is required, the rate at which the depth of the 
water is changing, i.e. dh/dt. 
The relation between Fand h is also given and therefore dV/dh can 
be obtained. 
dV dV dh 
Use can therefore be made of — = — which may be written 
dt dh dt 
. , „ dh dVldV 
m the form — = — / — • 
dt dt dh 
INTRODUCTION TO THE CALCULUS 15 
Since 
V = JaP (30-A), 
dV ^- = lOnh-nh2; dh 
dh dVldV 100 
" dt dtjdh 207th-nh29 
therefore when h = 5, 
dh _ 1 / 100 \ __4_m 
dt ~ 7i \ 100-25 ) " 3;r; 
i.e. the depth is increasing at 4/3TI cm per sec when the depth is 5 cm. 
Exercises 1.6 
1. Water is poured into a conical vessel, of semi-vertical angle 45° and axis vertical, 
so that the depth of the water in the vessel increases at a uniform rate of 4 cm per 
sec. Calculate the rate of increase of the volume of water in the bowl: (i) when the 
depth is 5 cm; (ii) after 2 sec. 
2. Gas is released from an elastic spherical vessel and the volume V ft8 of the 
vessel at time t sec is given by V = 3027r — nt2 for values of t < 14. Assuming that the 
vessel retains its spherical shape for this time, calculate the rate at which the 
243rc 
radius of the vessel is decreasing when its volume is —r— ft8. 
3. A particle moves along the curve y = xz so that its ^-coordinate is changing 
at the rate of 2 units per sec. Calculate: 
(i) the rate at which its ^-coordinate is changing when its x-coordinate is 2; 
(ii) the rate of change of the area of the rectangle formed by the axes, the ordinate 
and the abscissa through the particle when its jc-coordinate is 4. 
4. The relation between pressure and volume of a certain gas expanding "adiaba-
tically" (i.e. without addition or subtraction of heat) is given by the formula 
pv1** = k where k is constant. If a, b are the rates of change of/? and v respectively 
when v has the value vlt prove that 
— = -lAkvr2'4 o 
5. A circular metal disc is being heated and its radius is increasing at the rate of 
0*01 in. per min. 
(i) Calculate the rate at which its surface area is increasing when its radius is 
10-1 in. 
(ii) If the radius is initially 10 in., calculate the time at which the surface area is 
increasing at a rate of 0.41 n in* per min. 
16 SIXTH FORM PURE MATHEMATICS 
6. The coordinates of a point P on a certain curve may be expressed in the form 
x = 3/ — 1, y = 2/2+4 where / varies with the position of P on the curve. The tan­
gent to the curve at P meets the .y-axis at T and the line through P parallel to the 
jc-axis meets the j>-axis at N. Using the relation dy/dx = (dy/dt)/(dx/dt) to find the 
equation of the tangent at P, or otherwise, find the length of AT in terms of t. Hence 
find the rate of change of the length of NT compared with y when / = 1. 
1.7 Differentiation of a Function of a Function 
A function such as <\/(x2+1) cannot be differentiated directly by the 
rules considered thus far. It is differentiated as follows: 
Put 
Then 
u = x2+l. 
y = V"> 
dv 1 dt/ 
- T - = 2*; dw 2y/u dx 
dy dy du 1 
2x = * dx dudx 2y/u V(*2+l) 
By this method the general result which follows can be obtained. If 
y = Au) where u is a function ofx, then 
%-m£ a...) 
and thus the derivative of a function of a function can be written 
down at once.In the example given above u = x2+1, /(w) = <y/w and 
/ ' ( « ) = 1/(2 y/u). 
Example. Differentiate (JC3 +1)3: 
d(x3+l)3 
dx = 3(x
3+l)2.3x2 = 9JC2(JC»+ 1)» 
INTRODUCTION TO THE CALCULUS 17 
Exercises 1.7 
1. For each of the following, write down, but do not simplify, the expression 
for dy/dx: 
( i ) y = V(* 3 +x+l ) ; 
( i i ) , = (2**+l)«; 
2. For each of the following, calculate the value of dy/dx at the point whose 
x-coordinate is given: 
(i)y = (2x-l)*t ( * = 1 ) ; 
(ii)'=vdkir (* = 2); 
(iii) y = V(2**+9), (x = 3); 
(iv)y = (x-l)\ (* = 0). 
3. A ladder 10 ft long stands on a horizontal plane and leans in a vertical plane 
against a vertical wall. The ladder slips down the wall in a vertical plane and when 
the height of the top of the ladder is 6 ft its speed is 4 ft per sec. Calculate the speed 
of the foot of the ladder at this instant. 
4. With the usual notation, the s-t equation for a particle moving in a straight 
line is 
Calculate the velocity of the particle when t = 1. 
5. Find the equation of the tangent and the equation of the normal to the curve 
8 
y = 2 at the point (2, 1). Calculate the coordinates of the point at which the 
tangent meets the curve again and show that the normal does not meet the curve 
again. 
1.8 Maxima and Minima 
Increasing and Decreasing Functions. The function f(x) is said to be 
an increasing function of x at the point x = xi, if f(xi+e) >/(xi) 
for positive values of e however small e may be and/(x) is said to be a 
18 SIXTH FORM PURE MATHEMATICS 
X=Xx 
decreasing function of x at this point if /(xi+e) </(xi). Since, by 
definition, 
Ax1) = lim[/(x+£e>-/(^ 
then f(x) is increasing at x = xi if /'(*i) > 0 and /(x) is decreasing 
at x = Xi if f'(xi) < 0. 
If /'(*i) = 0, /(x) is neither increasing nor decreasing at x = Xi and 
is said to have a stationary value at this point. 
If both of/(xi ± e) > /(xi) however small e may be, fix) is said to have 
a minimum value at x = xi and if both of f(x±±e) </(xi) however 
small e may be 9 fix) is said to have a maximum value at x = xi. 
Fig. 5. 
In Fig. 5: 
(i) f{x) increases and/'(x) > 0 from A to B; 
(ii) /(x) is stationary and /'(*) = 0 at B; 
(iii) fix) decreases and/'(x) < 0 from B to C; 
(iv) fix) is stationary and/'(x) = 0 at C; 
iy)f\x) increases and/'(*) > 0 from C to 2). 
INTRODUCTION TO THE CALCULUS 
In Fig. 6: 
(i) f(x) decreases and/'(x) < 0 from P to Q; 
(ii) f{x) is stationary and/'(*) = 0 at Q; 
(iii) f(x) decreases and/'(*) < 0 from Q to R; 
(iv) /(x) is stationary and/'(x) = 0 at R; 
(v) f(x) increases and f'(x) > 0 from i? to S. 
Fig. 6. 
Fig. 7 shows three different kinds of stationary points. They are not 
the only types of stationary points but they are the only ones which 
will be considered here. At this stage we consider only functions for 
which f'(x) exists. A detailed discussion in the general case can be 
found in A Course of Mathematics for Engineers and Scientists, Vol. 1, 
Chap. V, by B. H. Chirgwin and C. Plumpton. In each case the graph 
off'(x) for corresponding values of xis shown below the graph of/(x). 
Fig. 7 shows that: 
(i) at a maximum point /'(*) = 0 and/'(x) is a decreasing function 
of x; 
(ii) at a minimum point f\x) = 0 and f'(x) is an increasing function 
of x; 
20 SIXTH FORM PURE MATHEMATICS 
(iii) at a stationary point which is also a point of inflexion/'(*) = 0 
and f'(x) is stationary. (Points of inflexion in general are defined in 
§ 14.1 of Vol. II. It is not necessary for a point of inflexion to be a station­
ary point.) 
( i ) Maximum (8) Minimum (iii) Stationary point which 
is also a point of inflexion 
Fig. 7. 
Maximum and minimum points represent the greatest and least values 
of the function respectively only within a limited region of the curve. 
If maximum and minimum points are used to determine greatest and 
least values of a function this limitation must always be considered. 
Example (iv), below, illustrates the case of a function of x which is 
defined only for a limited range of values of x, (0 -*s x < /)• The greatest 
value of y occurs when x = /, but this is not a maximum value as defined 
in this section. 
INTRODUCTION TO THE CALCULUS 21 
Examples, (i) Show that the function fix) = 4x3+6;c2+5JC+ 1 
increases for all values of x: 
f(x) = 12*2+12x+5 = 3(2x+l)2+2. 
Since 3(2* +l ) 2 +2 is positive for all values of x,f'ix) is positive for all 
values of x and therefore fix) increases for all values of x. 
(ii) Find the stationary points of the curve y = 3x5 — 5JC3+4 and find 
the nature of each. Sketch the curve. 
y = 3x5—5x3+4; 
.-. dj>/dx = 15JC4-15X2 = 15X2(JC-1)(;C+1); 
therefore there are stationary points atjc = 0 ; j c = l ; j c = — 1 . 
In the following table x = 0±e means x = ± a positive quantity 
however small. 
X 
- 1 - e 
- 1 + e 
0 - e 
0+e 
1 -e 
1 + e 
JC2 
+ 
+ 
+ 
+ 
+ 
+ 
x-1 
-
-
+ 
x+1 
+ 
+ 
+ 
+ 
+ 
/'(*) 
+ 
-
+ 
Conclusion 
f'(x) is decreasing at x = - 1 
f(x) has a maximum here 
f'(x) is stationary at x = 0 
/(JC) has an inflexion here 
/'(*) is increasing at x =* 1 
/(JC) has a minimum herd 
When x = 0, j = 4: there is an inflexion with the tangent parallel to 
the jc-axis at (0, 4). 
When x = - 1 , y = 6: there is a maximum at (— 1, 6). 
When x = 1,^ = 2: there is a minimum at (1, 2). 
7b sketch the curve: 
(a) When x is large and positive, y is large and positive, i.e. as 
X -*► + «>, jy -* 4- oo. 
(b) As JC — — oo, ^ -* - oo. 
22 SIXTH FORM PURE MATHEMATICS 
Fig. 8 (i) shows the information we have about the shape and other 
important features of the curve and Fig. 8 (ii) shows how this information 
is used to give a completed sketch of the curve. 
•(-1,6) max 
• 0,4 Inflexion 
(I) 
As x—►-oo 
*+co 
►+oo 
Fig. 8. 
It should be noted that, since the sign of f'{x) is constant between 
successive zeros of/'(*), it would be suflSicient to show (a) that/'(x) is 
positive for x < — 1; (b) that/'(x) is negative for — 1 < x < 0 and also 
for 0 < x •< 1; and (c) that/'(x) is positive for x > 1, in order to estab­
lish the nature of the stationary points. 
INTRODUCTION TO THE CALCULUS 23 
(iii) Calculate the greatest volume of a circular cylinder inscribed in a 
sphere of radius JR. 
The radius of the sphere is constant. The volume of the cylinder 
is a function of its radius and also of its height and it involves the con­
stant R. One of the two variables, however, depends upon the other so 
that the volume of the cylinder may be expressed as a function of only 
one of the height and radius of the cylinder. This function will involve 
the constant R. 
Fig. 9. 
If the height of the cylinder is h (Fig. 9), then the radius of the cylinder 
is \/CR2 — A2/4) and the volume V of the cylinder is given by 
V = jt(R2-h2/4)h (0 < h < 2R); 
therefore V is stationary when h = y/\ R. 
If h~ 
4
 n . <*v. 
—- R, then — is positive. 
/4 dV If h > /-— R, then —- is negative. \J 3 ah 
Therefore dV/dh is decreasing at h = V T ^ -
Therefore F is a maximum at h = <\/\ R. 
P.M. 1—B 
24 SIXTH FORM PURE MATHEMATICS 
The (V, h) curve within the region of possible values of h is therefore 
as shown in Fig. 10. Hence V has a greatest value at h = y ' y R. 
v 
Fig. 10. 
The greatest value of Fis 
(iv) The deflection y of a uniformly loaded cantilever of length / is 
given in terms of the distance x from one end by the equation 
24EIy = wx2(6P-4lx+x2) 
where E9 /and w are constants. Find the greatest value of y. 
24EI dy/dx = w(\2Px-12&*+4*»); 
/. dy/dx = 0 (and y has stationary values) 
where 
4x(x2-3lx+3P) = 0. 
Since the equation x2 — 3/x+3/2 = 0 has no roots, the only stationary 
point is at x = 0 (a minimum if there were no limitation on the value 
of x) and dy/dx is positive for all positive values of x. The value of y 
therefore increases with x when x is positive and the physical limitations 
of the problem confine x to the range 0 < x < /. 
The greatest value, yG, of y occurs therefore when x = /: 
.\ 24EIyG = wl2(6P-4P+P),i.e. yG = wl*/8EI. 
INTRODUCTION TO THE CALCULUS 25 
Exercises 1.8 
1. For each of the following, find whether the function is increasing, decreasing, 
or stationary at the point named. 
(i) 2x+—, 
(ii) x3+3xs+l, 
(iii) 1/(^+4), 
(iv) (x*+l)2, 
(v) x-l/x*, 
(x= - 1 ) ; 
(x = - 2 ) ; 
(x = 0); 
(x = - 2 ) ; 
(x = 1). 
2. Discuss the function JC4—2x2 —3, stating for what ranges of values of x it in­
creases, for what ranges of values of x it decreases, and for what values of x it is 
stationary. Sketch the graph of the function showing clearly its maximum and mi­
nimum points and the points where it crosses the axes. 
3. By sketching the graph of/'(x) show that/*(jt) = 2JC3+3X2+6JC+4 increases 
for all values of x. 
4. Find the stationary points, if any, for each of the following functions. In each 
case, find the nature of the stationary points and sketch the graph of the function: 
(i) * 2 - 6 * + l ; 
(ii) x2+x+6; 
(iii) 2xs-15;c2+24;t+l; 
(iv) 3**-4*3+1; 
(v) x*+x+l; 
(vi) x2+2/x. (Note that as x -+ 0 from the positive side y -> 4- «>, and as x -*- 0 
from the negative side y -► — «>. There is no point of the curve where x = 0 and 
the curve has two branches.) 
5. A piece of wire 12 in. long is to be cut into two pieces, one of which is to be 
bent to form a square and the other to form a circle. Calculate, in terms of n, the 
length of the side of the square when the combined area enclosed by the two figures 
has its least possible value. 
6. A solid circular cylinder is to be made from 100 cm3 of material. Calculate, 
correct to 2 significant figures, the least possible surface area of such a cylinder. 
7. The coordinates of any point P on the curve y = 12—x* may be taken as 
(f, 12 -12) where t varies. Perpendiculars from P to the coordinate axes meet these 
axes at A and B respectively. Find the value of t in the range 0 -< / < + V12 for 
which the area OAPB is a maximum. Discuss the variation of the area with / in 
the cases- V12 < / < 0 and t > + V12. 
26 SIXTH FORM PURE MATHEMATICS 
8. In the usual notation the distance-time equation of a particle moving in a 
straight line is 
s = 2t*-15t*+36t. 
Calculate the minimum velocity of the particle. Describe its motion with particular 
reference to its position at t = 2 and t = 3 and illustrate your answer with a sketch 
of the s-t curve. 
9. It was estimated that for a particular job the cost per mile of a certain type of 
heavy transport was partly constant, partly varied as the average speed, and partly 
varied inversely as the average speed. Experiment gave the results. 
Speed in m.p.h. Cost per mile in £ 
20 2-68 
30 265 
40 2-74 
On the above assumptions, calculate correct to 1 m.p.h. the speed for which the cost 
per mile would be the least possible. 
10. A curve y — x*+ax2 + bx+c passes through the point (0,1) and has station­
ary values at the points where x = 2 and x = — 2. Find the equation of the curve. 
1.9 Second Derivative 
The derivative of /'(*) is called the second derivative of f(x) and is 
denoted by /"(*)• The differential coefficient of dy/dx with respect to 
x is written d2y/dx2 (dee two y by dee x squared) and is called the second 
derivative of y with respect to JC. Thus, in the usual notation, the acceler­
ation of a particle moving in a straight line is dv\dt or d2s/dt2. (Note 
dv ds dv v dv that -7- = -—— = — and therefore a second, and very useful, ex-dt dt ds ds 
pression for the acceleration is v dv/ds.) 
Second derivative as a test for the nature of stationary points. If/'(*i)=0 
and/"(*i) is positive, f(x) is stationary at x = Xi and f'(x) is increasing 
at x = JCI. It follows that/(x) has a minimum value at this point. In the 
same way if/'(*2) = 0 and/"(*2) is negative, f(x) has a maximum value 
at x = JC2. The converses of these statements are not, however, true. 
There are points which satisfy all the other criteria for maximum or mini­
mum points and for which /'(*) = 0 and also/"(*) = 0; e.g., y = x4 
has a minimum at the origin although d2yldx2 = 0 at the origin and, 
by contrast, y = x3 has neither a maximum nor a minimum at the origin. 
INTRODUCTION TO THE CALCULUS 27 
Such cases occur when the solution of the equation of the tangent with 
the equation of the curve gives more than two equal roots. If, therefore, 
we use these criteria for maximum and minimum points we must carry 
out a more detailed investigation in the cases in which d2y/dx2 = 0. 
Example. We repeat the second worked example of § 1.8: 
fix) = 3JC5-5JC3+4; 
.\ / ' (* )= 15JC2(JC-1)(JC+1), 
/"(JC) = 60JC3-30JC; 
therefore there are stationary points at x = 0, x = 1, JC = —1. 
/"( — 1) = — 30; therefore maximum at x = — 1; 
/" (+! ) = +30; therefore minimum at x = 1; 
/"(0) = 0. 
A further investigation is therefore needed concerning the nature of 
the point at JC = 0 and this is carried out in the manner of the example 
in§ 1.8. 
This forms a useful alternative method of testing the nature of station­
ary points but it is by no means the better one in all cases. The method 
first discussed is the better one in nearly all the cases in which the obtain­
ing of the second derivative is a complicated process. 
1.10 Parameters 
The coordinates of a point on a curve can usually be expressed in 
terms of a third variable which is called a parameter. For example 
(f3, t2), where t varies, are the coordinates of a point on the curve yz = x2. 
The advantage in many cases of the use of parameters lies in the fact 
that the relation between the coordinates of a point on the curve is 
implied by the parametric form of those coordinates. The y = /(JC) 
(cartesian) form of the equation of the curve need not be introduced 
into the calculation. In some cases the parametric equations of a 
curve can be obtained in a comparatively simple form, whereas the 
cartesian equation would have an involved form. e.g. x = a(0+sin 0), 
y = tf(l—cos 0). In such cases the properties of the curve can be 
28 SIXTH FORM PURE MATHEMATICS 
obtained easily from the parametric equations, but with difficulty, if at 
all, from the cartesian equation. 
Examples, (i) Find dy/dx and d2y/dx2 for the curve x = ct, y = c/t 
(where c is constant). 
dx 
d 7 ~ 
dx 
d*y 
dx2 
C' dt 
dy jdx 
'"dtjdt~ 
d© 
dx ~ 
c 
~ - 7 . ; 
- * / " 
dt 
1 
jdx 
U 
dt[ t*)\c t*(C ct* 
Note that, whereas dyjdx has some of the properties of a fraction, 
d2yjdx2 has not. 
(ii) Find the equation of the tangent at the point whose parameter 
is ti on the curve x = t2-hl9 y = 1 — t* and find the parameter of the 
point in which it meets the curve again. 
x = f2+l; dx/dt = 2t; 
y = 1 _ * 3 ; dy/dt = _„3,2 . 
/. dy/dx = -3t/2; 
therefore the tangent at (f J+1, 1—tf) is 
( y - l + ff) = - * * ( * - * } - 1 ) , 
i.e. 2y+3t1x-(tf+3t1+2) = 0. 
This meets the curve at points whose coordinates are (f2+l, l — /8): 
/ . 2(l-f3)+3f1(*2+l) - (*?+3*i+2) = 0; 
/. 2t3-3ht2+q = 0. 
But (t — ti) is known to be a repeated factor of the left-hand side of this 
equation and so 
INTRODUCTION TO THE CALCULUS 29 
(t2-2th+tf)(2t+h) = 0, 
The tangent meets the curve again at the point whose parameter is 
- |-/i, i.e. at the point [(*J/4) + 1,1+ (rJ/8)]. 
Exercises 1.10 
1. The coordinates of a point on a curve are given, in each of the following cases, 
in terms of a parameter /. In each case find dy/dx and d2y/dx2 in terms of r: 
(i) x = 5f2, y = 10/; 
(ii) JC= 1 + /3, y = 3 f 2 - l ; 
(iii) * = 1/f, >> = f2; 
(iv) * = r2, >> = /+1/ / . 
2. Find the values of dy/dx and of d2j>/d*2 at each of the points named for the 
curve whose parametric equations are given below: 
(i) x = t\ 
(ii) x = A-t\ 
(iii) x = l/t\ 
(iv) x = t*+t+l, 
y = 4//, 
>- = 4f, 
J- = 2//, 
y = 5t, 
it = l); 
(*= - i ) ; 
( ' = - i ) ; 
(/ = 2). 
3. In the usual notation the v, s equation of a particle moving in a straight line 
is given by v = (s+1)2. Find the acceleration when s = 1. (Use/ = vdv/ds.) 
4. Find the equation of the tangent and the equation of the normal to the curvex = 3/2, y = t3 at the point whose parameter is tt. Find the parameter of the point 
at which the tangent meets the curve again. 
5. Find the stationary value of y for the curve whose parametric equations are 
x = t1—At, y = tz—9tl+24t and state whether it is a maximum or a minimum. 
Miscellaneous Exercises I 
1. A particle moves so that its distance x ft from a fixed point O is given by 
x = 9f2—2/3 where f is the time in seconds measured from O: 
(i) Find the speed of the particle when / = 3. 
(ii) Find the distance from O of the particle when / = 4 and show that it is 
then moving towards O. (N.) 
30 SIXTH FORM PURE MATHEMATICS 
2. The curve y — ax2+bx+c passes through the point (1, 2) and its gradient 
is zero at the point (2,1). Find the values of a9 b and c. (N.) 
3. Find the gradients at the points on the curve y = x(x — l) (x—2) where it 
crosses the x-axis. Write down the equations of the tangents at these points. Show 
that two of these tangents are parallel and find the coordinates of the points where 
these two tangents intersect the third. (N.) 
4. Find the coordinates of the point P on the curve Sy = 4—x* at which the 
gradient is j . Write down the equation of the tangent to the curve at P. 
Find also the equation of the tangent to the curve whose gradient is — | and 
the coordinates of its point of intersection with the tangent at P. (N.) 
5. A curve whose equation is of the form y = ax*+bx+c crosses the >>-axis at 
right angles at the point where y = 12 and also passes through the point (3, -15). 
Find 
(i) the values of a, b and c; 
(ii) the coordinates of the points at which the curve crosses the jc-axis; 
(iii) the gradients of the curve where it crosses the x-axis. (N.) 
6. Find the equation of the tangent and the equation of the normal to the curve 
y = (3 -2x2)2 at the point where x = 1. (N.) 
7. A particle is moving in a straight line so that at / sec from the start its distance 
s ft from a fixed point P is given by 
s = / 3 _ 6 / J + 9 / + 5 6 # 
Find 
(i) the velocity of the particle when / = 4; 
(ii) the acceleration of the particle when / = 4; 
(iii) when the velocity of the particle is zero; 
(iv) how far the particle is from P when its velocity is zero. (N.) 
8. If the abscissae of three points P, Q, R on the curve y = x2+x+l are x, 
x-h9x+h respectively, show that the chord QR has the same gradient as the tan­
gent to the curve at P. 
9. The total running cost of a ship per hour is £ (13+v2/52) where v knots is the 
speed. If £ C is the cost per nautical mile, write down an equation expressing C 
in terms of v; hence find the speed at which the cost per nautical mile is least. 
The ship makes a voyage of 800 nautical miles at a speed of 13 knots. How much 
money would have been saved if the journey had been made at minimum cost ? (N.) 
10. If y = 2xz+ax2+bx+3 has stationary values when x — 1 and when x = 2, 
find a and b, and determine the nature of these stationary values. 
Find also the equation of the tangent to the curve at the point where x — \. 
(N.) 
11. A block of ice in the form of a cube, whose edge is 2 m, begins melting, and 
its volume decreases at a constant rate, the block remaining cubical. If the rate of 
melting is such that the edge measures 1 m after 28 hours, find 
(i) the length of the edge after 16 hours; 
(ii) the rate at which the length of the edge is decreasing at that time. (N.) 
INTRODUCTION TO THE CALCULUS 31 
12. Find the maximum and minimum values of the function }(4x 3 -3;t - l ) . 
Sketch the curve y — f ( 4 x 8 - 3 * - l ) and find the greatest and least values of y 
for values of x in the range —1*5 to +1*5 inclusive. (N.) 
13. Find the equation of the tangent and the equation of the normal to the curve 
y = 2 + 5x—x2 at the point where x = 3. 
Show that the normal meets the curve again at the point where x = 1. (N.) 
14. A vessel has the form of an inverted circular cone whose height is equal to 
its base diameter. Water is poured steadily into it at the rate of 1 ft3 in 3 min. Find 
the speed in feet per minute at which the water level is rising when the depth of the 
water in the cone is 2 ft. (N.) 
15. A firm manufacturing baked beans sells them in closed cylindrical tins. The 
tins are to be of two sizes, of 24 in8 and 12 in8 capacity respectively. For economy 
in mass production, it is necessary that the two sizes of tin shall have the same ra­
dius. If half the volume of beans produced is sold in large tins, show that when 
a total volume of n in3 is produced the amount of tinplate used is rm(r2+ 16/?rr)/8 in2 
where r in. is the radius of a tin. Show also that the amount of tinplate used is least 
if each large tin has a height three times its radius. (N.) 
16. Sketch the curve whose equation is y = *2(3—x) and find the equation of the 
tangent at the point ( - 1 , 4). Show that the tangent meets the curve again at the 
point (5, -50). (L.) 
17. The distance s moved in a straight line by a particle in time / is given by 
s = at2+bt+c where a, b and c are constants. If v is the velocity of the particle after 
time /, show that 4a(s-c) = v2-b2. (L.) 
18. A manufacturer's costs are a fixed sum F shillings together with a fixed amount 
C shillings for each article sold. The number n of articles sold varies as the square of 
the amount by which the selling price of an article, x shillings, is less than 2C 
shillings. Show that his total profits on n articles are 
{kx(2C- x)2 - F-kC(2C- x)2} shillings 
where A: is a constant. Hence, or otherwise, show that these profits will be greatest 
when the selling price of an article is 4C/3 shillings. (N.) 
19. Show that the equation of the tangent to the curve y = x3 at the point 
P(a, a3) is 3a2x-y-2a3 = 0. 
The tangent at P meets the jc-axis at Q and intersects the curve again at R. The 
tangent at R meets the *-axis at S, and O is the origin. Prove that SO = 20Q. (L.) 
20. A container is in the form of a circular cylinder with a hemispherical cap at 
one end and a conical cap at the other end. The cylinder, the hemisphere and the 
base of the cone are of equal radius r, the height of the cone is also r, and the whole 
is symmetrical about the axis of the cylinder. If the capacity of the container is Kand 
its external surface area is 5, show that 
2V 
S = 7tr2V2 + —. 
r 
Find the ratio of the length of the cylinder to its radius if Kis fixed and it is required 
to make S a ininimum. (N.) 
32 SIXTH FORM PURE MATHEMATICS 
21. Show that the line x —3y+4 = 0 is a normal to the curve y = xt — lx-\-\2 
at one of the points in which it cuts the curve and find the equations of the tangent 
and normal at the other point of intersection. (L.) 
22. Find the equation of a tangent to the curve x = t,y = 1/r2. 
If the tangent at a point P on the curve cuts the x-axis at Q and the ;y-axis at R 
and meets the curve again at S, prove that 
(a) RP = 2PQ, (b) SR = PQ. (L.) 
23. A thin metal plate is in the form of a rectangle with a semicircle described 
externally on one of the sides of the rectangle as diameter. If the sides of the rectangle 
vary in such a way that the perimeter of the plate remains constant and equal to P, 
show that the maximum area of the plate is P2/2(4+TI). (C.) 
24. The volume of water in a vessel is %nx\9—x) in3 when the depth of the water 
is x in. If water is poured in at a constant rate of 4 in3 per sec, calculate, correct to 
two significant figures, the rate at which the level of the water is rising when the 
depth is 2 in. (N.) 
25. Find the equation of the tangent to the curve y = a3/x2 at the point P of 
which the coordinates are (a/t, at2). 
The tangent to the curve at P meets the axes Ox, Oy at A, B respectively. Express 
ABS in terms of r, and hence find the values of t for which the length AB is least. 
CHAPTER II 
METHODS OF COORDINATE GEOMETRY 
2.1 The Straight Line 
The equation of a straight line with a given gradient and through a 
given point was considered in Chapter I. The equation Ax+By4-C = 0,where A, B and C are constant, is the equation of a curve with a con­
stant gradient (d.y/djc = —A/B) and it is therefore the equation of 
a straight line. This form of the equation of a straight line includes all 
possible cases and it is called the general equation of a straight line. 
It should be noted that, if the coordinates of two of the points on the 
line are known, the ratios A:B:C can be determined uniquely. If 
A = 0, the line is parallel to the x-axis; if B = 0 the line is parallel to 
the j>-axis; if C = 0, the line goes through the origin. 
The line y = mx+c. This form of the equation of a straight line does 
not include the case x = constant. The gradient of the line is m, and c 
is the intercept on the >>-axis. 
The equation of the straight line through two given points. The gradient 
of the line joining the points A(xi, y{), Bfa, J2) is 0>2—>>i)/(*2—*i) 
(Fig. 11). The equation of the line is therefore 
X2 — X\ 
This is more conveniently written and remembered in the equivalent 
form 
y-yi = x—xx 
The equation 
(2.1) 
y2-yi x2-x! 
a b 
33 
34 SIXTH FORM PURE MATHEMATICS 
represents the line which passes through the points (a, 0), (0, b) on Ox and 
Oy respectively. The equation of any straight line not passing through O 
and not parallel to one of the coordinate axes can be expressed in this 
intercept form. 
+-x 
Fig. 11. 
The distance between two points whose coordinates are (*i, yi)9 (x2, ̂ 2). 
By Pythagoras' Theorem (Fig. 11), 
AB2 = AP2+PB2; 
.\ AB = V{(*i-*2)*+0>i->>2)2}. (2.2) 
Area. The area of a rectilinear figure can be obtained in special cases 
by various methods. The method illustrated below is the most convenient 
one when the coordinates of the vertices of the rectilinear figure are 
known. 
Example. Find the area of the quadrilateral ABCD where A is 
(2, 2), *is ( - 1 , 4), Cis ( - 3 , 1), D is (1, -1 ) . 
In Fig. 12: 
area ABCD = area PQRS- (area 5gC+area CRD+area 
DSA + area APE) 
= [25-(3 + 4 + l | + 3)] = 13|sq. units. 
METHODS OF COORDINATE GEOMETRY 35 
The area of a triangle whose vertices are(xi, >>i), (x2,y2), (*3, j^j.There 
are many cases of which only one is illustrated in Fig. 13. The formula 
obtained below is correct in all cases and is proved by another method 
in Volume II: 
area of A ABC = area ^Pg5-(area APC+ area BQC) 
= \PQ{AP+BQ)-^(AP.PC+BQ.QC) 
= \(x1-x2){y1-yz+y2-yz) 
- \iyi - y*)(*i - *3> - i(y2-yz)(x3 - x2) 
= Hxiy*—x&i+x2y* -xzy2+x3y! -^1^3). 
Q 
C(-3,l) 
K 
B(-l.4) 
^ ^ 0 
f 
0(1,-1) 
p 
A(2,2) 
•8 
Fig. 12. 
Because the sign of such an expression as (xi—x2) depends upon the 
relative positions of A and B9 the sign of the area obtained by this for­
mula will depend upon the way round the triangle in which (*i, j>i), 
(*2, y*)> (*3, ^3) are taken. Area is defined, for most purposes, as a posi­
tive quantity and the result is therefore written: 
Area of A = \\ ( x ^ - x ^ x + ^ 2 ^ 3 - X&2+*SJ'I-*:LV3)I • 
36 SIXTH FORM PURE MATHEMATICS 
{Definition: \x\ = + x if x is positive, 
| x | = —x if x is negative, 
| x | is called modulus x or worf x.) 
y 
B(x2,y2) 
Q 
-
\ / ° 
1 V 
C(x5.y3) 
, 
r 
A(K„y,) 
P 
Fig. 13. 
Example. Show that if R, S, T and V are fixed points and if a point P 
moves so that the sum of the areas of triangles PRS and PTV is constant, 
then the locus of P consists of segments of four straight lines. 
In such an example it is important to choose the positions of the 
origin and the coordinate axes so as to reduce the computation as far 
as possible. Let R be the origin, S be (a, 0), T be (A, k) and V be (/, w), 
where a, h, k, I and m are constants (Fig. 14). Let P be (x, y). 
Then 
area ISSP+area TVP = constant = ^4(say); 
.'. \-sya\ + \i(ly-mx+xk-hy+km-kl)\ = A. 
This gives four cases according as each of ya and 
ly — mx+xk—hy + hm — kl 
METHODS OF COORDINATE GEOMETRY 37 
is positive or negative. The equation is in each the equation of a straight 
line, but since neither of \\ya\ and \±(ly—mx+xk—hy+hm—kl)\ 
can be greater than A only a segment of each line is included in the locus. 
The segments are cut off by the lines \ya = ± A and 
i(ly—mx+xk—hy+hm—kl)= ±A. 
VU.m) 
1. Write down the gradient, and the coordinates of the points of intersection with 
the axes of each of the following lines: 
(i) 3jc-4y+5 = 0; 
(ii) x = 5y-6; 
(iii) 6*+15;y = 9; 
(iv)x/3 ~W5 = 1; 
(v) Ix+my = 1. 
2. Find the equations of the lines through each of the following pairs of points: 
(i) ( - 1 , 3), (4, 2); 
(ii) (3, -4 ) , (5, - 1 ) ; 
(iii) (2, -6 ) , ( - 5 , - 2 ) ; 
(iv) (a, b\ (2a, 3b); 
(v) {at\ 2at\ {a\t\ -2a//) . 
38 SIXTH FORM PURE MATHEMATICS 
3. Calculate the distance between each pair of points named in question 2. 
4. Prove that the points (9, 11), ( - 5 , 3), (3, - 1 ) are the vertices of a right-
angled triangle. 
5. Calculate the coordinates of the point of intersection of the line joining (5, - 4 ) 
to ( -3 ,4 ) with the line joining ( -13, 4) to ( - 5 , 1). 
6. Show that the points (3, -5 ) , (4, 1), (6, 13) are collinear. If the point (5a, la) 
is also on the line, find the value of a. 
7. A is (6, 1), B is ( - 4 , -4 ) , Cis ( - 1 , 5). Find 
(i) the coordinates of the foot of the perpendicular from C to AB; 
(ii) the area of the triangle ABC. 
8. Find the area of each of the triangles whose vertices are as given below: 
(i) (2, 1), ( - 4 , -3 ) , (3, 4); 
(ii) (3, 1), ( - 2 , -6 ) , (4, 5); 
(iii) (0, 0), (3, 6), ( - 2 , - 6 ) ; 
(iv) (ct, c/r), (-cf, -c/t), (ct\ let). 
9. A is (2, 3), B is ( -1 ,4) , C is (3,0) and D is ( - 4, - 2). Find the locus of a point 
P which moves so that area PAB — area PCD. Show the four points and the com­
plete locus in a diagram 
10. A is (1, 0), B is (3, 0), C is (0, 1), D is (0, 3). The point P moves so that the 
sum of the areas PAB and PCD is 4 sq. units. Find the complete locus of P and draw 
a diagram to illustrate your answer. 
2.2 The Division of a Line 
The coordinates of the mid-point of the line joining A(x^ y{) to B(x2, J2). 
Let M(x, y) be the mid-point of AB (Fig. 15) and let ML and MN be the 
perpendiculars from M on the ordinate through B and the abscissa 
through A respectively. 
Fig. 15. 
METHODS OF COORDINATE GEOMETRY 39 
Then 
AN I ML = AM I MB = 1; 
X— X± = X2 — X\ 
.'. x = Uxi+x*)* y = Kyi+J>2). (2.3) 
This result is a particular case of the more general one which gives the 
coordinates of the point which divides AB in the ratio m:n. If P(x, y) 
A W | , | U ^ — 
S 
P(x,y^ 
r 
h 
: 
i 
0 
f 
B ( x 2 ^ 
r 
r 
L 
Q(X,YL 
r R 
S 
Fig. 16. 
(Fig. 16) is the point which divides AB internally in the ratio m : n and if 
Q(X, Y) is the point which divides AB externally in the ratio m : n, then 
PN__AP__m_ 
BL~ PB~ nl 
which reduces to 
Similarly 
. y-yi = m 
J>2-;> n 
_ ityi+/nj>g 
x = /wci + ntX2 m+n 
40 SIXTH FORM PURE MATHEMATICS 
Also 
which reduces to 
Similarly 
QS 
QR 
Y ■ 
X 
AQ m 
BQ n 
Y-yi _ m 
Y-y2 ~~ n 
my2-ny1 
m — n 
mx2—nxi 
m—n 
If external division in the ratio p : q is defined as division in the ratio 
p : —q, these results can be reduced to the single form: The coordinates 
of the point which divides the line joining the points (JCI, ji), (x2, y2) in the 
ratio m : n are 
(nxx+mx2 nyx+my*^ 
\ m+n ' m+n ) 
Examples, (i) Find the coordinates of the points P and Q which divide 
the line joining A(—9, 14) to 5(5*4, —5*2) internally and externally re­
spectively in the ratio 5:1. Show that PQ subtends a right angle at each of 
the points L(10, -9 ) , M(\0% -4 6),N(129 - 7 4 ) . 
The x-coordinate of P is 
l(-9)+5(5-4) 
5+1 
The .y-coordinate of P is 
l(14)+5(-5-2) 
= 3. 
6 
The x-couiuinate of Q is 
(-1X-9)+5(5-4) 
5 - 1 
The ̂ -coordinate of Q is 
(-l)(14)+5(-5-2) 
= - 2 . 
= 9. 
5 - 1 = - 1 0 ' 
METHODS OF COORDINATE GEOMETRY 41 
-2 + 9 
Therefore the gradient of PL is = — 1 and the gradient of QL 
3-10 
. - 1 0 + 9 % is = 1: 
9 -10 
therefore ^ PLQ is a right angle. 
Similarly the gradient of PM is — -|- and the gradientof QM is 3: 
therefore -̂ PMQ is a right angle. 
The gradient of PNis 3 and the gradient of QNis —-§-. 
Therefore ^ PNQ is a right angle. 
(ii) The vertices of a triangle ABC are ,4(5, 2), £ ( - 4 , -1) , C(2, 8). 
The point P divides BC in the ratio 1 : 2 and the point Q divides AB in 
the ratio 2:1. AP and CQ intersect at H. If BH produced intersects AC 
at R9 calculate the ratio AR:RC. 
P is the point 
i [2(-4)+1(2)], i [2(-1)+1(8)], i.e. ( -2 , 2); 
j2 is the point 
i [l(5)+2(-4)], i [ l ( 2 ) + 2 ( - l ) ] , i.e. ( -1 ,0); 
Therefore the equation of AP is y = 2 and the equation of Cg is 
x+1 ,y 
2+1 8 ' 
i.e. 8x-3.y+8 = 0. 
The coordinates of H are given by the solution of the simultaneous 
equations 
y = 2, 8JC-3^+8 = 0, 
i.e. x = - £ , y = 2. 
Therefore the equation of 2?# is 
y+1 *+4 
2 + 1 - - 1 + 4 ' 
i.e. 4 X - 5 J > + 1 1 = 0. 
42 SIXTH FORM PURE MATHEMATICS 
The equation of AC is 
x-2 _y-S 
5-2 " 2 - 8 ' 
i.e. 2x+j>-12 = 0. 
The coordinates of R are given by the solution of the simultaneous 
equations 
4x- 5y+11 = 0, 2x+y-12 = 0, 
i.e. x = 3y, y = 5. 
If R divides AC in the ratio /: m, then 
2m+S7 _ 
«+/ " 
Hence / = m and i* is the mid-point of AC. 
Exercises 2.2 
1. Three of the vertices of a parallelogram are (—4, 2), ( - 3 , 2), (2,1). Calculate 
in all possible cases the coordinates of the fourth vertex. (Use a property of the 
diagonals of a parallelogram.) 
2. Use coordinate methods to verify that the sum of the squares on any two sides 
of a triangle is equal to twice the square on half the third side plus twice the square 
of the median which bisects that side. (Chose your axes in the most convenient posi­
tion in relation to the triangle.) 
3. Find the coordinates of the points which divide AB in the required ratio in 
each of the following cases: 
(i) A(l9 3), £ ( -4 ,8 ) , 4:1 internally; 
(ii) ^ ( - 2 , -4 ) , £(3, 1), 1:2 internally; 
(iii) A(4, 6), £ ( - 2 , -3 ) , 3:1 externally; 
(ivM(2,4), B ( - 2 , - 6 ) , 3:2 externally; 
(v)^(4,6), 5(2,8), -2 :1 . 
4. In each of the following cases find the ratio in which C divides AB: 
(iM(3,4), 5 (8 , -1 ) , C(7,0); 
(ii) A(-2, - 5 ) , 5(4, -3 ) , C(7, - 2 ) ; 
(iii) A(6, - 2 ) , 5 ( - 6 , 2 ) , C(9, - 3 ) ; 
(iv) A(-3,7), 5 ( - l l , 5), C(-9,5£). 
5. The medians of a triangle are concurrent at the point which divides each 
median from the vertex to the mid-point of the opposite side in the ratio 2:1. Calcu-
METHODS OF COORDINATE GEOMETRY 43 
late the coordinates of the point of intersection of the medians (the centroid) of a tri­
angle whose vertices are the points (xlt yt)t (x2i y2)> (xi9 y3). 
6. The line joining the origin to a point P(at2, 2at) on the curve y2 — Aax is 
divided at Q in the ratio k:l. Find the coordinates of Q and, by eliminating r, show 
that the equation of the locus of Q as / varies is y2 = -j—- . 
7. The lines 2x+y-2 = 0, 2x+5y-6 = 0 and JC+J>-3 = 0 form the sides 
AB, BC, CA respectively of a triangle ABC. A fourth straight line divides AB in the 
ratio 2:1 and AC in the ratio 3:1. Calculate the ratio in which this line divides BC. 
8. The straight line of gradient 1 through A(-2, —1) meets the curve 
x*+yt-4x-6y+ll = 0 at P and fi. Calculate the ratio AP.AQ. 
2.3 The Equation of a Circle 
If P(x9 y) is a point on a circle, radius a and centre Q(p, q) (Fig. 17), 
then QN2+NP2 = a\ 
:. (x-p)2+(y-q)2 = a2 (2.5) 
and this is therefore the equation of the circle. In the special case when 
the centre of the circle is at the origin, the equation of the circle is 
x2+y2 = a2. (2.6) 
Fig. 17. 
4 4 SIXTH FORM PURE MATHEMATICS 
The General Equation of a Circle. The equation 
x2+y2+2gx+2fy+c = 0 
can be written in the form 
(x+g)2+(y+f)2 = g 2 + / 2 - c . 
(2.7) 
(2.8) 
If g2+P—c is positive, equation (2.8) is the equation of a circle whose 
centre is (—g, —f) and whose radius is <\/(g2+f2—c). 
y 
P(x,.y,) 
-►x 
Fig. 18. 
The length of a tangent to this circle from the point (*i, yi) outside the 
Circle. From Fig. 18, 
PT2 ^Ptf-CT2; 
:. PT2 = (*!+*)■+(yi+f)2-(g2+f2-c) 
= xl+yl+lgXi+lfyi+c. 
Therefore length of tangent is 
Vte+J*+3?*i+2#i+c). (2.9) 
Examples, (i) Obtain the equation of the circle through the points 
^(2,3) ,5(6,-3) ,C(4,- l) . 
METHODS OF COORDINATE GEOMETRY 45 
The mid-point of AB is (4, 0); the gradient of AB is - 3/2. 
The mid-point of BC is (5, —2); the gradient of BC is — 1. 
Therefore the equation of the perpendicular bisector of BC is 
0>+2) = (x-5), i.e. x-y = 7. 
The equation of the perpendicular bisector of AB is 
y = £ (*-4), i.e. 2x-3y = 8. 
These two lines meet at the centre of the circle at the point given by the 
solution of the simultaneous equations 2x—3y = 8, x—y = 7. 
Therefore the centre of the circle is (13, 6). The radius r is the distance 
of (13, 6) from (2, 3), i.e. r = V(ll2+32) = ^130. 
The equation of the circle is 
(x-13)2+0>-6)2= 130, 
i.e. x2+j>2-26x-12^+75 = 0. 
This method provides a check at the stage when the radius is calculat­
ed. It is easy to verify that the distance of the centre from each of the 
given points is the same. An alternative method, using the general 
equation of the circle, does not provide so simple a check. 
(ii) Find the equations of the circles which pass through the points 
(—5, 9), (2, 8) and which touch the x-axis. 
The centre must be on the positive side of the x-axis. 
The j-coordinate of the centre is r where r is the radius of the circle. 
Let the centre be (d, r); then 
(</+5)2+(r-9)2 = (rf-2)2+(r-8)2 = r2; 
.'. <P+10rf+25 + r*-18r+81 = </2-4rf+4+r2- 16r+64 = r2; 
.'. 14</-2r+38 = 0 
and 
</2-4rf-16r+68 = 0, 
.-. </2-4</-̂ 8(14<M- 38)4-68 = 0; 
.\ rf2-116*/-236 = 0; 
/. (</+2)(</-118) = 0. 
46 SIXTH FORM PURE MATHEMATICS 
Hence 
rf=-2, r = 5 or rf=118, r = 845. 
The two circles are therefore 
(*+2)2+0>-5)2 = 25 and (x- 118)2+0>-845)2 = 8452. 
Exercises 2.3 
1. Write down the coordinates of the centre and the length of the radius (in surd 
form if necessary) of each of the following circles: 
(i) x2+y2+5x+6y-15 = 0; 
(ii) 16x2+16y2-16x+Sy-59 = 0; 
(hi) x2+y* = 8x+4; 
(iv) (ax+p)*+(ay-p)2 = a*; 
(v) a2b2x2+a2b2y2-2ab2x-2a2by+b2+a2-a2b2 = 0. 
2. Find the equation of the circle through the three given points in each of the 
following cases: 
(i) (4, 6), ( - 2 , 4), (8, - 6 ) ; 
( i i ) ( - 3 , -4 ) , (5, -2 ) , (7, -10 ) ; 
(iii) ( - 5 , - 4 ) , ( - 3 , - 2 ) , ( - 9 , 0 ) ; 
(iv) (0, 0), (6, 2), ( -14, -21); 
(v) (a, 2a)f ( - a , -a ) , (3a, a). 
3. In each of the following cases write down the equation of the circle which 
satisfies the given conditions: 
(i) centre (1, 4), radius 2; 
(ii) touches both axes, radius 1; 
(iii) centre (2, 3), touches the x-axis; 
(iv) centre (3, 4), passes through (7, 0); 
(v) centre on the line x+y = 1, touches the x-axis at (3, 0). 
4. Find the equation of the circle which touches the ^-axis at (0, 4), and passes 
through (2,2). 
5. (i) Find the equation of the circle of which one diameter is the line joining 
(2, 4), ( - 5 , -3 ) . 
(ii) Find the equation of the circle of which one diameter is the line joining 
(*i> J>i)> (*2> ̂ 2)- (Use the "angle in a semicircle" property.) 
6. Calculate the length of, each of the tangents drawn to the following circles 
from the points named: 
(i) x2+y2+4x-6y+5 = 0, (1, - 1 ) ; 
(ii) 3x2 + 3y2-5jc+ll.y-54 = 0, (-12, -3); 
(iii) x2+y2 + lax+2by - 6ab = 0, (a, b); 
(iv) (x-x)2+(y-P)2 = a2, (2a, 2p). 
METHODS OF COORDINATE GEOMETRY 47 
7. Find the equations of the circles which touch the >>-axis and intersect the 
jc-axis at (1, 0) and (9, 0). 
8. Prove that the points (9, 6), (4, -4 ) , (1, - 2 ) , (0, 0) are concyclic. 
9. A tangent of length 3 units is drawn from a point B on the line x+y = 1 to 
the circle x*+yi+10x+4y—2 = 0. Calculate the coordinates of the possible posi­
tions of P. 
10. The length of the tangent from a point P to the circle x2+y2 = 5 is twice 
the length of the tangent from P to 
xi+y*-$x-6y+24 = 0. 
Show that P must lie on the circle3jc2 + 3;'2-32jc-24y+101 = 0. 
11. Show that the circles 
x2+y2-16x-12y + 75 = 0, 
5x2+5y*-32x-24y+75 = 0 
touch each other and find the coordinates of their point of contact. 
12. Find the equation of the circle which touches x2+y2—6x+2y-{ 5 = 0 a 
(4, - 3) and passes through (0, 7). 
2.4 The Intersection of Lines and Circles 
The equation 
AiX+Biy+d+XiAzx+Bzy+Cz) = 0 
where A is a constant is the equation of a straight line. It is satisfied by 
the coordinates of the point of intersection of 
Axx+Bxy+d = 0 and A2x+B2y+C2 = 0. 
It is therefore the equation of a straight line through the point of inter­
section of these two lines. This technique forms the basis of a useful 
method of coordinate geometry which is illustrated by the following 
examples. 
Examples, (i) Find the equation of the straight line through the inter­
section of 
5 x - 1 7 j + l l = 0 and 1 3 J C + 1 9 J - 5 = 0 
which is parallel to the line x—2y+1 = 0. 
48 SIXTH FORM PURE MATHEMATICS 
5x-l7y+U + X(\3x+ l9y-5) = 0 
is a straight line through the intersection of the given lines. It is parallel 
to x—2y+l = 0 if its gradient is y, 
. .r -5-13A 1 
»•* l f 317+191 = I' 
This gives A = 7/45 and hence the required line is 
5^-17^ + 11 + & (13^+19^-5) = 0 
which simplifies to 
79*-158^4-115 = 0. 
(ii) Find the equation of the common chord of the circles 
x2+y2-4x+4y-42 = 0; x2+j>2+8x-8>>-18 = 0. 
Find the equation of that circle through the points of intersection of 
these two circles which passes through the origin. 
The equation 
J C 2 + J 2 - 4 X + 4 ^ - 4 2 - ( J C 2 + ^ 2 + 8 J C - 8 ^ - 1 8 ) = 0 
is the equation of a straight line which is satisfied by the coordinates 
of any point which lies on both circles. If the two circles intersect it is 
therefore the equation of the straight line through their common points. 
The distance between the centres of the circles is 6 y/2 and the radius 
of each is 5<\/2. The circles therefore intersect. 
Therefore the equation of the common chord is x—y+2 = 0. 
The equation 
x2+y2_4x+4y_42 + X(x2+y2+8x-Sy-l8) = 0, 
where A is a constant, is the equation of a circle which is satisfied by 
the coordinates of any point which lies on both circles. Since the two 
circles intersect it is therefore the equation of a circle through their 
common points. This circle goes through the origin if — 42 —18A = 0, 
i.e. if A = -7 /3 . 
Therefore the required circle is 
x2+y*-4x+4y-42-l(x2+y2+Sx-8y-lS) = 0 
which reduces to 
x2+y2+\7x-\7y = 0. 
METHODS OF COORDINATE GEOMETRY 49 
Exercises 2.4 
1. Find the equation of the line joining the origin to the point of intersection of 
2JC-11 .K+17 = 0 and 3x-9y+5 = 0. 
2. Find the equation of the line which is parallel to the line 2x+3y+5 = 0 and 
which passes through the intersection of 
JC+7>>-5 = 0 and 2 J C - 3 ^ + 1 = 0 . 
3. The equations of the sides of a triangle are 
x-y-15 = 0, 2 J C - 3 J > - 4 0 = 0, x-2y-24 = 0. 
Find the equation of each altitude and show that the three altitudes are concurrent 
at the origin. 
4. Find the equation of the common chord of the circles 
x*+y*-2x+4y-4 = 0, x2+y2 +4x-6y-\2 = 0. 
5. Find the equation of the circle through the intersection of the two circles 
X2+;K2 + 2 ;C+4}>-11 = 0, x2+y2-4x-10y+4 = 0 
which also passes through (3, 4). 
6. The line x -y +1 = 0 intersects the circle x2+y2 - 5x+2y - 1 2 = 0 at A and B. 
Find the equation of the circle on AB as diameter. 
7. Prove that the two points of intersection of the circles 
x2+y2-\2y+\0 = 0 and x2+y2-Sx-4y+10 = 0, 
and the points (2, 3), (10, — 5) are concyclic. Find the equation of the circle on which 
these points lie. 
8. Find the equations of the two circles which pass through the intersection of 
x2+yt-4x-\$y + $4 = o, x2 + y2-2x-\6y + 64 = 0, 
and which touch the jc-axis. 
2.5 The Parabola x = at2, y = 2at, a > 0 
(i) The cartesian equation of this curve is y2 = 4ax. 
(ii) There is no part of the curve for which x is negative. 
(iii) The curve goes through the origin. 
(iv) The curve is symmetrical about Ox. 
(v) dy/dx = 2a/2at = 1/r. The curve has no stationary points. 
(vi) The tangent at the origin is the j>-axis. 
(vii) The curve is therefore as shown in Fig. 19. 
50 SIXTH FORM PURE MATHEMATICS 
The Focus-Directrix Property of the Parabola. The distance of 
P{at\ 2at) from S(a, 0) is ^/{(at2-a)2+4a2t2} = at2-{-a. The distance 
of P from the line x+a = 0 is also at2+a. Every point on the parabola 
is therefore equidistant from (a, 0) and the line x+a = 0. It can easily 
be proved, conversely, that if a point is equidistant from (a, 0) and 
x+a = 0 it must lie on the parabola. The parabola is therefore the 
locus of points which are equidistant from the point (a, 0), which is 
called the focus, and the line x+a = 0, which is called the directrix. 
► x 
Fig. 19. 
A curve, which is the locus of points which are such that the distance 
of each from a fixed point, the focus, is in a constant ratio to its distance 
from a fixed straight line, the directrix, belongs to a family of curves 
known as conic sections. Each can be shown to have the shape of a 
section of a right circular cone. The equations of the conic sections are 
all of the second degree and all equations of the second degree are 
equations of conic sections. Included in this family of curves in 
METHODS OF COORDINATE GEOMETRY 51 
addition to the parabola, are the circle, the ellipse, the hyperbola and 
pairs of intersecting straight lines. The last three of these will be 
discussed in subsequent chapters. 
Tangent and Normal to the Parabola. The tangent at P(at2,2at) to4h*/ 
parabola is 
The normal at P is 
y-2at = j(x-at2), 
i.e. x-ty+at2 = 0. 
y-2at = -t(x-at2), 
i.e. tx+y-2at-at3 = 0. 
(2.10) 
(2.11) 
► x 
Fig. 20. 
In Fig 20 the tangent at P meets the x-axis at Tand the normal meets the 
x-axis at G. Therefore Tis (-at\ 0): 
.'. ST = a+at2 = PM = PS; 
:. ^ STP = ^ SPT = ^ MPT. 
52 SIXTH FORM PURE MATHEMATICS 
Therefore the tangent to the parabola bisects the angle between the 
line joining P to the focus and the line through P parallel to the x-axis. 
Therefore PG is the external bisector of ^ SPM. 
Therefore the normal PG bisects ^ SPL. 
This is the principle of the parabolic reflector. Rays from a source 
of light at the focus of the parabolas which are sections through the 
axis of the reflector are all reflected parallel to the axis of the parabolas 
and so form a parallel beam. 
The converse principle is used in the giant reflecting radio telescope 
at Jodrell Bank. Radiation waves from such great distances that they 
provide parallel "rays" of radiation are all reflected by the parabolic 
bowl to the receiving aerial at the focus. 
The chord joining the points (at2, 2at), (as2, las) is 
x—as2 y—2as 
at2—as2 ~~ 2at—2as 
which simplifies to 
2x-(t+s)y+2ats = 0. (2.12) 
This chord passes through the focus (a, 0) if 2a+2ats = 0, i.e. i, 
s = — \/t. Parameters at the ends of a, focal chord of this parabola are 
therefore of the form t and — 1/f. 
Example. Prove that the tangents at the ends of a focal chord of a 
parabola intersect at right angles on the directrix. 
If P (at\ lot) andj2(tf/'2, -2a /0 are the ends of the focal chord PQ 
the gradient of the curve at P is \\t and the gradient at Q is — t. The 
tangents at P and Q are therefore at right angles. 
The tangent at P is 
x-ty+at2 = 0 (1) 
and the tangent at Q is x+y/t+a/t2 = 0, 
i.e. t2x+ty+a = 0. (2) 
These tangents meet at the point whose coordinates are the solution 
of (1) and (2), i.e. where x = — a. The tangents therefore meet at right 
angles on the directrix. 
METHODS OF COORDINATE GEOMETRY 53 
Change of Origin. If the origin is moved to the point 0'(a, b) with the 
axes remaining parallel to their original directions, and if P(x, y) becomes 
P(X9 Y) referred to the new axes 0 % OT through (a, b\ (Fig. 21), then 
X = x—a and Y = y—b; 
:. x = X+a and y = Y+b. (2.13) 
Hence the equation of the curve y = f(x) is transformed into 
Y+b=f(X+a).Thus: 
i 
0 
5 
r 
tojbj 
P(x,y)(X.Y) 
►X 
Fig. 21. 
1. The equation (y+b)2 = 4k(x+a) is transformed by changing the 
origin to the point (—0, —b) into the equation 
(Y-b+b)2 = 4ifc(;r-a+a), 
i.e. into the equation Y2 = 4fcJf. 
0>+b)2 = 4fc(x+0) is therefore a parabola with its vertex at (—a, — b) 
and its focus at (—a+k, —b). 
2. (y+b)2= — 4k(x+a) is transformed by a similar change of origin 
into the equation 
Y2 = -AkX 
and is therefore a parabola with its vertex at (—a, —b) and its focus at 
( - a - * , -b). 
54 SIXTH FORM PURE MATHEMATICS 
3. The equations x = 2at, y = at2, which are the parametric equa­
tions of the curve x2 = 4ay, represent a parabola with its vertex at the 
origin, its axis along the j>-axis, and its focus at (0, a). 
Examples, (i) Show that y = ax2+bx+c is the equation of a para­
bola and state the coordinates of its vertex and of its focus. 
The equation 
y = ax2+bx+c 
reduces to the form 
/ b\2 (b2-4ac) 
y = a\X+2i) ST"' 
/ A. b \ 2 l I _i_b2-4ac\ 
/ b -(b2-4ac)\ 
which is the equation of a parabola with vertex at I — -tr-> TZ 1 
/ b_ ~ 6 2 + 4 A C + 1 \ 
\ 2a9 4a ) ' 
b - 6 2 +4ac+ l 
and focus at 
(ii) The normal at P(at29 lat) to a parabola meets the jc-axis at G, and 
GP is produced to Q so that PQ = 2GP. Prove that the locus of Q is a 
parabola and state the coordinates of its vertex and of its focus. 
From equation (2.11) the equation of PG (Fig. 22) is 
tx+y-2at-at* = 0. 
Therefore the coordinates of G are (2a+at29 0). 
The coordinates of Q which divides GP externally in the ratio 3 : 2 are 
[3at2-2(2a+at2), 6at], i.e. (at2-4a, 6at). 
If Q is (x, y), then 
" ( £ ) - * 
which reduces to y2 = 36a (x+4a) which is the equation of a parabola 
with vertex at (—4a, 0) and focus at (5a, 0). 
The locus of the intersections of perpendicular tangents to the parabola 
x = at2, y = 2at. The tangent at (at2, 2ai) is x-ty+at2 = 0. Since 
1/f is the gradient, the equation of the tangent which is perpendicular 
METHODS OF COORDINATE GEOMETRY 55 
to this is obtained by writing — l/t for t. Hence the equation of the 
perpendicular tangent is t2x+ty+a — 0. 
These tangents meet where 
x(f2+l) + a(f2+l) = 0, i.e. where x = 
The required locus is therefore the directrix. 
-a. 
1. With the data of Fig. 20, prove the following: 
(i) ST = SG = SP; 
(ii) ^ RSP is a right angle; 
(iii) SM is perpendicular to PT and intersects PT on the tangent at the vertex; 
(iv) NG = 2a. 
2. Find the equation of the locus of the feet of the perpendiculars from the focus 
on to tangents to the parabola x = at2, y = 2a/. 
P.M. 1—C 
56 SIXTH FORM PURE MATHEMATICS 
3. Find the coordinates of the vertex and the coordinates of the focus for each 
of the following parabolas: 
(i) y2 = 3ax; 
(ii) x2 = Say; 
(iii) x = at2, y = at; 
(iv) (y+a)2 = 2a(x-a); 
(v) y = 2x 2 -5x+2; 
(vi) x2-4x+4y+4 = 0; 
(vii) y2-2x + 4y + 6 = 0; 
(viii) x = 3at2 + 5a, y = 2at; 
(ix) x = af—A, ^ = 4af2-0. 
4. PQ is a focal chord of a parabola and a line through Q parallel to the axis of 
the parabc1^ meets the directrix at K. Prove that PK goes through the vertex. 
5. The tangent at P(at2, 2at) to a parabola meets the ^-axis at R and the x-axis 
at T. If O is the origin and ORQT is a rectangle, prove that the locus of Q is a para­
bola. 
6. The parabolas x = at2, y = 2at and x = 2as, y = as2 intersect at P. The tan­
gent at P to the first parabola meets the x-axis at Tand the tangent at P to the second 
parabola meets the x-axis at R. O is the origin. Prove that OT = 20R. 
7. The chord PR of the parabola x = at2, y = 2at cuts the x-axis at (na, 0); 
PQ is a focal chord. RQ produced cuts the x-axis at K. Prove that RK:QK = ml. 
8. Find the coordinates of the point at which the normal at P(at2,2at) to a para­
bola meets the curve again. If the normal at P cuts the axis at G(3a, 0) and meets 
the curve again at Q, prove that GQ = 3GP. 
9. Prove that the normal chord through P(4a, 4a) on the parabola x = at2, 
y = 2flf subtends a right angle at the focus. 
10. The tangent at P(at2, 2at) to a parabola meets the >>-axis at Q and the normal 
at P meets the x-axis at G. M is the mid-point of PQ, N the mid-point of PG. Show 
that the locus of M is a parabola and the locus of N is a parabola and in each case 
state the coordinates of the focus and of the vertex. 
2.6 The Rectangular Hyperbola x = ct, y = c/t, c > 0 
(i) The cartesian equation of this curve is xy = c2. 
(ii) As JC — 0 from the positive side, y — + «>, and as x -•* 0 from the 
negative side, y -* — <*>. In such a case as this, in which the shape of the 
curve tends to a straight line as one variable or the other tends to infinity 
(the value of y given by the equation xy = c2 can be made as large as 
we please by taking the value of x sufficiently small), the straight line 
is called an asymptote to the curve. A second concept of an asymptote, 
that of "a tangent at infinity" to a curve will be discussed in Chapter VI. 
METHODS OF COORDINATE GEOMETRY 57 
In the case of this rectangular hyperbola, both axes are asymptotes to the 
curve. 
(iii) There are no stationary values of y. Also dy/dx -** 0 as x -*■ ± «>. 
Further dy/dx is everywhere negative and so y = c2/x is a decreasing 
function of x for all values of x. As x -*■ 0, dy/dx — — ». 
(iv) The curve is therefore as shown in Fig. 23. 
Fig. 23. 
The equation of the tangent at (ct, c/i). 
dx/dt = c, dy/dt = -c/t2. 
.-. dy/dx = - l / * 2 ; 
therefore the tangent is 
( y -c /0= - ? ( * - t f ) , 
i.e. x-ff2.y-2cf = 0. 
77ie normal at (ct, c/t) is 
0>-c/0 = *2(*-c0, 
i.e. t*x-ty-c(t*-l) = 0. 
(2.14) 
(2.15) 
58 SIXTH FORM PURE MATHEMATICS 
Examples, (i) Prove that the area of a triangle bounded by the axes 
and a tangent to the curve is constant. 
The tangent at P(ct, cjt) is 
x+ t2y-2ct = 0. 
This line meets the x-axis at M(2ct9 0) and the j-axis at N(0, left). 
Therefore the area of AOMN is 2c2 which is constant. 
(ii) By changing the origin of coordinates show that 
xy—ax+ay—5a2 = 0 
is the equation of a rectangular hyperbola. If the normal to the curve 
at a point P meets the curve again at Q, find the equation of the tangent 
to the curve at Q in relation to the original axes and in terms of para­
metric coordinates of P. 
The equation 
xy—ax+ay—5a2 — 0 
may be written 
(x-\-a)(y—a) = 4a2 
and if the origin is moved to (—a, a) so that with regard to new axes 
O'X and O' Y parallel to Ox9 Oy we have x = X—a and y = Y+ a, the 
equation becomes 
XY = 4a2 
which is the equation of a rectangular hyperbola. Referred to the new 
axes the equation of the normal at (2at, 2ajt) on this curve is 
(Y-lalt) = t2(X-2at), 
i.e. / 3Ar-rF+2a(l -^) = 0 
and this meets the curve again at the point Q(2as, 2a/s) if 
2at3s-2atfs+2a(l-t4) = 0. 
This gives a quadratic equation for s in the form 
t*s2+(l-t*)s-t = 0, 
i.e. (s-t)(t*s+l) = 0; 
.\ s = t or -1/f3; 
/. eis(-2a/f3 ,-20f3); 
METHODS OF COORDINATE GEOMETRY 59 
therefore the tangent at Q is 
(Y+2at*) = -t%X+2a/t*)9 
i.e. t6X+Y+4at* = 0. 
Referred to the original axes, the parametric coordinates of P are 
(2at—a, 2a/t+a) and the equation of the tangent at Q is 
t*(x+a) + (y-a) + 4at* = 0, 
i.e. t*x+y+(at*+4atz-a) = 0. 
Fig. 24. 
(iii) Prove that the orthocentre of a triangle inscribed in a rectangular 
hyperbola lies on the curve . 
If P, Q, R, S are points on the rectangular hyperbola 
x = ct, y — c/t 
of parameters tl912, h> h respectively, the equation of PQ is 
x-ch = y-c/h 
ct2—cti ~~ c\H—c\t\ 
60 SIXTH FORM PURE MATHEMATICS 
and this simplifies to 
x+htiy-cih+td = 0. 
The equation of RS is 
x+titiy-c{tz+td = 0; 
Therefore RS is perpendicular to PQ if the product of the gradients of 
these lines is — 1, i.e. if t\HUt± = — 1. 
But this is also the condition that QS i_ RP and that PS _L QR. 
Hence, if /if2*3*4 = — 1, S is the orthocentre of triangle PQR. The 
orthocentre of triangle PQR therefore lies on the curve (Fig. 24). 
2.7 The Semi-cubical Parabola x = at2, y = at3, a > 0 
(i)