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A2-Level Maths: Core 4for Edexcel

C4.2 Coordinate geometry

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Parametric equations of curves

Converting from parametric to Cartesian form

The parametric equations of some standard curves

The area under a curve defined parametrically

Examination-style question

Co

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Parametric equations of curves

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Parametric equations of curves

All of the curves we have looked at so far have been defined by a single equation in terms of x and y.

For example, using the parameter t a curve is defined by:

We can plot this curve for –3 < t < 3 using a table of values:

Curves can also be defined by writing x and y in terms of a third variable or parameter.

x = t2 – 3

y = 2t

6420–2–4–6

61–2–3–216

y = 2t

x = t2 – 3

3210–1–2–3t

(6, –6) (1, –4) (–2, –2) (–3, 0) (–2, 2) (1, 4) (6, 6)

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y

x0

Parametric equations of curves

Each value of t gives us a coordinate that we can then plot on a set of axes.

t = –3

t = –2

t = –1

t = 0

t = 1

t = 2

t = 3

In this example, drawing a smooth line through these points gives us a parabola.

In most cases, a graphical calculator or a graph-plotting computer program can be used to produce curves that have been defined parametrically.

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Parametric equations of curves

This diagram shows a sketch of the curve defined by:3

= + 23

tx

2= 9y t

y

xA BFind the coordinates of the points A and B where the curve meets the x-axis.

The curve meets the x-axis when y = 0, that is when:2 9 = 0t

2 = 9t= 3t

This point is called a cusp.

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Parametric equations of curves

When t = –3:

33= + 2

3x =11

When t = 3:

3( 3)= + 2

3x

= 7

So the coordinates of A are (–7, 0) and the coordinates of B are (11, 0).

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y

x0

Parametric equations of curves

This diagram shows a sketch of the curve defined by:

The curve meets the line x = 1 at points A and B.Find the exact length of the line segment AB.

When x = 1: 2 1=1t

2 = 2t

= 2t

A

B

x = 1x = t2 – 1

y = t3 – 4t

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Parametric equations of curves

3= ( 2) 4 2y

3= ( 2) + 4 2y

When t = :2

When t = – :2

= 2 2 4 2= 2 2

= 2 2 + 4 2= 2 2

So the coordinates of A are (1, ) and the coordinates of B are (1, – ).

2 22 2

The length of line segment AB = 2 2 2 2

= 4 2

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Parametric equations of curves

Converting from parametric to Cartesian form

The parametric equations of some standard curves

The area under a curve defined parametrically

Examination-style question

Co

nte

nts

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From parametric to Cartesian form

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Find the Cartesian equation for the following pair of parametric equations:

x = 3t + 1

y = 5 – 2t

In many cases, a curve that has been defined parametrically can be expressed in Cartesian form by eliminating the parameter. For example,

In examples of this type, we make t the subject of one of the equations and then substitute this expression into the other equation.

The Cartesian form of an equation only contains the two variables x and y.

Converting from parametric to Cartesian form

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Substituting this into the second equation gives:

This Cartesian equation represents a straight line graph.

1= 5 2

3

xy

2 + 25 =

3

xy

3 15 = 2 + 2y x

3 =17 2y x

If x = 3t + 1 then1

=3

xt

Converting from parametric to Cartesian form

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Find the Cartesian equation for the following pair of parametric equations:

x = 5 – t2

y = 3t2 – 4

Substituting this value of t2 into the second equation gives:

= 3(5 ) 4y x

If x = 5 – t2 then:2 = 5t x

=15 3 4y x

=11 3y x

The second equation is written in terms of t2 so we can leave this as it is.

Converting from parametric to Cartesian form

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Find the Cartesian equation for the following pair of parametric equations:

x = 3 + 2 sin θ

y = 1 + 2 cos θ

We can eliminate the parameter θ using the identity sin2 θ + cos2

θ = 1.

Squaring and adding these equations gives:

x – 3 = 2 sin θ

y – 1 = 2 cos θ

2 2 2 2 2 2( 3) + ( 1) = 2 sin + 2 cosx y 2 2= 4(sin + cos )

= 4

Converting from parametric to Cartesian form

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This is the equation of a circle of radius 2 centred at the point (3,1).

The Cartesian equation is therefore2 2( 3) + ( 1) = 4x y

Find the Cartesian equation for the following pair of parametric equations:

x = 2 cos θ

y = cos 2θ

Using the double angle formulae we can write:

y = 2 cos2 θ – 1

x2 = 4 cos2 θ and so the Cartesian equation is:

2

= 12

xy

Converting from parametric to Cartesian form

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Parametric equations of curves

Converting from parametric to Cartesian form

The parametric equations of some standard curves

The area under a curve defined parametrically

Examination-style question

Co

nte

nts

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Parametric equations of standard curves

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Parabolas

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Rectangular hyperbolae

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Circles centred at the origin

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Circles centred at the point (a, b)

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Ellipses

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Parametric equations of curves

Converting from parametric to Cartesian form

The parametric equations of some standard curves

The area under a curve defined parametrically

Examination-style question

Co

nte

nts

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The area under a curve defined parametrically

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The area under a curve defined parametrically

We know that the area under the curve y = f(x) between the limits x = a and x = b is given by:

Suppose, however, that we wish to find the area under a curve that is defined in terms of a parameter t.

We can write the area in terms of the parameter as:

where t1 and t2 are the limits x = a and x = b rewritten in terms of the parameter t.

=b

aA y dx

2

1

=t

t

dxA y dt

dt

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The area under a curve defined parametrically

For example, consider the curve defined by the parametric equations:

Suppose we want to find the area under this curve between x = –2 and x = 4.

x = 2t y = t 2 + 3

Since t = , these limits can be written in terms of t as:

2x

t = –1 and t = 2

Also, = 2dx

dt

y

x–2 4

A

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The area under a curve defined parametrically

The area, A, is given by:2

1

=t

t

dxA y dt

dtSubstituting t1 = –1, t2 = 2, and y = t2 + 3 gives: = 2

dx

dt2 2

1= 2( + 3)A t dt

2 2

1= 2 + 6t dt

23

1

23= + 6t x

16 23 3= ( +12) ( 6)

= 24

So the required area is 24 units2.

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Parametric equations of curves

Converting from parametric to Cartesian form

The parametric equations of some standard curves

The area under a curve defined parametrically

Examination-style question

Co

nte

nts

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Examination-style question

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Examination-style question

a) Find the coordinates of the points A and B.

b) Show that the Cartesian equation of the curve C is x2 – y2 = 16.

1= + 4x t

t

1= 4y t

t

The diagram shows part of the curve C, defined by the parametric equations:

The line y + 3x = 12 cuts the curve C at points A and B.

x

y

0 A

B

y + 3x = 12 C

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Examination-style question

a) Substituting and into y + 3x = 12 gives:1

= 4y tt

1= + 4x t

t

1 14 + 3 + 4 =12t t

t t

1 34 + +12 =12t t

t t

4+ 8 =12t

t

21+ 2 = 3t t

22 3 +1= 0t t

(2 1)( 1) = 0t t

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Examination-style question

When t = 1:

B is the point (5, –3).

1= + 4(1)

1x = 5

1= 4(1)

1y = 3

A is the point (4, 0).

When t = :12

121

2

1= + 4( )x = 4 1

212

1= 4( ) =y = 0

The line and the curve intersect when t = and when t = 1.12

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Examination-style question

b) Squaring the parametric equations of the curve gives:

Subtracting y2 from x2 gives:

22 1

= + 4x tt

22 1

= 4y tt

2 22

1= + 8 +16x t

t2 2

2

1= 8 +16y t

t

2 2 2 22 2

1 1= + 8 +16 + 8 16x y t t

t t

2 2 =16x y

and

= 16