A Level Fur ther Mathematics for AQA

Brighter Thinking

A Level Fur therMathematics for AQAStudent Book 1 (AS/ Year 1)Stephen Ward and Paul Fannon

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Contents

ContentsIntroduction ............................................................. ivHow to use this book ............................................... v

1 Complex numbers

1: Definition and basic arithmetic of i ...................002: Division and complex conjugates ....................003: Geometric representation .................................004: Locus in the complex plane ..............................005: Operations in modulus–argument form ..........00

2 Roots of polynomials

1: Factorising polynomials .....................................002: Complex solutions to polynomial

equations ............................................................003: Roots and coefficients .......................................00 4: Finding an equation with given roots ..............005: Transforming equations .....................................00

3 Ellipses, hyperbolas and parabolas

1: Introducing the ellipse, the hyperbola and the parabola ........................................................00

2: Solving problems with ellipses, hyperbolas and parabolas .....................................................00

3: Transformations of curves .................................00

4 Rational functions and inequalities

1: Cubic and quartic inequalities ..........................00

2: Functions of the form = ++y ax b

cx d ......................00

3: Functions of the form y ax bx cdx ex f

= + ++ +

2

2 ..............00

4: Oblique asymptotes ..........................................00

5 Hyperbolic functions

1: Defining hyperbolic functions ...........................002: Hyperbolic identities .........................................003: Solving harder hyperbolic equations ...............00

6 Polar coordinates

1: Curves in polar coordinates ..............................002: Some features of polar curves ..........................003: Changing between polar and Cartesian

coordinates .........................................................00

Focus on … Proof 1 ............................................. 00

Focus on … Problem solving 1 ........................... 00

Focus on … Modelling 1 ..................................... 00

Cross-topic review exercise 1 .............................00

7 Matrices

1: Addition, subtraction and scalar multiplication ...........................................00

2: Matrix multiplication ..........................................003: Determinants and inverses of ×2 2 matrices ...004: Linear simultaneous equations .........................00

8 Matrix transformations

1: Matrices as linear transformations ....................002: Further transformations in 2D ...........................003: Invariant points and invariant lines ...................004: Transformations in 3D ........................................00

9 Further vectors

1: Vector equation of a line ...................................002: Cartesian equation of the line ..........................003: Intersections of lines ..........................................004: Angles and the scalar product ..........................00

10 Further calculus

1: Volumes of revolution ........................................002: Mean value of a function ...................................00

11 Series

1: Sigma notation ...................................................002: Using standard formulae ...................................003: Method of differences .......................................004: Maclaurin series ..................................................00

12 Proof by induction

1: The principle of induction .................................002: Induction and series ..........................................003: Induction and matrices ......................................004: Induction and divisibility ...................................005: Induction and inequalities .................................00

Focus on … Proof 2 ............................................. 00

Focus on … Problem solving 2 ........................... 00

Focus on … Modelling 2 ..................................... 00

Cross topic review exercise 2 .............................00

Practice paper 1 ....................................................00Formulae .................................................................00Answers to exercises .............................................00Glossary ..................................................................00Index .................................................................... ..00Acknowledgements ...............................................00

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© Cambridge University Press 2017 The third party copyright material that appears in this sample may still be pending clearance and may be subject to change

Before you start…

GCSE You should be able to use the quadratic formula.

1 Solve the equation − + =3 6 2 02x x , giving your answers in simplified surd form.

GCSE You should be able to represent a locus of points on a diagram.

2 Draw two points, P and Q, 5 cm apart. Shade the locus of all the points that are less than 3 cm from P and closer to Q than to P.

AS Level Mathematics Student Book 1, Chapter 5

You should be able to solve simultaneous equations.

3 Solve these simultaneous equations.

+ =+ =

2 5

102 2

x y

x y

Finding the square roots of negative numbersUntil now, every number you have met has been a real number. You will have been told that it is not possible to find the square root of a negative number and that therefore no number squares to give a negative answer. However, in this chapter, you will learn that there is a number that squares to give −1, and so a whole new branch of numbers appears: complex numbers. Remarkably, these numbers have many useful applications in the real world, from electrical impedance in electronics to the Schrödinger equation in quantum mechanics!

Section 1: Definition and basic arithmetic of iThere is no real number that is a solution to the equation = −12x but mathematicians have defined there to be an imaginary number that solves this equation. This imaginary number is given the symbol i.

In this chapter you will learn how to:

• work with a new set of numbers called complex numbers • do arithmetic with complex numbers • use the fact that complex numbers occur in conjugate pairs • interpret complex numbers geometrically • interpret arithmetic with complex numbers as geometric transformations • represent equations and inequalities with complex numbers graphically.

1 Complex numbers

1

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In all other ways, i acts just like a normal constant.

Simplify these terms.

a i3 b i4 c i5

a = ×i i i3 2 Use the normal laws of indices.

= −= −

i

i

( 1) = −i 12 .

b i i i= × = −=

( 1)1

4 2 2 2

c i i i i

i

= × = ×=

15 4

WORKED EXAMPLE 1.1

Remember that the imaginary part of the complex number

i= +z x y is the real number y, not yi .

Common error

Your calculator may have a ‘complex’ mode, which you can use to check your calculations.

Tip

= −i 12

or equivalently, = −i 1

Key point 1.1

You need to be familiar with some common terminology.

• A complex number is one that can be written in the form +x yi where x andy are real. Commonly, z is used to denote an unknown complex number and is used for the set of all complex numbers.

• In this definition, x is the real part of z and it is given the symbol Re( )z ; y is the imaginary part of z and it is given the symbol Im( )z . So, for example, −Re(3 i) is 3 and −Im(3 i) is −1.

You can now do some arithmetic with complex numbers.

WORKED EXAMPLE 1.2

Given that = +z 3 i and = −w 5 2i find the value of each expression.

a +z w b −z w c zw d ÷w 2

a + + − = + + −= −

i i i i

i

(3 ) (5 2 ) (3 5) ( 2 )

8Group the real and the imaginary parts.

b + − − = − + += − +

i i i i

i

(3 ) (5 2 ) (3 5) ( 2 )2 3

Continues on next page

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A Level Further Mathematics for AQA Student Book 1

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The original purpose of introducing complex numbers was to solve quadratic equations with negative discriminants.

See Focus on… Problem solving 1 to find out how some cubic equations can be solved by using a formula.

Focus on...

a Find the value of −4. b Hence solve the equation − + =4 5 02x x .

a − = −4 4 1 Use the standard rules of square roots: =ab a b .

= i2 = −i 1.

b ( 4) 16 4 1 5

2= − − ± − × ×x

= ± −4 42

Use the quadratic formula.

= ± i4 22 Using part a.

= ± i2

WORKED EXAMPLE 1.3

c + × − = × + × − + × + × −= − + −

i i i i i i

i i i

(3 ) (5 2 ) (3 5) (3 2 ) ( 5) ( 2 )

15 6 5 2 2

Multiply out the brackets.

= − − × −i15 (2 1)

= − i17

Remember, = −i 12 .

d − = −

= −

5 22

52

22

2.5

i i

i

Use the normal rules of fractions.

You may be sceptical about the idea of ‘imagining’ numbers. However, when negative numbers were introduced by the Indian mathematician Brahmagupta, in the 7th century, there was just as much scepticism. For example, in Europe, it took until the 17th century for negative numbers to be accepted. Mathematicians had worked successfully for thousands of years without using negative numbers, treating equations such as + =x 3 02 as having no solution. But once negative numbers were ‘invented’ it took only a hundred years or so to accept that equations such as + =3 02x also have solutions.

Did you know

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1 Complex numbers

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When you look at a new number system, one of the most important questions to ask is: ‘When are two numbers equal?’

Although this may seem obvious, in mathematics it pays to be careful. If two rational numbers a

b and cd are equal, it does not mean that =a c and =b d.

Despite its apparent simplicity, this result has some remarkably powerful uses. One is to find square roots of complex numbers.

Notice that there are two numbers that have the same square; this is consistent with what you have already seen with real numbers.

Solve the equation = −8 6i2z .

Let = + iz x y.

= −+ = −

i

i i

8 6

( ) 8 6

2

2

z

x y

z is a complex number, so write = +z x yi , where x and y are real.

+ + = −i i i2 ( ) 8 62 2x xy y Expand the brackets.

+ − = −i i2 8 62 2x xy y = −i 12

: 8 (1)2 2x yRe − =

: 2 6 (2)xyIm = −

Now equate the real parts and equate the imaginary parts.

From (2): = −y x3

Substituting into (1):

− =9 822x

x

Use substitution to solve these simultaneous equations.

− =− − =

9 8

8 9 0

4 2

4 2

x x

x x

Multiplying through by 2x leads to a disguised quadratic.

− + =− = + =

= ±

( 9)( 1) 0

9 0 or 1 03

2 2

2 2

x x

x x

x

There are two possible values for 2x . However, x is real so + =1 02x is impossible.

= −

= −

y x3

1 or 1

Find the value of y for each x value.

∴ = − − +3 or 3i iz Write the answer in the form = +z x yi .

WORKED EXAMPLE 1.4

If two complex numbers are equal, then their real parts are the same and their imaginary parts are the same.

Key point 1.2

4


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EXERCISE 1A

1 State the imaginary part of each complex number.

a i − +3 5i ii −8 2i b i +6 i ii −19 i

c i −2i 8 ii −7i 2 d i 15 ii −3i

e i i2 ii + −(1 i) i

f i + + − ∈a b a b1 i i for , ii − − − ∈b a a b2 4i ( i ) for ,

2 Simplify each expression, giving your answers in the form +x yi .

a i +2i 3i ii −i 9i b i 5i2 ii −i2

c i −( 3i)2 ii (4i)2 d i + − −(4i 3) (6i 2) ii − − −2(2i 1) 3(4 2i)


a i +i(1 i) ii −3i(2 5i) b i + +(2 i)(1 2i) ii + +(5 2i)(4 3i)

c i + −(2 3i)(1 2i) ii + −(3 i)(5 i) d i +(3 i)2 ii −(4 3i)2

e i +

−

3

212 i 3

212 i ii + −(3 2i)(3 2i)


a i +6 8i2 ii −9 3i

3 b i +5 2i10 ii −i 4

8

c i + +3 i2 i ii − −9i 6 4i

2

5 Evaluate each expression, giving your answers in the form +x yi .

a i −4 ii −49 b i −8 ii −50

c i − −4 363 ii − + −1 25

3 d i + −2 16 256 ii − −5 2 4 9

4

6 Solve the equations, simplifying your answers.

a i + = 9 02x ii + =36 02x b i = −102x ii = −132x

c i − + = 2 5 02x x ii − + =10 02x x d i + =3 20 62x x ii + = −6 5 5 2x x

7 Evaluate, simplifying your answers.

a i i3 ii i4 b i −( 2i)4 ii −( 5i)3

c i −(1 3i)3 ii +( 3 i)3 d i +

3

212

i3

ii − +

3

212

i3

8 Find real numbers a and b such that:

a i + − = +a b( i)(3 2i) 5i 1 ii + + =a b(6 i)( i) 2

b i + + = −a b( 2i)(1 2i) 4 i ii + + = +a b(1 i)(1 i) 2i

c i + + = − −a b a b( i)(2 i) 2 ( 1)i ii + = −a b ai( i) 6i

9 By writing = +z x yi , solve these equations.

a i = −4i2z ii = 9i2z b i = +2 2 3i2z ii = +5 i2z

5

1 Complex numbers

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10 Find the exact values of , a b ∈ such that + − = −a b(3 i)( i) 4i. Give your answers in the form k 3.

11 Find the exact values of a b, ∈ such that + + = + −a b b a(1 i)(1 i) 9i .

12 By writing = +z x yi , solve the equation + = −z zi 2 i 3 .

13 a Find values x and y such that + + = −x y( i )(2 i) i.

b Hence express − +3i

2 i in the form +x yi .

14 By writing = + iz x y, solve the equation = − −3 4i2z .

15 Solve the equation = i2z .

16 Use an algebraic method to solve the equation = −12 5i2z .

If = +z x yi , then the complex conjugate of z , z x y* i= −

Key point 1.3

Section 2: Division and complex conjugatesIn Section 1, you saw that quadratic equations may have two complex roots. In Worked example 1.3, the roots were +2 i and −2 i; the imaginary part arises from the term after the ± sign in the quadratic formula. These two numbers form a complex conjugate pair. They differ only in the sign of the imaginary part.

So for example, if = +z 2 i then = −* 2 iz , or if = −w 3 4i then = +w* 3 4i.

At first, the concept of conjugates may not appear particularly useful, but you will need them when you are dividing complex numbers.

Write +−

3 2i5 i in the form +x yi .

ii

i) ( i)i) ( i)

3 25

3 2 55 5

+− = + +

− +((

Multiply the numerator and the denominator by the complex conjugate of the denominator, +5 i.

= + + +−

i i ii

15 10 3 25

2

2 2Use the difference of two squares in the denominator.

= +− −

= +

13 1325 ( 1)

0.5 0.5

i

i

i 12 = − .

WORKED EXAMPLE 1.5

In Chapter 2 you will use the fact that complex roots occur in conjugate pairs to factorise and solve cubic and quartic equations.

Fast forward

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Worked example 1.5 shows the general procedure for dividing by a complex number.

You can prove that this procedure always results in a real number in the denominator.

Worked example 1.6 shows how to use the idea of equating real and imaginary parts to solve equations involving complex conjugates.

To divide by a complex number, write as a fraction and multiply the numerator and the denominator by the conjugate of the complex number in the denominator.

Key point 1.4

This procedure should remind you of rationalising the denominator when working with surds, which is also based on the difference of two squares.

Tip

PROOF 1

Prove that *zz is always real.

Let = +z x yi , where x and y are real.

Then = −z x y* i .

Writing z and *z in terms of their real and imaginary parts is often the best way to prove results about complex conjugates.

So: ( )( )* i i= + −zz x y x y

= − + −i i i2 2 2x xy yx y

( )2 2

2 2

= − −= +

x y

x y= −i 12 .

which is real. x and y are real numbers.

Find the complex number z such that + = +3 2 * 5 2i.z z

Let = + iz x y.

Then: + = + iz z*3 2 5 2

+ + − = +i i ix y x y3( ) 2( ) 5 2

+ = +i ix y5 5 2

As before, the best way to describe complex conjugates is in terms of their real and imaginary parts.

: 5 5z xRe =: 2z yIm =

You can now equate real and imaginary parts.

= =1, 2x y

So = +1 2iz

WORKED EXAMPLE 1.6

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1 Complex numbers

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Being able to divide complex numbers means that you can solve more complicated equations.

EXERCISE 1B

1 Find the complex conjugate of each number.

a i −2 3i ii +4 4i b i −i 3 ii +3i 2

c i 3i ii −i d i −45 ii 9

2 Write in the form +x yi . Use a calculator to check your answers.

a i −+

3 2i1 2i ii −

4i3 5i b i 4

i ii − 1i

c i +−

4 i4 i ii +

−2i 12i 1 d i +

−(1 i)

1 i

2

ii −+

(i 2)i 2

2

3 Solve these equations.

a − = − +z z2 3 4 3(i ) b + = −z z2i 1 4i( 3)

4 Solve these simultaneous equations.

a − =+ + = −

z w

z w

2 3i 5

(1 i) 3 4i

b + + − =− + =z w

z w

(1 i) (1 i) 1

(1 i) 2i i

Solve these simultaneous equations.+ + − = ++ + = − +

z w

z w

(1 i) (2 i) 3 4i

i (3 i) 1 5i

+ + − = ++ + + + = + − +

(1 ) (2 ) (3 4 )(1 ) (1 )(3 ) (1 )( 1 5 )

z w

z w

i i i i i i

i i i i i i

+ + + = − ++ + + = − +

(1 ) (2 1) 4 3(1 ) (2 4 ) 6 4

z w

z w

i i i i

i i i i

The best method to use here is elimination: multiply the first equation by i and the second by +(1 i) …

Subtracting:− − = −i iw( 1 2 ) 2

… and then subtract them to eliminate z.

= −− − × − +

− +

= + =

w 21 2

1 21 2

51 4

ii

ii

i i

Divide by − −( 1 2i): multiply top and bottom by the complex conjugate.

Substituting into the second equation:

+ + = − +i i iz w(3 ) 1 5

+ + = − +i i i iz (3 )( ) 1 5

+ − + = − +i i iz ( 1 3 ) 1 5

=i iz 2=z 2

Substitute back to find z: use the second equation as it looks simpler.

So =z 2, = iw .

WORKED EXAMPLE 1.7

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5 Find the complex number, z, if:

a − =2 * 1 4iz b + =3 * 2 9iz

6 By writing = +z x yi , solve these equations.

a + = −2 * 2 7iz z b + = − −2 i * 3 iz z

7 If x and y are real numbers find the complex conjugate *z when:

a i = + +z x y3 ( i ) ii = − −z x y(2 i ) b i = + + −z x y( 3i ) (2 i) ii = + − −z x y(3 3i) ( i )

c i + + +x y x yi 1i ii + − +x y x yi 1

i d i + − −x

x yx

x yi i ii + + −x

x yx

x yi i

8 Write −−

3 5i2i 1 in the form +a bi. Show all your working.

9 a Let = +z x yi . Find, in terms of x and y , the real and imaginary parts of +3i 2 *z z .

b Find the complex number z such that + = −3i 2 * 4 4iz z .

10 Find real numbers x and y such that + = +3i 4i *x y z z , where = +z 2 i.

11 By writing = +z x yi , prove that =( *) ( )*2 2z z .

12 Solve + =3 * iz z .

13 Solve + = −i 1 *z z .

14 If = +z x yi , find the real and the imaginary parts of +z

z 1 in terms of x and y , simplifying your answers as far as possible.

Section 3: Geometric representationYou can represent real numbers on a number line.

10 2 3 4–1–2–3–4

To represent complex numbers you can add another axis, perpendicular to the real number line, to show the imaginary part. This is called an Argand diagram.

Im

ReO 1 2 3 4

1

2

3

4

–1

–2

–3

–4

–1–2–3–4

Using real and imaginary axes was first suggested by the land surveyor Caspar Wessel. However, it is named after Jean-Robert Argand who popularised the idea.

Did you know

9

1 Complex numbers

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Operations with complex numbers have useful representations on Argand diagrams.

Consider two numbers, = + i1 1 1z x y and = + i2 2 2z x y . On the diagram they are represented by points with coordinates ( , )1 1x y and ( , )2 2x y . Their sum, + = + + +( ) i( )1 2 1 2 1 2z z x x y y , has coordinates + +( , )1 2 1 2x x y y . But you can also think of coordinates as position vectors:

+

=

++

x

y

x

y

x x

y y1

1

2

2

1 2

1 2

So you can add complex numbers geometrically in the same way as you add vectors.

Im

ReO

z1

+ z2

z1

z2

You can represent subtraction similarly, by adding the negative of the second number.

Im

ReO

z1

z2

z1– z

2

–z2

Represent −3 2i on an Argand diagram.

Im

ReO

3

z = 3 – 2i–2

WORKED EXAMPLE 1.8

The point representing 3 - 2i has coordinates (3, -2).

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Taking a complex conjugate results in a reflection in the real axis.

Im

ReO

z

z *

Two complex numbers, z and w, are shown on this Argand diagram.

Im

ReO

w

z

Represent these complex numbers on the same diagram.

a −w z b *z

Im

ReO

w

z

z*–z

w – z

a To find −w z , first label −z and then draw a parallelogram.

b *z is the reflection of z in the real axis.

WORKED EXAMPLE 1.9

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1 Complex numbers

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EXERCISE 1C

1 Represent each number on an Argand diagram. Use a separate diagram for each part.

a i = +z 4 i and =w 2i ii = − +z 3 i and = −w 3i

b i Let = +z 4 3i. Represent z, −z and zi . ii Let = − +z 2 5i. Represent z , −z and zi .

c i Let = +z 2 i. Represent z, z3 , − z2 and − zi . ii Let = −z 1 3i. Represent z, z2 and z2i .

2 Solve each quadratic equation and represent the solutions on an Argand diagram.

a i − + =4 13 02z z ii − + =10 26 02z z

b i + + =2 17 02z z ii + + =6 13 02z z

3 Each diagram shows two complex numbers, z and w. Copy the diagrams and add the points corresponding to *z , −w and +z w .

a i ii Im

ReO

w

z

Im

ReO

w

z

b i ii Im

ReO

w

z

Im

ReO

w

z

See Focus on… Modelling 1 to find out how complex numbers can be used in electronics.

Focus on...Modulus and argument

When a complex number is represented on an Argand diagram, its distance from the origin is called the modulus. There are lots of complex numbers with the same modulus; they form a circle centred at the origin. You can uniquely describe a particular complex number by giving its angle relative to the real axis; this is called the argument. The argument is conventionally measured anti-clockwise from the positive real axis.

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The symbols used for these are:

• z or r for the modulus• zarg or θ for the argument.

Im

ReO

|z|

z

arg z

The notation θ[ ]= ,z r is sometimes used for a complex number z with modulus r and argument θ .

This description of a complex number is called the modulus–argument form.

The description in terms of the real and imaginary parts, = +z x yi , is called the Cartesian form.

Note here that the argument is not usually stated in degrees but in an alternative measure of angle called radians.

Radians

The radian is the most commonly used unit of angle in advanced mathematics.

You can deduce the sizes of other common angles: for example, a right angle is one quarter of a full turn, so it is π ÷ = π2 4 2 radians.

Although sizes of common angles measured in radians are often expressed as fractions of π, you can also use decimal approximations: for example, a right angle measures approximately 1.57 radians.

° = π360 2 radians

Key point 1.5

Radians are covered in more detail in A Level Mathematics Student Book 2, Chapter 7.

Rewind

13

1 Complex numbers

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a Convert °75 to radians. b Convert 2.5 radians to degrees.

a =75360

524

°75 is 75360 as a fraction of a full turn.

× π = π524 2 5

12Calculate the same fraction of π2 .

∴ ° = π75 512 radians This is the exact answer. By evaluating π5

12 on a

calculator you can also give a decimal answer: 75°= 1.31 radians (3 s.f.).

b π2.52

2.5 radians is π2.52 as a fraction of a full turn.

π × = …2.52 360 143.24

∴ = °2.5 radians 143 (3 s.f.)

Calculate the same fraction of °360 .

WORKED EXAMPLE 1.10

It is worth remembering some common angles in radians.

Degrees 0° 30° 45° 60° 90° 180°

Radians 0 6π

4π

3π

2π π

Tip

When you need to evaluate trigonometric functions of angles in radians, make sure that your calculator is set to radian mode.

Tip

EXERCISE 1D

1 Express each angle in radians, giving your answer in terms of π.

a i °135 ii °45 b i °90 ii °270

c i °120 ii °150 d i °50 ii °80

2 Express each angle in radians, correct to 3 decimal places.

a i °320 ii °20 b i °270 ii °90

c i °65 ii °145 d i °100 ii °83

3 Express each angle in degrees.

a i π3 ii π

4 b i π56 ii π2

3

c i π32 ii π5

3 d i 1.22 ii 4.63

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Converting between the modulus–argument and Cartesian forms

To convert between the modulus–argument and Cartesian forms you can use a diagram and some trigonometry.

Im

ReO

|z|

z = x + iy

θx = |z| cos θ

y = |z| sin θ

To convert from modulus–argument form to Cartesian form:

θ= cosx z

θ= siny z

To convert from Cartesian to modulus–argument form:

= +2 2z x y

θ =yx

tan

Key point 1.6

The argument can be measured either between 0 and π2 or between −π and π. It will be made clear in the question which is required.

Always draw the complex number on an Argand diagram before finding the argument. It is important to know which angle you need to find.

Tip

A complex number z has modulus 3 and argument π6 . Write the number in Cartesian form.

= π =x 3 63 3

2cos

= π =y 3 632sin

Use θ=x z cos and θ=y z sin .

∴ = +z 3 32

32 i The Cartesian form is = +z x yi .

WORKED EXAMPLE 1.11

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WORKED EXAMPLE 1.12

Write -2 3i in modulus–argument form, with the argument between 0 and π2 .

Give your answer correct to 3 s.f.

= + -

== .

2 ( 3)

133 61 (3 s.f.)

2 2z Use = +2 2z x y .

Before attempting to find the argument, draw a diagram to see where the complex number actually is.

α

α

=

∴ =

32

0.983

tan Calculate the angle α. Remember to use radian mode on your calculator.

θ = π -=

2 0.9835.30 (3 s.f.)

From the diagram, θ α= π -2 .

So = z 3.61 and arg z 5.30= (3 s.f.)

Im

ReO

|z|

z = 2 – 3i

θ

α

2

–3

Find the argument of = - -z 5 3i, where -π < πzarg ø .

Which is the correct solution? Identify the errors made in the incorrect solutions.

Solution 1 Solution 2 Solution 3

( ) =arctan 35 0.540

π - =0.540 2.60

∴ = -zarg 2.60

)(= -

= -

zarg arctan 35

0.540

( )-- =arctan 3

5 0.540

= π -=

zarg 2 0.540

5.74

WORK IT OUT 1.1

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There is another nice way of expressing complex numbers which, although strictly in Cartesian form, explicitly shows the modulus and argument.

θ θθ θ

= +

= +

= +

z x y

z z

z

i

cos i sin

(cos i sin )

A complex number z with modulus r and argument θ can be written as

θ θ= +z r(cos i sin )

Key point 1.7

In Further Mathematics Student Book 2 you’ll see another way of writing complex numbers that involves the modulus and argument.

Fast forward

When writing a number in this form, it is important to notice that there must be a plus sign between the two terms and the modulus must be a positive number; otherwise you need to draw a diagram to find the argument.

Write ( )= − π − πz 2 cos 3 i sin 3 in the form r(cos i sin )θ θ+ , with the argument between −π and π.

= − π + π2 3 2 3cos i sinz In order to draw the diagram, you need to identify the signs of the real and imaginary parts.

= − π <

= π >

z

z

Re( ) 2 cos 3 0

Im( ) 2 sin 3 0

From the diagram:

θ= = = πz z2, 23arg

2 23

23z cos i sin )(∴ = π + π

θ = π − π = π3

23

⎧ ⎨ ⎩

⎧⎨⎩ θ

Im

Im

ReO

Re

π3

z

WORKED EXAMPLE 1.13

An Argand diagram can help to identify the modulus and argument of a negative and complex conjugate of a number.

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WORKED EXAMPLE 1.14

A complex number (cos i sin )θ θ= +z r is shown in the diagram. The argument is measured between −π and π.

θ

Im

Re

r

z

On the same diagram, mark the numbers −z and *z . Hence write −z and *z in modulus–argument form.

To find −z, make both real and imaginary parts negative.

*z is the reflection of z in the real axis.

z z r− = =* All three points are the same distance from the origin, so all have the same modulus.

θθθ

− = − π −= − π

= −

( ) ( )

( )

arg

arg

z

z*

As the argument is measured between −π and π, both numbers on the diagram have a negative argument.

( ( ) ( ))( ( ) ( ))

*

cos i sin

cos i sin

θ θθ θ

∴ − = − π + − π= − + −

z r

z rWrite both numbers in modulus–argument form.

θ

Im

ReO

r

z

z*–z

EXERCISE 1E

1 Find the modulus and the argument (measured between −π and π) for each number.

a i 6 ii 13 b i −3 ii −1.6

c i 4i ii 0.5i d i − i2 ii −5i

e i +1 i ii +2 3i f i − −1 3i ii −4 4i

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2 Find the modulus and the argument (measured between 0 and π2 ) for each number.

a i +4 2i ii −4 3i

b i −i 3 ii +6i 2

c i − −3 i ii − +3 2i

3 Given the modulus and the argument of z, write z in Cartesian form.

a i = = πz z4, arg 3 ii = = πz z2 , arg 4

b i = = πz z2, arg 34 ii = = πz z2, arg 2

3c i = = − πz z3, arg 2 ii = = −πz z4, arg

4 Write each complex number in Cartesian form without using trigonometric functions. Display each one on an Argand diagram.

a i π + πcos 3 i sin 3 ii π + πcos 34 i sin 3

4

b i 3 cos 2 i sin 2 )( π + π ii ( )( ) ( )− π + − π5 cos2

i sin2

c i 4(cos 0 i sin 0)+ ii π + πcos i sin

5 Write each complex number in the form (cos i sin )θ θ+z .

a i =z 4i ii = −z 5

b i = −z 2 2 3i ii = +z 3 i3

6 Find the modulus and the argument of each number.

a i ( )π + π4 cos 3 i sin 3 ii ( )π + π7 cos 37 i sin 3

7

b i π + πcos 5 i sin 5 ii ( ) ( )− π + − πcos 4 i sin 4

c i ( )π − π3 cos 8 i sin 8 ii ( )π − π7 cos 45 i sin 4

5

d i ( )− π + π10 cos 3 i sin 3 ii ( )− π + π2 cos 6 i sin 6

e i 6 cos 10 i sin 10( ) ( )− π + − π

ii

7 a Write ( )π + π3 cos 74 i sin 7

4 in Cartesian form in terms of surds only.

b Write −4i 4 in the form (cos i sin )θ θ+z .

8 Let = +z 1 3i and = −w 3 3 3i.

a Find the modulus and the argument of z and w.

b Represent z and w on the same Argand diagram.

c Find the modulus and the argument of zw. Comment on your answer.

12 cos 3 i sin 3( ) ( )− π − − π

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Section 4: Locus in the complex planeYou have met various examples of representing equations and inequalities graphically. You can represent an equation involving two variables, such as − =x y2 5, on a graph, show an inequality such as

< x2 5ø on the number line and shade a region corresponding to an inequality in two variables, such as 3 2y xù , on a graph.

O 5

y

x

x – 2y = 5

–2.5

1 2 3 4 5

O

y

x

y = 3 x 2

Now that you can represent complex numbers as points on an Argand diagram, there is a way to show equations and inequalities involving complex numbers graphically.

Locus involving the modulus

In Section 3, you defined the modulus of a complex number, z, as its distance from the origin, O, on an Argand diagram. This means that all complex numbers with the same modulus, r, form a circle around the origin, with radius r.

You can extend this idea to measure the distance between any two complex numbers on an Argand diagram.

If you write two numbers in Cartesian form, = + i1 1 1z x y and = + i2 2 2z x y , then:

− = − + −( ) i( )1 2 1 2 1 2z z x x y y

so:

− = − + −( ) ( )1 2 1 22

1 22z z x x y y

Notice that the last expression is the distance between the points with coordinates ( , )1 1x y and ( , )2 2x y .

Im

ORe

|z| = r

z

9 If (cos i sin )θ θ= +z r , write in terms of r and θ , simplifying your answers as far as possible.

a + *z z b *zz c *

zz

10 If θ θ= +z cos i sin , express the real and imaginary parts of −+

zz

11 in

terms of θ , simplifying your answer as far as possible.

Part c requires a double angle formula from A Level Mathematics Student Book 2, Chapter 8.

Rewind

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The distance between points representing complex numbers 1z and 2z on the Argand diagram is given by −1 2z z .

Key point 1.8

You can now find complex numbers that satisfy equations or inequalities involving the modulus by thinking about the geometric interpretation, rather than by doing calculations.

Show on an Argand diagram a set of points satisfying the equation − =z 2 3.

The points will lie on a circle, centre 2, radius 3. The equation says ‘the distance between z and 2 is 3’.

Im

ReO 2 5–1

WORKED EXAMPLE 1.15

The equation − =z a r represents a circle with radius r and centre at the point a.

Key point 1.9

Shade on an Argand diagram the set of points satisfying the inequality − + <z 2 3i 2.

− + <z 2 3 2i

− − <z (2 3 ) 2i

You need to write the inequality in the form − <z a 2.

The points will lie on a circle, centre −2 3i, radius 2. The inequality can be interpreted as saying, ‘the distance between z and

−2 3i is less than 2’.

Remember that the dashed line means that the circle itself is not included.

WORKED EXAMPLE 1.16

Im

ReO

2–3i–3

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You can also describe the inequality in Worked example 1.16 by using x y( , ) coordinates and set theory notation. The circle with centre −(2, 3)

and radius 2 has equation − + + =( 2) ( 3) 42 2x y , so the set of all points satisfying the inequality is x y x y{ i : ( 2) ( 3) 4}2 2+ − + + < .

Sketch the locus of points in the Argand diagram that satisfy − = −z z4 2i .

The equation says ‘the distance from z to 4 is the same as the distance from z to 2i’.

This is the perpendicular bisector of the line segment joining the points 4 and 2i.

O

Im

Re

2i

4

WORKED EXAMPLE 1.17

Locus involving the argument

The argument measures the angle that the line connecting z to the origin makes with the real axis. So the three points shown on the diagram all have the same argument, θ.

Since the argument is measured from the positive x -axis, any number on the dashed part of the line would have argument θ + π (or θ − π if the argument is negative).

If you replace z by −z a, the line shifts from the origin to the point a.

A locus is a set of all the points that satisfy a given condition.

Tip

A half-line is a line extending from a point in only one direction.

Tip

The locus of points satisfying arg( )z a θ− = is the half-line starting from the point a and making angle θ with the positive x-axis.

Im

ReO

θa

Key point 1.10

θ

θ – π

Im

ReO

z1

z2

z3

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Sketch the locus of points on the Argand diagram that satisfy arg( 3 2i) 4z − + = π .

− + = π( 3 2 ) 4arg iz

− − = π( (3 2 )) 4arg iz

You need to write the expression in brackets in the form θ− =arg( )z a .

The half-line starts from the point −3 2i and makes a °45 angle with the horizontal.

Im

ReO 3

–2

π4

WORKED EXAMPLE 1.18

WORKED EXAMPLE 1.19

On a single diagram, shade the locus of points that satisfy the inequalities < + − < π0 arg( 1 i) 23

z and z 3ø .

< + − < π0 ( 1 ) 23arg iz

< − − + < π0 ( ( 1 )) 23arg iz

The first inequality represents the region between two half-lines, starting from the point − +1 i and making

angles 0 and π23 with the vertical axis.

The second inequality represents the inside of the circle with radius 3 and centre at the origin.

The half-lines should be dashed and the circle solid.

You can indicate the required region by shading.

Im

Re

–1 + i

23

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EXERCISE 1F

1 Sketch each locus on an Argand diagram. Shade the required regions where relevant.

a i − =z 2i 5 ii − =z 3 5 b i + =z 4 1 ii + =z 3i 2

c i − i 2z < ii + >z i 3 d i − + >z 3 i 2 ii + −1 2i 1z <

2 The set of points from Question 1 part ai can be described as x y x yi : ( 2) 252 2{ }+ + − = .

Write the sets of points for the rest of Question 1 using similar notation.

3 Sketch each locus on an Argand diagram. Shade the required regions where relevant.

a i − = +z z2 2i ii + = −z z1 i

b i − < +z z3i i ii + > −z z2 3

4 Sketch each locus on an Argand diagram.

a i = πzarg 3 ii = πzarg 4 b i = − πzarg 6 ii = − πzarg 34

c i = πzarg 2 ii = πzarg

5 On an Argand diagram, shade the region where < −z1 3i 3ø .

6 On an Argand diagram, sketch the locus of points where − = +z z3i 6 .

7 On an Argand diagram, shade the region where π < < πz6 arg 23 .

8 a On the same diagram, sketch the loci of + = −z z1 3 and = πzarg 4 .

b Hence find the complex number z that satisfies both equations.

9 On an Argand diagram, shade the region where − − <z 3 i 3 and < πzarg 3.

10 On an Argand diagram, shade the locus of points that satisfy Re( ) 1z ø , <Im( ) 3z and < πzø0 arg 34 .

11 Find the complex number z that satisfies − = + − = −z z z z3i i and 1 i .

12 z is a complex number satisfying − − =z 2 2i 2. Find the maximum possible value of zarg .

Section 5: Operations in modulus–argument formAddition and subtraction are quite easy in Cartesian form but multiplication and division are more difficult. Raising to a large power is even harder. These operations are much easier in modulus–argument form, thanks to the following result.

When you multiply two complex numbers you multiply their moduli and add their arguments.

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• =zw z w• = +arg( ) arg argzw z w

Key point 1.11

The proof requires the use of compound angle formulae from the A Level Mathematics course:

+ = +A B A B B Asin ( ) sin cos sin cos

+ = −A B A B A Bcos ( ) cos cos sin sin

PROOF 2

Let: θ θ= +cos sin( i )1 1 1 1z r and

θ θ= +cos sin( i )2 2 2 2z rStart by introducing variables.

θ θ θ θ= + +cos sin cos sin( i )( i )1 2 1 2 1 1 2 2z z r r

θ θ θ θθ θ θ θ

= −+ +

cos cos sin sin

sin cos sin cos

(i ( ))

1 2 1 2 1 2

1 2 2 1

r r

Multiply them together and group real and imaginary parts.

θ θ θ θ= + + +cos sin( ( ) i ( ))1 2 1 2 1 2r r Now use the compound angle formulae given above.

So: =1 2 1 2z z r r

θ θ= +arg( )1 2 1 2z z

This is in the form (cos i sin )θ θ+z so, by comparison, you can state the modulus and argument of 1 2z z .

i.e. =1 2 1 2z z z z and = +arg arg arg( )1 2 1 2z z z z

Finish the proof with a conclusion.

A similar proof gives the result for dividing complex numbers.

When you divide two complex numbers you divide their moduli and subtract their arguments.

• =zw

zw

• arg arg argzw z w( ) = −

Key point 1.12

Compound angle formulae are covered in A Level Mathematics Student Book 2, Chapter 8.

Rewind

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a If ( )= π + πz 3 cos 3 i sin 3 and ( )= π + πw 5 cos 4 i sin 4 , write zw in the form (cos i sin )θ θ+r .

b Write )

)((

π + π

π + π

6 cos 23

isin 23

3 cos6

i sin6

in the form +x yi .

a = × =zw 3 5 15

= π + π = πzw( ) 3 4712arg

15 712

712cos i sin )(∴ = π + πzw

Multiply the moduli and add the arguments.

b = =r 63 2

θ = π − π = π23 6 2

= π =x 2 2 0cos

= π =y 2 2 1sin

cos i sin

cos i sini

))

((∴

π + π

π + π =6 2

323

3 6 6

Then convert to Cartesian form: θ θ= =x r y rcos , sin .

WORKED EXAMPLE 1.20

Find the argument of ( ) ( )= π + π × π + πz 3 cos 95 i sin 9

5 5 cos 43 i sin 4

3 .


= π × π

= π

zarg 95

43

125

2

π + π = π95

43

4715

So: = π − π

= π

zarg 4715 2

1715

= π + π

= π

zarg 95

43

4715

WORK IT OUT 1.2

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EXERCISE 1G

1 Simplify each expression, giving your answers in the form (cos i sin )θ θ+r .

a i ( ) ( )π + π × π + π3 cos 6 i sin 6 7 cos 5 i sin 5 ii ( ) ( ) ( )− π + − π

× π + πcos 9 i sin 9 4 cos 3 i sin 3

b i ++

8 (cos 6 i sin 6)2 (cos 2 i sin 2) ii

( )( )

π + π

π + π

15 cos 7 i sin 7

5 cos 2 i sin 2

2 Write each of these expressions in the form cos i sinθ θ+ , where arg z−π < πø .

a i ( ) ( )π + π π + πcos 3 i sin 3 cos 34 i sin 3

4 ii ( ) ( )π + π π + πcos 25 i sin 2

5 cos 4 i sin 4

b i ( ) ( )π + π π − πcos 3 i sin 3 cos 4 i sin 4 ii ( ) ( )π + π π − πcos 23 i sin 2

3 cos 25 i sin 2

5

c i π + π

π + πcos 3

5 i sin 35

cos 4 i sin 4

ii π + π

π + πcos 3 i sin 3

cos 34 i sin 3

4

d i π + π

π − πcos 5 i sin 5

cos 4 i sin 4

ii π + π

π − πcos 4 i sin 4

cos 25 i sin 2

5

3 Write ( )

( )π + π

π + π

6 cos 23 i sin 2

3

2 cos 4 i sin 4

in the form θ θ+r(cos i sin ).

4 Write ( )( )

π + π

π + π

8 cos 56 i sin 5

6

4 cos 23 i sin 2

3

in the form + ix y.

5 Let cos 4 i sin 4= π + πz and cos 3 i sin 3= π + πw .

a Write zw in the form +x yi , without using trigonometric functions.

b Hence find the exact value of πtan 712 .

6 Use trigonometric identities to show that θ θ θ θ θ θ+ = − + − = π − + π −1cos i sin cos( ) i sin( ) cos(2 ) i sin(2 ).

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• A complex number is of the form = +z x yi , where x and y are real and = −i 1.• Apart from the fact that = −i 12 , the arithmetic of complex numbers is the same as for real numbers.• To divide by a complex number, multiply top and bottom by its complex conjugate = −( * i )z x y .

• It is useful to represent complex numbers geometrically using an Argand diagram. Addition can be represented on a diagram in the same way as adding vectors.

• There are two ways of describing numbers in the Argand diagram: Cartesian form +x y( i ) and modulus–argument form θ θ+r(cos i sin ). The two forms are linked by:

• = +2 2r x y

• θ = yxtan

• θ=x r cos• θ=y r sin

• Multiplication and division are easier in modulus–argument form:• = +zw z warg( ) arg arg

• arg arg argzw z w( ) = −

• =zw z w

• =zw

zw

• Equations and inequalities can be represented on an Argand diagram:• − =z a r represents a circle with centre a and radius r• − = −z a z b represents the perpendicular bisector of the line segment connecting points a and b• arg( )z a θ− = represents a half-line starting from a and making angle θ with the positive real axis• vertical and horizontal lines are represented by equations of the form =Re( )z c or =Im( )z c, respectively.

Checklist of learning and understanding

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Mixed practice 1 1 If = − +z 2 3i, find an expression for zarg . Choose from these options.

A ( )−tan 32

1 B ( )−−tan 3

21 C ( )π − −tan 3

21 D ( )π + −tan 3

21

2 What is ( *)*zz is equivalent to? Choose from these options.

A z B *z z C (Re )2z D 2z

3 Let = −1 i1z and = +3 5i2z .

Showing all your working clearly, find, in the form +x yi :

a 1 2z z b 1

2

zz

4 Express = −+

z 3i 2i 3

in the form +x yi .

5 a For the complex number = − +z 2 5i, find:

i z ii zarg , where −π < πzarg ø .

b State the modulus and argument of *z .

6 a Solve the equation + + =14 53 02z z .

b Represent the solutions on an Argand diagram.

7 Find the complex number z such that − = −3 5 * 4 3iz z .

8 On an Argand diagram, illustrate the locus of points z that satisfy the inequality < − +z3 3 4i 5ø .

9 It is given that = +z x yi , where x and y are real numbers.

a Find, in terms of x and y , the real and imaginary parts of − −(1 2i) *z z .

b Hence find the complex number z such that − − = +(1 2i) * 10(2 i)z z .

[©AQA 2010]

10 Find, in terms of w, the complex number that satisfies both − =z w 2 2 and − = πarg( )4

z w .

Choose from these options.

A +w 2 2 B + +w 1 i C + +w 2 2i D + +w 2 2i

11 Given that = +w 1 3i and = +z 1 i show that +−

=z w

z wRe 2

20.

12 Let z and w be complex numbers satisfying +− = +

−ii

11

ww

zz

.

a Express w in terms of z.

b Show that, if =Im( ) 0z , then =Re( ) 0w

13 Represent on an Argand diagram the region defined by the inequalities < − π0 arg( 2i) 34

z ø and −z 2i 2ù .

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14 Two complex numbers, a and b, are shown on the Argand diagram.

a Add the points representing the numbers +a b and *a to the diagram.

b Sketch the locus of points z that satisfy − = −z a z b .

15 Two loci, 1L and 2L , on an Argand diagram are given by:

L z: 6 5i 4 21 + − =

L z: arg( i) 342 + = π

The point P represents the complex number − +2 i.

a Verify that the point P is a point of intersection of 1L and 2L .

b Sketch 1L and 2L on one Argand diagram.

c The point Q is also a point of intersection of 1L and 2L . Find the complex number that is represented by Q.

16 Let z and w be complex numbers such that = −w z1

1 and = 12

z . Find the real part of w.

17 If z cos i sinθ θ= + prove that θ−+ =1

1i tan

2

2zz .

[©AQA 2013]

18 If z and w are complex numbers, solve the simultaneous equations:

+ = +z w3 9 11i− = − −w zi 8 2i

19 Let = +6 i 221z , and = −1 i2z .

a Show that = π + πcos 512

i sin 512

1

2

zz

.

b Find the value of 1

2

zz

in the form +a bi, where a and b are to be determined exactly in surd form.

Hence find the exact values of cos 125π and πsin 5

12 .

20 By considering the product + +(2 i)(3 i), show that + = πarctan 12 arctan 1

3 4.

21 If θ< < π0 2 and θ θ= + −(sin i(1 cos ))2z find zarg in its simplest form.

Im

ReO

a

b

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Before you start…

AS Level Mathematics Student Book 1, Chapter 4

You should be able to use the factor theorem to factorise cubic polynomials.

1 a Show that 2x = is a root off( ) 2 9 7 63 2x x x x= - + + .

b Hence factorise xf( ) completely.

Chapter 1 You should be able to perform arithmetic with complex numbers and solve quadratic equations with complex roots.

2 a Expand and simplify (3 2i)(5 i)- + .b Solve the equation 2 5 02x x- + = .

Chapter 1 You should be able to work with complex conjugates.

3 Given that = -z 3 5i, find:a z z+ * b zz*

Using complex numbers in factorisingThis chapter draws together ideas from the AS Mathematics course about factorising cubics and quartics with the theory of complex numbers from Chapter 1 of this book to enable you to find complex roots of these polynomials. It also looks at the relationship between the coefficients of a polynomial and its roots.

Section 1: Factorising polynomialsYou are already familiar with the link between factorising and real roots of a polynomial: for example, if x = 2 is a root of the equation p(x) = 0, then (x - 2) is a factor of the polynomial p(x).

This is the factor theorem (which you met in AS Level Mathematics Student Book 1, Chapter 4), and it also applies to polynomials with complex roots.


• factorise polynomials and solve equations which may have complex roots • link the roots of a polynomial and its coefficients • use substitutions to solve more complicated equations.


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You can use the same method to factorise polynomials of higher degree.

a Solve the equation 4 40 02x x- + = .b Hence factorise 4 40.2x x- +

a

i

i

x4 ( 4) 4 40

24 144

24 12

22 6

2

=± - - ×

= ± -

= ±

= ±

Use the quadratic formula.

b Therefore:

i i

i i

x x x x

x x

4 40 ( (2 6 ))( (2 6 ))( 2 6 )( 2 6 )

2 - + = - + - -= - - - +

Make the link between roots and factors.

WORKED EXAMPLE 2.1

Let f( ) 153 2x x x x= - - - .

a Show that f(3) 0= .b Hence write xf( ) as a product of a linear factor and a real quadratic factor.c Solve the equation xf( ) 0= and write xf( ) as a product of three linear factors.

a f(3) 3 3 3 1527 9 3 150

3 2= - - -= - - -=

Substitute x 3= into xf( ).

b x x x x x Bx C15 ( 3)( )3 2 2- - + = - + + ⇒ = =B C2, 5

f x x x x( ) ( 3)( 2 5)2∴ = - + +

As 3 is a root, x( 3)- is one factor. You can find the quadratic factor by long division or comparing coefficients.

c =xf( ) 0

x x x

x x x

( 3)( 2 5) 0

3 or 2 5 0

2

2

- + + == + + =

If x x2 5 02 + + = :

i

i

x 2 2 4 52

2 162

2 42

1 2

2= - ± - ×

= - ± -

= - ±

= - ±

Use the formula to solve the quadratic.

WORKED EXAMPLE 2.2


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This means that you can now factorise some expressions that were impossible to factorise using just real numbers. A particularly useful case is this extension of the difference of two squares identity to the sum of two squares.

( i)( i)2 2a b a b a b+ = + -

Key point 2.1

3 or 1 2x x i∴ = = - ± State all three roots of xf( ).

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

( 3)( 1 2 )( 1 2 )

x x x x

x x x

f i i

i i

∴ = - - - + - - -= - + - + +

Each root z has a corresponding factor x z( )- .

Factorise 81 164z - .

z z z81 16 (9 4)(9 4)4 2 2- = - + This is a difference of two squares.

= - + - +z z z zi i(3 2)(3 2)(3 2 )(3 2 ) The first factor is a difference of two squares again. The second factor is a sum of two squares.

WORKED EXAMPLE 2.3

EXERCISE 2A

1 Solve the equation xf( ) 0= and hence factorise xf( ).

a i f( ) 2 22x x x= - + ii f( ) 6 252x x x= + +

b i f( ) 3 42x x x= + + ii f( ) 2 52x x x= + +

c i f( ) 3 2 102x x x= - + ii f( ) 5 4 22x x x= + +

2 Factorise each expression into linear factors.

a i 42z + ii 252z +

b i 4 492z + ii 9 642z +

c i 14z - ii 16 814z -

3 Given one root of the cubic polynomial, write it as a product of a linear factor and a real quadratic factor, and hence find the other two roots.

a i 2 143 2x x x+ - - ; root x 2= ii 3 7 53 2x x x+ + + ; root x 1= -

b i 2 5 63x x- + ; root x 2= - ii 3 23 2x x- - ; root x 1=

4 f( ) 13 56 783 2x x x x= - + -

Show that x( 3)- is a factor of xf( ) and find all the solutions of the equation xf( ) 0= .

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5 a Given that x(2 1)+ is a factor of 2 16 63 2x ax x+ + + , show that a 9= .

b Factorise 2 9 16 63 2x x x+ + + completely.

6 f( ) 4 4 21 94 3 2x x x x x= - - -

a Show that x( 3)- is a factor of xf( ).

b Factorise xf( ) and hence solve the equation xf( ) 0= .

7 f( ) 3 13 104 3 2x x x x x= + - - -

a Show that x( 1)+ and x( 2)- are factors of xf( ).

b Write xf( ) as a product of two linear factors and a quadratic factor. Hence find all solutions of the equation xf( ) 0= .

8 Find all solutions of the equation 21 100 04 2x x+ - = .

This result is not true if the polynomial has complex coefficients. For example, the equation i 2 02z z- + = has solutions i- and 2i.

Did you know?

Complex solutions of real polynomials come in conjugate pairs. If zf( ) 0= then z =f( *) 0.

Key point 2.2

Given that 1 i 6- is a root of the polynomial f( ) 213 2x x x x= + + + , find the other two roots.

1 6i- is a root so 1 6i+ is also a root. Complex roots come in conjugate pairs.

So ( (1 6 ))x i- - and ( (1 6 ))x i- + are factors of xf( )

Therefore ( (1 6 ))( (1 6 ))x xi i- - - + is a factor.

Link roots to factors (using the factor theorem).

WORKED EXAMPLE 2.4

See Focus on… Proof 1 for a proof of this result.

Focus on...

Section 2: Complex solutions to polynomial equationsYou know from Chapter 1 that if a quadratic equation has two complex solutions, then they are always a conjugate pair: for example, the equation 6 25 02x x- + = has solutions = +x 3 4i and = -x 3 4i. This happens because of the ± in the quadratic formula.

If you look at the cubic and quartic polynomials you factorised in Exercise 2A, you will see that the complex roots were either real or a complex conjugate pair. It can be proved that this result generalises to any real polynomial (a polynomial in which all of the coefficients are real numbers).


This result can be very useful when you are factorising polynomials and solving equations. If you know one complex root, then you can immediately write down another one.

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i i

i i i i

i i

x x

x x x

x x x x x

x x

( (1 6 ))( (1 6 ))

(1 6 ) (1 6 ) (1 6 )(1 6 )

6 6 7

2 7

2

2

2

- - - += - + - - + - += - - - + += - +

Expand and simplify.

x x x x k x x( 21) ( )( 2 7)3 2 2+ + + = + - +By inspection, k = 3.So the solutions are = ±x 1 6i and x = -3.

Use polynomial division to find the third factor.

In Worked example 2.4, in order to find the quadratic factor of xf( ), you needed to expand x x( (1 i 6 ))( (1 i 6 ))- - - + , which is a product of the form x z x z- -( )( *). You get this type of expression whenever you are expanding two brackets corresponding to a pair of complex conjugate roots, and so it is useful to know a shortcut.

x z x z x z x z- - = - +( )( *) 2Re( )2 2

Key point 2.3

In the case of x x( (1 i 6 ))( (1 i 6 ))- - - + , z 1 i 6= - and so:

• =Re( ) 1z

• 1 ( 6 ) 72 2 2z = + =

( (1 i 6 ))( (1 i 6 )) 2 1 7 2 72 2x x x x x x∴ - - - + = - × + = - +

Given that one of the roots of the polynomial f( ) 5 26 46 68 is 3 5i4 3 2x x x x x= - + + - - , find the remaining roots.

3 5i- is a root so 3 5i+ is also a root. Complex roots come in conjugate pairs.

So ( (3 5 ))x i- - and ( (3 5 ))x i- + are factors of xf( ). Link roots to factors (using the factor theorem).

Therefore:i ix x x x

x x

( (3 5 ))( (3 5 )) 2 3 34

6 34

2

2

- - - + = - × += - +

is a factor.

Multiply out the brackets (using the shortcut from Key point 2.3).

If = +z 3 5i, then:zRe( ) 3= and 3 5 342 2 2= + =z .

WORKED EXAMPLE 2.5


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Find a cubic polynomial with roots 3 and 4 i+ .

Give your answer in the form 3 2x bx cx d+ + + .

3 is a root of xf( ), so -x( 3) is a factor. From the factor theorem, if z is a root then x z( )- is a factor.

4 i+ is a root, so 4 i- is another root. Complex roots occur in conjugate pairs.

So xf( ) also has factors ( (4 ))x i- + and ( (4 ))x i- -

Hence: f i ix x x x

x x x

x x x

( ) ( 3)( (4 )) ( (4 ))

( 3)( 2 4 17)

( 3)( 8 17)

2

2

= - - + - -= - - × += - - +

x x x11 41 513 2= - + -

Use the result of Key point 2.3 to expand the brackets corresponding to the complex roots. If z 4 i= + then:

zRe( ) 4= and 4 1 172 2 2z = + =

Expand and simplify.

WORKED EXAMPLE 2.6

The polynomial f( ) 4 3 2x x bx cx dx e= + + + + has roots 3i and 5 i- . Find the values of the real numbers b, c, d and e.

The other two roots are 3i- and 5 i+ . Write down the remaining roots and use them to find factors.

Hence the two quadratic factors are:i ix x x( 3 )( 3 ) 92+ - = +

and:i ix x x x( (5 ))( (5 )) 10 262- - - + = - +

The complex conjugate pairs will give quadratic factors.

WORKED EXAMPLE 2.7

Factorising:x x x x x x x x5 26 46 68 ( 6 34)( 2)4 3 2 2 2- + + - = - + + -

Divide xf( ) by 6 342x x- + (using polynomial division or comparing coefficients).

For the roots of the second factor:x x

x x

x

2 0( 2)( 1) 0

2or 1

2 + - =+ - =

= -

You already know the roots of the first factor. Find the roots of the second factor by factorising.

So the roots of xf( ) are: -1, 2 and 3 5i± .

You can also find a polynomial with given roots.


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Find a cubic polynomial with real coefficients, given that two of its roots are 2 and -2i 1.



The other complex root is 2i 1+ , so the polynomial is:

x x x( 2)( (2i 1))( (2i 1))- - - - + ( 2)( 4i 5)2x x x= - - -

(2 4i) (8i 5) 103 2x x x= - + + - +

The other complex root is - -2i 1, so the polynomial is:

+ - - - - -x x x( 2)( (2i 1))( ( 2i 1)) ( 2)( 2 5)2x x x= + + +

4 9 103 2x x x= + + +

The other complex root is 1 2i- - , so the polynomial is:

x x x( 2)( ( 1 2i))( ( 1 2i))- - - + - - - ( 2)( 2 5)2x x x= - + +

103x x= + -

WORK IT OUT 2.1

Hence: f x x x x

x x x x

( ) ( 9)( 10 26)

10 35 90 234

2 2

4 3 2

= + - += - + - +

Multiply the two factors to get the polynomial.

So 10 35 90 234b c d e, , , .= - = = - =

EXERCISE 2B

1 Find the real values of a and b such that the quadratic equation 02x ax b+ + = has the given roots.

a i 5i and 5i- ii 3i- and 3i b i 3 4i- and 3 4i+ ii 1 2i+ and 1 2i-

2 Given one complex root of a cubic polynomial, factorise the polynomial and write down all its roots.

a i 11 43 653 2x x x- + - ; root x 3 2i= + ii 7 153 2x x x- - + ; root x 2 i= +

b i 3 7 53 2x x x- + - ; root = -x 1 2i ii 2 14 403 2x x x- - + ; root x 3 i= -

3 Given one complex root of the quartic polynomial, find one real quadratic factor and hence find all four roots.

a i 2 14 8 404 3 2x x x x- + - + ; root x 1 3i= + ii 6 11 6 104 3 2x x x x- + - + ; root x 3 i= -

b i 2 8 14 394 3 2x x x x- + + + ; root x 2 3i= - ii 3 27 21 584 3 2x x x x- + + + ; root x 2 5i= +

c i 2 10 8 244 3 2x x x x+ + + + ; root 2i- ii 4 21 64 804 3 2x x x x- + - + ; root 4i

4 Given that 5 i- is one root of the equation 8 6 52 03 2x x x- + + = , find the remaining two roots.

5 p( ) 3 16 483 2x x x x= + + +

a Show that =p(4i) 0. b Hence solve the equation xp( ) 0= .

6 2 5i+ is a root of the equation 4 30 4 29 04 3 2x x x x- + - + = .

a Write down another complex root of the equation.

b Find the remaining two roots.

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Section 3: Roots and coefficientsWhen you first learned to factorise quadratic equations you were probably told to look for two numbers that add up to the middle coefficient and multiply to give the constant term. For example, 10 212x x- + factorises as x x( 3)( 7)- - because - + - = -( 3) ( 7) 10 and ( 3) ( 7) 21- × - = . This means that the roots, 3 and 7, add up to give the negative of the middle coefficient, 10, and multiply to give the constant term, 21.

However, there are infinitely many other quadratic equations with the same roots, because you can multiply the whole equation by a constant. For example, another equation with roots 3 and 7 is 3 30 63 02x x- + = .

The two roots still add up to 10, which is 303 , and multiply to give 21,

which is 633 . This is a particular example of a very useful general result.

7 Two roots of the equation 8 21 32 68 04 3 2x x x x- + - + = are 2i and 4 i- . Write down the other two roots and hence write 8 21 32 684 3 2x x x x- + - + as a product of two real quadratic factors.

8 f( ) 5 4 44 3 2z z z z z= + + + +

a Show that =f(2i) 0.

b Write zf( ) as a product of two real quadratic factors.

c Hence find the remaining solutions of the equation zf( ) 0= .

9 Find a quartic equation with real coefficients and roots 4i and 3 2i- .

If p and q are the roots of the quadratic 02ax bx c+ + = then:

• p q ba+ = -

• pq ca=

Key point 2.4

PROOF 3

If a quadratic equation has roots p and q then:

x p x q

x p q x pq

( )( ) 0

( ) 02

- - =- + + =

Write the equation in factorised form.

ax a p q x apq( ) 02 - + + = You can multiply the whole equation by a number, and it will still have the same roots.

Hence: ( )b a p q= - + and =c apq Compare coefficients with 2ax bx c+ + .

So: + = -p q ba and =pq c

a

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You can also find other functions of roots. This requires some algebraic manipulation.

The equation 5 3 1 02x x+ + = has roots p and q. Find the value of:

a p q1 1+ b 2 2p q+ .

p q ba+ = - = - 3

5

pq ca= = 1

5

You could actually find the roots by using the quadratic formula, but Key point 2.4 provides a much quicker way to answer the question.

a

( )( )

+ = +

=-

= -

p qp q

pq1 1

35

15

3

It is often helpful to combine fractions into a single fraction.

b p q p pq q( ) 22 2 2+ = + +

p q p q pq( ) 2

35 2 1

5

2 2 2

2( ) ( )⇒ + = + -

= - -

You can get 2 2p q+ by squaring p q+ .

125

= -2 2p q+ is a sum of two squares, so you may

think that you made a mistake because the answer is negative. However, p and q can be complex numbers, so this is in fact possible!

WORKED EXAMPLE 2.8

The Greek letters α, and β are often used instead of p and q for the roots.

The equation 3 ( 1) 02x kx k+ - + = has roots α and β.

a Write down expressions for α β+ and αβ in terms of k.b Find 3 3α β+ in terms of k.

a 3( 1)

3

α β

αβ

+ = -

=- +

k

k

Apply Key point 2.4:

α β

αβ

+ = -

=

ba

ca

WORKED EXAMPLE 2.9


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Cubic and quartic equations

For cubic equations there are three relationships between roots and coefficients. Just as in the proof of the relationships between coefficients and roots of a quadratic equation, if the roots of the equation

03 2ax bx cx d+ + + = are p q, and r then you can write:

( )( )( )3 2ax bx cx d a x p x q x r+ + + = - - -

Expanding and comparing coefficients then gives the results.

EXERCISE 2C

1 The equation 2 02x kx k- + = has roots α and β. Find the value of each expression in terms of k.

a α β+ b 2αβ c 2 2α β+

d 1 1α β+ e 3 3α β+ f 1 1

2 2α β+

2 The equation 3 02 2ax x a+ - = has roots p and q. Find the value of each expression in terms of a.

a p q3 3+ b 2 2p q c 2 2p q+ d ( )2p q-

b ( ) 3 3

3 ( )

3 3 2 2 3

3 3

α β α α β αβ βα β αβ α β

+ = + + += + + +

You can find 3 3α β+ from the binomial expansion of

3α β( )+ .

( ) 3 ( )3 3 3α β α β αβ α β⇒ + = + - +

k k k

k k k

k k k

27 3 13 3

27( 1)

39 927

3

3

3 2

( ) ( )= - - - + -

= - - +

= - + +

You need to try to express everything in terms of α β+ and αβ .

If p, q and r are the roots of the cubic 03 2ax bx cx d+ + + = then:

• p q r ba+ + = -

• pq qr rp ca+ + =

• pqr da= -

Key point 2.5

You can again combine these with algebraic identities to find other combinations of roots.

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The equation 3 4 7 03 2x x x- + + = has roots , α β and γ .

Find the value of 2 2 2α β γ+ + .

( ) ( ( ))

2 ( ) ( )

2 2 2

2( )

2 2

2 2

2 2 2

2 2 2

α β γ α β γα α β γ β γα αβ αγ β γ βγα β γ αβ βγ γα

+ + = + += + + + += + + + + += + + + + +

You can find 2 2 2α β γ+ + from the expansion of ( )2α β γ+ + .

α β γ

αβ βγ γα

+ + = - - =

+ + = =

31 3

41 4

Now use Key point 2.5

baα β γ+ + = -

caαβ βγ γα+ + = .

α β γ∴ + + = -=

(3) 2(4)1

2 2 2 2

WORKED EXAMPLE 2.10

Similar relationships between roots and coefficients can be found for polynomial equations of any degree. The expressions get increasingly complicated.

If p, q, r and s are the roots of the quartic 04 3 2ax bx cx dx e+ + + + = , then:

• p q r s ba+ + + = -

• pq pr ps qr qs rs ca+ + + + + =

• pqr pqs prs qrs da+ + + = -

• pqrs ea=

Key point 2.6

To help you to remember these equations, notice that the pattern is always the same.

• The sum of all the roots is related to b

a.

• The sum of all possible products of two roots is related to c

a.

• The sum of all possible products of three roots is related to d

a.

The signs alternate between - and +.

Tip

The equation 2 3 5 04 3x x x- + - = has roots p q r s, , and .

Find the value of p q r s1 1 1 1+ + + .

p q r spqr pqs prs qrs

pqrs+ + + = + + +1 1 1 1 Combine into a single fraction.

))

((=

-

-

=

1252

15

Use Key point 2.6:

pqr pqs prs qrs da+ + + = -

pqrs ea=

WORKED EXAMPLE 2.11

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Remember that an arithmetic sequence is one in which the term goes up by a constant number.

Rewind

The roots p q r, and of the equation 3 03 2x x cx d- + + = form an arithmetic sequence.

Show that c d 2+ = .

2- = -

⇒ + =q p r q

p r qIf p q r, and form an arithmetic sequence then q p r q- = - .

p q r+ + = - - =31 3

You know three equations relating the roots to the coefficients.

The most useful one seems to be p q r ba+ + = - .

2 31

q q

q

∴ + ==

Substitute from p r q2+ = .

pq qr rp c

p r rp c

+ + =+ + =

To involve c and d you need to use the other two relationships

as well. Start with pq qr rp ca+ + = with q 1= .

rp c+ =2rp c⇒ = - 2

You know that + = =2 2p r q .

pqr d

pr d

= -= -

Finally use pqr da= - with q 1= .

c d

c d

2

2

∴ - = -

⇒ + = , as required.Substitute from rp c 2= - .

WORKED EXAMPLE 2.12

You can use information about the roots to prove facts about the coefficients of an equation.

EXERCISE 2D

1 The equation 3 6 12 4 03 2x x x+ + - = has roots p, q and r. Find the value of:

a p q r+ + b p q r1 1 1+ + c 2 2 2p qr pq r pqr+ + d 2 2 2p q r+ + .

2 , , α β γ and δ are the roots of the given quartic equation. In each case, find the value of the given expressions.

a 2 4 5 3 1 04 3 2x x x x- + + - = ; αβγδ and αβγ αβδ αγδ βγδ+ + + .

b 3 2 1 04 2x x- + = ; α β γ δ+ + + and αβ αγ αδ βγ βδ γδ+ + + + + .

c 5 3 8 04x x- - = ; α β γ δ+ + + and αβγδ .

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You can already do this by writing down the factors and expanding brackets (see Worked example 2.7), but this method is more direct.

Rewind

3 The equation 4 2 4 1 03 2x x x- + + = has roots p q, and r. Find the value of 2 2 2p qr pq r pqr+ + .

4 The equation 3 2 5 03 2x x x+ - + = has roots a b, and c. Find the value of:

a a b c1 1 1+ + b ab bc ca

1 1 1+ + .

5 When two resistors of resistances 1R and 2R are connected in series in an electric circuit, the total resistance in the circuit is 1 2R R R= + . When they are connected in parallel, the total resistance satisfies 1 1 1

1 2R R R= + .

Two resistors have resistances equal to the two roots of the quadratic equation 3 12 4 02R R- + = . Find the total resistance in the circuit if the two resistors are connected:

a in series b in parallel.

6 The cubic equation 3 5 3 03x x- - = has roots α, β and γ .

a Write down the value of αβγ . b Show that 1α β αβ+ = - .

7 A random-number generator can produce four possible values, all of which are equally likely. The four values satisfy the equation 9 26 29 10 0.4 3 2x x x x- + - + =

If a large number of random values are generated, estimate their mean.

8 The equation 04 3 2x bx cx dx e+ + + + = has roots p p p, 2 , 3 and p4 . Show that 125 310

4e b= .

9 The equation 2 3 2 03 2 2x ax a x+ + + = , where a is a real constant, has roots p q, and r.

a Find an expression for 2 2 2p q r+ + in terms of a.

b Explain why this implies that the roots are not all real.

Section 4: Finding an equation with given rootsYou can use the relationships between roots of polynomials and their coefficients to find unknown coefficients in an equation with given roots.

The quadratic equation 5 02x bx c+ + = has real coefficients and one of its roots is 4 7i+ . Find the values of b and c.

The other root is 4 7i- . Complex roots occur in conjugate pairs.

Then: (4 7 ) (4 7 ) 5bi i+ + - = -

b

b

= -

= -

8 540

Use p q ba+ = - .

WORKED EXAMPLE 2.13


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A quartic equation 14 18 04 3 2x ax x x b+ + - + = has real coefficients and two of its roots are 3i and 1 2i- . Find the values of a and b.

The four roots are: 3i, 3i- , 1 2i+ , 1 2i- As you know two of the complex roots, you can find the other two (their conjugates).

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

2 12

a

a

a

i i i i+ - + - + + = -

= -

= -

Use p q r s ba+ + + = - , being careful

to note how the coefficients have been labelled in this question (a is the coefficient of 3x here).

(3 )( 3 )(1 2 )(1 2 ) 1bi i i i- - + = Use pqrs e

a= .

b

b

9(1 2 )45

2 2+ ==

Use =* 2zz z .

WORKED EXAMPLE 2.14

(4 7 )(4 7 ) 5ci i+ - = Use pq c

a= .

c

c

4 7 5325

2 2+ =

=Remember: zz z=* 2 .

You can also find a new equation with roots that are related to the roots of a given equation in some way, and you can do this without solving the equation. The strategy is to use the sum and product of roots of the first equation to find the sum and product of roots of the second equation.

The quadratic equation 3 4 7 02x x- + = has roots p and q. Find a quadratic equation with integer coefficients and roots 2p and 2q .

p q+ = - - =43

43

pq = 73

You don’t need to find p and -q – just their sum and product.

Let the equation be ax bx c 02 + + = .

Then: p q ba

2 2+ = - and p q ca

2 2 =

The coefficients of the new equation are related to the roots 2p and 2q .

WORKED EXAMPLE 2.15


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Set a = 1.

Then: b p q( )2 2= - + and c p q2 2= .

All equations with the required roots are multiples of each other, so you can set a 1= .

c p q

pq( )

2 2

2

==

You need to relate 2 2p q+ and 2 2p q to p q+ and pq. The second one is easier.

73

499

2( )=

=

Substitute pq 73= .

p q p pq q

p q p q pq

( ) 2

( ) 2

2 2 2

2 2 2

+ = + +⇒ + = + -

You can square p q+ to get 2 2p q+ .

b p q

p q pq

( )

(( ) 2 )

43 2 7

3

269

2 2

2

2( ) ( )

= - += - + -

= - -

=

Substitute p q 43+ = and pq 7

3= .

The equation is: x x269

499 02 + + = Substitute into 02ax bx c+ + = .

x x9 26 49 02⇔ + + = You want the equation with integer coefficients, so multiply through by 9.

The equation 3 2 03 2x x- + = has roots p q r, and . Find a cubic equation with roots p3 , q3 and r3 .

For the equation x x3 2 03 2- + = :

+ + =+ + =

= -

p q r

pq qr rp

pqr

30

2

Use Key point 2.5 for the original equation.

If the equation with roots 3 , 3p q, and r3 is x bx cx d 03 2+ + + = then:

Now use the roots to find the coefficients. You can set the coefficient of 3x to be 1.

p q r b+ + = -3 3 3 Using the first part of Key point 2.5 for the new equation.

WORKED EXAMPLE 2.16


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You will see in Section 5 that you could also find the equation from Worked example 2.16 by using the substitution =u x3 .

Fast forward

3( )3(3)

9

p q r b

b

b

+ + = -= -= -

p q r 3+ + =

+ + =p q q r r p c(3 )(3 ) (3 )(3 ) (3 )(3 )

+ + ===

pq qr rp c

c

c

9( )9(0)

0

Using the second part of Key point 2.5 for the new equation.

pq qr rp 0+ + =

= -p q r d(3 )(3 )(3 )= -

- = -=

pqr d

d

d

2727( 2)

54

Using the third part of Key point 2.5 for the new equation.

pqr 2= -

The equation is x x9 54 03 2- + = .

The equation 3 5 3 03 2x x x- + - = has roots , and α β γ . Find a cubic equation with roots αβ βγ γα, and .

For the equation x x x3 5 3 03 2- + - = :

531333

α β γ

αβ βγ γα

αβγ

+ + =

+ + =

=

Use Key point 2.5 for the original equation.

If the equation with roots and , αβ βγ γα is x bx cx d 03 2+ + + = then:

You can take the coefficient of 3x to be 1.

b

b

b

13

13

αβ βγ γα+ + = -

= -

= -

By Key point 2.5, the sum of the roots of

the new equations is b1- .

( )( ) ( )( ) ( )( ) cαβ βγ βγ γα γα αβ+ + = Use the second part of Key point 2.5.

WORKED EXAMPLE 2.17


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( ) cαβγ α β γ+ + = Factorise.

( ) ( ) =

=

c

c

33

53

53

( )

d

d

d

d

( )( )( )

33

1

2

2

αβ βγ γααβγ

( )

= -= -

= -

= -

The product of the roots is d1- by Key

point 2.5.

So the equation is

x x x13

53 1 03 2- + - = .

EXERCISE 2E

1 Given the roots of the equations, find the missing coefficients.

a i 3 3 03 2x ax x b- - + = ; roots 1, 1, 2- - ii 2 10 03 2x ax x c+ + + = ; roots 1, 1, 2

b i 4 03 2x ax x b- + - = ; roots 3, 2i, 2i- ii 9 03 2x bx x d+ + + = ; roots 2 i, 2 i, 1+ -

2 Find the polynomial of the lowest possible order with the given roots.

a i 5 2i, 5 2i+ - ii 3 i, 3 i- + b i 1, 3i, 3i- ii 5, i, i-

c i 2, 1, 1 2i, 1 2i- + - ii 3, 1, 2 i, 2 i- + - d i 2, 4 3i, 4 3i- + - ii 1, 2 3i, 2 3i- + - -

3 A cubic equation has real coefficients and two of its roots are 2 and 4 i- .

a Write down the third root.

b Find the equation in the form 03 2x bx cx d+ + + = .

4 The quartic equation 04 3 2x ax bx cx d- + - + = has real coefficients, and two of its roots are 3i and 3 i- .

a Write down the other two roots. b Hence find the values of a and d .

5 The equation 3 4 1 03 2x x x- + + = has roots p q, and r. Find a cubic equation with roots p q3 , 3 and r3 .

6 The equation 4 3 5 03x x- + = has roots , α β and γ . Find a cubic equation with integer coefficients and roots α β γ- - -2, 2 and 2.

7 The quadratic equation 5 3 2 02x x- + = has roots p and q. Find a quadratic equation with roots p1 and q

1 .

8 The equation 3 2 04 3x x x- + + = has roots a b c, , and d. Find a quartic equation with roots a b c2 , 2 , 2 and d2 .

9 Let p and q be the roots of the equation 5 3 2 02x x- + = .

a Find the values of pq and 2 2p q+ .

b Hence find a quadratic equation with integer coefficients and roots 2p and 2q .

You can also write this equation as 3 5 3 03 2x x x- + - = .

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Section 5: Transforming equationsIn Section 4, you saw how to use relationships between roots and coefficients of a polynomial equation to find another equation with roots that are related to the roots of the first one.

In some cases, however, there is an easier way to find this equation – by using a substitution.

Suppose the original equation, with roots p and q, is 5 2 1 02x x- + = .

If you set x u3= you get a new equation.

53

23

1 0

5 6 9 0

2

2

u u

u u

( ) ( )- + =

⇔ - + =

Since u x3= this equation has roots p3 and q3 .

10 The equation 3 5 03 2x x- + = has roots , α β and γ .

a Find the value of 1 1 1αβ βγ γα+ + .

b Find a cubic equation with roots 1 , 1α β and 1

γ .

11 The equation 4 3 7 03x x- + = has roots , and p q r. Find a cubic equation with integer coefficients and roots , and pq qr rp.

12 a Expand 2αβ βγ γα( )+ + .

b If , and α β γ are the roots of the equation 2 6 03 2x x x- + + = find a cubic equation with roots , and 2 2 2α β γ .

Given an equation in x that has a root x p= , if you make a substitution u xf( )= , then the resulting equation in u has a root u pf( )= .

Key point 2.7

The equation 3 2 5 03 2x x x- + + = has roots p q r, and .

a Find a cubic equation with roots - - -p q r2, 2 and 2.b Hence find the value of p q r( 2)( 2)( 2)- - - .

a u x= - 2 x u⇒ = + 2

When x p= , u p 2= - so make the substitution u x 2= - .

WORKED EXAMPLE 2.18


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If you were just asked to find - - -p q r( 2)( 2)( 2), you could expand - - -p q r( 2)( 2)( 2) and use the relationships between roots and

coefficients of the original equation. You can decide for yourself whether this or the substitution method is simpler.

Tip

The roots of the quadratic equation 3 5 02x x- + = are p and q. Find a quadratic equation with roots p5 and

q5 .



From the first equation: p q 1+ = and pq 5=

So for the second equation: p q5 5

15+ = and

p q5 5

525

15× = =

Hence the new equation is: 15

15

02x x- + =

5 1 02x x⇔ - + =

Let u x5= ; then

p5 and

q5 are the

roots for u.

Make the substitution x u5= .3(5 ) (5 ) 5 0

15 1 0

2

2

u u

u u

- + =⇔ - + =

Replace x by x5 :

35 5

1 02x x( ) ( )- + =

3 5 25 02x x⇔ - + =

WORK IT OUT 2.2

u u u

u u u u u u

u u u

3( 2) ( 2) 2( 2) 5 0

3( 6 12 8) ( 4 4) 2 4 5 0

3 17 34 29 0

3 2

3 2 2

3 2

+ - + + + + =+ + + - + + + + + =

+ + + =

Substitute for x .

b ∴ - - - = -p q r( 2)( 2)( 2) 293

This is the product of the roots of the equation in u.

A substitution is particularly useful if it transforms a difficult equation into one that you can solve easily.

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Use the substitution x u 1= - to solve the equation 4 6 4 80 04 3 2x x x x+ + + - = .

u u u u( 1) 4( 1) 6( 1) 4( 1) 80 04 3 2- + - + - + - - = Make the given substitution.

- + - +u u u u4 6 4 14 3 2

+ - + -u u u4 12 12 43 2

+ - +u u6 12 62 + - - =u4 4 80 0

It’s a good idea to line up terms when expanding lots of brackets.

i

u

u u

u u

81 0

( 9)( 9) 03, 3

4

2 2

- =- + =

= ± = ±

You can solve the equation in u easily by factorising.

x x x x2, 4, 1 3 , 1 3∴ = = - = - + = - -i i Using x u 1= - .

WORKED EXAMPLE 2.19

EXERCISE 2F

1 Use the given substitution to transform the equation in x into a polynomial equation for u.

a i 3 1 03x x- + = ; x u2= . ii 2 5 03 2x x+ + = ; x u3= .

b i 3 4 03x x- + = ; x u 2= - . ii 2 1 03 2x x+ + = ; x u 1= + .

c i 3 4 15 03 2x x x- + + = ; x u2 1= + . ii 6 10 03 2x x x+ - + = ; x u3 2= - .

2 The equation 3 4 2 03x x- + = has roots p q r, and . Use a suitable substitution to find a cubic equation with roots p q r1, 1 and 1- - - .

3 The equation 2 2 5 04x x+ + = has roots a b c d, , and . Find a quartic equation with integer coefficients and

roots a b c d2 , 2 , 2 and 2 .

4 a Show that the substitution x u 2= - transforms the equation 6 21 26 03 2x x x+ + + = into the equation 9 03u u+ = .

b Hence solve the equation 6 21 26 03 2x x x+ + + = .

5 a Find the value of c so that the substitution x u c= + transforms the equation 12 45 54 03 2x x x- + - = into the equation 3 03 2u u- = .

b Hence find all the solutions of the equation 12 45 54 03 2x x x- + - = .

6 The equation 3 2 5 02x x- + = has roots α and β. Using the substitution x u3 1= + , or otherwise, find the

value of 13

13

α β( )- -

.

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• Complex roots of real polynomials occur in conjugate pairs: if zf( ) 0= then z =f( *) 0. You can use this fact to factorise cubic and quartic polynomials.

• Coefficients of a polynomial can be expressed in terms of its roots.• For a quadratic equation 02ax bx c+ + = with roots p and q:

o p q ba+ = -

o pq ca =

• For a cubic equation 03 2ax bx cx d+ + + = with roots p, q and r:

o p q r ba+ + = -

o pq qr rp ca+ + =

o pqr da= -

• For a quartic equation 04 3 2ax bx cx dx e+ + + + = with roots p, q, r and s:

o p q r s ba+ + + = -

o pq pr ps qr qs rs ca+ + + + + =

o pqr pqs prs qrs da+ + + = -

o pqrs ea=

• You can use these relationships to find:• a polynomial with given roots• a polynomial with roots related to the roots of another polynomial.

• The second of these types of problem can sometimes also be solved by making a substitution.


7 The equation 3 9 1 02x x- + = has roots α and β. Using a suitable substitution, or otherwise, find an equation with roots 2α and 2β .

8 The substitution x u k= - transforms the equation 4 11 14 10 04 3 2x x x x+ + + + = into the equation of the form 04 2u bx c+ + = .

a Find the value of k.

b Hence solve the equation 4 11 14 10 04 3 2x x x x+ + + + = .

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Mixed practice 2

1 Find the sum of the roots of the equation 3 4 7 04 2x x x+ - - = .


A 43 B 4

3- C 73- D 0

2 The quadratic equation 2 02 2x kx k- + = is transformed into an equation in u by the substitution

x u 12= - .

Find the root of the transformed equation in terms of the constant k.


A k 12- B k 1

2- - C k2 1+ D k2 1- -

3 Given that z 3i= is one root of the equation 2 9 18 03 2z z z- + - = , find the other two roots.

4 f( ) 3 2z z az bz c= + + + where a b c, , are real constants. Two roots of zf( ) 0= are z 1= and z 1 2i= + . Find a b, and c.

5 One of the roots of the polynomial x x x x= + - +g( ) 3 7 153 2 is 1 i 2+ .

a Write down another complex root and hence find a real quadratic factor of xg( ).

b Solve the equation xg( ) 0= .

6 Find a quartic equation with real coefficients given that three of its roots are 2i, 1- and 5.

7 The quadratic equation 5 7 1 02x x- + = has roots α and β.

a Write down the values of α β+ and αβ.

b Show that 395

αβ

βα+ = .

c Find a quadratic equation with integer coefficients, which has roots 1α α+ and 1β β+ .

[©AQA 2012]

8 Given that z 1 2i= + is one root of the equation 15 03 2z z z+ - + = , find the other two roots.

9 Two roots of the cubic equation 03 2z bz cz d+ + + = ∈R( , , )b c d are 2- and 2 3i- .

a Write down the third root.

b Find the values of b c, and d .

10 The polynomial 653 2z az bz+ + - has a factor of z( 2 3i)- - . Find the values of the real constants a and b.

11 a Show that ( ) 3 ( )3 3 3α β α β αβ α β+ = + - + .

b Let α and β be the roots of the quadratic equation 7 2 02x x+ + = . Find a quadratic equation with roots 3α and 3β .

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12 The equation 3 4 7 1 03 2x x x- + + = has roots p q, and r. Use the substitution u x 3= - to find a cubic equation with roots p 3- , q 3- and r 3- .

13 The equation 5 9 4 03x x- + = has roots , α β and γ . Use a substitution of the form u kx= to find a cubic

equation with roots 2 , 2α β

and 2γ

.

14 The cubic equation 03z pz q+ + = has roots ,α β and γ .

a i Write down the value of α β γ+ + . ii Express αβγ in terms of q.

b Show that 33 3 3α β γ αβγ+ + = .

c Given that α = +4 7i and that p and q are real, find the values of:

i β and γ ii p and q.

d Find a cubic equation with integer coefficients which has roots 1 , 1α β and 1

γ .

[©AQA 2012]

15 a A cubic equation 03 2ax bx cx d+ + + = has roots , ,1 2 3x x x .

i Write down the values of 1 2 3x x x+ + and 1 2 3x x x in terms of a b c, , and d .

ii Show that 1 2 2 3 3 1x x x x x x ca

+ + = .

b The roots α, β and γ of the equation 2 16 03 2x bx cx+ + + = form a geometric sequence.

i Show that 2β = - .

ii Show that c b2= .

16 a Show that:

i ( ) 2( )2 2 2 2p q r p q r pq qr rp+ + = + + - + +

ii p q q r r p pq qr rp pqr p q r+ + = + + - + +( ) 2 ( ).2 2 2 2 2 2 2

b Given that the cubic equation 03 2ax bx cx d+ + + = has roots p q, and r:

i write down the values of p q r+ + and pqr in terms of a b c, , and d

ii show that pq qr rp ca+ + = .

c The equation 2 5 2 03x x- + = has roots 1x , 2x and 3x .

i Show that 512

22

32x x x+ + = .

ii Find the values of 12

22

22

32

32

12x x x x x x+ + and 1

222

32x x x .

iii Hence find a cubic equation with integer coefficients and roots , 12

22x x and 3

2x .

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Before you start…

A Level Mathematics Student Book 1, Chapter 3

You should be able to use the discriminant to determine the number of solutions of a quadratic equation.

1 Find the set of values of k for which the equation 2 5 02 − + =kx x has exactly two real roots.


You should be able to interpret solutions of simultaneous equations as intersections of graphs.

2 Find the value of c for which the line 3= +y x c is a tangent to the graph of

52= +y x .


You should be able to recognise the

graph of =y kx

, and know that it has

asymptotes.

3 Sketch the graph of 3=y x and state the equations of its asymptotes.


You should be able to recognise the relationship between transformations of graphs and their equations.

4 Find the equation of the resulting curve after the graph of 32= −y x x is:a translated 3 units in the positive

x -directionb stretched vertically with scale

factor 3.


You should be able to write down the equation of a circle with a given centre and radius, and to find the centre and radius from a given equation.

5 Find the centre and radius of the circle with equation 4 10 72 2+ + − =x x y y .

Conic sectionsIn A Level Mathematics Student Book 1 you met two curves with equations that involve squared terms: the parabola, such as 3 2=y x , and the circle, such as 52 2+ =x y . In this chapter you will meet some other curves with similar equations; for example: 32 =y x , 2 52 2+ =x y and

52 2− =x y .


• recognise and work with Cartesian equations of ellipses, hyperbolas and parabolas

• solve problems involving intersections of lines with those curves and find equations of tangents

• recognise the effects of transformations (translations, stretches and reflections) on the equations of those curves.

3 The ellipse, hyperbola and parabola

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All such curves can be obtained as intersections of a plane with a cone; hence they are known as conic sections. As well as having many interesting mathematical properties, they have applications in modelling planetary orbits and the design of satellite dishes and radio telescopes.

Section 1: Introducing the ellipse, hyperbola and parabolaThe ellipse

You already know that the equation 252 2+ =x y represents a circle with radius 5, centred at the origin. Similarly, the equation 4 4 252 2+ =x y can

be rewritten as 254

2 2+ =x y , so it represents a circle with radius 52

. But

what happens if the coefficients of 2x and 2y in this equation are not equal?

Use graphing software to investigate equations of this form. Here are some examples.

Circle

Ellipse

Parabola

Hyperbola

O–

x

y

4x2 + 9y2 = 25

5–2

5–2

5–3

– 5–3

O x

y

x2 + 16y2 = 1

1–1 – 1–4

1–4

O x

y

5–5

3

–3

+ = 1x2

–25

y2

–9

In each case there is a closed curve centred at the origin. The axes intercepts are most easily seen if the equation is written in the form

12

2

2

2+ =xa

yb

.

55

3 The ellipses, hyperbola and parabola

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The equation 12

2

2

2+ =xa

yb

represents an ellipse centred at the origin, with

x-intercepts ( , 0)±a and y-intercepts (0, )±b .

O x

y

a–a

b

–b

Key point 3.1

It can be shown, using the universal law of gravitation and Newton’s laws of motion, that planets follow elliptical orbits.

Earth

Mars

Jupiter

Saturn

Uranus

Neptune

Ea

iter

Venus

Mercury

Did you know?

Sketch these ellipses, showing the coordinate axes intercepts.

a 49 25 1

2 2

+ =x yb 48 3 122 2+ =x y

a

O x

y

7–7

5

–5

Comparing this equation to Key point 3.1, 7=a and 5=b .

b x y

x y

x y

+ =

+ =

+ =

48 3 12

4 14 1

4 1

2 2

2 2

2

14

2

To compare this to Key point 3.1, the right-hand side needs to equal 1. So divide the whole equation by 12.

WORKED EXAMPLE 3.1


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The hyperbola

What happens if, in the equation of an ellipse, the 2y term, is negative? Here are some examples.

x y

x y

x y

+ =

+ =

+ =

48 3 12

4 14 1

4 1

2 2

2 2

2

14

2Multiplying by 4 is the same as dividing by 1

4.

O x

y

1–2

1–2

2

–2

–

You can now see that this is an ellipse with 12=a

and 2=b .

O x

y

4x2 – 9y2 = 25

–2.5 2.5 O x

y

–1 1

x2 – 16y2 = 1

O x

y

–5 5

– = 1x2

–25

y2

–9

Each of these graphs is a curve called a hyperbola. As with the ellipse, it

is convenient to write the equation in the form 12

2

2

2− =xa

yb

.

You can see several common features in all the graphs.

• The curve has the x and y -axes as lines of symmetry. This is because both x and y terms in the equation are squared.

• You can find the x -intercepts, which are also the vertices of the curve,

by setting 0=y ; so, for example, the curve with equation x y2

25

2

91− =

crosses the x -axis at (–5, 0) and (5, 0). • There is a range of values of x for which the curve is not defined.

For example, the curve with equation 25 9 1

2 2

− =x y is not defined for

x5 5− < < . • For large values of x andy , the curve seems to approach a straight line.

Because of the symmetry of the curve, this line passes through the origin. Investigate the gradient of the line by looking at a few more examples.

To see why x y2

25

2

91− = is‘nt

defined for x5 5− < < , rewrite the equation in the form

25 1 92

2

= +

xy

. Since 2y is

never negative, 2x can’t be smaller than 25(1 0) 25+ = , so x can’t be between –5 and 5 .

Tip

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The equation 12

2

2

2− =xa

yb

represents a hyperbola with x-intercepts ( , 0)±a

and asymptotes = ±y ba x.

O x

y

y = xb–ay = – xb–a

a–a

Key point 3.2

Sketch these hyperbolas, showing the axis intercepts and stating the equations of the asymptotes.

a 36 25 12 2

− =x yb 9 25 42 2− =x y

a

O x

y

y = x5–6y = – x5–

6

6–6

– = 1x2–36

y2–25

Comparing this to Key point 3.2, 6=a and 5=b .The x -intercepts are ( 6, 0)± .

The asymptotes are 56= ±y x . The asymptotes are given by = ±y b

a x.

WORKED EXAMPLE 3.2


O x

y

– = 1x2–25

y2–9

y = x3–5y = – x3–

5

O x

y

– = 1x2–16

y2–49

y = x7–4 y = – x7–

4

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You can derive the equations of the asymptotes of a hyperbola (rather than just guessing them from the graph as you did in Worked example 3.2).

b x y

x y

x y

− =

− =

− =

9 25 4

94

254 1

1

2 2

2 2

2

49

2

425

First write the equation in the form 12

2

2

2− =xa

yb

.

Multiplying by 94 is the same as dividing by 4

9 .

O x

y

y = x3–5

y = – x3–5

2–3

– = 19x2

–4

25y2

–4

2–3

–

You can now see that this is a hyperbola with 23=a

and 25=b .

Hence the x -intercepts are 23 , 0( )± .

Asymptotes:

35

2523

= ±

= ±

y x

y x

The asymptotes are given by = ±y ba x.

PROOF 4

Prove that the asymptotes of a hyperbola with equation xa

yb

12

2

2

2− = are y ba x= ± .

The asymptotes pass through the origin, so their equations are y mx= ± .

First, by symmetry of the curve, the asymptotes must pass through the origin so the y-intercept is zero. The two asymptotes must also be reflections of each other in the x-axis so their equations are of the form y mx= and y mx= − .

Consider a line of the form y mx= . Whether or not this line crosses the hyperbola depends on the value of m.

The hyperbola never intersects its asymptotes, so consider the intersection of the curve with the lines of the form y mx= ± .

xa

mxb

b x a m x a b

b a m x a b

( ) 1

( )

2

2

2

2

2 2 2 2 2 2 2

2 2 2 2 2 2

− =

− =− =

To find any possible intersections, substitute y mx= into the equation of the hyperbola.

Multiply both sides by the common denominator a b( )2 2 to clear fractions.


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You have actually met a special case of the hyperbola before: 1=y x (or,

more generally, =y kx ). It can be shown that this is indeed a hyperbola

but rotated so that the asymptotes are the coordinate axes. If k then it can be written as 2=k c . To emphasise the symmetry between x and y the

equation 2

=y cx is usually written as 2=xy c .

Solutions only exist if b a m

a m b

m ba

ba

m ba

02 2 2

2 2 2

22

2

− >⇔ <

⇔ <

⇔ − < <

Therefore for m 0> if m ba

> the line doesn’t cross the hyperbola and if m b

a< it does cross. The limiting case,

when m ba

= , gives the gradient of the asymptote.

The RHS is positive and x2 can’t be negative. This is only possible if the expression in the bracket on LHS is positive.

You can divide both sides by a2 without affecting the direction of the inequality because a2 is a positive number.

O x

y

m � b–a– = 1x2

–a2(mx)2

–b2

m � b–a

m � b–a

Similarly for m < 0 to give the gradient of the other

asymptote as m ba

–= . Hence the equations of the

asymptotes are y ba

x= and y ba

x= − .

The equation 2=xy c represents a hyperbola with vertices at ( , )c c and (– , – )c c

and coordinate axes as asymptotes.

O x

y

(c, c)

(–c, –c)

Key point 3.3

You may be surprised that the equations 2=xy c and

12

2

2

2− =xa

yb

represent members

of the same family of curves. In A Level Further Mathematics Student Book 2 you will see how to prove that the curves are related by rotation.

Fast forward

A graph shows how the limiting case gives the asymptotes.

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The parabola

In the equations of an ellipse and a hyperbola, both x and y terms are squared. You already know that 2=y kx is an equation of a parabola with a vertex at the origin. If the y term is squared instead, the curve still has the same shape, but is ‘sideways’.

Sketch the curve with equation 4 5=xy , labelling the coordinates of the vertices.

xy

xy

=

=

4 554

To use Key point 3.3 the equation needs to be in the form 2=xy c .

O x

y

( , )–��5–2

–��5–2

( , )��5–2

��5–2

This is a hyperbola with coordinate axes as asymptotes.

It has 54

2 =c so 52=c .

WORKED EXAMPLE 3.3

The equation 42 =y ax represents a parabola with m vertex at the origin, tangent to the x-axis.

O x

y

Key point 3.4

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A curve has equation 122 =y x .

a Sketch the curve.

b Let P be a point on the curve with coordinates ( , )p q . Show that the distance of P from the point (3, 0)F equals the distance of P from the line 3= −x .

a

O x

y

y2 = 12x

This is a parabola with vertex at the origin.

b

O x

y

y2 = 12x

x = –3

P (p, q)

F (3, 0)

d1

d2

Sketch a graph showing the required distances.

d p p

d p q

p p q

p p p

p p

p

= − − = +

= − + −= − + += − + += + += +

( 3) 3

( 3) ( 0)

6 9

6 9 12

6 9

( 3)

1

22 2 2

2 2

2

2

2

The distance from a point to a vertical line equals the difference between the x -coordinates.

You can use Pythagoras’ theorem to find the distance between two points.

Point P lines on the parabola, so 122 =q p .

Hence 32 1= + =d p d , as required.

WORKED EXAMPLE 3.4

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The property illustrated in Worked example 3.4 can also be taken as the defining feature of a parabola: it is the locus of the points that are equal distances from a given point ( )F and a given line ( )l .

The point F in Worked example 3.4 is called the focus of the parabola, and has an important special property. Horizontal rays reflected off a parabola with equation

42 =y ax all pass through the point F with coordinates ( , 0)a .

Satellite dishes usually have a parabolic shape.

Did you know?

F

l

F

EXERCISE 3A

1 Sketch these ellipses, showing the coordinates of any axis intercepts.

a i 4 9 1

2 2

+ =x yii

25 4 12 2

+ =x yb i

9 122

+ =xy

ii 16 1

22+ =x y

c i 9 16 12 2+ =x y ii 25 9 12 2+ =x y d i 4 25 162 2+ =x y ii 49 16 362 2+ =x y

2 Write down the equation of each ellipse.

a i

O x

y

5–5

3

–3

ii

O x

y

2–2

4

–4

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b i

O x

y

��3–��3

��5

–��5

ii

O x

y

��7–��7

��10

–��10

c i

O x

y

1–2

1–3

– 1–3

1–2

–

ii

O x

y

1–5

1–4

– 1–4

1–5

–

d i

O x

y

5–2

7–3

– 7–3

5–2

–

ii

O x

y

2–3

7–4

– 7–4

2–3

–

3 Sketch these hyperbolas, stating the coordinates of the vertices and the equations of the asymptotes.

a i 4 9 1

2 2

− =x yii

25 4 12 2

− =x yb i

9 122

− =xy

ii 16 1

22− =x y

c i 9 16 12 2− =x y ii 25 9 12 2− =x y d i 4 25 162 2− =x y ii 49 16 362 2− =x y

4 Sketch these hyperbolas, indicating the coordinates of the vertices.

a i 25=xy ii 16=xy b i 4 1=xy ii 49 1=xy

c i 16 20=xy ii 3 25=xy

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5 Find the equation of each hyperbola, giving your answers in the form 12

2

2

2− =xa

yb .

a i

O x

y

y = x3–5

y = – x3–5

–5 5

ii

O x

y

y = x2–3

y = – x2–3

–3 3

b i

O x

y

y = – x��2–��5

–��5 ��5

y = x��2–��5

ii

O x

y

y = – x��7–��3

–��3 ��3

y = x��7–��3

c i

O x

y

y = – x2–��3

y = x2–��3

1–2

1–2

–

ii

O x

y

y = – x3–��5

y = x3–��5

1–3

1–3

–

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d i

O x

y

2–3

2–3

–

y = x5–2

y = – x5–2

ii

O x

y

3–4

3–4

–

y = x2–3

y = – x2–3

6 Sketch each of these curves, showing the coordinates of any axes intercepts and vertices, and equations of any asymptotes.

a i 4 25 1

2 2

− =x yii

100 36 12 2

− =x yb i 82 =y x ii 202 =y x

c i 9 4 1

22+ =x y ii 25 9 12

2

+ =xy

d i 9=xy ii 144=xy

7 What is the equation of the curve in the diagram?

O x

y

10–10

7

–7

+ = 1x2

–100

y2

–49


A 100 49 1

2 2

− =x yB 100 49 12 2+ =x y

C 100 49 1

2 2

+ =x yD 49 100 12 2− =x y

8 What are the equations of the asymptotes of the hyperbola with equation 4 9 12 2− =x y ?Choose from these options.

A 49= ±y x B 9

4= ±y x

C 23= ±y x D 3

2= ±y x

9 An ellipse has equation 25 9 12 2

+ =x y and ( , )P p q is a point on the

ellipse. Points 1F and 2F have coordinates ( 4, 0)− and (4, 0). Show that the sum of the distances 1 2+PF PF does not depend on the value of p.

The property illustrated in Question 9 can be used as a defining feature of an ellipse: it is the locus of points whose sum of distances from two fixed points is constant.

Did you know?

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Section 2: Solving problems with ellipses, hyperbolas and parabolasIn A Level Mathematics Student Book 1 you learnt how to find intersections of lines with parabolas and circles. You also learnt how to use the discriminant to determine the number of intersections. You can use those methods to solve problems involving ellipses and hyperbolas.

You can also use the discriminant to solve problems about tangents and normals.

The line 2= +y x c intersects the hyperbola with equation 25 4 12 2

− =x y at two points.

Find the range of possible values of c.

25(2 )

4 1

4 25(4 4 ) 100

96 100 (100 25 ) 0

2 2

2 2 2

2 2

− + =

− + + =+ + + =

x x c

x x cx c

x cx c

To write an equation for the intersection points, substitute 2= +y x c into the equation of the hyperbola.

Clear the fractions: multiply both sides by the common denominator.

This is a quadratic equation in x . Make one side equal to zero.

There are two solutions:(100 ) 4(96)(100 25 ) 0

10000 9600 38400

96

2 2

2 2

2

− + >− >

>

c c

c c

c

For two intersection points, the discriminant needs to be positive.

O x

y

96

–4��6 4��6

4 6 or 4 6∴ < − >c c

This is a quadratic inequality for c, so you can sketch the graph to see that there are two solution intervals.

WORKED EXAMPLE 3.5

See A Level Mathematics Student Book 1, Chapter 3, for a reminder of quadratic equations and inequalities.

Rewind

It is also possible to use differentiation to find gradients of tangents and normals, but so far, you don’t know how to differentiate equations that have both x and y on the same side. These are called implicit equations and they are covered in A Level Mathematics Student Book 2.

Fast forward

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The line 25= +y mx , with 0>m , is a tangent to the ellipse with equation 400 225 1

2 2

+ =x y.

a Find the value of m.

The line touches the ellipse at point Q.

b Find the equation of the normal to the ellipse at Q.

a 400

( 25)225 1

225 400( 50 625) 90 000

9 16( 50 625) 360

2 2

2 2 2

2 2 2

+ + =

+ + + =+ + + =

x mx

x m x mx

x m x mx(9 16 ) 800 6400 02 2+ + + =m x mx

You need to find the value of m for which there is only one intersection point between the line and the ellipse. Start by writing an equation for the intersections: substitute 25= +y mx into the equation of the ellipse.

As there is one solution: (800 ) 4(9 16 )(6400) 0

640 000 (36 64 )(6400) 0

100 36 64 0

36 361

2 2

2 2

2 2

2

− + =− + =

− − ==

∴ =

m m

m m

m m

m

m

This is a quadratic equation for x , so make one side equal to zero.

For the line to be a tangent, this equation should only have one solution. This happens when the discriminant is zero.

b Coordinates of Q:

25 800 6400 0

32 256 0

( 16) 016

2

2

2

+ + =+ + =

+ =∴ = −

x x

x x

x

x

2516 25 9

= += − + =

y mx

You first need to find the coordinates of Q. The x -coordinate is the solution of the quadratic equation with the value of m from part a.

Gradient of normal: 1 11 = − = −m m

Equation of normal: 9 1( 16)− = − +y x

7 0+ + =x y

The normal is perpendicular to the tangent, so their gradients multiply to 1− .

Use ( )1 1 1− = −y y m x x for the equation of the normal.

WORKED EXAMPLE 3.6

The question says that m is positive.

A normal to the curve is perpendicular to the tangent at the point of contact. See A Level Mathematics Student Book 1, Chapter 6, for a reminder of perpendicular lines.

Rewind

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EXERCISE 3B

1 Find the value of m for which the line 2= +y mx is tangent to the parabola with equation 52 =y x .

2 A hyperbola has equation 16 25 12 2

− =x y.

a Sketch the hyperbola, stating the equations of the asymptotes.

b The line with equation 2= +y x c is tangent to the hyperbola. Find the possible values of c.

3 The line y mx 5= − is a tangent to the ellipse 4 13 12 2

+ =x y.

Find the possible values of m.

4 A parabola has equation 62 =y x .

a Show that the line with equation 32= +y x is tangent to the parabola and find the coordinates of the

point of contact, T .

b The normal to the parabola at T intersects the parabola again at Q. Find the coordinates of Q.

5 The line 3 5= +y x is tangent to the ellipse 2 7 12 2

+ =x y at the point P .

a Find the coordinates of P .

The normal to the ellipse at P intersects the ellipse again at Q.

b Find, to 3 significant figures, the coordinates of Q.

6 A curve has equation 16 12

2+ =x y .

a Sketch the curve, showing the intercepts with the coordinate axes.

b A line with equation 2= +y mx intersects the ellipse at two distinct points. Find the range of possible values of m.

A parabola has equation 202 =y x . A line passes through the point (0, )c , with 0≠c , and has gradient m.

Given that the line is tangent to the parabola, express m in terms of c.

= + ⇒ = −y mx c x

y cm

For the intersection with the parabola:

202 = −

y

y cm

20 202 = −my y c

20 20 02 − + =my y c

Write the equation for the intersection of the line and the parabola: substitute x from the equation of the line into the equation of the parabola.

One solution: 20 4 20 0

80 4005

2 − × ==

=

m c

mc

m c

If the line is a tangent to the parabola, this quadratic equation for y should have only one solution so the discriminant is zero.

Note that this gives a single value of m for every 0≠c .

WORKED EXAMPLE 3.7

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7 A hyperbola has equation 4 16 12 2

− =x y.

a Write down the equation of the asymptote with a positive gradient.

b Show that every other line parallel to this asymptote intersects the hyperbola exactly once.

8 A hyperbola has equation 12 2− =x y .

a The line 1= +y mx is a tangent to the hyperbola. Find the possible values of m.

b Show that this hyperbola has no tangents which pass through the origin.

Section 3: Transformations of curvesIn Sections 1 and 2 you only looked at ellipses and hyperbolas centred at the origin, and parabolas with vertex at the origin. You can use your knowledge of transformations of graphs to translate these curves.

Translations

You know how to apply a translation to a curve.

The hyperbola with equation 10 6 12 2

− =x y is translated by the vector

27−

.

Find the equation of the resulting curve and state the coordinates of its vertices.

( 2)10

( 7)6 1

2 2− − + =x y Replace x by x( 2)− and y by ( 7)+y .

− − + =3( 2) 5( 7) 302 2x y

− + − + + =(3 12 12) (5 70 245) 302 2x x y y

x x y y3 12 5 70 2632 2− − − =

Clear the fractions: multiply by the common denominator (30).

The original hyperbola has vertices ( 10 0),− and ( 10 0), ,

The vertices of a hyperbola 12

2

2

2− =xa

yb

are at ( , 0)±a .

so the new hyperbola has vertices

(2 10 7),− − and (2 10 7),+ −The curve has been translated by

27−

.

WORKED EXAMPLE 3.8

Replacing x by ( – )x p and y by ( – )y q results in a translation of the curve by

the vector

p

q.

Key point 3.5

Transformations of graphs are covered in A Level Mathematics Student Book 1, Chapter 5.

Rewind

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In A Level Mathematics Student Book 1 you learnt how to complete the square to find the centre and radius of a circle when the equation is written in an expanded form. You can apply the same technique to ellipses, hyperbolas and parabolas.

a Show that the equation 4 16 – 9 90 2452 2+ + =x x y y represents a hyperbola, and find the coordinates of its vertices and the equations of its asymptotes.

b Hence sketch the curve.

a 4 16 4 ( 4 )

4 ( 2) 4

4( 2) 16

2 2

2

2

[ ]

+ = += + −= + −

x x x x x

x

x

You want to write the equation in the form ( ) ( )

12

2

2

2

− − − =x pa

y qb

, so start by completing

the squares (separately for the x and y terms).

y y y y

y

y

9 90 9( 10 )

9 ( 5) 25

9( 5) 225

2 2

2

2

[ ]

− + = − −= − − −= − − +

Remember to take out a factor of –9 from each term (rather than just 9).

Hence:

+ − + =+ − − − + =+ − − =

+ − − =

4 16 9 90 245

4( 2) 16 9( 5) 225 245

4( 2) 9( 5) 36

( 2)9

( 5)4 1

2 2

2 2

2 2

2 2

x x y y

x y

x y

x y

Now put these back into the original equation.

This is a hyperbola with equation 9 4 12 2

− =x y

translated by the vector 25−

.

The vertices of the original hyperbola are at ( 3 0),− and (3 0), , so the vertices of the new hyperbola are ( 5 5),− and (1 5), .

The original hyperbola was translated 2 units to the left and 5 units up.

The asymptotes of the original hyperbola

are 23=y x and 2

3= −y x .

The asymptotes of a hyperbola are given by

= ±y ba x.

The new asymptotes are:

− = +( 5) 23 ( 2)y x

= +23

193y x

and

− = − +( 5) 23 ( 2)y x

= − +23

113y x

The translation replaces x by ( 2)+x and y by ( 5)−y .

Divide both sides by 36 to get the equation in the required form.

WORKED EXAMPLE 3.9

See A Level Mathematics Student Book 1, Chapter 3, for a reminder of completing the square.

Rewind


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Stretches

You also know how to apply horizontal and vertical stretches to curves.

Replacing x by xp and y by

yq results in a horizontal stretch with scale factor p

and a vertical stretch with scale factor q.

Key point 3.6

Show that the ellipse with equation 4 25 92 2+ =x y can be obtained from a circle with radius 1 by applying a horizontal stretch and a vertical stretch.

State the scale factor of each stretch.

4 25 92 2+ =x y

+ =49

259 1

2 2x y

You want to see the connection between the given ellipse and the circle 12 2+ =x y , so start by making the RHS of the equation equal to 1.

( ) +

=2

353 1

2 2x y This is the equation of the circle, with x

replaced by 23( )x and y replaced by

53

y.

Hence the ellipse is obtained from the circle 12 2+ =x y by a horizontal stretch with scale

factor 32 and a vertical stretch with scale

factor 35 .

Remember that a stretch with scale factor p replaces x by x

p .

WORKED EXAMPLE 3.10

b a

O x

y

4(x + 2)2 – 9(y – 5)2 = 36

y = – x – 2–3

19–3

y = – x + 2–3

11–3

(–5, 5) (1, 5)

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Reflections

A parabola has equation 122 =y x. Find the equation of the resulting curve when this parabola is:

a reflected in the y-axisb reflected in the line =y x.

a 122 (– )=y x

= −122y xReflection in the y-axis replaces x by −x.

b 122 =x y

= 112

2y x

Reflection in the line =y x swaps x and y.

WORKED EXAMPLE 3.11

Find the equation of the hyperbola shown in the diagram.

O x

y

y = 2xy = –2x

(0, –3)

(0, 3)

WORKED EXAMPLE 3.12

• Replacing x by x− reflects a curve in the y-axis.• Replacing y by y− reflects a curve in the x-axis.• Swapping x and y reflects a curve in the line =y x.

Key point 3.7


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Reflect the hyperbola in the line =y x .

O x

y

(–3, 0)

y = x1–2

y = – x1–2

(3, 0)

You can obtain this hyperbola from a standard hyperbola by reflecting it in the line =y x .

First find the equation of the reflected hyperbola by using its vertex coordinates and equations of asymptotes.

The asymptotes are:

2 12= ± ⇔ = ±x y y x

The reflection swaps x and y ; you can use this to find the asymptotes of the reflected hyperbola.

Equation of the reflected hyperbola:

12

2

2

2− =xa

yb

From the vertices:3=a

The vertices are at ( , 0)±a .

From the asymptotes:

12

32= ⇒ =b

a b

The equations of the asymptotes are = ±y ba x.

The reflected hyperbola has equation:

( )− =

31

2

2

2

32

2x y

− =94

9 12 2x y

The original hyperbola has equation

94

9 12 2

− =y x

Reflection in =y x swaps x and y .

EXERCISE 3C

1 Find the equation of each curve after the given transformation. Give your answers in expanded and simplified form.

a i 6 2 1

2 2

+ =x y; translation with vector

32−

ii 3 10 12 2

+ =x y; translation with vector 1

4−

b i 16=xy ; translation with vector 5

3−

ii 100=xy ; translation with vector 1010

c i 182 =y x ; vertical stretch with scale factor 3 ii 122 =y x ; horizontal stretch with scale factor 2

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d i 6 9 1

2 2

− =x y; reflection in the line =y x ii

10 5 12 2

− =x y; reflection in the line =y x

e i 42 =y x ; reflection in the y -axis ii 42 =y x ; reflection in the x -axis

2 Each of these curves is a translation of a standard ellipse, hyperbola or parabola. Identify the translation vector, and hence sketch each curve.

a i 9 – 18 4 8 – 23 02 2+ + =x x y y ii 2 25 – 100 76 02 2+ + + =x x y y

b i 9 – 18 – 4 8 – 31 02 2 + =x x y y ii 4 – 24 – – 2 31 02 2 + =x x y y

c i – 2 – 8 25 02 + =y y x ii 4 – 12 16 02 + + =y y x

d i – 3 – – 1 0=xy x y ii – 2 – 11 0+ =xy x y

3 An ellipse with equation 5 3 1

2 2

+ =x y is translated by the vector

p

q. The resulting curve has equation

3 – 18 5 10 17 02 2+ + + =x x y y .

Find the values of p and q.

4 A hyperbola with equation 3 12 2 2− =x b y is translated by the vector 14

. Given that the translated hyperbola passes through the origin, find the positive value of b.

5 The curve 1C with equation 16=xy , is translated by the vector 3

8−

. The resulting curve is 2C .

a Sketch the curve 2C , showing the coordinates of the vertices and equations of any asymptotes.

b Find the equation of 2C in the form f( )=y x .

6 A parabola with equation 42 =y ax is stretched parallel to the x -axis. The equation of the resulting curve is

52 =y ax . Find the scale factor of the stretch.

7 A parabola with equation 42 =y ax is reflected in the line =y x . The two curves intersect at the origin and at one more point R . Find, in terms of a, the coordinates of R .

8 Sketch the curve with equation 2 – 16 – 9 18 12 02 2 + + =x x y y , stating the coordinates of the vertices and equations of any asymptotes.

9 The curve 1C has equation 3 2 1

2 2

− =x y.

a Sketch the curve 1C , showing the intercepts with the coordinate axes.

The curve 2C is obtained from 1C by a stretch with scale factor k parallel to the y -axis.

b Given that the point (3, 5) lies on 2C , find the value of k.

10 The curve with equation 12

2

2

2+ =xa

yb

is stretched parallel to the x -axis with scale factor 23 . The resulting

curve is the circle with equation 42 2+ =x y . Find the positive values of a and b.

11 The line =y mx, where 0>m , is tangent to the curve with equation 4

2 5 12 2( )− + =x y

.

a Find the exact value of m.

b Hence find the equations of the tangents to the curve 2 5 1

2 2

+ =x y which pass through the point ( 4, 0)− .

12 The hyperbola with equation 4 16 12 2

− =x y is translated by the vector

3−

p. One asymptote of the new

hyperbola is 2 – 5=y x . Find the equation of the other asymptote.

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13 The ellipse 1C has equation 48 36 12 2

+ =x y. 1C is obtained from a circle 2C , with radius 6, by a horizontal

stretch.

a State the scale factor of the stretch.

The point (6, 3)P lies on 1C and the point Q lies on 2C . P is the image of Q under the stretch.

b Find the coordinates of Q.

c Find the equation of the tangent to 2C at Q.

d Hence find the equation of the normal to 1C at P .

14 The diagram shows the hyperbola with equation ( 3) – 252 2+ =x y . The line =y mx is a tangent to the hyperbola, and crosses the asymptotes at the points A and B .

O x

y

(x + 3)2 – y2 = 25

A

B

y = mx


b Write down the equations of the asymptotes of the hyperbola.

c Find the exact distance AB.

• The equation 12

2

2

2+ =xa

yb

represents and ellipse with axis intercepts ( , 0)±a and (0, )±b .

• The equation 12

2

2

2− =xa

yb

represents a hyperbola with vertices at ( , 0)±a and asymptotes = ±y ba x.

• The equation 2=xy c represents a hyperbola with vertices ( , )c c and (– , – )c c and the coordinate axes as asymptotes.

• The equation 42 =y ax represents a parabola with the vertex at the origin and the x -axis as the line of symmetry.

• You can use the discriminant of a quadratic equation to solve problems about the number of intersections of a line with a parabola, hyperbola or an ellipse, including identifying tangents.

• You can apply transformations to curves by changing their equations.• Replacing x by ( – )x p and y by ( – )y q results in a translation with vector

p

q.

• Replacing x by xa and y by

yb results in a horizontal stretch with scale factor a and a vertical stretch with

scale factor b.• Replacing x by −x results in a reflection in the y -axis.• Replacing y by −y results in a reflection in the x -axis.• Swapping x and y results in a reflection in the line =y x .


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Mixed practice 3 1 What is a possible equation of the curve shown in the diagram?


A 3 4 1

2 2

− =x yB 42 =y x

C 3 4 1

2 2

+ =x yD 5=xy

2 What is the equation of the ellipse shown in the diagram?


A 4 162 2+ =x y B 2 4 1

2 2

+ =x y

C 4 16 12 2+ =x y D 4 16 642 2+ =x y

3 A parabola with equation 32 =y x is translated by the vector 4

3−

.

What is the equation of the resulting curve?


A – 3 3 42 = +y x B ( – 3) 3 42 = +y x C 3 3 – 42( ) ( )+ =y x D ( – 3) 3 122 = +y x

4 The line –4= +y x c is a tangent to the ellipse 3 2 1

2 2

+ =x y. Find the possible values of c.

5 The line – 5=y mx is tangent to the hyperbola 4 13 1

2 2

− =x y. Find the possible values of m.

6 The parabola 1C has equation 122 =y x .

a Find the value of m for which the line 2= +y mx is tangent to 1C , and find the coordinates of the point of contact.

The parabola 1C is reflected in the y -axis to form a new curve, 2C .

b Write down the equation of 2C .

c Find the equation of the tangent to 2C at the point 43 , 4( )− .

7 Point (3, 4)P lies on the circle with equation 252 2+ =x y .

a Show that the tangent to the circle at P has equation 3 4 25+ =x y .

The circle is stretched horizontally to form an ellipse with equation 4 1002 2+ =x y .

b Find the scale factor of the stretch.

c Find the equation of the tangent to the ellipse at the point (6, 4).

O x

y

O x

y

2–2

4

–4

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8 The ellipse with equation 3 1

22+ =x y is translated by the vector

32−

. What is the equation of the resulting curve?


A – 6 3 – 12 18 02 2+ + =x x y y B – 6 3 12 18 02 2+ + + =x x y y

C 6 3 – 12 18 02 2+ + + =x x y y D 6 3 12 18 02 2+ + + + =x x y y

9 The parabola 1P has equation 82 =y x . The line 1l with equation 2= +y mx is tangent to the parabola at the point A.


b Find the coordinates of A.

The parabola 1P is translated by the vector 2

3−

to form a new parabola, 2P .

c Find the coordinates of the points where the parabola 2P crosses the coordinate axes.

d Sketch the parabola 2P , showing the coordinates of its vertex.

e Find the equation of the tangent to 2P at the point (0, 7).

10 The parabola 1C , with equation 52 =y x , is reflected in the line =y x to form a new parabola, 2C . The two parabolas intersect at the origin and another point, A.

a Write down the equation of 2C .

b Find the coordinates of A.

c Use differentiation to find the equation of tangent to 2C at A.

d Hence find the equation of the tangent to 1C at A.

11 Point ( , )P p q lies on the parabola with equation 42 =y ax . For all values of p, the distance of the point P from the line 5= −x is the same as its distance from the point (5, 0). Find the value of a.

12 A curve 1C has equation 9 16 12 2

− =x y.

a Sketch the curve 1C , stating the values of its intercepts with the coordinate axes.

b The curve 1C is translated by the vector 0

k, where 0<k , to give a curve 2C .

Given that 2C passes through the origin (0, 0), find the equations of the asymptotes of 2C .

[© AQA 2015]

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13 An ellipse is shown at right.

The ellipse intersects the x -axis at the points A and B . The equation

of the ellipse is 4

4 12

2( )− + =xy .

a Find the x -coordinates of A and B .

b The line ( 0)= >y mx m is a tangent to the ellipse, with point of contact P .i Show that the x -coordinate of P satisfies the equation

(1 4 ) 8 12 02 2+ − + =m x x .ii Hence find the exact value of m.

iii Find the coordinates of P .

[©AQA 2013]

14 The hyperbola with equation x y4 9 362 2− = is translated by the vector 51

. What are the asymptotes of the resulting curve?Choose from these options.

A x y4 9 1− = and 4 9 29+ =x y B x y2 3 7− = and 2 3 13+ =x y

C x y4 9 11− = − and x y4 9 29+ = − D x y2 3 13− = and x y2 3 7− =

15 A hyperbola with equation 122

2− =xyb , where b is a positive constant, is reflected in the line =y x .

Given that that the original hyperbola and its image intersect, find the range of possible values of b.

16 a Show that the line = +y mx c is a tangent to the ellipse 12

2

2

2+ =xa

yb

when 2 2 2 2+ =a m b c .

b The line l is a tangent to both the ellipse 25 21 1

2 2

+ =x y and the ellipse

16 57 12 2

+ =x y. Find the possible

equations of l.

17 An ellipse E has equation

16 9 12 2

+ =x y

a Sketch the ellipse E , showing the values of the intercepts on the coordinate axes.

b Given that the line with equation = +y x k intersects the ellipse E at two distinct points, show that –5 5< <k .

c The ellipse E is translated by the vector

ab

to form another ellipse whose equation is

9 16 18 642 2+ + − =x y x y c. Find the values of the constants ,a b and c.

d Hence find an equation for each of the two tangents to the ellipse 9 16 18 642 2+ + − =x y x y c that are parallel to the line =y x .

[© AQA 2014]

O x

y

A

P

B

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A Level Fur ther Mathematics for AQA

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