IB Diploma Haese Mathematics: Further Mathematics HL: Linear Algebra and Geometry FM Topic 1 FM Topic 2 1921972351, 9781921972355

Table of contents :
Linear algebra
- Systems of linear equations
- Gaussian elimination
- Matrix structure and operations
- Matrix multiplication
- Matrix transpose
- Matrix determinant and inverse
- Solving systems of linear equations using matrices
- Elementary matrices
- Vector spaces
- Linear transformations
- Geometric transformations
- Eigenvalues and eigenvectors
- Theory of knowledge: representing space

Geometry
- Similar triangles
- Congruent triangles
- Proportionality in right angled triangles
- Circle geometry
- Concyclic points, cyclic quadrilaterals
- Intersecting chords and secants theorems
- Centres of a triangle
- Euclid's angle bisector theorem
- Apollonius' circle theorem
- Ptolemy's theorem for cyclic quadrilaterals
- Theorems of Ceva and Menelaus
- The equation of a locus
- The coordinate geometry of cicles
- Conic sections
- Parametric equations
- Parametric equations for conics
- The general conic equation

Citation preview

FUNLTE 3

for use with

e m m a r g o r P a IB Diplom

¥

HAESE

MATHEMATICS

Specialists in mathematics publishing

Mathematics

for the international student

Further Mathematics HL:

Linear Algebra and Geometry

FM Topic 1 FM Topic 2 Catherine Quinn Peter Blythe Robert Haese

Michael Haese

for use with IB Diploma Programme

MATHEMATICS

FOR THE INTERNATIONAL STUDENT

Further Mathematics HL: Linear Algebra and Geometry Catherine Quinn Peter Blythe Robert Haese Michael Haese

B.Sc.(Hons), Grad.Dip.Ed., Ph.D. B.Sc. B.Sc. B.Sc.(Hons.), Ph.D.

Haese Mathematics

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978-1-921972-35-5

© Haese & Harris Publications 2014

Published by Haese Mathematics Pty Ltd. 152 Richmond Road, Marleston, SA 5033, AUSTRALIA First Edition Reprinted

2014 2015

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FOREWORD Further Mathematics HL: Linear Algebra and Geometry has been written to provide students and teachers with appropriate coverage of these two Further Mathematics HL Topics, to be first examined in 2014. The Linear Algebra topic introduces students to matrices, vector spaces, and linear transformations. Useful preparation for this topic including the introduction to matrices is provided online with our MYP5 Extended book. The Principle of Mathematical Induction is useful for proofs in this topic, but is not essential in its preliminary study. The Geometry topic aims to develop students’ geometric intuition and deductive reasoning, particularly in plane Euclidean geometry. Most of this topic can be done using prior knowledge from the MYP5 Extended course. However, the final sections on conic sections require calculus from the HL Core course. Detailed explanations and key facts are highlighted throughout the text. Each sub-topic contains numerous Worked Examples, highlighting each step necessary to reach the answer for that example. Theory of Knowledge is a core requirement in the International Baccalaureate Diploma Programme, whereby students are encouraged to think critically and challenge the assumptions of knowledge. Discussion topics for Theory of Knowledge have been included on pages 124 and 129. These aim to help students discover and express their views on knowledge issues. Graphics calculator instructions for Casio fx-9860G Plus, Casio fx-CG20, TI-84 Plus and TI-nspire are available from icons in the book. Fully worked solutions are provided at the back of the text. However, students are encouraged to attempt each question before referring to the solution. It is not our intention to define the course. Teachers are encouraged to use other resources. We have developed this book independently of the International Baccalaureate Organization (IBO) in consultation with experienced teachers of IB Mathematics. The text is not endorsed by the IBO. In this changing world of mathematics education, we believe that the contextual approach shown in this book, with associated use of technology, will enhance the student's understanding, knowledge and appreciation of mathematics and its universal applications. We welcome your feedback. Email: Web:

[email protected] www.haesemathematics.com.au

CTQ RCH

ACKNOWLEDGEMENTS The authors and publishers would like to thank all those teachers who offered advice and encouragement on this book.

PJB PMH

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124 01 ]c=[ 21 ]p:[ 20 ] a:[_ 4=|2 L21 _”] 12 32 31 1l8|-2lcl+4lD|=7

o 1

Rows: 2|5 Columns: 2

-

Textiout [Expression input

1B-2iCI+4PD|

3 -m

6

TABLE OF CONTENTS

TABLE OF CONTENTS

-

LINEAR ALGEBRA

CARTCTIZOT@OUOQwW>

SYMBOLS AND NOTATION USED IN THIS BOOK

Systems of linear equations Gaussian elimination Matrix structure and operations Matrix multiplication Matrix transpose Matrix determinant and inverse Solving systems of linear equations using matrices Elementary matrices Vector spaces Linear transformations Geometric transformations Eigenvalues and eigenvectors

11 15 25 32 39 41 49 55 60 83 100 113

THEORY OF KNOWLEDGE

124

(REPRESENTING SPACE)

GEOMETRY

DUWOoOZIOCAT"ZTOQT@DUOTP

THEORY OF KNOWLEDGE

127 (EUCLID’S POSTULATES)

129

Similar triangles Congruent triangles Proportionality in right angled triangles Circle geometry Concyclic points, cyclic quadrilaterals Intersecting chords and secants theorems Centres of a triangle Euclid’s angle bisector theorem Apollonius’ circle theorem Ptolemy’s theorem for cyclic quadrilaterals Theorems of Ceva and Menelaus The equation of a locus The coordinate geometry of circles Conic sections Parametric equations Parametric equations for conics The general conic equation

133 138 141 145 156 163 167 172 176 178 182 190 194 200 216 219 223

WORKED

231

INDEX

SOLUTIONS

342

Q

SYMBOLS AND NOTATION

USED IN THIS BOOK

is approximately equal to

>

is greater than

>

is greater than or equal to


n).

is an underspecified system of linear equations,

2 equations in 3 unknowns. If an underspecified system of linear equations is consistent then it will have infinitely many solutions.

since there are

LINEAR ALGEBRA

AUGMENTED

(Chapter 1)

13

MATRICES

A system of m x n linear equations can be written as a rectangular array of m by (n + 1) numbers by leaving out the + and = signs, and the variables. We call this an augmented matrix (AM). The general system

a1171 + a12T2 + oo + ATy = by a2171 + AT + ... + A2 Ty = bo Am1%1 + Gma®a + ..o + Q@

a1 azr

a2 G2

.. ...

ain | b G2, | b2

Am1

Gm2

o

Gmn | b

has AM

= b

.

For example, the system

2wy + 3xs 1 — Ty

—

—

1

=

X4

—

- 3I3

has AM

=10

x3+2x4

x2+ 2Z2

x4=5

2 1

3 -1

0 1

1|5 |10 2

1

-2

3

-1

1

An augmented matrix is a matrix of coefficients.

EXERCISE 1A 1

Explain why each of the following is not a linear equation:

a 2

8—y=3

T1=7—./Z2

b

z;— 215+ 23=10

€

T1+x9—

213+ 14 =2

1 1.0 0f5 ‘00116 00 2 0|8 00 0 1|2

b11224 21 3 —-1|3

X

+

To

Ty

—To+

2z1 +x9

—

I3

=

4

T3

=

8

b

1+

does the system

o solutions

For what value(s) of k£ € R

{

3z

— 323 =0

For what value(s) of a € R a

6

¢

Write down the augmented matrix for the system of equations: a

5

—z—22=0

Write down the system of linear equations corresponding to the augmented matrix, and state if the system is underspecified or overspecified:

a[lli—?’zl 9 1|-1 4

bz

Find the solution set for: a

3

2xy +To+ 2374 =3

b .

1 +To—x3—x4 =05

a0 —203=7 +

C

x3=2

=4

Tty 3x+3y=a

is the system

20 —4y =8

+ax3+a4=1

dry —x4

have:

infinitely many solutions r—2y==F

3o

¢ .

exactly one solution?

consistent?

=06

14

LINEAR ALGEBRA (Chapter 1) z+y+

7

8

Find the relationship between p, ¢, and r given that the system

a

Explain why the system

r+y= 20—y =

z=p

T +2z=g¢q 2e+y+3z=r

isconsistent.

is inconsistent.

3r+y=28

b

9

a b

s this system underspecified or overspecified?

Determine whether the system

r14+xe—

x3—T7=0

{ oy — 2+ 205 — 9 = 0

Under what conditions is the system

1+

T

— T3

isis h homogeneous . =a

{ %y — 9+ 5 — 8 = b

h 0mogeneous ?

LINEAR ALGEBRA

(Chapter 1)

EINIT GAUSSIAN ELIMINATION In previous years, we have solved

2 x 2

15

systems of linear equations by elimination.

2r+y=—1 {z C gy =17,

Consider the system

From using the method of elimination, we know we can:

e

interchange the equations, called swapping 20 +y=-1

o3y e

has

=17

Luti

z—3y=17

as the same solution as

%ty — 1

replace an equation by any non-zero multiple of itself, called scaling

2 =-1 {xw:;z 17 e

th

has the same solution as

{

—62

—3y =3 z _ 35 _17

Itiplying by —3 {multiplying by —3}

replace an equation by a multiple of itself plus a multiple of another equation, called pivoting. If we replace the second equation by “twice the second equation, minus the first equation”, we have 2x — 6y = 34 -2+ y=-1) —T7y =35

2 =-1 {xx:;z 17

. has the same solution as

2c+y=—1 { 1773 _ 35

The principles of swapping, scaling, and pivoting are applied to augmented matrices as elementary row operations. We can hence: e

i

.

e

interchange rows replace any row by a non-zero multiple of itself

e

replace any row by itself plus a multiple of another row.

For example, the system e

2

=1

{;:;Z

17

has AM

(?

of the system.

_

713 | 171 )

If we interchanged rows 1 and 2, we would write:

21—1N1—317 vl T2 1] means “which has the same solution as”

e

Elementary row operations do not change the solution

indicates rows 1 and 2 have been interchanged

If we multiplied row 1 by —3, we would write:

1

=317

1

3|17 indicates row 1 has been replaced by “—3 x row 17

2L

16

LINEAR ALGEBRA (Chapter 1) o

If we replaced row 2 by “twice row 2 minus row 1”, we would write:

2 1|1\ _ (2 1]|-1 1 3|17 07735

indicates row 2 has been replaced by “twice row 2 minus row 17

In the process of row reduction, we use elementary row operations to eliminate variables from selected rows of an augmented matrix. This allows us to systematically solve the corresponding system of linear equations.

SOLVING

2 x 2

SYSTEMS OF LINEAR EQUATIONS

To solve a 2 x 2 system of linear equations by row reduction, we aim to obtain a 0 in the bottom left corner of the augmented matrix. This is equivalent to eliminating z; from the corresponding equation.

Use elementary row operations to solve:

.

{

2c+3y =4 br+4y =17

.

2

In augmented matrix form, the system is

( 5 o

L

3|4

4 ‘ 17)

2

3|4

0

—7|14

Check your solution by substitution into the

y=-2

Substituting into row 1,

2R275R1*>R2

2z + 3(—2) =4

e

o

2z =10 =5

the solution is

=z =5,

y = —2.

In previous courses, you should have seen that

ax + by = ¢ where a, b, ¢ are constants, is a line in the

Cartesian plane. Given two such lines, there are three possible cases which may occur: Intersecting lines

Parallel lines

Coincident lines

-—

-—

one point of intersection

no points of intersection

infinitely many points of

a unique simultaneous

no simultaneous solutions

intersection

solution

For example:

{2z +3y=1 z—2y=28§

For example:

2743y =1

{2x 1 3y="7

infinitely many simultaneous

solutions

For example: {2x773y:1

4z 4+ 6y =2

LINEAR ALGEBRA

(Chapter 1)

17

EXERCISE 1B.1 1

2

Consider the system of linear equations Y 4

r—3y=2

{ 2r+y=-3

a

Write the system as an augmented matrix.

b

Replace the second row with “the second row minus twice the first row”.

¢

Hence solve the system.

By inspection, decide whether the pair of lines is intersecting, parallel, or coincident, and state the number of solutions to the system.

4oy = a{:cy

2% —y=—1 b{xy

3z + 6y =

5

6

r+y=a,

a€R

Use elementary row operations to solve:

5

r—3y=-8 4z 4+ 5y =19

b

r+ Ty =-17 2c —y =11

.

2z +3y = -8 r+4y=-9

d

3r—y=9 4o+ 3y =-1

.

o

Explain why there are infinitely many solutions.

O

Try to solve the system using elementary row operations. Explain what happens.

an

{2z+6y=8.

Let

y=t,

t € R.

Solve the system in terms of ¢.

Q

Consider onsider the th system 1

z+3y=4

Let

z =s,

s € R.

Solve the system in terms of s.

®

4

z+y=4

d

Tz —5y=28 2z = 10y + 14

c

3

r+4y =13

Explain why your solutions in ¢ and d are equivalent.

Consider the system

T —5y =28 { 2z — 10y =a

where

a € R.

a

Write the system as an augmented matrix, and perform an elementary row operation to make the bottom left corner element 0.

b

Explain what the second row means for the cases where the system have in this case?

¢

Find all solutions for the case where

Discuss the solutions to

r+3y=4 { 2e+ay=>

a = 16.

for

oral

all

beR.

a,5€

a # 16.

How many solutions does

18

LINEAR ALGEBRA (Chapter 1)

SOLVING

3 x 3

SYSTEMS OF LINEAR EQUATIONS a1y + ajpze + azrs = di

The general

3 x 3

systems of linear equations

a21T1 + A20%o + ag3w3 = do

can be written as

az1z1 + azaTe + aszrs = ds az1 az;

a3 | di

g azz

>0

oo

o

We can use elementary row operations to reduce the matrix to the form

o

o

azs | da asz | ds Qo

the augmented matrix

a2

in which

=

ann

there is a triangle of zeros in the bottom left corner. We call this row echelon form. From the row echelon form, we can see that: e

If

h#0,

the third row means

hxs =i.

We can therefore solve for x3, and hence for 25 and z;

using rows 2 and 1 respectively. The system has a unique solution.

s

z+3y—2z2=15

Solve using elementary row operations:

The system has AM

~10 Using row 3,

1

3

-1]15

1

-1

-2]0

1

3

—-1]

15

0

-5

3

—23

R2

-

2R1

0

—4

—1|-15

Rg

—

R1

1

3 =5 0

-1] 3 —17|

5R3 — 4Ry — R3

2

~

26 +y+z2="7 r—y—2z=0

0

1

1|7

15 |-23 17

—17z =17 Loz=-1

Substituting into row 2,

—5y+ 3(—1) = —23

Substituting into row 1,

x4+ 3(4) — (—1)

the solutionis

z =2,

—5y=—20

y=4,

z=—1.

=15

— —

R2 R3

LINEAR ALGEBRA e

If h=0 and i # 0, the third row means 0z; + Oxz + Oz is no solution and the system is inconsistent.

oo~ cCo~

2

W=

The system has AM

Row 3 means that

where

113

-5

—112

R2

-

2R1

—

R’_}

—-10

—-21(9

R3

—

3R1

-

R3

R3

72R2

0

In this case there

1|3 18 118

2

2 -5

i # 0.

19

T+ 2y + z=3 20 —y+2=28 3z —4y + z=1

Solve using elementary row operations:

2 -1 —4

=4

(Chapter 1)

1 (3 —-1(2 0

5

Oz + Oy + 0z = 5,

HR3

which is absurd.

there is no solution, and the system is inconsistent.

If

=0

solutions.

and

i = 0,

the last row is all zeros.

We let

x3 =t

where

¢t € R

In this case the system has infinitely many

and write z; and x5 in terms of ¢. In this case the solution

is a parametric representation with parameter ¢. We call z; and z basic variables and z3 a free

variable.

N

Solve using elementary row operations:

Cow

OO

wWH

The system has AM

2

e

-1 1 -3

15 —-1(2 3|8

-1

1

5

3

-3

-1

2R2

-

R1

-3

3

1

2R3

-

3R1

-1 3

0

1 5 -3|-1

010

z =t¢,

then using row 2,

R2 —

Ry +Rs — Rs

Row 3 indicates there are infinitely many solutions. If we let

—

3y —3t=—1

Jy=3t—1

R3

20

LINEAR ALGEBRA (Chapter 1)

Substituting into row 1,

2z —(t—3)+t=

EXERCISE 1B.2 1

Solve each system of linear equations by row reduction to echelon form: r+4y+11z2=7

a

20 —y+

z+6y+172=9 r+4y+8z=4

b

Consider the system


j.

7 < j.

0 2 -1

100 010 0 0 1 100 00 0 0 0 4

-

The elements

LINEAR ALGEBRA

(Chapter 1)

EQUALITY OF MATRICES < means

Two matrices are equal if they have the same order and the elements in corresponding positions are equal. A=B

&

[T

bi]‘

“if and only if”.

for all 4, ]

For example,

a (c

b [ d)—(y z)

S

a=w,

b=z,

c=y,

and

d = z.

MATRIX ADDITION Carla has three boxes of sports equipment: A, B, and C. The boxes contain bats, balls, and cones according to the matrix shown.

Carla has ordered more equipment for the boxes.

10 bats, 20 balls,

and 15 cones will be added to each. The new equipment is given by

A 12 32 26

Box B 15 25 28

10

10

20

15

the matrix shown.

20 15

When the new equipment is added to the boxes, we have the matrix addition:

12 32 26

15 25 28

11 21 20

|+

10 120 15

10 20 15

10 22 20 ) =52 15 41

25 45 43

21 41 35

To add two matrices, they must be of the same order, and we add corresponding elements.

A + B = (aij) + (biy) = (as; + bij) MULTIPLES OF MATRICES A cake recipe requires 3 cups of flour, 2 cups of sugar, and 6 eggs.

We can represent these ingredients using the matrix

3

C = | 2

6

If we make two cakes using this recipe, we will need 6 cups of flour, 4 cups of sugar, and 12 eggs. We can represent this using the matrix Notice that to get 2C by 2.

2C=C+

C = |

4

12

from C, we multiply each element of C

C 11\ 21 | 20/ 10

20 15

bats balls cones

27

28

LINEAR ALGEBRA (Chapter 1)

Similarly, to make three cakes using the recipe, the ingredients needed are given by

3x3 9 3C=(3x2|=1|6 3x6 18 If we made a cake using only half the ingredients, we would need

If A = (a;;)

hasorder

m x n, and k is a scalar, then

%C =

(3

31 X6

3

% x2|=11

kA = (ka;;).

We use capital letters for matrices and lower-case letters for scalars.

So, to find kA, we multiply each element in A by k. The result is another matrix of order

bxay

m X n.

NEGATIVE MATRICES The negative matrix A, denoted —A, is actually —1A.

—A = (-1 x ai5) = (—aij) —A is obtained from A by reversing the sign of each element of A.

A

For example, if

A:,

=3

then

*A:(,Q

1

,4)'

MATRIX SUBTRACTION To subtract two matrices, we define

A—B=A+

(—B).

For example, suppose that over the next 6 months, Carla’s sports equipment is lost or broken according to the matrix

3

2

4

15

12

7

4

0

3

The equipment remaining is given by 22 52 41

25 45 43

21 41 ) — 35

3 |15 4

2 12 0

4 7 | =| 3

19 37 37

23 33 43

17 34 32

To subtract matrices, they must be of the same order, and we subtract

corresponding elements. That is,

A — B = (a;; — b;;).

3T T Let

R

3

A—(4

2 1

5

6)’

B—(1

2

0

7

A+B

-3

71),

and

C—(4

3

(3

21

“\a56)T\1 (5 ~\5

2 12

9

0

Find:

A—-B

—1iB

A+B

.o

2).

A+ C

3A

2A — 3B

A + C cannot be found as A and C have

-3

different orders.

7 21

=2 5

A—-B

3A

_ (3 “\4

2 5

(1 “\3

2 -2

1)y 6

(2

1

0 7

=3 -1

(9 —\12

6 15

3 18

4 7

7%B

2A — 3B

-1

=

1

2

0

1

T2

71

4 10

(6 _ “\8

% 2

—\5

(0

4 —11

(6

2\ 12

0 21

3

-9 -3

11 15

A=

T

Let

N

and

=W

=

EXERCISE 1C.1 C=

4A C-A

Consider two

m x n matrices

A+B=B+

-1 2 -3 5 0 2 -2C —2A + 4C A = (a;;)

and

Find:

A +2C 1A B = (b;;).

Prove that:

kA + kB = k(A + B)

A

B—-—A=—-(A—-B)

(a+b)A =aA + bA

A+A+A+....+A=FkA,

keZ"

k of these

Find z and y such that:

( )=( ZIQ

3

4

y 3

9 y+7

)

(

r

y

2y

—y Ty

forall

a,beR

30 4

LINEAR ALGEBRA (Chapter 1) Two teachers are comparing the grades their students have scored in recent exams. Last year Keith’s students obtained 9 As, 12 Bs, and 7 Cs. This year his students obtained 8 As, 14 Bs, and 5 Cs. Tatiana’s students obtained 12 As, 6 Bs, and 13 Cs last year.

This year they obtained 9 As, 9 Bs, and 10 Cs. a

Write the results of Keith’s students in a

b

Write the results of Tatiana’s students in a 3 x 2

¢

Find

d

Explain the significance of the matrices found in c.

5

a

Let

K+ T

and

A_(0

-3

6

matrix T.

2

4),

B_(3

2

—4

71),

0

and

C—0

forall

and

A™ exists provided A is square and n € Z* (AB)C = A(BC) provided A, B, and C are of appropriate order. {associative law}

ne€R

a, b, ceR {associative law}

Note that in general, A(kB) = k(AB) # kBA. We can change the order in which we multiply by a scalar, but we cannot reverse the order in which we multiply matrices.

eErrs0°0°0°0%0%0 Expand and simplify:

a

b (A — 2B)?

(2A +1)?

a

(2A+1)2

(2A +1)(2A +1)

(2A +1)2A + (2A + 1)1 = 4A2% 1 21IA + 2A1 + I2 =4A% +2A +2A +1

{X? = XX

by definition}

{distributive law} {distributive law}

{AI=1A=A

and P =1}

‘We cannot simplify b further, since in general BA # AB.

= 4A% 4 4A +1 b

( A —2B)? (A —2B)(A — 2B) (A — 2B)A + (A — 2B)(—2B) = A% — 2BA — 2AB + 4B?

{X? = XX

by definition}

{distributive law} {distributive law}

EXERCISE 1D.3 1

Expand and simplify where possible:

a

X(2X+1)

e B-—A)B+A)

b (31 + B)B

f (A2

¢ D(D?+3D + 2I)

g

(51 — 2B)?

d (A + h (A+

B)(C — D)

38

LINEAR ALGEBRA (Chapter 1)

Suppose

A% =2A — I.

kA + I

where

k,l € Z.

Write A® and A* in the form

A% = A x A?

A=A x A3

=A(2A 1)

= A(3A —2I)

=2A% — AI

= 3A2% — 2AI

=202A—-1) - A =4A -2l — A =3A -2 2

Suppose

a 3

+ 2I.

Write in the form

A?

b

A(2A +3I)

where

b

k, [ € Z:

c A8 A? = I.

Simplify:

(A-1)?

¢

A(A + 51)2

Show using counter-examples that the following are not true in general:

a

A2=0

=

A=0

b

5

Find all 2 x 2 matrices A for which

6

Find the error in the argument:

= 7

kA + [I

At

Suppose A is a matrix with the property

a 4

A% =3A

=3(2A—-1)—2A =6A — 31— 2A =4A - 31

=

A2=A = A=0orl

A? = A. A% =2A

A’-2A=0

AA-21)=0 = A=0or2l

Explain why the binomial expansion for real numbers can also be used to expand

n € 7+,

but cannot be used to expand

(A + B)".

(A + kI)",

LINEAR ALGEBRA

N

(Chapter 1)

MATRIXTRANSPOSE

The transpose AT of matrix A is the matrix obtained by writing the rows of A as the columns of A

For example, if

2

A =

matrix, then AT = (a;;) 1

(0

31

4

),

then

AT=1[1

isan n x m 2

o

isan m xn

4

=W

If A= (a;)

matrix.

The square matrix A is:

o

symmetric if AT = A

o

skew-symmetric if AT = —A.

For example:

°

( ,3 1

PROPERTIES

_02)

(g

o

is symmetric

_21)

is skew-symmetric.

OF TRANSPOSE

Provided that the orders of the matrices are appropriate for the operations to be performed:

o

(ANT=A

o

(sA)T =sAT

e A+BT=AT+B"

and

(A—B)' =AT — BT

for any scalar s € R

e (AB)T = BTAT

Prove that

(AB)"T = BTAT

for matrices A, B of appropriate

order so that multiplications may be performed.

If

C

=

AB

then

Cij =

(AB)T = CT

where

If we let D = B'AT,

k E a,mbn]‘ n=1

k

cj; = Z: Qjnbni n=1

then

d;;

n

z

bniajn

Z

ajnbni

=i k

= =cj;

Since

¢j; =d;;

forall

(AB)T = BTAT

i, j, C' =D

39

40

LINEAR ALGEBRA

(Chapter 1)

EXERCISE 1E 1

For

A:(1

3 4

0

2) 1

a (ADT f (A—B)T 2

For

A=

b g 1 2

-1

4

4

B:(2

b

3), -1

find:

¢ AT + BT h (—2B)T

and

1

0 -2

1

(A+B)T AT — BT

01 1 3

a AB 3

and

-1 0

B =

2 3

4

(AB)T

d

1 2|,

0

5

(3A)7 i —2B”

3AT

find:

¢ ATB"

d

BTAT

Assuming the matrices are of suitable shape for the operations being performed, prove that:

a

(ANT=A

b

(A+B)=AT

a

Prove that

b

Generalise the result in a using

+ BT

c

(sA)T =sAT

(A1A2A3)T = AJASA Ay,

Ag,

Ag,

..., Ay,

Use mathematical induction to prove your generalisation

is true for all n € Z7. 5

e

For help with mathematical induction see the HL Core course.

Prove that:

a if A is symmetric, then AT is symmetric b

if A and B are symmetric, then

¢

if A and B are symmetric,

6

Prove that

7

Suppose A is a square matrix.

8

9

A = (a;;)

a

Show that

b

Is

A — A"

AAT,

A + B

AB is symmetric

is skew symmetric

ATA,

symmetric?

Give examples of 3 x 3

is symmetric

and


(

-1

3

Y

) =

< 10 )

-1

(10 -1)

An n x n system of linear equations can be written in the form AX = B where A is an n X n square matrix of coefficients, X is an n x 1 column matrix of variables, and B is an n x 1 column matrix of constants. Consider such a system of the form

AX = B.

If the square matrix A is invertible, then the system has a unique solution which can be found as follows:

AX =B

invertible if A=

{premultiplying by A™'}

. ATHAX)=A"'B (AT'A)X =A"'B IX=A"'B X=A"'B

The matrix A is

exists.

If the square matrix is not invertible, then the system does not have a unique solution. 3 €1 TR

Solve using matrices:

.

3z — 2y =10 { —z+3y=-1 .

3

(_1

In matrix form, the system is

-2 3

a7

) (y)

10

= (_1>,

A= (2 7)) ten A= @@ - (2D-7 .

(3

_17132

Premultiply by the inverse matrix on both sides.

50

LINEAR ALGEBRA

(Chapter 1)

EXERCISE 1G 1

Solve using matrices: s

2z +4y = —6

b

r—3y=13

br—y="7

2

3

4

5

br+2y=3

—3r—2y=>5

d

—2z+5y=4

—xr—3y=15

3z —2y =20

-3 =17 { 61123 _ 7

Consider the system a

Write the system in the form

b

Find

|A|.

.

AX =B

where

X = (z)

What does this tell us about the system?

Consider the system

2x —ky =4 w43y = 4

{

a

Write the system in the form

b

Find

|A|.

AX = B

where

X = (Z)

What does this tell us about the system?

.

.

.

Consider the matrix equation

(1

3

=2

5

4 ) X =

(_3

0

14).

a

Does this equation represent a system of linear equations? Explain your answer.

b

Find X using an inverse matrix.

Show that if X; and X, are solutions of AX = B of AX=B, forall teR.

then

X3 =¢X;+ (1 —1¢)Xo

is also a solution

Explain the significance of this result. 6

For what values of k does the system have a unique solution? r+2y—32=5

a

20 —y—2z2=28 kr+y+2z=14

2 —y—42=28 3r—ky+z=1

b

S —y+kz=-2

Example 20

T—y—z=2

In matrix form, the system is:

1

-1

-1

B

1 9

1 -1

3 -3

y | = z

7 -1

X

B,

The system has the form

x

1 -1

-1\

z

9

-3

vl=[1

z=0.6,

using matrix methods and a graphics calculator.

rc+y+3z="7 9z —y—3z2=-1

Solve the system

1

-1

y=-5.3,

3

A

'/

z=39

2

7 | =

-1

0.6

=53 3.9

GRAPHICS CALCULATOR INSTRUCTIONS

2

=

so X=A"'B

LINEAR ALGEBRA 7

(Chapter 1)

Use matrix methods and technology to solve: 3r+2y—z=14

a

r—y—2z=4

r—y+2z=-8 2c+3y—2=13

5r +y+2z=—6

b

3z —4dy —2=17 T+2y—2=23

r—y+3z=-23 Tx+y—4z Oz —y+4z=-9

e

T+ 3y — 5z =89

f

132z— 17y + 232z = —309 8

=62

1.3z +2.7y — 3.1z =8.2 2.8z — 0.9y +5.62 = 17.3 6.1z + 1.4y — 3.2z = —0.6

Describe the limitations of using matrix methods for solving systems of linear equations.

A rental company has three different makes of car for hire: P, Q,and R. These cars are located at yards A and B on either side of a city, or else are being rented. In total they have 150 cars. At yard A they have 20% of P, 40% of Q, and 30% of R, which is 46 cars in total. At yard B they have 40% of P, 20% of Q, and 50% of R, which is 54 cars in total. How many of each car type does the company have? Suppose the company has

z of P,

It has 150 cars in total, so

y of Q,

= +y+

and

2z =150

BOBS CoRAENTAL &

z of R. . . (1)

Yard A has 20% of P + 40% of Q + 30% of R, and this is 46 cars. 2

4

T 10Y

0%

T

3,

107

_=46

22 + 4y + 32 =460

. . (2)

Yard B has 40% of P + 20% ofQ + 50% of R, and this is 54 cars. 40% t 10y 2 + 15 5, =_254

-3

4z + 2y + 5z =540

We need to solve the system:

r+y+2z=150 2z + 4y + 3z = 460

[AI-T[E]

4x 4+ 2y + 5z = 540 In matrix form, we write:

11 1\ [z 150 2 4 3|y ] =460 4 2 5/\% 540 z

11

1\ /150

vl=[2 4 3 z 425

45

460 | = [ 55 540 50

the company has 45 of P, 55 of Q, and 50 of R.

{using technology}

51

52

LINEAR ALGEBRA

(Chapter 1)

Managers, clerks, and labourers are paid according to an industry award. Xenon employs 2 managers, 3 clerks, and 8 labourers with a total salary bill of €352 000. Xanda employs 1 manager, 5 clerks, and 4 labourers with a total salary bill of €274 000. Xylon employs 1 manager, 2 clerks, and 11 labourers with a total salary bill of €351 000. Let x, y, and z represent the salaries (in thousands of euros) for managers, clerks, and labourers respectively. a

Write the above information as a system of three equations.

b

Solve the system of equations.

¢

Determine the total salary bill for the company Xulu which employs 3 managers, 8 clerks, and 37 labourers according to the industry award.

A mixed nut company uses cashews, macadamias,

and Brazil nuts to make three gourmet mixes. The table alongside indicates the weight in hundreds of grams of each kind of nut required to make a

Cashews Macadamias

kilogram of mix.

Brazil nuts

1 kg of mix A cost $12.50 to produce, 1 kg of mix B costs $12.40, and 1 kg of mix C costs $11.70. a

Determine the cost per kilogram of each of the different kinds of nuts.

b

Hence, find the cost per kilogram to produce a mix containing 400 grams of cashews, 200 grams of macadamias, and 400 grams of Brazil nuts.

Susan and Elki opened a new business in 2007. Their annual profit was £160 000 in 2010, £198 000 in 2011, and £240000 in 2012. Based on this information, they believe that their annual profit can be predicted by the model

(&

P(t)=at+b+ P

pounds

where ¢ is the number of years after 2010. a

Find the values of a, b, and ¢ which fit the profits for 2010,

b

The profit in 2009 was £130000. Does this profit fit the model

¢

Susan and Elki believe their profit will continue to grow according to this model. Predict their profit in 2013 and 2015.

t=0 gives the Es

2011, and 2012.

in a?

@

k

N\

If Jan bought one orange, two apples, a pear, a cabbage, and a lettuce, the total cost would be $6.30. Two oranges, one apple, two pears, one cabbage, and one lettuce would cost a total of $6.70. One orange, two apples, three pears, one cabbage, and one lettuce would cost a total of $7.70. Two oranges, two apples, one pear, one cabbage, and three lettuces would cost a total of $9.80. Three oranges, three apples, five pears, two cabbages, and two lettuces would cost a total of $10.90. a

Write this information in the form AX = B where A is a quantities matrix, X is the cost per item column matrix, and B is the total costs column matrix.

b

Explain why X cannot be found from the given information.

¢

If the last line of information was replaced with “three oranges, one apple, two pears, two cabbages, and one lettuce cost a total of $9.20”, can the system be solved now? If so, what is the solution?

LINEAR ALGEBRA

(Chapter 1)

ACTIVITY Cryptography is the study of encoding and decoding messages. Cryptography was first developed for the military to send secret messages. Today it is also used to maintain privacy when information is transmitted on public communication services such as the internet. To send a coded message, it must first be encrypted into code called ciphertext. When the recipient wishes to read the message, the ciphertext must be deciphered. A simple method for encrypting messages is to use matrix addition or multiplication. The messages are then deciphered using either matrix subtraction or an inverse matrix. Suppose the letters of the alphabet are assigned integer values, with Z assigned 0 as shown below:

A|B|C|D|E|F|G|H|I 1123|456 | 7|89

JIK|L ]|10f11]12

N|JO|P|[Q|R|S|T|U|V|W|X|Y 14 | 15|16 [ 17 | 18 [ 19 [ 20 | 21 | 22| The word

SEND

.

can be written as the string of numbers

19

matrix form

23 | 24 | 25

19 5 14 4 which we can write in 2 x 2

5

(1 4

4).

Now suppose we encrypt the message by adding the matrix

( 2T

13

5

)

195+27_2112 14 4 13 5/ \27 9 Before this matrix can be transmitted all of its numbers must be written in modulo 26, or mod 26.

This means that any number not in the range 0 to 25 is adjusted to be in it by adding or subtracting multiples of 26. The matrix to be sent is therefore

the string says

ULAI,

21 ( 1

12 ), 9

and this is sent as the string

21 12 1 9.

By itself,

which has no apparent meaning.

The message SEND MONEY is then encoded.

PLEASE

could be broken into groups of four letters, and each group

SENDIMONEI|YPLEIASEE R,

*%

%

7%

R1*R2*>R1

3

-3

-1

L

3

-5

3 3

1 101

-2

31

14

1f-5

=A™

A=

0 |1 3

1

1003

102

for

R+

1|-%2

1|-5

00

A"l

1 0 0 O -3 1 0

001§ 00

~|lo

1 O 0 0 -3 1

0

~lo1o|d

A=

-1

R2R1

%

3

1 ' ENE]

1

3|0

0

0

0

10|

1

0

1

%

00

~l0

2|1 0 3|0 1 1|0 0

3(0 2|1 -6]1

4

(I|A™1).

A =W~}

= (ExEjp_y ... EoE)) 7}

0

(WA |W).

&

121

Ry —3R3 — R,

Check this result

using technology.

3

12

L

12

1

6

¥

58

LINEAR ALGEBRA (Chapter 1)

EXERCISE 1H State the elementary a

(4

%Rg

—

b

R3

f

a

1 00 010 00 3

d

1 0 0

Matrix

0 -2 0

A= |

Ry

R3+

%Rl

—

R3

¢

Ry

—2Ry

—

Ry

d

Ri1


Y 2

where

T

=

y

T+ 2y z Y

T

T+y ={,_ y

z

T : R® s R?>

is a linear transformation.

where

x

T

y

z

transformation. 4

Prove that if T : R™ +— R™

T(k1uy

5

6

+ kous

+ ... + kpuy) = k1 T(wr)

+ k2T(u2)

ot 7(( ) T:R*+—

R?

P

.

.

isalinear transformation where

ro (1))

=

Tz

(yz)

is not a linear

is a linear transformation, then

T:R?+— R? is a linear transformation where

2

. . . is a linear transformation.

T

+ ... + k. T(u,).

T(((l)))

= (é)

1

1

0

=13

and

and

T

T((?))

= (é)

1

-1

9

=\

_3)

86

LINEAR ALGEBRA

7

(Chapter1)

T:R?+— R? is alinear transformation where

wr((3)

8

T:R3— and

R3

is a linear transformation where 0

T

0 1

=11

1

0 0

=1

and

1

=(-1], 2

T( (?))

T

0

1

0

-3

=(

0

=1

2

3

4

T

a

3 0

b

T

b c

Determine which of these transformations are linear:

-

a

T:R?—R?

where

T((

b

T:R?—

where

T((z))

R?

¢

T:R¥®—R

d

T:R?>—R?

where

Y

))

y

T =|y—=z z2

= (z—y)

T

T|

[y

2

=(z+y—22)

z

T

_

T

[ty

\x—4z

N

where

e

8

9

a

T

5

1

Find:

T((l))

2

KERNEL AND RANGE For the linear transformation

T : R™ — R™:

o o

the domain of T is R" the co-domain of T is R™

o

the kernel (or null space) of T, denoted

ker(T),

is the set of all vectors u in the domain of T

such that T(u) =0 e

the range of T, denoted R(T), w = T(v) for some v € R™. domain R”.

is the set of all vectors w in the co-domain of T such that

co-domain R™

LINEAR ALGEBRA

(Chapter 1)

Theorem on kernel and range: If T:R"™+— R™

Proof:

is a linear transformation, then:

e

ker(T)

is a subspace of the domain

e

R(T)

is a subspace of the co-domain

For

ker(T)

(1)

ker(T)

is non-empty as T(0) =0

(2)

Forall

uy, us € ker(T),

=

R™

and

R™.

O € ker(T).

T(u; +uz) = T(u;) + T(uz) =0+0

{addition property} {uy, uy € ker(T)}

=0

=

Thus

(3)

ker(T)

u; + us € ker(T)

is closed under vector addition.

For u € ker(T),

keR,

T(ku)=kT(u)

{scalar multiplication property}

=k0

{u € ker(T)}

=0

Thus

ker(T)

= ku € ker(T) is closed under scalar multiplication.

From (1), (2), and (3),

For

R(T)

(1)

R(T)

(2)

Forall

ker(T)

is a subspace of the domain

is non-empty as T(0) =0 wy, wo € R(T),

w1 +wp =

Thus (3)

Forall

Thus

R(T)

W;

=

R™.

0 € R(T).

=T(vq) +T(v2)

forsome

=T(v1 +v2)

{addition property}

+ Wg

vy, vo € R"

€ fl(T)

is closed under vector addition.

we R(T),

R(T)

keR,

kw=£kT(v) forsome veR" =T(kv) {scalar multiplication property} = kw e R(T)

is closed under scalar multiplication.

From (1), (2), and (3),

R(T)

If T:R™—

is a subspace of the co-domain

R™

e

nullity (T)

e

rank (T)

R™.

is a linear transformation, then:

is the dimension of ker(T), is the dimension of R(T).

and

87

Consider T:R2—R?

where

ker(T)

Y

T((*))=|2+y|. Y

R(T)

Find:

z—y

nullity (T) Consider

o -{0)

z=y=0

0 . m)hn{(l), =z|1|+y| 1

0

1

1

1 -1

( )} -1

rank (T) =2

nullity (T) =0

Consider T :R?— R?

where

ker(T)

T((g))

R(T)


+dz +ey+

The points A, Py, B, P, lie on a

.

and P; PPy

Py

PB

k 7 1 was given in Apollonius’ circle theorem.

75P

A

—=3 =

v

b

‘

PA

P2

line,

is a right angle.

a

Prove that

b

Hence,

f =0

APy _ APy

BP;

BPy’

a = (.

prove

circle theorem.

the converse

of Apollonius’

178

GEOMETRY

(Chapter 2)

REM FOR ATERALS If a quadrilateral is cyclic, then the sum of the products of the lengths of the two pairs of opposite sides is equal to the product of the diagonals. AB.CD + BC.DA = AC.BD

Proof:

We first draw [AH], 61 = 05 as shown. Now in As ABH

e

0 =0y

e

a3 =as

where

H

lies on

[DB]

such

that

and ACD:

{construction}

{angles subtended by the same arc}

The triangles are equiangular and therefore similar.

ABAC _ CDBH

BH=2BD AC Also, in As ADH

e e

and ACB:

AHD=a+6

{exterior angle of AABH}

and

65 =6,

AHD = ABC = o + ¢

(3, =[,

{angles subtended by the same arc}

The triangles are equiangular and therefore similar.

HDBC _ ACDA

HD_BCDA

o

AC

Using (1) and (2), BD = BH+ HD — A'f\'SD + Bi'CDA . Hence,

(1

BD=

AB.CD

+ BC.DA

AC

AB.CD + BC.DA = AC.BD

{angles subtended by the same arc}

GEOMETRY

(Chapter 2)

179

Find AC given that [BD] has length 12 cm.

AB.CD + BC.DA = AC.BD L

{Ptolemy’s theorem}

6x9+5x10=ACx12 . 104 =AC x 12

. AC=8%cm

THE CONVERSE FOR PTOLEMY’S THEOREM FOR CYCLIC QUADRILATERALS If the product of the lengths of the diagonals of a quadrilateral equals the sum of the products of the lengths of its pairs of opposite sides, then the quadrilateral is a cyclic quadrilateral. If AB.CD

B

A

+ BC.DA

= AC.BD

then ABCD is a cyclic quadrilateral.

Proof:

Let CAD — a, ADB = 3, and BAC = 4. Choose a point H on the same side of AC

BAH = o and ACH = 3. Now

CAH=DAB=a+60

and

ACH = ADB = 8 As ACH and ADB

CH AC _ AH DB

AD

——

Rearranging (1),

CH =

AC.DB

=

()

as B, such that

() — @

AB

are equiangular and therefore similar.

180

GEOMETRY

(Chapter 2)

Given (2) and that BAH = DAC = a, in As BAH and DAC we have two sides in the same ratio and the included angle between them equal. As BAH and DAC are similar.

BHCD _ ABAD

BH _ ABCD AD

@

Using (3) and (4),

CH—BH

AC.DB

=

AD

AB.CD AD

_ AC.BD — AB.CD AD _ BC.AD

{by the assumption of the converse}

=BC BC + BH = CH,

which can only be true if B, C, and H are collinear.

- ACB=2 .

ACB = ADB

ABCD

are equal angles subtended by [AB].

is a cyclic quadrilateral.

s N

In the given quadrilateral ABCD,

BD = 8.2 cm.

LR 53 cm

Is ABCD

AC = 10.5 cm

and

a cyclic quadrilateral?

3.4 cm

D

9.2 cm

AB.CD + BC.DA =74x%x92+4+53x%x34 =86.1

and

AC.BD =10.5x8.2 = 86.1

Hence, by the converse of Ptolemy’s theorem, ABCD is a cyclic quadrilateral.

EXERCISE 2) 1

The side lengths of a cyclic quadrilateral, in clockwise order, are 6 cm, 9 cm, 7 cm, and 11 cm. If one diagonal is approximately 12.0 cm long, find the length of the other diagonal.

2

Three consecutive sides of a cyclic quadrilateral have lengths 6 cm, 5 cm, and 11 cm. Its diagonals have approximate lengths 10.1 cm and 9.54 cm. Find the length of the fourth side of the cyclic quadrilateral.

GEOMETRY

3

181

In each figure, determine whether PQRS is a cyclic quadrilateral: a

P

7Tcm

b

— Q

9em

S

L

(Chapter 2)

72mm

Q

11cm

13cm

110 mm

R

Consider a cyclic quadrilateral ABCD with the dimensions given. Diagonal [AC] has length m, and diagonal [BD] has length n. a

Write down an equation connecting the variables.

b

Suppose we construct rectangles on sides [AB], [BC], and [AC], with widths y, 2z, and n respectively. What can be deduced about the shaded areas?

¢

Suppose ABCD is a rectangle. What formula does Ptolemy’s theorem

give in this case?

5

a

Use the given figure and the Cosine Rule to deduce

b

If the other diagonal has length y units, show that

that

22 — (ac + bd)(ab + cd) (bc + ad) :

5> _

o

¢

6

B

A

(ac+bd)(ad + bc)

ab+cd

’

Hence, prove Ptolemy’s theorem.

[AC] is a diameter of a circle with centre O and radius 1 unit.

C

BAC = a and DAC = s. Use Ptolemy’s theorem to prove the addition sin(a + ) = sinacos 8 + cos asin .

formula

182

GEOMETRY

(Chapter 2)

[7% IENTHEGREMS OF CEVA AND MENELAUS We have seen previously that:

e

three or more lines are concurrent intersect at a common point

if they

e

three or more points are collinear if one straight line passes through all of them.

\‘\B‘\C‘\ A

Points A, B, and C are collinear.

These lines are concurrent at P.

The converse theorems of Ceva respectively.

and Menelaus

enable us to establish concurrency and collinearity,

CEVA'S THEOREM Any three concurrent lines drawn from the vertices of a triangle divide the sides (produced if necessary) so that the product of their respective ratios is unity.

A

A

Z B

Y

AZ BX CY ZB'XC'YA

B

C

Nz NxA Ny may help you to

A

write down the correct ratios.

2

X ©

Proof of Ceva’s theorem:

We use the theorem that if two triangles have the same base, then the ratio of their areas is the same as the ratio of their altitudes. In AABC,

[AX], [BY], and [CZ] intersect at O.

We draw altitudes [BP] for AAOB

and [CQ] for AAOC.

In As BXP and CXQ:

e BPX = CQX e

BXP = CXQ

{vertically opposite angles}

The triangles are equiangular and BX

BP

area of AAOB

CX

CQ

areaof AAOC

..

similar. !

@

{as As have common base [AO]}

GEOMETRY .

.

Likewise,

hvg

YRR AY

f AB

arca 0l =HOG

e

area of ABOA

Multiplying (1), (2), and (3) gives 3 CT T

(2)

AZ

and

/DO

7B XC YA

f AA

o2 = LB o BZ

area of ABOC

areaofABOC

e

(Chapter 2)

183

(3)

X 2wa of AROR

X Ac=a of ABOC _

_arcaof AAOC

_a.:ea—eme’il

IERT

A

P

B

P divides [AB] in the ratio 2:1 and Q divides [BC] in the ratio 3: 7.

Q

Find the ratio in which R divides [CA].

C

P divides [AB] in the ratio

Q divides [BC] in the ratio

But

2P BQCR_,

3:7

=

=

AP:PB=2:1

AP

BQ 3 &=7

2 1

If P divides [AB] in the ratio r : s then AP:PB=r:s.

{Ceva’s theorem}

PB QC RA

%

S

212 %l

2=X =X 3 CR 177

2:1

I

C

6

R divides [CA] in the ratio

7 : 6.

THE CONVERSE OF CEVA'S THEOREM If three lines are drawn from the vertices of a triangle to cut the opposite sides (or sides produced) such that the product of their respective ratios is unity, then the three lines are concurrent. Proof:

Let [BY] and [CZ] meet at O.

Suppose [AO] produced meets [BC] at point X’.

But

BX' AZ = CY 22 X/C YA ZB BX CY AZ —.—.—=1 XC YA 'ZB BX' _ BX X'C ~ XC

{Ceva’s) theorem}

. {given}

X and X’ coincide {as B, X, X/, and C are collinear} [AX], [BY], and [CZ] are concurrent.

184

GEOMETRY

(Chapter 2)

Example 17 LMN is a triangle. X is on [LM], Y is on [MN], and Z is on [NL]. NZ=5cm, YN=6cm, ZL =2cm, and XM =5 cm. Prove that [MZ], [NX], and [LY] are concurrent.

[MZ], [NX], and [LY] are concurrent.

{converse of Ceva’s theorem}

Let the medians

respectively.

of AABC

be

[AP],

[BQ],

and

[CR]

Ry O S RB PC QA =

[AP], [BQ], and [CR] are concurrent.

{converse of Ceva’s theorem}

EXERCISE 2K.1 1

A

T

B

T divides [AB] in the ratio 3: 7. S divides [BC] in the ratio 5: 3. Find the ratio in which R divides [AC].

C

2

In AABC, D lies on [BC] such that BD = 1BC.

A

E lies on [AC] such that CE = 2CA. }

E

[BE] and [AD] intersect at O, and [CO] produced meets [AB] at F. Find:

F

; B

D

C

a

AF:FB

b

area of AAOB

: area of ABOC.

GEOMETRY 3

C

In the diagram,

Y

7

X L

BZ : ZC =2:1

and

(Chapter2)

AY : YC =3:2.

a

Find the ratio in which X divides [AB].

b

Find the ratio in which S divides [AZ].

B

P, Q, and R lie on sides [AB], [BC], and [CA] of triangle ABC respectively, such that

BQ = 2BC,

185

and

CR = 1CA.

AP = %AB,

Prove that [AQ], [BR], and [CP] are concurrent.

5

Use the converse of Ceva’s theorem to prove that the angle bisectors of a triangle are concurrent.

6

The inscribed circle of triangle PQR has tangents [QR], [RP], and [PQ] which touch the circle at A, B, and C, respectively. Prove that [PA], [QB], and [RC] are concurrent.

7

Use the converse of Ceva’s theorem to prove that the altitudes from the three vertices of a triangle

are concurrent.

MENELAUS’ THEOREM So far, when we have considered ratios of lengths of line segments, we R i have used the length or magr}ltude only.. To state Menelaus. tl?eorem, we need to use sensed magnitudes, which means that a ratio is taken to be positive or negative depending on whether the line segments are written as vectors with the same direction.

We only use sensed magnitudes L —, ey

For example: °

C B /é//

AB

.

.

Bc 18 positive, since AB and BC have the same direction.

/CA//V .

.

-

=

.

.

.

.

Bc 8 negative, since AB and BC are opposite in direction.

If a transversal is drawn to cut the sides of a triangle (produced if necessary), then using sensed magnitudes, the product of the ratios of alternate segments is minus one.

transversal

186

GEOMETRY

Proof:

(Chapter 2)

(for Case 1)

We draw perpendiculars from A, B, and C to the transversal. As BQX and CRX are similar

B

=

CX

_

BX

£

CR

BQ

XC ~ CR As CYR and AYP are similar

CIEECR

=

AY

cY

YA As BQZ and APZ are similar

You should show that this proof also

CR

PA

A

=

BQ BZ AZ _ PA

holds for Case 2.

ZB

BQ

Q

=-1

Example

AP

19

P divides [AB] in the ratio [AC] in the ratio 5: 2.

2: 3,

In what ratio does R divide [BC]?

AQ _5 QC

2

Co_-ac_gc_ (a0) AQ \qQcC QA

—-AQ

R divides [BC] externally in the ratio

15 : 4.

i

}

BR

Since —

does

RC not lie

< 0, R

on [BC],

but rather on [BC] produced.

and Q divides

GEOMETRY

(Chapter2)

187

THE CONVERSE OF MENELAUS’ THEOREM If three points on two sides of a triangle and the other side produced (or on all three sides produced) are such that the product of the ratios of alternate segments is equal to minus one, then the three points are collinear.

Az i BXCY XC'YA'ZB

A

transversal

5

then X, Y, and Z are collinear.

B

Proof:

(for the illustrated case)

Let XYZ' be a straight line.

A

. transversal

........

|

BX

CY

AZ

XC

YA

Z'B

—.—.——

But

!

= —1

{Menelaus’ theorem}

BXOYAZ_

XC'YA'ZB

_______ X

L !

7'B

ZB

{A, 7', Z, and B are collinear} 7' and Z coincide. X, Y, and Z are collinear. 3 €1 IR 1]

In a triangle two angles are bisected internally, and the third angle is bisected externally. Prove that the points where the angle bisectors meet the triangle’s sides are collinear. A

Let the triangle bisectors at A and respectively. Let by [CZ] where Z

J

B A

be ABC, and the internal angle B meet [BC] and [AC] at X and Y’ the external angle at C be bisected lies on [AB] produced.

By the angle bisector theorem, as [AX] bisects BAC, AB

@

@

AC

BX

e

XC

=

BX

AB

CX

AC

===

.. (1)

Likewise, as [BY] bisects AfiC,

BA_AY _ CY_ BC BC CY AY BA CY

-

BC

@

Also, as [CZ] bisects the external angle,

_CA_AZ CB BZ

_

=

AZ_CA

7B

CB

AZ AC 22_2% ZB BC

0

188

GEOMETRY

(Chapter 2)

From (1), (2), and (3),

AB

—

BC

AC

G

=-1 =

X, Y, and Z are collinear.

{converse of Menelaus’ theorem}

EXERCISE 2K.2 1

Transversal XYZ of triangle ABC cuts [BC], [CA], and [AB] produced, at X, Y, and Z respectively. If BX: XC=3:5 and AY : YC = 2: 1, find the ratio in which Z divides [AB].

2

A

Prove Menelaus’ theorem by constructing [AW] to [XB], to meet the transversal at W. Hint: Look for similar triangles.

B

X

C

3

ABC is a triangle in which D divides [BC] in the ratio Find the ratio in which [BE] divides [AD].

4

In the figure alongside, P and Q are the midpoints of sides [AB] and [AC] respectively. R is the midpoint of [PQ]. [BR] produced meets [AC] at S, and [AR] produced meets [BC] at M.

5

a

Show that M is the midpoint of [BC].

b

Find the ratio in which S divides [AC].

¢

Find the ratio in which R divides [BS].

Common

drawn

for

external

the

tangents

three

pairs

parallel

are

of

illustrated circles. The circles have different radii a, b, and ¢ units. Use the converse of Menelaus’ theorem to prove that X, Y, and Z are collinear. DEMO

2 : 3, and E divides [CA] in the ratio

5 : 4.

GEOMETRY

(Chapter 2)

189

A, B, and C lie on a circle. The

tangents

at

A,

B,

and

C

meet

[CB] produced, [AB] produced, and [AC] produced at D, E, and F respectively.

7

Consider two lines.

One line contains the

distinct points A, B, and C. The other line contains the distinct points D, E, and F.

Suppose [AE] and [BD] meet at X, [AF] and [CD] meet at Y, and [BF] and [CE] meet at Z. Pappus of Alexandria discovered that X, Y, and Z are always collinear. Prove Pappus’ theorem.

Hint:

Prove that DB : DC = AB? : AC?.

b

Prove that D, E, and F are collinear.

F

E

D

a

Produce [EA] and [FB] to meet at G. Let [DC] intersect [GF] at 1. Apply Menelaus’ theorem to each of the five transversals of triangle GHI. GEOMETRY PACKAGE

190

GEOMETRY

II

(Chapter 2)

THE EQUATION OF A Locus

A locus is a set of points satisfying a particular equation, relation, or set of conditions. The plural of locus is loci.

If P(z,y) represents any point in a locus, the Cartesian equation connecting = and y is the equation of the locus. Example 21

Consider the locus of all points which are equidistant from a

Find the equation of the locus.

a

b

A(—1, 0)

and

B(5, 4).

Describe the locus.

AP = BP, . AP? = BP?

@)+ A2+ 1+

= (2 -5+ (y—4)° =2 — 102+ 25+ 4% — 8y + 16

122 + 8y — 40

A(-1,0)

3z + 2y = 10

b

The locus is the line bisector of [AB]. Check:

.

3z + 2y = 10,

q

_

4-0

.

.

gradient of perpendicular is

The midpoint of [AB] is

P y)

which is the perpendicular

The gradient of [AB] = 5o

.

/

4

53

2

3

—3.

4 967 + 144 = 9y° + 967 + 256

162 2 +L 72 7y

=

=112

(or

7)2

2

LY T +¥%

=1

EXERCISE 2L 1

2

Find the distance from:

a

(3,2)

¢

(2,-1)

to y=3z—2

b

(-1,4)

d

(-1,-3)

b

ar+by+c; =0

to 4z —3y =141 to mex+y=>5

meR.

Find the distance between the parallel lines: a

3

to 2z+5y+6=0

3x+2y=>5and

3x+2y+1=0

Find the value of k if:

a

the distance from

b

A(1, —2)

(k, —3)

to 3z —2y+6 =0

is equidistant from

z+y =%k

and

is /13 units x —y+7=0.

and

ax+by+c2=0

GEOMETRY

4

Find the equation of the locus of all points which are parallel to the line

(Chapter 2)

=z —y

= 4

193

and are

24/2 units from it.

5

Find the locus of all points

P(z, y)

6

Suppose

Bis (3, 0).

A is (—3,0)

and

which are equidistant from

N(—1, 8)

Find the locus of all points

and

P(z, y)

S(5, 4). such that APBisa

right angle.

7

8

Find the Cartesian equation of the locus of P(z, y) a

P is the same distance from

b

P is equidistant from the lines

a

A(—1,0) i

ii b

and

B(3,0)

(2, 1)

if:

asitis from

3z —4y =3

are given points.

2z —y =25

and

5z — 12y = 4.

P(x, y)

moves such that

%

=2.

Find the Cartesian equation of the locus of P.

Describe the locus of P. Give reasons for your answer. AP

Repeat a for the case where

3

%

e

]

The distance from the point P(z, y) to A(—1, 3) is a half of the distance from P to the line 4 2y = 7. Find the Cartesian equation of the locus of P.

AP = 12 x Ps distance from z +2y—7=0

@7+ (9P = 3 et

|z+2y—7]|

4oy

(x+1)2+(y—3)%= W 2022

+2c+14+y2 — 6y +9)

—

72

{squaring both sides}

=2 + 4% 4+ 49 + 4oy — 28y — 14z {(a+b+c)?=a?+b%+c%+ 2ab+ 2be + 2ac}

202 + 20y? + 40z — 120y + 200 = x2 + 4y? + 49 + 4zy — 28y — 14x 1922 — 4zy + 16y + 5dx — 92y + 151 =0 9

10

Find the Cartesian equation of the locus of R(z, y) a

equal to its distance from the line

b

half its distance from the line

¢

1.5 times its distance from

Suppose

that:

A is (2,0)

a AQ+BQ=6

and

if R’s distance from

A(3, 0)

is:

= = —3

= = 12

4 z = 3.

Bis (—2, 0). Find the Cartesian equation of the locus of Q(z, y)

b AQ — BQ=2.

such

194

GEOMETRY

(Chapter 2)

CIRCLES A circle is the set of all points which are equidistant from a point called its centre.

THE CENTRE-RADIUS FORM OF THE EQUATION OF A CIRCLE The equation of a circle with centre

(h, k)

and radius r is

(x—h)?2+(y—k)2=r2 The proof is a simple application of the distance formula.

Example 26

Find the equation of a circle with centre

The equation is

and radius /7 units.

(z —2)? + (y — —3)* =

whichis

(. —2)*+ (y+3)?

If we expand and simplify we obtain

(2, —3)

(x—2)%+(y+3)?2= 22 — 4z

This equation is of the form

2

4 4 + 32 + 6y

+y?

+ dx + ey

In fact, the equation of any circle can be put into this form. The general form of the equation of a circle is

2 + y2 +dr+ey+

f=0.

We are often given equations in general form and need to find the centre and radius of the circle. We can do this by ‘completing the square’ for both the x and y terms.

For the equation of a circle to be in general form, the coefficients

of 22 and y? should both be 1.

GEOMETRY €.

(Chapter2)

195

108F)

Find the centre and radius of the circle with equation

22 + y? + 6x — 2y — 6 = 0.

22+ +6x—2y—6=0

22 + 6z +y2 -2y =6 2?4+ 62 +3%+y? —2y+12=6+32+12

(3P4

(y—1)2=16=42

the circle has centre €1

{completing the squares}

(—3, 1)

and radius 4 units.

10801]

The point (m, 2) lies on the circle with equation Find the possible values of m.

Since

(m, 2)

lies on the circle,

(z —2)% + (y — 5)? = 25.

(m —2)%+ (2 —5)? =25

S

(m—22+9=25

EXERCISE 2M.1 1

Find the centre and radius of the circle with equation:

a 2

3

4

(-2

+((y-3)2=4

22+ (y+3)2=9

¢

(z—-2%4+y2=7

Write down the equation of the circle with:

a

centre

(2, 3)

and radius 5 units

¢

centre

(4, —1)

and radius v/3 units

b

centre

(—2,4)

d

centre

(—3, —1)

and radius 1 unit and radius v/11 units.

Find, in centre-radius form, the equation of the circle with the properties:

a

centre

(3, —2)

and touching the z-axis

¢

centre

(5,3)

d

(—2,3)

and

e

radius /7 and concentric with

and passing through (6, 1)

b

centre

(—4, 3)

and touching the y-axis

(4, —1)

are end-points ofa diameter

(z 4 3)%+ (y — 2)% = 5.

Describe what the following equations represent on the Cartesian plane:

a @422+ @y-72=5 5

b

b

(z+22+(y-772=0

Consider the shaded region inside the circle, centre (h, k), radius 7 units.

a

Let

P(z,y)

Show that b

be any point inside the circle.

(z — h)? + (y — k)% < r%

What region is defined by the inequality (x—h)2+ (y — k)2 > r??

¢ (z+224+@Yy-72=-5

196

6

GEOMETRY

(Chapter 2)

Without sketching the circle with equation

(z + 2)2 4 (y — 3)? = 25,

determine whether the

following points lie on the circle, inside the circle, or outside the circle:

a A(2,0)

7

8

10

¢ D(3,0)

d E(4,1)

Find m given that:

a

(3, m)

b

(m, —2)

¢

(3, —1)

lies on the circle with equation (z +1)% + (y —2)? =25 lies on the circle with equation lies on the circle with equation

(z + 2)? + (y — 3)? = 36 = 53.

(z + 4)? + (y +m)?

Find the centre and radius of the circle with equation:

a ¢ e 9

b B(L 1)

2?2 +y?+6x—-2y—3=0 22+ +4y—1=0 22+9y? -4z —-6y—3=0

b 22 +y>—6x—2=0 d 2?2 +y? +4r-8y+3=0 f 22+y?—-8x=0

Find £ given that: a

22 +y?—

122 +8y+ k=0

b

224 y?+ 62 — 4y =k

¢

2?2 +y?+4r—2y+ k=0

is a circle with radius 4 units

is a circle with radius /11 units represents a circle.

1In general form, a circle has equation

22+ y? + dz + ey + f = 0.

a

. . Show that its centre is

b

Hence, find the centre and radius of the circle with equation

¢

Comment on the locus with equation z? + y? +dz +ey+ f =0 i

d 4f.

3x2 + 3y? + 6z — 9y +2 = 0.

in the case:

d?+e? — 8z —4

=

0

at the

—8x—4=0

+y2 =4 cox? 8z Lox?—8r 44yt =4442 (-4

4+ =20

the circle has centre

(4, 0).

—2-0_ -2 4 The gradient of [CP] is =74 "2 8—-4

P(8,-2)

the gradient of the tangent is %

..

the equation of the tangent is

2z —y = 2(8) — (—2)

whichis

[3'€1

2z —y = 18.

110

Find the equations of the tangents from the external point

224+ 9% — 10z — 2y + 16 = 0.

22 +1y? — 10z — 2y + 16 =0 -

2% — 102

22—

+y?—2

=-16

10z + 52 + 42 — 2y + 12

s

(@=5)2+@-12%=10

which is a circle with centre

(5, 1)

16 + 5% 4+ 12

and radius v/10 units.

Let m be the gradient of a tangent from P.

it has equation

But

(0, —4)

y = max + ¢ for some constant c.

lies on the tangent, so ¢ = —4

the equation is

y = mxz —4

whichis

maz —y —4=0.

The centre of the circle is v/10 units from each tangent.

Im®) 4] _ V10

21

o

{point to a line formula}

+ 1) =+/10(m2?

5m =5

25m?% — 50m + 25 = 10m? 4 10 15m? —50m 4+ 15=10 3m?2—10m+3=0

(m=3)(8m—1)=0 ..

the tangents are

m=3

or

Wl

o

o

y =3z —4

and

y = %x —4.

P(0, —4)

to the circle with equation

198

GEOMETRY

(Chapter 2)

EXERCISE 2M.2 1

2

Find the equation of the tangent to the circle with equation:

a

22+ y?+6z—10y+17=0

b

22+ y?+ 6y =16

at the point P(—2, 1)

at the point P(0, 2).

The boundary of a circular pond is defined by the equation

2?2 +y? — 242— 16y + 111 = 0.

A straight path meets the edge of the lake at grid reference

A(3, 4). a

Given that the grid units are metres, find the diameter

b

Find the equation of the straight path.

of the circular pond.

(2, 3) and radius 4 units. P(8, 7)

A

3

A circle has centre

4

Find the equations of the two tangents from the origin O to the circle with centre radius 2 units.

of the two tangents from P to the circle.

is external to the circle. Find the equations

A circle has centre

(3, 4).

has equation y = 3.

(4, 3)

and

One tangent from the origin O

Find the equation of the other tangent.

6

7

A circle with centre

(3, —2)

has a tangent with equation

a

Find the equation of the circle.

b

Find the tangent’s point of contact with the circle.

Consider the circle a

a tangent

x? 4 3? — 42 + 2y = 0. b

3z — 4y + 8 = 0.

Find the value(s) of k for which

a secant

¢

C(r, 0)

3z +4y =k

is:

an external line.

is the centre of a fixed circle with radius r.

A is a point which is free to move on the circle, and M is

the midpoint of [OA]. a

Find the Cartesian equation of the locus of M.

b

Describe the locus of M.

GEOMETRY

(Chapter 2)

199

Line segment [AB] has fixed length p units. A can only move on the z-axis, and B can only move on the y-axis. M is the midpoint of [AB].

10

Suppose

a

Find the Cartesian equation of the locus of M.

b

Describe the locus of M.

A is (1, 0), Bis (5, 0), and k is a constant.

P(z, y)

is a point such that

g =k

for

all positions of P. Find the equation and nature of the locus of P if:

a k=3 1

Suppose

b k=3 A is (2, 0)

and

B is (6, 0).

The point

¢ k=1 P(z, y)

moves such that

%

= 2

for all

positions of P. a

Deduce that P lies on a circle, and find the circle’s centre and radius.

b

The circle in a cuts the z-axis at points Py and Py, where Py is to the right of P;. coordinates of P; and Ps.

Deduce the

AP AP, _ AP, . — = —= BP BP; BP,

¢

Show that

d

Hence, deduce that [PP] bisects AfiB, and [PP5] bisects the exterior angle A§B, for all positions of P.

200

GEOMETRY

(Chapter 2)

CINcowicsicrions Consider a right-circular cone, which means the apex is directly above the centre of the base. Suppose you have a second identical cone which you place upside-down on the first. Now suppose the cones are infinitely tall.

We call the resulting shape a double inverted right-circular cone. When a double inverted right-circular cone is cut by a plane, 7 possible intersections may result. 1

a point when the plane meets the double-cone where the apexes touch,

2

and at no other points

4

a circle when the plane is perpendicular to the axis of symmetry, and not through X axis of symmetry

a line when the plane is tangential to the double-cone, for

3

example (AB)

5

X

an ellipse when the double-cone is cut such that o > 0 axis of symmetry

a line-pair when the plane contains an axis of symmetry of the

double-cone

6

a parabola when the double-cone is cut such that a =3 axis of symmetry

GEOMETRY

7

an hyperbola when the

(Chapter 2) DEMO

1,2, and 3 are called degenerate conics and 4 to 7 are called the

double-cone is cut such that a < 3.

non-degenerate conics. axis of symmetry

FOCUS-DIRECTRIX DEFINITION OF AN ELLIPSE, HYPERBOLA, AND PARABOLA Suppose P(z,y) moves in the plane such that its distance from a fixed point F (called the focus) is a constant ratio e of its distance to a fixed line (called the directrix). The locus of P is a conic which is o

anellipseif

e

aparabolaif

e

an hyperbola if e > 1.

0 0}

the equation of a rectangular hyperbola can be z? — y? = a? 2 or y? — 2% 2 = a?. 2

Since

0% =a%(e*—1),

e*—1=1 e=+2

{since e >0}

So, every rectangular hyperbola has eccentricity v/2.

In the HL Core course, we saw rectangular hyperbolae with equations of the form

xy = k

where k is

a constant.

In fact,

xT

— a

2

Y

2

— =; =1 a

.

under a rotation of

T

+7 T

becomes

zy =

a 2

5

where a is a constant. These are

the only rectangular hyperbolae for which y can be written as a function of x.

Sketch

422 — 9y = 36

=

by finding the axes intercepts and asymptotes.

::3

the graph cuts the z-axis at (3, 0) ) 42 22

The asymptotes are

ol

and

which are

(-3, 0). )

y = 5z

.

212

GEOMETRY

(Chapter 2)

3 €T

An hyperbola has foci

(42, 0)

and directrices

@ = £3.

At what points does it cut the axes?

The foci lie on the z-axis and the centre is (0, 0). Now

ae=2 a? =2

and

& = e

1 x 5=1

{as

a=1

a >0}

the hyperbola cuts the z-axis at (£1, 0) but does not cut the y-axis.

Example 39 Each circle in a set touches the z-axis, and the y-axis cuts off a chord of length 2 cm from each circle. Find the nature of the locus of the centres of all such circles.

Let the centre of one of the circles be

C(z, y),

and

let M be the midpoint of [AB].

Now

CM =z

Loy=z%+1

and

CA =y

the locus of C is

{radius of circle}

{Pythagoras}

> — 2> = 1

equation of a rectangular hyperbola

which is the

all centres lie on a rectangular hyperbola.

EXERCISE 2N.4 For each hyperbola:

iv

2

Find the axes intercepts. Find the foci and the equations of the corresponding directrices. Find the equations of the asymptotes. Sketch the graph.

a

25z% — 16y> = 400

b

4y —22=16

¢

22—y

d

y2-22=9

=4

Find the equation of the hyperbola with the following properties: a

vertices

¢

foci

(£4,0),

(+12,0),

e=

1%

directrices

z

wleo

1

b

centre O,

d

vertices

y-intercept —2, (::%, 0),

directrix

directrices

x

y :g

2 V3

|[PF — PF/| = 2, foci

3

4

6

y = +2x,

f foci (0, +£2), directrices y = 4

g

asymptotes

h

transverse axis 4 units long on y-axis,

vertices

213

(£4, 0) ¢ = %

For any hyperbola:

a

Prove that

b

Show that b can be interpreted as the shortest distance from a focus to an asymptote.

|PF — PF/| = 2a.

Find the equations of the tangent and normal to 422 —9y? = 36

Use implicit differentiation!

at the point:

a 5

(£3,0)

(Chapter 2) wiloo

GEOMETRY

(30

b

(3v2 —2)

Find the equation of the tangent and normal to z? —y% =9 the points on the curve where

Consider the hyperbola the hyperbola.

2

at

x = 5.

2

z_2 — 2_2 = 1. a

Suppose

P(z1, y1)

lies on

a

Show that the equation of the normal to the curve at P is a*y1x + b2z1y = (a® + V)39

b

Find the equation of the tangent to the curve at P.

¢

Suppose [PT] is a tangent to the curve and T lies on the asymptote with positive gradient. (b:vl +ayr s bxy +ay1)

Show that T has coordinates

7

a

y

Each circle in a set touches the z-axis, and the y-axis cuts off a chord of length 4 cm from each circle. Find the

T

nature of the locus of the centres of all such circles.

4cm

AN

TRANSLATING CONICS In general, if a conic is translated Under the translation

o

(Z),

we replace z by

= —h,

and y by

y — k

in its equation.

(Z)

the parabola y? = 4az

becomes

(y — k)%= 4da(xz— h)

v _ 1 becomes e the ellipse ? + z= 2

? ST

Eg

e

the hyperbola

o

the rectangular hyperbola

1 becomes

(@=m? -

2

S

(=k? o _ 1

@=hn> _ (-k? @ = 1

xy = c?> becomes

(x — h)(y — k) = c2.

214

GEOMETRY

(Chapter 2)

If a conic has an equation which can be put into one of these forms, we can then sketch it and find details of any foci and directrices. Example 40

@+2?

Sketch the ellipse _1)2

(z +2)2 9

i

-1

4

=1

Now

a? =9

and

Since

b% = a*(1 — €?),

T

-1)? =1

comes from

%

2

and give details of its foci and directrices. 2

S yI =1

)

under the translation

(

»? =4.

1—e?=

e’ = e= Now

{e >0}

ae=+/5

2

and

2

% 4 yI =1 .

.

(2+2?

< = €

has foci

and directrices

Hence

a

(+/5, 0)

3 x = £—=. V5

(-1?

+

4

=1

has foci

(—2++/5, 1) and directrices = = —2 4

EXERCISE 2N.5 1

2

3

4

Sketch each conic, giving details of any foci and directrices:

@=12 —

@+3? L1_

b (4P =-8+2)

¢ (@+22- W=

Find the Cartesian equations of the ellipse with:

a

foci (—5,2)

b

focus

(—3,4)

¢

foci

(—1, —3)

and (1, 2), and eccentricity% with corresponding axis extremity and

(5, —3),

and one directrix

(—5, 4),

and eccentricity £

= = 13.

Find the equation of the hyperbola with: a

centre

(2, —1),

b

focus

(2, 3)

¢

foci

(2, —2)

focus

(1, —1),

with corresponding directrix

and

(6, —2),

Consider the curve with equation

a

and eccentricity 2

Write the equation in the form

where

= = —1,

and eccentricity 2

|[PF—PF/|=2.

zy — 2x + 3y — 10 = 0.

(x — h)(y — k) = 2.

b

Identify the curve and sketch its graph.

¢

Check the position of the graph by finding the axes intercepts from the original equation.

GEOMETRY

5

6

Consider the curve with equation

Write the equation in the form

b

Identify the curve and sketch its graph.

¢

Find the axes intercepts from the original equation.

d

Find the coordinates of any foci and the equations of any directrices.

(y — k)% = 4a(z — h).

For each conic:

ii iii

7

y? — 8z + 6y + 22 = 0.

a

i

Write the equation in a suitable form so that the curve can be identified and graphed. Sketch the graph of the curve. Find the coordinates of any foci and the equations of any directrices.

a

22 +4y> —6x+32y+69=0

b

42? —9y? + 162+ 18y =9

Explain why

(Chapter 2)

322 +y? — 62 — 4y +40 =0

does not have a graph.

Complete the square YA CLE anianlcy

215

216

GEOMETRY

(Chapter 2)

LEIEEPARAMETRIC EQUATIONS Parametric equations are equations where both = and y are expressed in terms of another variable called the parameter. The parameter takes all real values unless otherwise specified, and often represents an angle 6, or a time ¢.

For example,

© =2cosf,

for the circle

y =2sinf

x? +y? = 4.

is a parametric representation

The parameter is # where

0 € R

is

the angle measured anticlockwise from the positive z-axis to the point P(z, y).

The identity

)

x =

notice that

However,

2sinf,

2cosf

y =

would

cos20 +sin?60

is very useful.

also be a

suitable parametric representation for this locus. In this case # would have a different meaning.

=1

There may be infinitely many parametric representations for the one Cartesian equation. For example, for the line with equation r=t

y=8—t

or

and so on.

y=t,

x +y

= 8

we could use

t=8—t or x=1—t,

'

y="7+1t,

Example 41

Find the Cartesian equation of the curve with parametric equations:

a

z=sinf,

y=2cosf 2 t

, y=2—3t

where

b

z =sinf —cosf,

y=-sin26

b

We use the identities cos? and sin 260 = 2sinf cos 6.

t#0.

Now

+ sin?6 = 1

z? = (sinf — cos6)> =sin?6 — 2sinf cos @ + cos? 0 =1-—sin26

=1—y y=1

¢

Since

z—1=

%

and

y — 2 = —3¢,

we can eliminate ¢ by multiplying.

(z-1(y-2)= (%) (—3t) = —6 Ty

—2z—y+2+6=0 zy—2z—y+8=0

—z?

or alternatively,

GEOMETRY

(Chapter 2)

Find a suitable parametric representation for:

a

zy=

Ty

a

If

b

=5

=1t

y

y2=1

¢

Oz

5

then

y=—,t9é0.

b

If y=t

then

27 ¥ 5 B=

t2

2

C

w__y_zl

4

9

In cases such as a and b there are several sensible answers.

() - ( - (3) - () 8wl

2

But

sec?f =tan?6 +1

we let

gzsece

z =2secl,

for all 6.

and

%:tanfi

y=3tan6

are the parametric equations.

PARAMETRIC DIFFERENTIATION Consider a curve with parametric equations .

.

d;

Using the Chain Rule,

A

dt

dy

= = g(t),

y = h(t).

d.

dx at

W) =g —= = —

or

——

is the gradient of the tangent at any point with parameter ¢ on the curve.

A curve has parametric equations

x =t? — ¢, y =2t — 3.

Find the equation of the tangent to the curve which has gradient %

Bdt _9t—1 and Ydt =2 dy

dr .

2

2t—1

.

s

Since the gradient of the tangent is £,

2

o

3

= When

¢t =3,

(6, 3)

=6

and

y =3

1is the point of contact.

Thus, the equation of the tangent is which is

2z — 5y = 2(6) — 5(3) 2z — by = —3.

217

GEOMETRY

218

(Chapter 2)

EXERCISE 20 Find the Cartesian equation of the curve with parametric equations:

a

r=t,

y=%

b

z=t

y=1-5t

d

z=t

y=t>-1

e

z=1t2 y=13

c

x=1+4+2t,

f

=12

y=3—1

y=4t

Find the Cartesian equation of the curve with parametric equations:

a

x=2cosf,

d

z=sinf,

y=3sinf y=cos26

b

=2+cosf,

e

xz=tanf,

y=sinf

y=2sech

¢

x=cosf,

y=cos20

f

z=cosf,

y=sin20

y? =9z

Find a suitable parametric representation for:

a

z+4y=5

b

zy=-8

¢

d

22 +y2=9

e

4z’ +y* =16

foa?=—dy

g

2 2 =15 — 3z°+5y°

h

2 r_ 1

1+9v

2

2 2 i|16Z_Y_ 9

1

Consider the curve represented by z = 2t2, y = t. a b

Where does the line

=+ y =3

meet the curve?

Check your answer by first converting the parametric equations into Cartesian form.

Find the equation of the tangent to:

a

x=3t

y=t>—-3t

¢

z=secl,

y=tanf

at t =2

b

z=2cosf,

b

z=1-t

y=>5sinf

at §=1%

at 0= F.

Find the equation of the tangent to:

a

x=1—1t2

y=4t

with gradient 4

Find the coordinates of the points where the line

equations

= = 1+sinf,

y=1—

A curve has parametric equations

cosé.

x =t + %,

y=t>

= + 2y = 3

y=1—

%,

the Cartesian equation of the curve

b

the equation of the normal to the curve at the point where

(1, 0).

meets the curve with parametric

t # 0.

a

passing through

Find:

¢ = 2.

The illustrated ellipse has equation

2

2

% + 311_6 =1.

Ais

(0,4).

Find the nature of the locus of the midpoints of all chords from A to the ellipse. Hint: Write the coordinates of B in terms of parameter 6. Then write the coordinates of M in terms of 6.

GEOMETRY

(Chapter 2)

I3 PARAMETRIC EQUATIONS FOR CONICS

219

The standard parametric equations for the non-degenerate conics are:

x =acosf,

o

x=at?

y=asinf

for the circle

y=2at

for the parabola

y? = dax y?

22

.

.

e

x =acosf,

y=>bsin@

for the ellipse

= S5 z

e

z=ct,

%,

for the rectangular hyperbola

zy = c

e

x =asecl,

y=

t#0

y=>btan6

for the hyperbola

(acosb, asinf) (acosf,

bsind)

=

1

-—==1

The parameter 6 is called the eccentric angle. z°

5

4+ y*

5

= a”

5

. . . is called the auxiliary circle.

SUMMARY OF TANGENTS AND NORMALS TO CONICS In the following Exercise we will prove these equations of tangents and normals for the non-degenerate conics:

Circle

2?1y’ =a

2

at (acosd, asinf) Parabola y? = 4daz at (at?, 2at)

Tangent

Normal

(cosf)z + (sinf)y = a

y = (tanf)z

z —ty = —at?

tr +y = at’ + 2at

(bcosO)z + (asin@)y = ab | (asinf)z — (bcos)y = (a® — b?)sinf cos Rectangular hyperbola t2x—y=ct3—§

alp

Hyperbola

a

b2

at (asec6, btan@)

bz — (asinf)y = abcosf

(asin@)z + by = (a® + b%) tan

220

GEOMETRY

(Chapter 2)

Example 44

Find the equation of the normal to

dz _ asecOtanf do

2

2

2—2 = Z—z =1

at the point

(asec6, btan@).

and

bsec? 0

asectanf _

bsech atanf

E(

1

a

)(cosQ)

\cos@

sin6

b asinf

the gradient of the normal is

", the equation of the normal is

_asind

(asin @)z + by = (asin®)(asecd) + b(btanh) = (J,QSil’l9
=80

b

822+ 287y — 13y +40 =0

in standard form.

Hence, identify the conic and sketch its graph.

2

3

Write each conic in the form

A(2’ — h)? + B(y' — k)? + f = 0.

conic.

a ¢

2?—ay+y*-20+y—-3=0 322 —6zy —5y? + 3z + 9y = 10

b d

a

Use questions 1 and 2 to complete the following table: a+c | Ay and Ao | A+

“

b

15

5, 10

15

Hence, identify and sketch the

22 +dxy — 29> + 252 — By —5=0 222 —dzy+ 5y +dr—2y =1 Ao | Sign of M2 | Tipe of conic

>0

ellipse

Use the table to make conjectures dealing with the eigenvectors A\; and As. ABC is a set square which is right angled at A. [AB] and [AC] have lengths 2 units and 1 unit respectively. The set square is free to move so that A always lies on the z-axis and B on the y-axis. [AC] makes an angle 6 with the z-axis as shown.

a

Find the coordinates of C in terms of 6.

b

Find the Cartesian equation of the locus of C.

¢

Find possible transformed equations for the locus of C. Use your observations from question 3 to identify the

d

conic. Give reasons for your answer.

GEOMETRY

(Chapter 2)

229

INVESTIGATION 5 For the conic section with equation

ax? + 2bxy + cy? + dz + ey + f = 0,

the value

b* — ac

is called the discriminant. In this Investigation we will observe how the value of the discriminant allows us to quickly identify the conic type.

What to do: 1

2

We have seen that the conic section with equation

ax? + 2bzy + cy? +dx +ey+ f =0

can

be written in the form

xTAx + vIx + f = 0

where

and

v =

det A

(j)

Show that

Suppose the conic

x =

(x)’ Y

A=

(a

b

b),

¢

is the negative of the discriminant of the conic.

az? + 2bxy + cy® +dz +ey+ f =0

a'z' 2+2b'z'y +c'y 2+ d'x’ +e'y’ + f =0

is rotated through angle 6 to produce

such that the constant f is equal in both equations.

Show that:

3

a

the trace of matrix A is conserved, so

b

the determinant of matrix A is conserved, so a'c’ — b 2 = ac — b?

¢

the discriminant of the conics are the same, so b’ 2 — a/c’ = b* — ac.

Consider the following proof that if line-pair: Suppose

> —ac

> 0

az? + 2bxy + cy? +dz +ey+ f =0

dr'2+ 2y

+ Y 2+ dd

We choose 6 so that

Using2¢,

a’ +c

+ ey + f=0.

=a+c

then the conic is either a hyperbola or a is rotated through angle 6 to obtain

b = 0.

b?—ac=02—adc =-dcd —a'd >0 a/ and ¢’ are opposite in sign

The rotated conic is

1

! (z

... (1)

o'z’ 2 +cy' 2 +d'a’ +e'y +f=0 2! ’ 2

o

d'

d, } CL’z

This equation is of the form

’ 2

;Y

1 } a’

2

yo

€’

ad

y,

€ c’y

Tl f a'cd

i(z’ —h)2+ fi(y' -k)?=C

which is a hyperbola if C' # 0 or a line-pair if C' = 0.

{using (1)}

Prove that:

a

if b2 —ac < 0 then the conic is an ellipse, a circle, a point, or else has no graph

b

if > —ac =0 graph.

then the conic is a parabola, a line, a pair of parallel lines, or else has no

230

WORKED

Worked

Solving the first two equations, not satisfy the third equation the system is inconsistent.

Solutions

1

S————

a

It contains a product of two variables,

b

It contains the squared term

a

Llet

z3z4.

by =7

z1 +z2

—

3

=7

r] —x2

+223

=9

and

The system is

—,/Zz.

bs =9

21 + x2 — 23 = a

2wy —x2

x=t

It is homogeneous if

8t—y=3

y=28t—3 the solution setis

b

Let

zo =5 x1

and

= =t,

z3 =t

the solution set is where s, t € R.

Let

1

=28 —t+

za =71,

1

xz3=3s,

and

10,

x2 = s,

x3 = s, 3

a

b

wq =t,

z1+2x2=3 z0 = —4 221 4+ mp = —1 z1+x2+

x4

x2

a

b ¢ 5

a b

¢ 6 7

7,5, t€R.

—t

—2,

z9

=

underspecified as it has more unknowns (4) than equations (2).

x4 =3

This system is neither underspecified nor

=8 2y =2

The system has AM

The system has AM

overspecified as it has the same number of equations and unknowns. 1 1

1 -1

2

(1) 11 0o 3 0o 0

The system has AM No solutions exist when parallel).

—11]4 118

a #

a € R

2

|

2

)

1|3

7y = -7

The lines are neither coincident nor parallel they intersect in exactly one point. the system has a unique solution. The lines are not coincident, but they have the same gradient they are parallel and never meet. the system has no solutions. If a =4,

..

the lines are coincident.

the system has infinitely many solutions.

If a #4, a €R, the lines are parallel. the system has no solutions.

The system has AM

-3|0

4

712 ‘ Z)

(1

—3|-8

-1 1 4

12,

0

—-1]|5 1|1 —116

a € R a = 12

= —3(—1) =2

The second equation is a multiple of the first. the lines are coincident and have infinitely many points of intersection. the system has infinitely many solutions.

-3|-8

Infinitely many solutions exist when coincident).

There is no value of exists.

-3

2 |-7

1

1

(;

b+8=0 b=-8.

r=-1 the unique solution is =z = —1, y = —1.

7,

This system is

= T3+ x4 =

23

4

—r+2s

223+ 224 =4

(1

Substituting into row 1,

This system is overspecified as it has more equations (3) than unknowns (2).

2¢1 + @2+ 3w3 —

¢

where

and and

cy=-1

Ty =-r+25s—t—2 =

3] 7

Using row 2,

x4 =t

z3

(1 0o

z3 = t,

1 +r—2s+t=-2

the solution setis

a =0 a=0

The system has AM

10

= 2s —t+

+a23=b+8

EXERCISE 1B.1 W

t € R.

—2s+t=10

1

¢

y=8t—3,

equations than

the system is not homogeneous.

—2w32.

¢ It contains the square root term 2

with

231

= 3, y = 0 but this does

The system is overspecified as it has more unknowns. The system is

EXERCISE 1A

SOLUTIONS

5

17|

Using row 2,

19

51

(the lines are

for which exactly one solution

The system is consistent if it has at least one solution k=4 Adding the first two equations gives 2z +y + 3z =p+ the third equation is 2z +y + 3z =7

¢ and

the system is consistent if p+ g = 1.

(If p+q # r, the two planes are parallel and no solutions would exist.)

— Ry

17y =51

y=3

(the lines are

Ry — 4Ry

Substituting into row 1,

= — 3(3)

the unique solutionis

=z =1,

=1 y = 3.

The system has AM 1 2

7| -1

1 0

7 —15|

Using row 2,

—17 11 —-17 45

Ry — 2Ry

—

Ry

—15y =45 cLooy=-3

Substituting into row 1,

=+ 7(—3) = —17 z =4

the unique solutionis

=z =4,

y = —3.

232

WORKED

SOLUTIONS

The system has AM

2 1

3-8 4|-9

(2 0

3| 5|

5

-8 -10

Using row 2,

a

(1

2R2 — R1 — R2

5y = —10

b

coy=-2

Substituting into row 1,

2z + 3(—2) = —8 ¢

=-2

L2

r=-1

the unique solution is The system has AM (3 4

- ( 3 0

= —1,

y = —2.

6

3Ry — 4R, — R

Substituting into row 1,

3z — (—3) =9

the unique solution is = =2,

y = —3.

represent coincident lines, which points. has infinitely many solutions. AM

meet

=4

_ 4(a—6)—3(b—8) a—6

the solution setis

4-s

(s',

y=1t,

t€R.

3

b#8,

=4

—3t,

0x+ 0y # 0

y=1t,

t € R.

which is not possible.

the lines are parallel and the system has no solutions. EXERCISE 1B.2

y =

= =s,

4—s

,

1

sER.

a

.

y =

4—s 3

74—5'

3

!

3)

z=4—-4+5s ’

NS

The system has AM

be a point in the second solution set.

_

a—6

.

teR.

the solution setis

4=

In the first solution set, when

2:473

The second equation is a multiple of the first, so we obtain a

Let

which

= —5t=38

1

If27> a#6,

at

which

tER.

The second row gives

=2

(1 0

Ry —2R; — R

When a # 16, the second row gives Oz + Oy # 0 is not possible. there are no solutions; the lines are parallel. When a = 16, the second row gives Oz + Oy = 0 is true for all z, y € R. .. there are infinitely many solutions to the system.

2

3r =6

2

8 |a—16

The system has AM

Cooy=-3

1

0

-5 0

r=>5+8 the solution setis « =5t+8,

13y = —39

The equations infinitely many the system The system has

5|8 —10|a

Usingrow 1,

-1 | 9 ) 13| -39

Using row 2,

1 2

Let y=t,

9 ) |-1

—-11 3

The system has AM

~lo

1 1 1

4 6 4

11(7 179 814

1

4

11

0

0

-3|-3

2

Using row 3,

6|2

7

R3 — R1 —

R3

—3z = —3 soz=1

Substituting into row 2,

2y + 6(1) =2 2y=—4

the y-coordinates are the same, the z-coordinates

are also the same. any point in the second solution set is also in the first solution sct. the solution sets are equivalent.

Ry — Ry — Ry

Sy

Substituting into row 1, the unique solutionis

=-2

=+ 4(—2)+11(1) =7 z =4,

y=

r=4 -2, z=1.

WORKED b

e

The system has AM

~(0

2 2 3

-1 -2 2

2

-1

17

0

-1 7

2

—1

0

0

—61 [ —122

Using row 3,

—61z = —122

~[0

-8|-13 -5|-31 3

Ry —Ri — Ry 2R3 — 3R1 — R3

Ry —3R; — Ry Rs —5R; — R3

R3 + 7Rz

102 ~|lo0o -4 0o 0

-1

R3 — 2R3 — R3

17

-8 |

—13

— R3

f The system has AM

—y —8(2) = —13

2

Sooy=-3

Substituting into row 1,

2z — (—3) +3(2) = 17 2c =38 =4

the unique solutionis ¢

z =4,

y=

-3,

~|0

z=2.

The system has AM 2 5 8 2

0

3 6 9

~|0

4|1 7|2 10|4 3

-3

-3

-6

2

3

4

0 0

-3 0

2Ry —5R1

0

1 0

0

-6

1

-2

0o

3 0

4

0

5 -2

4|0 5

1

—22

a

-9(35

0

2424

LRy + Ry — Ry

24z =24

—22y — 9(1) = 35 22y = —44 =-2

2z +4(—2)+

(1) =1 2 =8

|1 10

Ry —2R1

R3 +3R1

~(0

— Ra

— R3

~0

—210 010

R3 +2R2

— R3

b ¢

3y —2t=20

2

2 1

-1 7

1

2

0 1 0

1

4|1 1|k

y=

-5

2|

5

2 -5 0

3

1

3

-5

Rz —2R1 — Ro

—2|k-3

Rz — Ry — R3

1 3 2| =5 O0|k-38

R3 + R2 — R3

Using row 3, the system has no solutions if &k # 8. The system has infinitely many solutions if the last row is all zeros. This occurs when k = 8. In this case we let z =t. using row 2, —5y+ 2t = —5

5y =2t+5

y=35t

y:§t+1

= — 2(%1&) +5t=1

Using row 1,

m:l—ls—lt

z =

1 — %t.

Yy =

x+2(%t+l)+t:3 m:lfgt

t,

wo

the solutions have the form

=z =4,

=4 -2, z=1.

The system has AM 1

3y =2t

z=t

1

the unique solutionis

1

Substituting into row 1,

2Ry — 3Ry — Ra 2R3 — 5R1 — R3

R3 — Rz — R3

Row 3 indicates there are infinitely many solutions. Let z=1. Using row 2,

2

1 1 -9|35 9 |-3

Substituting into row 1,

5 1 8 2 —11 | =3

-2 3

0

4 —22 6

Sy

1

—6|-1 0 1

-2 -1 0

2

=3[19 7 (1

R3 — 4Ry — R3

The system has AM 1 2 -3

=5 13

1

Substituting into row 2,

— R2

Row 3 means that Oz + Oy + 0z = 1 which is absurd. there are no solutions, and the system is inconsistent. d

3 5

1

z=1

1

—6|-1

0

4

Using row 3,

4

-1 4 4]|-5 0|1

Row 3 means that Oz + Oy + 0z = 1 which is absurd. there are no solutions, and the system is inconsistent.

z=2

Substituting into row 2,

233

The system has AM

12 -1 4 3.2 1|7 5 2 3|11 102 -1 4 ~lo0o -4 4|-5 0 -8 8 |-9

3 17 5|4 2 10

3

SOLUTIONS

the solutions have the form

teR.

z=t,

d

= =1

—

%t,

Y=

%t-f—l,

teR.

In row echelon From b and ¢ infinitely many the system

form, row 3 reads Oz + Oy + 0z = k — 8. the system has no solutions if &k # 8 or solutions if k = 8. never has a unique solution.

234

WORKED a

The system has AM

1

2

1

~10

0

|-k Ry — Ry R3 — R1

— —

5 —6

Ro R3

5t = —6

3y=5t+6 y:§t+2

Using row 1,

x:lfgt

¢ When

=z =1

— %t,

y=

%t + 2,

teR.

k # 13,

row 3 gives

(k— 13)z = —(k — 13) sooz=-1

Substituting into row 2,

—3y +5(—1) = —6

T=z la+2 the solutions have the form

z=1

EXERCISE 1B.3

k#13, a

a

~0

3 -1 -5

1

3

xo

0 1

0

=

—9,

-7 —-14

-5 9—2a a—-9|19-3a

3 -7

3 -5

¢

The basic variables are x1, z2,

variable.

z=—1

0

Ty = —5t—11

v

5t — 11 z+3(T)+3t:(—1)—1 —15t — 33 4

7

21t

[ T =

za=1t

x1+2t=05

—2t x7 =5

;] =5—2t —2t,

xo = 4 — 2t,

tER.

The system has AM

~[0

x9

3 1 4

1 -1 -2

1

0

0]

-1 1| 1

0

0

1|-6

1

b

=3,

12 -8 -8

1

0of

3

{using technology}

3

~[0

0o 1 11 0o 1

0 0 -1

1

0

0

2|5

0O

0

1

1(6

1

7

the solutions have the form 19 — 6t —5t — 11 = ,y=————, z=t, teR 7

0

2| 4 4|9 1|-2

2|4

{using technology}

The basic variables are x1, x2,

variable. Let

and x3, and x4

is the free

x4 = t.

r3+t=6 19 — 6t

1,

= —6.

Using row 3,

—14

z; =

The system has AM

—7y — 5t =9 —2(—1)

_—st—11

ot

xo =4

x3=6—t

z = ¢.

Substituting into row 1,

=4

the solutions have the form

The system has infinitely many solutions if the last row is all zeros. This occurs when a = —1.

using row 2,

zo +2t

x3=6—1

Rs — 2R> — R3

is the free

Using row 1,

By inspection, the system has the unique solution

a+1|a+1

In this case we let

Using row 2,

x3+t=26

forall

and x3, and z4

x4 = t.

Using row 3,

Ry —2R1 — R» R3 —3R; — R3

a—1 | 9—2a

= 3.

By inspection, the system has no solutions as row 3 means 0z 4+ Oy + 0z = 1 which is absurd.

3|la-—-1 1 7 a 16 a—1

x3

x; = 2,

b

a

3

NS

By inspection, the system has the unique solution

keR.

1 2 3

ac€R.

Is in reduced row echelon form.

Let

The system has AM

~[o0

b

%,

%a -2,

b

x + 2(%) —2(-1)=5 o oz=1 3 y=

y=

Not in reduced row echelon form as the row of all zeros is not at the bottom.

y=3% the unique solution is « = %,

a# -1,

z = %a +2,

a

3y=1

Substituting into row 1,

= +3(2a—2)+3(1)=a—1

¢ Not in reduced row echelon form as the leading 1 in row 3 should have zeros above it in column 4.

x+2(§t+2)—2t:5

the solutions have the form

Substituting into row 1,

Rs—3R.— Ry

The system has infinitely many solutions if the last row is all zeros. This occurs when &k = 13. In this case we let z =t. —3y+

y = %a -2

:c+$a—6+3:a—1

k—13|-k+13)

using row 2,

—7y —5(1) =9 — 2a Ty =2a—14

—2 5

0

z=t,

(a+1)z=a+1

Substituting into row 2,

—2 5 5 —6 k+2|—-k-5

2 -3

0

row3gives

z=1

-1

k

2 -3 -9

1 0

~|

3

-7

1

¢ If a# —1,

=215

-1

1

b

SOLUTIONS

3

Using row 2, zo +2t

=6—1

=4

xo =4

the solutions have the form x3=6—t

za=1t

Using row 1, x1+2t

—2t x; =5 —2t,

tER.

=5

x1 =5—2t xo = 4 —2t,

WORKED ¢

The system has AM 1 1 3

2 -1 3

6

9

10

~10

1

-3

0

0

0

1]

1

1

the solutions have the form 12:7%7%5,

7 -35 32

k=1

o

~10

8

23

=s

115 1

and

Using row 2,

x4

: technology} {using

b

~|l0

-2

3 2

-3 2

2 —1

1

0

0

0

0O

0

1

-2

1

0

321 4+ 2x0 + 23 =01

When

w4 =1,

where

s, t € R. ¢

©4 =s

and

1

Using row 3,

x3 —2s

Using row 2, Usingrow

1,

_

= —3t

=1

3

1 1

1

2

3

0

00

d

0O

1]3 5

3

1

0

0

0 0

0O o0

3z

1|46

L75

5

1

5

1

2

0

0]0.1

0|0

0

2714 0|53

0

of

0f—-H 4169 =T 1

0

0

1

0

0

0

4

{using technology}

3311

O

2714

2

{using technology}

_

4 4748 d2 4748

2711d+ 4169

2=

2173154 dB

27111

_ 4748

45 > T3

TTqg

_ 4169

s T4 T Tip -

when d =2, h(2 )z 37.2

at the point 2 km from the ocean the height of the hill is about 37.2 m above sea level.

1

2|5

41

h(d)

EXERCISE 1B.4

3|7

1

050

0

1|2 1|3

3

2

101 ~l0

2 -1

7T

TL="T35>

f The system has AM 1 -1

... (4

By inspection, the system has the unique solution

the solutions have the form z7 =1, xo = —3t, x3 =28, x4 =8, x5 =t, where s, t € R. 1 1

s

2

01

~

z2+3t=0 z1

1f12

1

= 28

©oxp

1

{using technology}

=10

x3

1

125

x5 = t.

+5za+23=0

From (1), (2), (3), and (4), the system has AM

3

0fO0

h/(d) =0

Do

The basic variables are x1, 22, and x3, and the free variables are x4 and xs5. Let

d =25,

... (3)

321(2.5)% + 222(2.5) + 3 = 0

913 6 |2

3f0

h/(d)=0.1

321(1)% + 222(1) + @3 = 0.1

z; = % + %s + 5t,

oOf1

.. (2)

with respect to d.

il When d=1,

3|1

-4 2

0

h(d)

=545 =g+ 38 + 5t

r3 =s,

1

... (1)

h'(d) = 3z1d® + 2z2d + x3

The system has AM

1

h=12

h(d) = 21d® + x2d? + z3d + 24

z1 — %5 — 5t = %

the solutions have the form

1

d=1,

%

1

e

7, s, t € R.

function

Ty = % — %s —t

IQ:%*%S*L

T5 =1,

i The gradient of the hill is the rate of change of height over distance. This is modelled by the derivative function R'(d), which is found by differentiating the height

= t.

x2 + %s +t=

Using row 1,

T4 =8,

When d = 2.5, h =46 z1(2.5)% + z2(2. 5)2 + x-g(? 5) + a4 = 46 o0y + 2ay + 2ag+as =46

The basic variables are x1 and x2, and the free variables are x3 and x4. Let

=71,

1 +x2+xz+xa=12

|3 5

5|2

When

r3

xlzg—r—és—t,

21(1)% + 22(1)? + 23(1) + 24 = 12

=}

1 =}

10

a

—-4]1 2 |2

o

13

where 4

The system has AM

1

3s+t=3 zlzgfrfésft

which is absurd.

-1 3

a1 +7+

{using technology}

0

0z1 + O0z2 + Oxz3 = 1

1 1 17

T2+ 58=—3

:czz—%—gs

By inspection, the system has no solutions as row 3 means d

2.

Using row 1,

0

0

0

Using row

9

Li1

235

The basic variables are x1 and x2, and the free variables are x3, x4, and 5.

3 4 4 7 10 | 15

19|

SOLUTIONS

a

b

NS

The second equation is a multiple of the first. So the lines are coincident, and there are infinitely many solutions.

the system has non-trivial solutions. The system is underspecified as it has more unknowns than equations. it has infinitely many solutions the system has non-trivial solutions.

236 2

WORKED a

SOLUTIONS

The system has AM

1 2

3 -1

~ (é Let

-—-1]0 5|0

(1)

x3

b

z =2,

Now

if z =21

zog —t =0 xo

1,

~[0

1

1 1

0 -1

1

0

0

z1 = —2¢t,

+ 2,

{using technology}

1]0

1

4x1 adA=|4%x6 4x5

the only solution is the trivial solution =0.

b

1 2 3

-1 1 -1

1 -1 2

110 =210 110

1

00

—%]o0

0

10

7§

001 Let

c

0

{using technology}

Using row 2,

zg+§t:0 znggt

tER.

p—2 1

1

0 0)

p—2

0 1-(p—2)?

(1-(-2%y=0

if (p—2)2#1,

y=0

So for non-trivial solutions

z1 = %t,

f

then and

y=y1

2

a

isasolution of

agz1

... (2)

Now if =cxzy; a1z + b1y

=0

and

= ay(cw1) + bi(cyr)

= c(ar@1 + b1y1) =¢(0) {from (1)} =0

y = cy1 and

ajxz+ by =0

asx + bay =0

2002y

then asx + bay

= az(cx1) + ba(cy1)

= c(azz1 + bay1) =¢(0) {from (2)} =0

y=cyr

-3 5

-3

2t

7

-10

-7

2

3

3

2

1

isasolution for all

¢ € R.

e

-6

10 | = 4 2 3 4

—2 |=|-9 -5

-3

1

-1

6

0 5

13 8 0 2 -2

1

5

,%

%

0

1

2

3

A+B

b

kA + kB

= (kai;) + (kbij)

= (aij + bij)

= (kai; + kbij)

= (bij) + (aiz)

=k(A+B)

B—A

= k(aij + bij)

d

(a+b)A

= (bij — aij) = (—(aij — bij))

= ((a+b)ai;) = (aas; + baij)

= —(as; — biz) =—(A—B)

= (aas;) + —aA +bA

A+A+A+

-4 -10 -4

4

=B+A
*+(au)

=(0)

+ aig) = ((maig) =(=A)+

EXERCISE 1C.2 I 1 a3A+4A=TA ¢ 2M - 2M =0 e

3(A+B)—B

=3A+3B-B —3A+2B

= (aij — aij)

1

Xl— 3B+ 3C

(-0 C=E3X+(-C)=A-1C+

o

s —8X+0=A-3C

=-2(a-£0)

R

A+ (—A) = (ai;) + (—ai;)

5

fC-3X=A-3C

A+0=0+A=4 and

—

.

=A

—0+A

A+ (—A)

3X — 3B =2B 4 C 3X - 3B+3B=2B+C+3B 3X+0=5B+C 1(3X) = 1(5B+C)

= (aiz)

= (0) + (aij)

b

3(X-B)=2B+C

A+0=(a;;)+(0) = (aij +0)

= (0+ai)

A +(=A)

X =4C - 2A

matrix is O = (0)

and

X=9C-3B

1

2(3X) =2(2C — A)

—A+(B+C) A+0=(aj;)+(0) = (aij +0)

o

_le

s 3Xt0=20-4A

= (aiz) + (bij + cij) m X n

X=4(C-B)

A)=2C IX+A (7)_ :"5+1+

= (aij + bij + cij)

a The zero

. .

3 o)

1Y) (-1 —4>*%

-1

74) _3

2 -6

237

2(A+B) — (A—B)

bB-X=C

14 21

=A-2A-C =-A-C

z =y

8

7

h

iA-2D-i(D-A)=A-2D1D+ 1A

2

y=x,

and

A—(2A+C)

and 4=y +7

y =0

9

aK=|[12

g

z)

2y==x,

y=—-y=2y

)

SOLUTIONS

238

WORKED

SOLUTIONS

EXERCISE 1D.1 B 1

Ais

1x3

and

Bis

‘(1

2

2x3

#

3

a b ¢

ABexists if n=m. If AB exists, it has order 3 x 2. The number of columns of B (2) does not equal the number of rows of A (3). BA cannot be found.

a

Ais

2x2

and

Bis

Z

6

Bis

1x2

and

Ais

2x2

(6 5)(% %) (6x3+5x1

= (23 4

a

Ais

BAis

Bis

1

3x1

r

and v

2

BA=

|3

|(3

1

=

6

8 12 4

2 3 1

(13

b

1 2x1 3x1 1x1

3

:(11

¢

2x3 3x3 I1x3

=19

504

Ais

2)(2

1

ABis

1x1

1

2 3 1 3

a

2

1

1x3

b

BAis

6 2 4 0

13 11 7T

c2|

d

0

1

193 284

8

2

3

1

0

3

-9

218 ) 257

in the first

2

6 4

8 5

4

13 11 T

12 12 9

9 0

4

6

31.32 20.88 20.88

1.3 1.8 —1.1

4 8 7

-3 —6.9 —-133

115 136 46 106

_ | -

12 4 3 12 8 | +3[2 97 3

04(

.

4 5 6 11

0 36.54 10.44

1

2

1

0

8.95 12.95 0.5

218 ) 257

7 6 2 8 0 7

5.22 10.44 41.76

= |

2x4 3x4 1x4

0 8 0 1

5222

3x3

4)

41.76 26.1 20.88

6 9 13

46.98 0 31.32

11 35 8 ) =28 17 23

|—-13[2

3

6 9 13

3

42 51 57

41 40 65

11 8 17

—12.7 —7.2 —19.3

125 315

.

Prices matrix =

405

375 0

2

3

—1

1

-3 2

—1x140x-3+1x2

—2x1+4+2x-3+—-1x2 OXx14+3x—-3+1x2

710

3

3

—6

HES

8

—2

-2

6

1

= $13937.45

= 3

6

6

¢ Total income = $6064.65 + $7872.80

= (13) Bis

193 284

4

7

NC gives the income for each month.

=(Bx2+1x3+4x1)

b

_03):

—4

_ ( 6064.65 —\ 787280

6x2+5x4)

3x1

4)|3

(1)

5

the restaurant had an income of $6064.65 month and $7872.80 in the second month.

v

AB=(3

5

(156 x 8.95 + 193 x 12.95 + 218 x 9.95 183 x 8.95 + 284 x 12.95 + 257 x 9.95

EXERCISE 1D.2

and

}1

7(2—14())

156 |, N:( 183

156 b NC = ( 183

1x2

32)

1x3

1

_

o

(g

8.95 a C=| 0.95 1295

v

BA

0

—1

AB cannot be found. b

1

3

1x2

¢

13)\, 2

d

The number of columns of A is not equal to the number of rows of B. AB cannot be found. 2

10)3}2§11

Monthly income =

50 65 120 42

42 37 29 36

18 25 23 19

65 82 75 72

125 315 405 375

51145 60655 61575 51285 total income = $51 145 4 $60 655 + $61 575 4 $51 285 = $224660

WORKED

t

g

h

b

= 3(3A 4 2I) + 2A = 9A + 61 + 24 = 11A + 61

A2=A

5

= = = = =

(I+ 5A)(A + 5I) (I+ 5A)A + (1 + 5A)51 1A + 5A% + 512 + 25A1 A+ 51+ 51 + 25A 26A + 101

—I-A—A+I = 21— 2A

2

2

_

2

= A

A=O0orl

_

where A is not O or I.

b d)' a (c

then

=A,

a®+bc=a

b a d)(c

b\ _(a d)i(c

ab+bd)

_

be+d? )

b d)

(a

b

\c

d

be=a(l—a)

ab+bd=b

. (3)

be=d(1—d)

b=c=0

then from (1) and (4),

If

0

0

o

0

0)

0

o

10

1)

band carenotbothO

then

.4

a=0orl and

As

(2

cla+d—1)=0

be+d?=d If

..(1)

ba+d—1)=0

ac+cd=c

o

d=0or1l

or

o)

1

o

0

1)

a+d—-1=0 d=1-a

A is of the form

= 11A% + 6AI

=A% 1A - AL+ 12

2

11 2 2\

Equating corresponding elements:

= A(11A + 6I)

= 1521(3A + 2I) + 858A + 858A + 4841 = 4563A + 30421 + 1716A + 4841 = 6279A + 35261 A(2A +3I) b (A-1)? = 2A% 4 3A1 =(A-I)(A-T) =(A-DA+(A-I)(-I) =2[+3A

A°

2

A # O

2

actecd

b AT =AxA3

= 11(3A + 2I) + 6A = 33A + 221 + 6A = 39A + 221

Af(c

where

i)

2

A2

#

a

11

a?+bc

6

a ( c

have but

#

A7 = 251 — 10B — 10B + 4B? = 251 — 20B + 4B> (A+B)? = (A+B)(A+B)(A+B) =[(A+B)A+ (A+ B)BJ(A+B) = (A2 +BA + AB + B?)(A +B) = (A% + BA + AB + B?)A + (A% + BA + AB + B%)B = A% + BA% + ABA + B?A + A%B + BAB + AB? + B®

= 3A% + 2A1

c

A =

2_[2

= (A —2I)(A —2I) = (A — 2D)A + (A —21)(—2I)

= A(3A + 2I)

a

Consider

(A-20)?

a A% = A x A?

9)=(0 0)=0

So, we have found A2 = O A2=0 % A=0

— AD — BD

= A% — 2IA — 2A1 + 412 =A% —2A —2A +4I =A% —4A 441

71)

2e=(1 )

=(A+B)C+ (A+B)(-D) = AC+ BC

A7(1

ol

e

(3I+B)B = 3IB + B? =3B+ B? (A+B)(C-D)

a Consider

239

-1

wl=

¢

X(2X+1) b =2X? + XI =2X2+X DMD*+3D+21) d =D® +3D% + 2DI =D3 4 3D% + 2D (B—A)(B+A) = (B—A)B+ (B—A)A =B? — AB + BA — A

4

1

vl

a

.

wl=

EXERCISE 1D.3 B

SOLUTIONS

ABf(O A # O

and

10

A=0

10

0)(0

0

or A=2L Bf

are

0

o)

"*

1

1o 1

1 0

0 1

1

1

I'ltol

1

o

¢ nullity (T) = dimension of ker(T) =0 d rank (T) = dimension of R(T) =3

sar((5))(=5 3)(%)

|0 10

oAt=

( 1 (3

-2}6

(1

0

-20

a basis for the row space of AT is 1

a basis for the column space of A is

an - ( )}

i (g) ¢ R(T)

L))~

0 0

0 0

0

o T((é)

0

x=y=2=0

Thus

ol =\

LAl

¢ nullity (T) = dimension of ker(T) d rank (T) = dimension of R(T) =2

L

kx1 ky1 kz1

0

[ 9-9 “\-18+18

{3} T+y+z

z+y+z

= kT(u)

Since the addition and scalar multiplication satisfied, T is a linear transformation.

2z

2%

_ [ kx1 +kya

ty2

T(v)

Y z—x

w=T(v)=

T(ku) =T\

Titz2ty1

_(

zZ—T

0

ker(T) = { 0

} .

wa (1)) = ()

.

{using technology} {( 1

-2)}.

{ ( _12 ) }

SOLUTIONS

267

_—

Il

—

-

,_.

—e—o

o = a

~~ ~

- o

S~—— SN—

o

WORKED

and

saA*

268

WORKED

B=|[S

Y

and

1 0 0

R

(2

[

T o S

Av = 0

0

0

)

Letting

y=s,

z=—t

and

where

0

1

1

-1

3

which is already in reduced row

w=1t

S

1

T 0

0

-1

1

-1

1

x y z

(ToS)

s,t€R,

0 0 1

0

-1

1

2

=

echelon form.

269

has standard matrix

T

has augmented matrix

1

0 1 0

S

0

SOLUTIONS

=

0 -1 -1

o

nw sy

o

= =s.

2 -1 3

where

s,t € R

a (SoT)(v) = S(T(v))

=

g

2

== Oo o

OO

=

ker(T) = lin

abasis for the row space of AT is Let S have standard matrix A and T have standard matrix B. A=|S

=l

1 0 1

nullity(T) + rank(T) =2 + 2

0

=

z+z—y ztz—y—2r z+zx—y+2z

3r—y+z

AB=|

1 0 1

=l

0o -2 -1

-1 0 -1

Let T have standard matrix A and S have standard matrix B.

o= ()]

((

(SoT)

0

0

1

0 0 1

has standard matrix

0O -1 0o

—r—y+z

0

T

T

y z

0 O 1

2 0o -1

0 -1 0

0 0 1

0 0 1 0

=l

-1

-2 -1

0

0 -1

0 1

N~

-«

S o T

0 -1 0

1

w

a (ToS)(v) = T(S(V)

b

2 0o -1

0

T

0

= dimension of the domain EXERCISE 1).3 B

0 0 1

1

B=|T

0 0 1

S

+

and

S

< Fe

=R2

d

0 -1 0o

0 1 0

Il

Now

1 0 0

&

) an =i

*. a basis for the column space of A is

b

/N

{(x 0). (0 1)}

Il

.

{using technology}

T

Y z

SOLUTIONS

2

-3

Y

T has standard matrix

Now

A~

?)

(73

2)

Y

(2

-2

=

BA is the standard matrix for

A =

!l= ( 72

((z))

iii AB is the standard matrix for To S

1/

( L2 ) 2 -3

AB # BA ToS # SoT,

6 (z)

)

T:RF - R™

( 7322_: yy

T has standard matrix

1

1

0

3

@

1 1 %ac — %z

#

-1

3

and

!

i

10

B

0

2

)z

T

0

-

i AB=(0

BA=10

0

0) {( 1

1

0

-1

1

oM

0 0

T

1

-1

-2

3

o

1

1

10

—-10

3

{(1t

1

o

1

0 0) (0

-10

0

001

020

1

0

0

0

0

0

0)=(0-10 100

)=100

00

00

0 0

110

0

-1

1 0|’

9 so

But

-1

—1 1

1

3

5

1

0 1]

1 0

0

3

column rank = 4.

R(T)

is the column space of A,

so rank (T) = column rank = 4.

v

s

0

o0l

0),

row rank = 4.

1

100

-100

(00010),(00001)},

A basis of the column space is

100

10

A basis of the row space is

O

10

((2))},

00

0

0

2

‘A:1717201~0100

so

020

{(;),

1

0

00

3

But R(T) is the column space of A, so rank (T) = column rank = 2. v

1

o

0

%)

column rank = 2.

2

0

)

3)N5

2

1

cos?a =

202"

3z — 14y =21

COSQ = ————=, V14 m?2

y:174z~>%y':174z'

o2y

—

1

y:%y’,

1

y=1'

3z —4y=6

¥ =3y Hence

and

275

3z —4y=6 — 3(22') — 4y’ =6 = 32’ — 1y =21

0 5 )

o 3

o =x a

x:%x’

e

1

b

SOLUTIONS

B’

>

T

v

(=1 4(=1)+ (=3))

v

—S8/(~1,-7)

and

C

y= —4x

b

Area ABCD

B

= area AABD + area ABCD

_1 =3

X9Xx3

+

1 3x9x3

x’:%m

and

y' =y

a T(-2,4) — T/(-7, 4)

8

a

||A|| =1,

arca A’B'C’'D’ = 27 units®

For a vertical stretch with

A

=

1

0

0

3

5

St

k& =

,

=z,

%, oy

)

=

Yy

wslen

~—

( 0

Since

o

The stretch has matrix

=

5

=

= 27 units?

also.

SOLUTIONS

The circle has the following axes intercepts, which transform to obtain the axes intercepts of the image:

(3,0) (0,3) (—3,0) (0, —=3)

— — — —

we

b

For

y=mz,

tana=m cos?a

(3,0) (0, 5) (-3,0) (0, —5)

1

=

A=

T

—

y=2y, Now,

=9,

and

z =

z,

_

m(k — hm)

-

a=3

and

14+ m?

_[m*(mh -

b=5

area = mab = 157 units?

Alternatively, |[|A|| = % . area of image =_53 X (7 x 3)2 = 157 units? 9

a

For

y=4a,

m

1+m2

1+m?2

1+ m?2

14 m?

2

2

fomatmy 14 m?

y = ma + ¢

2+

)

mh+m2‘75—k>4n275

1+ m?2

mh —k\2

-

14+ m?2

= k)% + (mh — k)? (14 m2)?

(1+m?2)2

[ (mh —k)? 1+

—

Imh V14 a

m2

] m2

units

For a horizontal stretch with b

A=|

°

a

0

0

1

b k = —,

b / n oal=—

a

/ Y=y

The circle has the following axes intercepts, which transform to obtain the axes intercepts of the image:

(_1,

3)

.

(—1+12’

17

—45:748)

)

[ (mh = k)21 +m?)

-

tana = 4

m

mh +m°k2

1+ m?2

z? +y?

1+m?

h+mk, |

0}

0

Ax'x

>0

_

(0 \0

b= —t

T2

x#0

-3

o .

MNad +ad + o+ a,2) >0

.

the cigenvalues of A are positive.

= /\Qxlsz ... (1)

1o

2a—3b=0 t#0,

an eigenvector is

(3 )

the eigenvalues are

_2

b—gt

(1) 2

3

{choosing

A\ =5,

Lo corresponding eigenvectors

A2 = 10

1 )

t,

t#0

t = 3} with

3 La)

g)

()G

D

(b)Y

o>

x;'Axa = x{' Aox2

z; =0}

t =1}

(0

)=

x=1|

N

x # 0

_

b

a =t¢,

il

n

asnotall

a\

3

_

Letting

{choosing

t£0

|[AM[-A|x=0

2

ccomes

.

(71)

A=10,

i PTIAP=

>0

{z2+z2+...4+22)>0

1

bp:(x1|xz>:(jl

1

#x2)

t#0,

.

For b

=" - @aB)'| = |n" - BTA"|

=A2(x1

a) b))

an cigenvector is

AB and BA have the same characteristic polynomial. they have the same eigenvalues.

10 Consider

-3 -2

.

| AL — AB|

forall

-3 -2

Lat+b=0

{from 5 a}

A2 4 4A has eigenvalue eigenvector X.

A>0

|[AM—-A|x=0

x:(jl)t,

(A% 4 4A)x = A%x + 4Ax = A%x + 4Xx = (A2 +4\)x

=|-aB)T|

A=5,

becomes

Ax = Xx

Consider

>\77’*O

A%~ 15A 450 =0

= XX + kx

8

then

-3

’72

are

(A + kI)x = Ax + kIx

b

[M—A[=0

A—8

linearly independent. a

and Xz are orthogonal.

EXERCISE 1L.2 I

A1 = A2 if the eigenvectors are linearly cigenvalues are not distinct.

=0

2)(43) o) o

c2

{BTAT = (AB)"}

=

Xg = _a

= (Ax1)"x> = (Mx1)"xa

has a non-trivial solution c

{A is symmetric}

ow

c1X1 + caxo = 0

= xlTATxg

il

6

WORKED

I A~~~/

282

D)

WORKED

e

(20 (1 ) (1)

—142p_1(2

7(2

-3

70 45

—3)(14

1

1

-15

(%

0

—\20

“\lo

9)(

6

(10

1

11

o

2

-1

2

3

If

100

)

A+1

0

then

|NI—A[=0 »

\sIf

becomes

For

A=—-2,

becomes

—0

LA=4V3

(

i Letting

4 —4

A=+3,

)

=

0 0

.

becomes

( 745)

becomes 1

5)@

(

1-3 o

t£0

{choosing

t =4}

-4

(75

a

_

(0

.

5)([))7(0)

a=t,

t#0,

.

)eigenvector

is

(

the eigenvalues are

(11

b=t

. {choosing

)

A\ = —2,

corresponding eigenvectors

A2 =7

( :L5 ) s

t£0

P:(_45 A% =PD?P~!

—

with

=P

p~1A%p = p~lppip~1D b

(-8 “\o

0 34

1

1+V3

o0




0 7\/5)

P

-1

1

330

T\1+VB

1-VB

0

:_;
1

)

330 4 330

:7L0}

a Pz, y), A(-2,0), and B(4,0)

6 = arccos (fl) 35

Now

V57

or=—D

=2

arccos

3

e

12z = 100

=

/9—4(1)(-12)

3457

100 — bz = Tz

o

{corollary of Apollonius’ circle theorem}

SoT= -

o_5__= 14 7 20—z

Let ABC =6

= (r +4) cm

3+

OA

CB

=4 cm

+ APy

By the angle bisector theorem, CA

AP;

=2

PA% = 4[PB?

(@ +2)° +9° =4[z — 9° +4°] x4

y® =4

- 8z + 16+ ¢

22 + 4z + 4+y? = 42 — 32z + 64 + 4y?

— 362 +60=0 327 +3y? oty — 120 4+20=0 PA

= 3

iv

o

PA? = 9[PB2] [

22 4z +4+y? =9[z* — 8z + 16 + 37 22 + 4z + 4+ 1y = 92 — 720 + 144 + 9y° 822 + 8y? — 76z + 140 = 0 zz+y27§z+3—;:0

WORKED b

When ",

.

k=1,

AP =PB

.

P lies on the perpendicular

.,

bisector of [AB], and this N traicht 1i .

18circle. @ straight

¢

A

EXERCISE 2) -

1

6 cm

Let the other diagonal be = cm.

fme, mot a

N By Ptolemy’s theorem,

9 em

12.00 ~ 6 x 7+ 11 x 9 12.0z ~ 141

o~ 118

If a circle has fixed cc?ntre C(p, q) and fixed radius r

the other diagonal is . approximately 11.8 cm long.

7cm

and P(z,y) moves on the CP = r.

2

6cm

Let the 4th side be

V

5cm

5z +6

P2 = g2 —2pr—2qy+p>+

. which is of the form d=—2p, e=—2q,

x 11 ~ 10.1 x 9.54

TR

@-p’+@y—a’=r"

+y?

g2 —r2=0

approximatel PP Y 6.07 cm long.

3

@

5

LoeRe0T th64th'51dflls

. with

22 + y? +dz +ey+ f =0 2 —r°.2 2 and f=p°+q°

cm.

By Ptolemy’s theorem,

zcm

2

307

NN

_

circle, then

SOLUTIONS

a

PQRS+SP.QR =7x134+9x11

and

PR.QS =14x12

=190

= 168

PQRS is not a cyclic quadrilateral.

b

72 mm

2

P A

Let

Py

AP =a,

BP =b,

BPP; = 8. As PiPPy = 90°, Now

PPy =x,

_

APP; =aq,

110

APy

area of ABP;P

2

PPy =y,

79 mm

R

BP;

of AAP,P

_ AR

area of ABPoP

BPy

In APQS,

cosa = 797

area of AAPP

area of AAP5P

area of ABP;P

area of ABPoP

{

1 . zorsina

1 . 5oy sin(90° + )

Lbrsing

Lbysin(90° — B)

sina

cosa

Snpg

cosB

sina_

s

APy BP;

a = (3,

=k,

or

=

AP, —— BP,

a positive constant

{as A, B, and Py are fixed}

79 x 118

p§R = 60° + arccos( }: 23‘11

BPy

In APSR

PR? = 792 +80? — 2 x 79 x 80 cos (60Q + arccos(1295% )

PR ~ 118.65

Now

PQ.RS+ SP.QR

=72 x 80+ 79

and

x 110

SQ.PR ~ 118 x 118.65

= 14450

~ 14001

PQRS is not a cyclic quadrilateral. 4

xzz+ wy = mn.

a

By Ptolemy’s theorem,

b

zz + wy = sum of areas of blue rectangles

mn = area of brown rectangle using Ptolemy’s theorem, blue area = brown area.

the angle bisector theorem applies.

AP APy _——=— BP BP; AP

BP

APy }

cosfB a=p

As

2 — 72 2 118 +118° — T

a = arccos( }g gf‘;},)

sing

tana = tan3 b

2

2x

{arca comparison theorem, altitudes are equal}

cosa

mm

BPPy = 90° — 3.

area of AAP;P

ang

,\




cd + ba

.

AF 1

{replacing d by a, a by b, ¢ by d, and b by ¢} (ac+bd)(ad + be)

v=

FB 2 AF:FB=1:2

ab + cd

2202

—

(ac + bd)(ab + cd)

=

be+ ad

b

(ac + bd)(ad + be)

’

ab+ cd

22y? = (ac + bd)? ac+bd = zy 6

area of ACBE .

area of AAOE

AC=2

area of ACOE

ABC = ADC = 90°

arca of AAOB

{angles in a semi-circle}

b . a — = sin 2 a = — cosa

{angles subtended by the same arc}

In AABD,

BD

sin(a+ )

sinf = g

= %

{equal altitudes}

= area of AABE

— area of AAOE

area of ACBE

area of ACOE

2

2

_ area of ABOC

c = — 2 B d cos 3 = — 2 A ... (1)

{cqual altitudes}

-

. sinff

ac+bd = 4cosasinB+4sinacosf

= %

_ area of ACBE

2

In AACD,

We have that AE: EC =1:2. We use the theorem that if two triangles have the same altitude, then the ratio of their areas is the same as the ratio of their bases. area of AABE

r=1

01 = 62

arca of AAOB

BD = dsin(a + 3)

2 BD = 2sin(a + 8)

2

3

a

By Ceva’s theorem,

: area of ABOC AX — XB

=1:2

BZ CY X=—=X—=1 zc YA AX —

XB

X divides [AB] in the ratio

sinf

— area of ACOE

=D

using the Sine rule,

d

=

fiXTxizl

(bd + ca)(bc + da)

v=

5: 7.

.. BD:DC=1:1 .. CE:EA=2:1

By Ceva’s theorem,

o

°

2 Xx2x=-=1

3 AX XB

3 : 4.

3 4

WORKED

SOLUTIONS

>

Let the triangle have vertices A, B, and C.

The altitudes from A, B, and C meet [BC], [AC], and [AB] at P,

~

o

= %a = %e {areas of triangles are

Q, and R respectively.

proportional to their

and so

o

RB PC QA As ABP cquiangular, similar.

f:%e

4(a+ f+e)=30b+c+d)

C

———Q

>

e+ 2e+25=23a+2a+2

BP

=

Also,

AR

We need to prove that

e+d+c—2f+a+b) a:%e

C

w

bases if altitudes are equal} Similarly,

AB

c

Likewise As are similar.

o

.17 _ 1—5fl+1—0l-fil

ow

Q AP

B

=1

BP CQ _ BP

RB'PC'QA

T717%

AB

BX CY AZ

AB

XC'YA'ZB

BC

AQ

{rearranging to fit

1. @, 3}

EXERCISE 2K.2 B 1

A

By the angle bisector theorem,

BX

AC

C

.03

So, by Ceva’s theorem, [AP], [BQ], and [CR] are concurrent.

By Menelaus’

BX CY AZ _

[BY] meets [AC] at Y. [CZ] meets [AB] at Z.

X

BQ

and

=1

[BC] at X.

B

ACR

T,CB AC| A8,

Let the triangle have vertices A, B, and C. The angle bisector [AX] meets

Y

AR

B BE' At

{converse of Ceva’s theorem}

7

QC

RB PC AQ

[AQ], [BR], and [CP] are concurrent. A

BC CY

XC'

BA

YA’

CA AZ — = —. CB 7B

and

CA

2

=1

31az_ 5278 AZ

C

X

ZB

Z divides [AB] externally in the ratio

..

z

AC BA CB

XC'YA'ZB

Y

B

theorem,

w

10 3

10 : 3.

A

by Ceva’s theorem, [AX], [BY], and [CZ] are concurrent.

By the tangents from an external point theorem,

P

g

RA = RB,

B

QA = QC,

Y

and PB = PC.

Thus Q A R

[PA], [QB], and [RC]

(2 @

C

AR

APBQER 5o 5,1

PBQCRA

o

AB

CR:RA=1:6

..

APC

PC

AC CR AR —=—=—

CR=1icCA C

B

=

BQ:QC=3:1

R

and

T

@

BQ = 3BC

.

AP

Finally, also As ABQ are similar.

>

o

BQC

Q

AC

AP =—22AB AP:PB=2:1

AP

BC _BQ _ QC —_—=—=..

=49:34

B

BP

=== 0 CB BR CR

>

AS:SZ = f+a b

P,

and CBR are and therefore

{Equal angles are marked.}

4(Ze+ Ge+e) =3(b+2b+ %e)

A

309

are concurrent.

2 QP8 _y AQ CP BR

C

B

RA

In As AWZ

and BXZ,

ap =az and (3 = By the triangles are similar AZ

BZ

=

AW

BX

()

{equal alternate angles}

X

310

WORKED SOLUTIONS In As AYW and CYX, a1 = a2, and 67 =602

[4

A

{vertically opposite angles}

the triangles are similar AW AY

x-w

?

AZ

BX

CY

AW

ZB

XC

YA

BX

BX

—

XC

BR 11 R RS~ 3°1 _ BR C RS

CcX

AW

B

M

XC =-1 —

=

E divides [CA] in the ratio

5:4

.

—

=

o

3 1

3: 1.

o

..

=

2:3

|

ol o

R divides [BS] in the ratio

D divides [BC] in the ratio

|

RS'AC'MB

[——

.

X

3

BR SA CM _

S

—.—.—=(-—).

Thuys

Also ARM is a transversal of ASBC, and by Menelaus,

A

Q

PN~

v

N B

fi D

C

Let [BE] and [AD] meet at X.

As APY and CQY

Now BXE is a transversal of AADC. by Menelaus’ theorem, AX

DB

AX(2>5_ 5)

4

1 2: 1.

AAPR

Q

&

PR BM

AABM

BM = 2(PR)

X

e

BZ

b

AY

CX

BZ

AY

CX

BZ

is similar

() (5)-(2) a

c

b

=-1

{converse of Menelaus’ theorem}

to

and @ is common}

... (I)

Likewise AARQ is similar to AAMC

and

MC = 2(RQ)

e (2)

Using (1) and (2), as PR = RQ BM

a

= MC

M is the midpoint of [BC].

b

AZ _a AZ

BX — 2 ox

and likewise

{a1 = a2, cqual corresponding angles,

c

AP 1 AB 2

¢

¢

Y, X, and Z are collinear

{midpoint theorem}

B

oY

YC'XB'ZA

[PQ] | [BC]

A

b‘

CQ

A = 2

e

X divides [AD] in the ratio a

a

X, Y, and Z are points on the three sides produced of AABC.

2

XD

AP

AY

Thus

|

AX

4

Y

CE

== XD BC EA

XD\

AY

are equiangular, and therefore similar.

A

BRS

AAPQ.

is

a

transversal

So, by Menelaus,

of

By the angle between a tangent and a chord theorem, a; =az =a3 and B = (. Consider As ABD, CAD: A

As ABD and CAD equiangular and therefore similar.

AS QR PB

SQ'RP'BA AS'1

0T

1

—1

2 AS 5Q 1

AS:SQ:QC=2:1:3 AS:SC=2:4 =1:2 S divides [AC] in the ratio

1 : 2.

are

D

area of AABD _ AB® arca of ACAD

CAZ

DB : DC = AB? : AC?

DB DC

. W

{area comparison theorem}

WORKED Likewise, As BCE CAE are equiangular therefore similar.

and and

2

a

(1,1)

lieson

B+

d b

BC?

_

BE

Finally, As BCF ABF are equiangular therefore similar.

and and

lieson

units

V13

az+by+c1 =0

Va2 +b2 . lea—eal units d = ——— + b2 Va?

Q)

area of ACAE ~ CA2 ~ AE

6

311

a(0) +b (‘—;1) +eo

d=

area of ABCE

+1

V9 +4

(0, 7761)

E

3z+2y=>5

SOLUTIONS

+6| =V13 —2(~3) [3(k) a ——————

3

V9+4

= 13 |3k + 12|

3k +12 = +13 3k=1or—25

area of ABCF

BC?

_

k:%orf%

CF

—_——=—=— area of AABF AB2 AF

AF CD BE

W

—

e

—

FC DB

EA

aB?

‘ca®

= ==

Be?

b IO+

..0 ®

ABZ

'Be?

vi+1

\—1—k|_\10\

CAZ

V2

from (3) from (1) from (2)

7

For transversal ABC, For transversal DEF,

EG

ZI

CH

HC

1B

GA

—.—.— CI BG AH HD

IF

k+1==£10 k=9or—11

theorem, D, E, and F are

AGHI has 5 transversals and for each we theorem. HX GB ID For transversal DXB, —.—.— = —1 XG BI DH HA GF 1Y For transversal AYF, ——— =1 AG FI YH HE GZ IC For transversal CZE, ——— =1

can use Menelaus”

—_—.— DI FG EH

&

r—y—4==4

r—y=0 or z—y=38 z—y=0 and x—y =28 are the two parallel lines. 5

PN =PS

V@ +12+

=1

oA

42+

HX GB P Al gr Iy uE'Gz de'uelas! XG BT \DH ,AG Bt YH ,B6 ZI |,CH | €T ;B6 G« up' ! GE!

1Y

—.—.—— XG ZI YH

AGHI

= —1

14y — 16y + 64 = 2% — 10z + 25

the locus is

3z —2y+6

+4% -8y +16

=0

gradient of [AP]

for points X, Y, and Z on two sides of

and one side produced.

X, Y, and Z are collinear.

{converse of Menelaus’ theorem}

EXERCISE 2L I

1

87 = /(257 + (y - 4)?

122 — 8y +24=0 6

GZ

is _\z—y_4" v1i+1

V2 le—y—4/=4

Multiplying all of these gives

HX

z—y—4=0

from

of P(z,y) The distance

|z —y —4] =22 —

= —1

GE

V2

|k+1| =10

=1

by the converse of Menelaus’ collinear.

-k -7 VI+T

_ 23 +5(2) +6] _

22

d="—"""__""T it ° 1+25 vz e |4(—1) — 3(4) —4| _ |—20] bd=" = = 4

619

5

cd=

13(2) — (=1) — 2|

4 g

ImED+(3) -5

9+1

=

5

Jio 1

’

s

)

unit

.t

\m2+8l

y—0

y

gradient dient of of [BP] [BP] = —— 23" =

units

Th

7.3

= ()(s) Yy

Yy

=1

sy

=2

soyP

=249

z?+y% =9 the locus is

2

+ y?=9

tmme=n =-1

-9)

312

7

WORKED SOLUTIONS

a

(172)2+(y71)2:%

(-2 + (-2 = Zmy=?’

522 —dz+4+y? —2y+ 1] =42® + 4> +25 —dxy + 10y — 20z

z =12}

522 =207 + 5y° — 10y +25 = 42° + y*> + 25

Let N be a point on the line

— dzy + 10y =207

22 + 4oy + 4y b

— 20y = 0

F 144 V25

|3z — 4y — 3] _

|5z — 12y — 4]

5

such that

AR = RN

o 4AR? =RN? oAz —3) 4y = (12— a)? 4[a? — 6z + 9+y?] = 144 — 24x +2

|3z74y73\:|5z—12y—4\

VI+16

= = 12

4a? =247 + 36 + 4y° — 2?

13

7 —144=0

322 4 4y = 108

13 |3z — 4y — 3| =5 |5z — 12y — 4| 13(3z — 4y — 3) = £5(5bz — 12y — 4) 39z — 52y — 39 = £[25x — 60y — 20| 39z — 52y — 39 = 25z — 60y — 20 or

39z — 52y — 39 = —25x + 60y + 20 14z +8y —19=0

or

T

64zr — 112y —59 =0

Let N be a point on the line

which is a line pair.

T=3

8

a

i

AR = $NR

AP = 2BP

AP? = 4BP?

4AR? = 9NR? Az —3)2 + 9 =9 — 3)? o A4fz® — 6z + 9+ 9% = 9f? — B+ 18]

(z+1)? +y2 =4(x - 3) + 97

22 4+ 20+ 1+ y? = 4[2® — 62 + 9 + 7] 40?4y — 242 +36 =22 +y? 422+ 1 322 +3y? — 262+ 35=0 ii The locus ofP is a circle.

b

i

{Apollonius’ circle theorem}

2AP = BP

4a? + 4y? =247 + 36 = 922 —247 + 16 522 — 4y? = 20

10

a

AQ+BQ=6 ViE—=22+y2+\/(z+2)2+y2=6

4AP? = BP? A

+2c 4+ 14y

Va2 —dz+4+y2=6—

=a® -6z +9+9y°

302 +3y* + 142 —5=0

9

/a2 +4z+4+y?

A dr A =36 —12¢/a% +de+4+ 12+ 25 + Az + A+ 47

42? +4y? +8c+4—a? -y +62-9=0

ii

such that

{squaring both sides}

Once again, by Apollonius’ circle theorem, the locus ofP is a circle.

so12y/a?

44z +4+y? =8z + 36

so3yat+dr4+4+y2=2x+9

a

oo 9(a? + 4z + 4+ y?) = 4a? + 36z + 81

{squaring both sides again}

922 + 367 + 36 + 9y? = 42> + 367 + 81 522 + 9y? = 45

b

AQ—-BQ =2 o

s Let N be a point on the line

@ = —3

A

such that

A

(z—3)% +y% = (z +3)?

—br Ity = 6+ 9 y? =12z

Aty

=/

22 tdz+ 4+ Y2 +2

eA

A2 f et Aty A l A

NR = AR

z—(=3)=+/(z—3)2

—de

AQ=BQ+2

4

{squaring both sides}

+y2 o

? Az +4+y2 Ay/a

Va2

tdr 44y

2?4 AF+4+y?

=8z

—4

=201

=42+ A4+ 1 {squaring both sides again}

3127y2:3

WORKED EXERCISE 2M.1 B

For D(3, 0),

a Centre b Centre

(2,3), r =2 units (0, —3), r =3 units

¢

( (2,0),

Centre

r=

b (x+2)2+(y—4)2=

a (@—2)%+(y—32=25

d (@+3)2+(y+1)2=11

c @42+ (y+1)2=3

(z+2)%+ (y—3)? =(4+2)2+(1-3)? =40 whichis > 25 E lies outside the circle.

D lies outside the circle.

lieson

(3,m)

As

313

d For E(4, 1),

(@ +2)%+ (y - 3)? =(3+2)2+(-3)? =52 43% whichis >25

V/7 units

SOLUTIONS

(y —2)? =25, (z+ 1)+

42 4 (m—2)%2=25 (m—2)2=9

(x—3)2+(y+2)2=4

m—2=43 m=5or—1

(m, —2),

As

lieson

= 36, % + (y —3)% (z+2)

)? (m+2 + 25 = 36 (m+2)2=11 m+2=+/11

(z+4)2+(y—3)%2=16

m=—-2++V11

As (3, —1)

lieson (z +4)> 4 (y+m)? =53,

72+ (m—1)%2=53 (m—-1)%2=4

m—1=+2 m=3or—1

r?=(5-4)2+(3+1)? sor?=1+16


0 forall

z,y € R

and

RHS < 0.

which is a circle, centre

(z—h)?+(y—k)?>r>

then

which is a circle, centre

PC%>r?2

=5 a

units. For

hascentre

C(—2,3)

(+2)°+(y—3)> =(2+2)2+(-3)?

7 = 4 units.

and

which is a circle, centre

(4, 0),

7 = 4 units.

2?2 +y?> 120 +8y+ k=0

b

For

2% — 12z + 36 + 3 + 8y + 16 = —k + 36 + 16

B(1,1),

(z+2)°+ (y—3)° =(1+27%+(1-3)2

=55

=13

=2

. A lies on the circle.

(2, 3),

(x—4)%+y*=16

Using 5 above:

A(2,0),

r = /17 units.

22 — 8z + 16+ y> =16

P lies outside the circle.

(z+2)2+ (y—3)?2=25

(-2, 4),

x2+y2782:0

PC>r

6

r = /5 units.

2? +y? —dz—6y—3=0 22 —dx+4+y>—6y+9=3+4+9 (x—2)%+(y—3)2 =16

< r?

(x—h)2+(y—k)? =2+9 (x—3)2+y?=11

(4,-1) d

(—3, 1),

whichis

0

the centre of the lake is at

k

d which is a circle centre | ——, 2

627

{3

f

b

equation is 3

r=y/4+2-2 =—

¢

i

/3T 15

When

equation

e=-3

.

the equation represents the point ii When

d? +e?

[ a2

< 4f,

TJr

-5

3

1

a

1

o M6 —48m + 36m? — 16m? — 16 =0 20m? — 48m =0

e

m(20m — 48) =0 tangents have equations y =7

y =7 and

and

Let a tangent have equation y = max. Now the distance of

(4, 3) to a tangent is 2 units.

Im-®) @) _ Vm?

2

tangents have equations

gradient of tangent = +

0

P(-2.1)

at P(—2,1)

y = 1—5210 +7—

12z — 5y = 61.

= — 4y = —6

1

4

to the line is 4 units}

z — 4y = (—2) — 4(1)

—2+3

the tangent is

_

+1 [4—m3| = 24/m2 + 1 (4m —3)% = 4(m? +1) 16m? — 24m +9 = 4m? + 4 12m? — 24m +5 =0

1-5

tangent has equation

= (3) +7—8m|

(4 —6m)? =16(m? +1)

4

C(—3, 5).

_4

y = mz + 7 — 8m

the equations are

e

Gradient of [CP] =

n

[4—6m|=4y/m2+1

— f is a non-real

22+ 4+ 62 —10y+17=0 22 462 +9+y? — 10y +25=9+25+17 (x+3)2+(y—5)2 =51

has centre

,

m=0or£12

complex number. the equation has no meaning.

EXERCISE 2M.2

(8,7) lies on it 7T=8m+c c=T7-8m

f=2}

d

P(8,7)

T

{distance of (2, 3)

and

r=0

.

y = mz + c.

=Ve

uni units

d?+e2=4f,

9z + 4y = 43

m(2)

{d=2

°

Let a tangent have

the tangent is

(—1, %)

12 -3

= %

9z + 4y = 9(3) + 4(4)

that is,

_< 2

8—4

gradient of tangent = —%

=c+7-/

d% 4 €% > 4f 322 +3y%+6z—-9y+2=0 @ +y?+20-3y+2=0 which is a circle centre

Gradient of [AC] =

e?

As

b

(12, 8), and the radius is

V97 m. diameter = 2v/97 ~ 19.7 m

10 22+ +detey+f=0 auz

(—

gradient of tangent is 0

2+ +4x—2y+k=0 e fdz+a+y® —2y+1=-k+4+1 (e+2%+(y-1)2=5-k

d\?

C(0, —3).

Gradient of [CP] = J 0-0 which is undefined.

E+13=11 L k=-2

5

m

2

V& + 13

k+13=V11

¢

P(0,2)

2?4 (z+3)2=25

(x+3)°+(y—2)°2=k+13

which is a circle with radius

22 +y? + 6y =16

m ~ 0.236

or

1.76

y ~ 0.236z

and

y ~ 1.76x.

9—56

WORKED 5

Let the other tangent have equation

y = ma.

Now the distance of (3, 4) to maz —y distance of (3,4) to =z —2y =0.

8

= 0

a

is equal to the

Let M have coordinates (X, Y). As M is the midpoint of [OA], A has coordinates But

Im@3) — (4] _ [B) —2(4)]

JiTd

VmE o1

SOLUTIONS

(2X, 2Y).

AC=r1r

JEX @ 0 =

©(2X —7)? +4Y2% =+2

o VEBm—4l=+/m2+1x5 - 5(3m —4)% = 25(m? 4+ 1)

315

{squaring both sides}

the Cartesian equation of the locus of M is (2¢ —1)2 + 4y? = r2.

b

5(9m? — 24m + 16) — 25m? — 25 =0 45m? — 120m + 80 — 25m? — 25 =0

(2¢ —7)2 +4y* =12

20m? — 120m + 55 =0 4m? —24m +11=0

(2m—-1)2m

—11)

=0

m=1Loril

the other tangent is y = %m

o B® 42+

6

Vo+16

=2 =5

344y —8=0

units

the circle has equation

(x—3)%+ (y+2)? =25

b

Let M have coordinates

B(0,2Y)

Now the gradient of the tangent is %.

Ais (2X,0) and Bis (0,2Y).

Let P be the point of contact of the tangent and the circle.

(X, Y).

[CP] has gradient 7%

[CP] has equation

4z + 3y = 4(3) + 3(—2)

thatis,

4z +3y

=26

P is at the intersection of 4z +3y =6 and 3z —4y Solving simultaneously gives = =0, y = 2.

= —8. But

point of contact is at (0, 2).

AB =p

VEX-02+0-2Y)2=p

z2+y274z+2y:0

L

2 —drt+a+yir2y+1=4+1

(=22 +@y+1)>=5

the centre is a

(2, —1)

and

P X 2 4+Y 2 —(—2)

-k ng 5 _B@+ac

2= (2), 2

V9 + 16

A sccant occurs when

‘

2—k

2

the Cartesian equation of the locus of M is

12—kl =5V5 2— k=455 k=2+5V5

b

4X? +4y? = p?

r = /5 units.

Tangent case is when

d

AAXZ 442 =p

b

2_

(P

2

This is the equation of a circle, centre

(0, 0), radius

d < v/5

o5

[2— k| < 5V5 —5V50

..

y? =8(1) —42V3

=- yi(z1 + 2a) yi1z z==x1+2a

x>0}

42

— y=

y1z + 2ay = z1y1 + 2ay1

(z1,y1) # (0,0),

2

The focal chord meets the parabola when

y1z + 2ay = y1(z1) + 2a(y1)

The normal cuts the z-axis when

{as

4-2

2V2z —y = 2v/2(2) — (0)

= o

is

(z1, y1)

2

2az — y1y = 2a(z1) — y1(y1)

equation

z>2a

4 -0 —\/_

_ 2a

The normal at

when

the focus is F(2, 0).

[FP] has gradient

which is which is

¢

..

..

the tangent at

.its

_ da=38 Soa=2

the gradient of the tangent at (x1, y1)

b

(4, 4v/2)

oo

Qis

v=

(1, —2\/5),

2v2 = _\/_

1

WORKED ii Using 3 a, the tangent at P(4, 4v/2) which is

4z + @

@ —Vv2y=—4

and the tangent at

Toeafer)

.. (1)

Q(1, 72\/5)

is

4z + 2v2y = —4

whichis

iii

2z +v2y= -2

... (2)

4z — \/g\/ay = —4x,

The gradient of the tangent at P is

VB8\Z1y = Az + 1) = = —2

Equating y in (1) and (2):

v2

(e 2=

(75) X (775) =—1

4z — y1y = —4z;.

[FP] has gradient

xr] —

24z

5

r=—2 the tangents at P and Q meet where the tangents meet on the directrix.

the focal chord through P has equation

yiz — (21— 2)y =2y (w1 =2y =n(z - 2)

The gradient of the tangent at P is

_yi(z—2)

S

(@1 -2)

and

(ylx(:”:;) )2 =8z

Szl(z—2)2

W:Sz

{as

y°

o

1

=8z}

a

i

z12? — (2 + 4)x + 4z

=0

(z1z —4)(x—21)

=0

y2

=

1

V2

2

Y 4+ =1

and

b2 =4

the z-intercepts are +3 and the y-intercepts are +2.

ii

r=— kgt

or I

1

32

—

z1

y==* 4 Qis | —,

x 0}

320

WORKED

SOLUTIONS

Woac—3x L— Ivaef3>=4

and

e

+ y? =1

b* =3

the z-intercepts are £2.

2 _

%

are

+v/3

and

the

y-intercepts

10 =10

3 1 1

=17

e= %

1 .iva=2x5=1

{as and

property}

e >0}

=5

2

—=—=4

1 2 The focus (0, 1) has corresponding directrix y = 4. The focus (0, —1) has corresponding directrix y = —4.

2

e

a

a=4

and

ae=3

e=3

the ellipse has equation

b = a2(1 — 52)

b2 2 _=16(1— )9

—

+ %

=1

f

Ay

ae=3

g

Ay

b=3

and

b=4

=7

42

—

7

b

=1

e=4

and

e=

ad) =

cooa=38

b2 =a?(1—e?)

%

3|

the ellipse has equation

—

+ —

= 1.

e:%

b2 = a?(1 —€?)

b’ =64(1- 1) b =48

T -3

and

9=d(1-1%)

a?=9x% coa?=12

v

the ellipse has equation

—

+ %

=1

WORKED ae=3

and

— =5

At (V2 2),

€

ae (E>

=15

€

which is

b? = a? — a%e?

.

.

T

the e cllipse ell k has equation 1]

3

Y

GQ;Z . fi a

b?=15-9 b’ =

vy

b

The gradient of the normal is

Wh en = 1—e

22z —y = 2v2(vV2) - (-2) 2v2z —y =6

whichis

b2

2

Z—Z = 74((:/25)) =22

the equation of the tangent is

= = ae.

_,

2v/2z + y=26

At (V2, -2),

—+—==1 = + 5

The latus rectum meets the ellipse when

j—z = 74((‘2/)5) = —2V2.

2V2z +y = 2v2(vV2) + 2

b2 =a?(1—e?)

2

321

the equation of the tangent is

coa?=15

2

SOLUTIONS

x 2 22 —+4+—==1 3+12

Yy =2,

i. 4x

2

w_2—1,l 3 3 cc272

b2 y=+—

4

LetNbe

(X,Y).

Ais

(3X,0),

22 —.

rectum is

At

a

2y Bis

z=4V2

(0, T)

At

\BX 024 (0- =) =k

-,

9x? 4 9y? k2 4k

02

So the locus of N is an ellipse with equation

oy

:—2 + %

=

7

a

b2

a

y

3

2

+ 2y

12 d

12

dz

LY 3

Wh en

S

z

de dy

a2 bz

de

a2y

dy

b211

the equation of the tangent is .

g

s

Veiz + a®yiy = bz (21) + a’y1(y1)

o

{implicit differentiation}

=z 3

de

vy

z =V2, =2,

.

P {implicit differentiation}

At (@1, (z1.y1), 91), —= = - =0

=1

y dy 12 dz dy

e (1)

=0

y dy

This works because of the focal-distance property. PPy s fixed, and MP; + MPy is a constant as the loop’s length is fixed.

6

@ + 2v2y = 3v2

2 2y d a_z + b_g fi

the string and pull the string taut as the pencil moves. The pencil

2

z+2V2y = (—V2) +2v2(2)

2 x2 v §+b_2,1

draws an ellipse.

2 T -+t

1

the equation of the normal is

which is

Make a loop with the string and place it over the pegs Py and Pa. Place a pencil on the inner side of

pencil

z—2V2y = (V2) - 2v2(2) = — 2v2y = —3V2

e

=3

4

string

1 = —.

(—+/2, 2), the gradient of the normal is

(2)

9y?2 9X2 4 —— =k?

Py

2

2, 2), the gradient of the normal isS5 (V/2.2). the gradien

whichis

3Y

Py

@

the equation of the normal is

AB=k

5

738

z? =2

°latus

the length of the

3

a2

z1T

T

4z

w2)? 3

ey a2 2 =_wd o

b

¢—=1 12 i

12

Y

v’ =4

2

a

3

=12x1

+ 2

+

2

olb2

{dividing by

22

a”b”}

.

{using (1)}

=

The end of the latus rectum in the first quadrant is

(ae, —>

—1-2

Y1y

2

{using 3}

The cquation of the tangent is

b2

wn

(5

whichis

=+ 2% =1

=

whichis

=

a

a

ex +y

= a.

322

WORKED ¢

SOLUTIONS

The gradient of the normal at

(z1, y1)

the normal has equation which is

is

2

ii a2=4

a”y1

and

b2y

b2 =a?(e? - 1)

16 = 4(e% — 1)

a’yiz — o1y = a®ya(z1) — b2@1(y1)

e —-1=4

a2ylz — bzzly = (a2 — bz)zlyl

=5

e=+5

EXERCISE 2N.4 B

a

i

ae=2V5

2522 — 16y% = 400

6

cuts the z-axis when

y =0

The focus

(0, —2v/5)

z2 =16 r=+4

vy=-"7

25

2

and

and

e>0}

22 S

and

(0, 2v/5)

the hyperbola cuts the z-axis at (4, 0) (—4, 0) but does not cut the y-axis. il a2=16

{as

The focus =2 y==

2?42

has corresponding directrix has corresponding directrix

iii The asymptotes have equations

y = :t%ac

b% =25

b2 =a?(e2 - 1)

iv

|

-

) S Il — o ©W

1

b =16

The focus (v/41, 0) has corresponding directrix z— 6 Vi The focus (—+/41, 0) has corresponding directrix g = —-16 Vat’ . b iii The asymptotes have equations y = +—x

i

2?2 —y? 2 y?

T

cuts the z-axis when

=4

y =0 22 =4

a

T =12

y = i%z

the hyperbola cuts the z-axis at (2, 0) (=2, 0) but does not cut the y-axis.

iv i

a2=4

and

and

b2 =4

b = (12(62 —1)

4=14(e® 1)

e2-1=1

2 =2

Ce=v2

{as

e>0}

ae =2V2

and b

i

2 cuts the y-axis when

= =0

yP=4

y =2

the hyperbola cuts the y-axis at (0, 2) and (0, —2) but does not cut the z-axis.

2

;:fi:fi

The focus

(2\/5, 0)

The focus

(—2+/2, 0)

z =12

a?

116

a

has corresponding directrix

has corresponding directrix

1:7\/5.

iii The asymptotes have equations

b y = +—z a

— 42 y—:tiac y =tz

WORKED iv

SOLUTIONS a=2

323

and

fi:§ e 5

2 8 e

5

o oe=8 b2 =a?(e® - 1)

b =4(21) b 2 _=4x 159

2_

b =3 2

9

g

ae =12 cuts the y-axis when

ane

« =0

y?=9

the hyperbola cuts the y-axis at (0, 3) (0, —3) but does not cut the z-axis.

and

and

b»>=9

the equation of the hyperbola is

9=9(2—-1)

Ay

e2-1=1 e2=2

ae=3v2

The focus

v="15

{as

e>0}

and

(0, 3v/2)

=1

z=cosf, y=cos20 We use the identity cos20 = 2cos260 — 1 y:2cos2t9—1 y:?zzfl

d

z=sinf,

y=cos20

We use the identity y=1-—2sin?0

cos20 = 1 — 2sin? 0

y:172z2 e

x =tan6,

y=2secl

We use the identity sec?

2

=W 4

2

16

: : directrices Hence

has foci

(iZ—”;s, 0)

9(y—1)2 16

i : and directrices

=1

2 —1 =1+t + tan’ 6

and

2 Y =1 1

322 4+ 9% — 6z — 4y +40 =0

322 — 2z +1%) +y? — 4y +2%2 = —40+3+4 3z —1)2 + (y—2)% = -33

which is not possible as the LHS is always positive.

the equation does not have a graph.

+z

2

4x2fy2:—4

has foci

@ _= -2+ 756

+ tan®6

Yy

6 = _= :tm.

(z+2)%

2v13 (=2 + =5=,1)

7

1

=1

f

o =cos,

y=sin20

We use the identities and Now

sin 20 = 2 cos @ sin 6 sin?0 =1 — cos? 0

3?2 =sin?20 = 4cos?@sin? 0

= 4cos? 0(1 — cos? ) y? = 4a?(1 - 2?)

4 2

WORKED r+4y =25 If y=t, then x=05—4t,

Gl

I —

z+4t=5 y=1t arethe parametric equations.

Ty = —8 If @ =1¢,

then

r=t

ty= 8

y=

-8

B

But

(t #0)

are the parametric

then

4

y=3tanf

t=-%orl

9

() ()-

tzfg,

When

z:2(7%)2:%

t=1,

z=2

and

they meet at (%,—%)

cos?0 +sin?0 =1 and

3

b

forall @

2 = 2t2,

and

yzfg

y=1

and

(2, 1).

y=t

z=2y?>

2 =sinf.

whichmeets

z+y=3

where

3

y=3sinf

2t2 +¢ =3

(2t+3)(t—-1)=0 When

2 =cosf

arc the parametric

22+t —-3=0

x Y —_t==1

z =3cosf, equations.

2 =tand. 3

a The line meets the curve when

2?2 +y? =9 2 2

welet

and

equations.

32 =09t

9

forall 0

= =secf 4

x =4sech,

y =3t (we do notneed y = £3¢, as t could be positive or negative) x =12, y=3t arc the parametric cquations.

But

sec?0 =1+tan?0 welet

equations. y? =9z If o =1t2,

SOLUTIONS

2y2:3—y

2% +y—3=

are the parametric

42? +y2 =16 r

2

+

4

But

¥y

2

_

16

1

() () -

When

a

cos?6 +sin?0 =1 T

welet

= =cosf 2

x =2cosf, equations.

dy

4

y=4sin@

When

2% =—4(—t%) = 4¢>

x =2,

y=—t>

() () 5

But

we let

z

—=

=cosf

V5

z = +/5cos 0,

y=

equations.

z

But

2

14+

4

+

y L

welet

z =2secl, equations.

= — 3y

V3sin6

=sinb.

dy

_

dxr

do

are the parametric

and (%,

9

%)

1

dy _ 5(_7'2)

tan20

forall 0 and

y=3tan6

2 =tané. 3

d.

©

—

dx

1

= 5. “

=12

y = 5sinf d

2sin0

dr

.

dy

@3y= (6)— 3(~2)

dy 3

2

= =secf 2

V3

.

x = 2cosf

forall§ —=

.

y =22 —3(2)=-2

is the point of contact and

do

and

z=23(2)=6

and

L

() -5 sec?0 =1+

dt

d

Y

3

t =2,

b

cos?6+sin?0 =1

_2t-3

dz

whichis

3

y=1t>-3t

Thus, the equation of the tangent is

arc the parametric cquations.

322 4+ 5y2 =15

2 v

F%

(6, —2)

=2t

v

Y93 dt

d

_

de

are the parametric

f a2 =4y

Ifwelet y=—t2,

d

Zdt

2 =sinf.

2=3-1=2

=3t

d

forall§ and

y=1,

_o(_L

-2(F)

—

=5cosl

5cosf —2sinf

y=5sin(%)

= %

is the point of contact.

-

2

Thus, the equation of the tangent is

5z + 2y = 5(%) +2(35)

are the parametric which is

5x+2y:%,

or

5\/5:c+2\/5y:20

329

330

WORKED
0,

and

SOLUTIONS

3-2V2>0,

so we have an ellipse. at+c=1+5=6

MtA=3+2V243-2/2=6

v

341

342

INDEX

INDEX allied angles alternate angles altitude

angle at the centre theorem angle between tangent and chord

angle bisector angle in a semi-circle theorem angles at a point

angles of a quadrilateral angles of a triangle angles on a line

angles subtended by the same arc Apollonius' circle theorem area comparison theorem

asymptote augmented matrix auxiliary angle

axiom axis of symmetry basic unit vectors

basic variable basis centre of an ellipse centre of an hyperbola centroid Ceva's theorem

characteristic polynomial check matrix chord chord of a circle theorem chord of an ellipse circle circumcentre co-domain column rank column space

column vector composition of transformations

concyclic points congruent triangle consistent system of equations corresponding angles

cyclic quadrilateral determinant of a matrix

diagonal matrix diagonalisable matrix

130 130 167 147 149 167 145 130 132 131 130 148 176 142 210 13 219 128,129 202 63 19,22 72 205 210 167 182 116 55 163 146 205 194, 200 168 86 78 78 25 93, 111 156 138 12 130 156 42,46 26 118

diameter of an ellipse

205

diameter of an hyperbola dimension

210 72

directrix

201

discriminant of a conic distance from a point to a line

229 190

domain

86

eccentric angle eccentricity

219 201

eigenbasis

115

eigenspace eigenvalue

115 113

eigenvector

113

elementary matrix elementary row operations elements of a matrix ellipse equal matrices Euclid's angle bisector theorem

55 13 25 200, 201, 204 27 172

exterior angle of a cyclic quadrilateral exterior angle of a triangle

158 131

external tangents

196

focal-distance property of ellipse

206 211

focal-distance property of hyperbola focus free variable Gaussian elimination general form of a circle equation homogenous hyperbola identity matrix image incentre inconsistent system of equations infinitely many solutions intersecting chords theorem intersecting circles theorem inverse of a matrix inverse transformation isosceles triangle theorem

201 19,22 22 194 12 200, 201, 209 26 100 168 12,19 19 163 150 41,42 94 131

kernel latus rectum of an ellipse

86 205

latus rectum of an hyperbola

210

line linear combination of vectors

200 62

linear equation

11

linear transformation linearly dependent vectors

83 70

INDEX linearly independent vectors

70

reduced row echelon form

line-pair

200

reflection

locus lower triangular matrix

190 26

rotation row echelon form

major arc major axis

145 205

row rank

major segment

145

matrix

25 27

matrix addition matrix multiplication

32,33

matrix subtraction median of a triangle

28 167

Menelaus' theorem

185

midpoint theorem minor

131 46

minor arc

145

minor axis minor segment

205 145

mutually orthogonal vectors

61

negative matrices no solution

28 19

non-singular matrix

42

null space nullity

77 77,87

row reduction row space row vector secant

secant-secant theorem secant-tangent theorem segment self-inverse matrix sense shear similar triangles singular matrix skew-symmetric matrix solution set spanning set

standard matrix

169

system of linear equations

range rank

rank-nullity theorem real Cartesian space rectangular hyperbolae

135 217 216 167 13 22 200 120 109 178 147 86 78, 87

89 60 211

163 145 43 106 108 133 42 39 11

107

orthocentre

parallel lines within a triangle parametric differentiation parametric equations perpendicular bisector pivoting pivots point powers ofa matrix projection Ptolemy's theorem for cyclic quadrilaterals radius-tangent theorem

164

standard parametric equations subspace symmetric matrix

200, 201, 202

25 163, 196

89 219

156 25

parabola

78

72

opposite angles of a cyclic quadrilateral order of a matrix

61 12

78 13

standard basis

stretch

orthogonal vectors overspecified

18

square matrix

100

45

103 102

66 26

object

orthogonal matrix

21,56

tangent tangents from an external point theorem on a known solution trace transformation transformation matrix transpose transverse axis trivial solution underspecified unique solution

65 39 11 196 149 97 116 83 100 39 210 23 12 18

unit vectors

63

upper triangular matrix

26

vector

60 72

vector cross-product vector dot product vertically opposite angles zero matrix

6l 130 26

343

344

NOTES