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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
© Boardworks Ltd 20069 of 29
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
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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
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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