C4.2 Coordinate geometry in the (x, y) plane

Report
A2-Level Maths:
Core 4
for Edexcel
C4.2 Coordinate
geometry
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Parametric equations of curves
Contents
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
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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.
Curves can also be defined by writing x and y in terms of a third
variable or parameter.
For example, using the parameter t a curve is defined by:
x = t2 – 3
y = 2t
We can plot this curve for –3 < t < 3 using a table of values:
t
x = t2 – 3
y = 2t
–3
–2
–1
0
1
2
3
6
–6
1
–4
–2
–2
–3
0
–2
2
1
4
6
6
(–2, 2)
(1, 4)
(6, 6)
(6, –6) (1, –4) (–2, –2) (–3, 0)
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Parametric equations of curves
Each value of t gives us a
coordinate that we can then
plot on a set of axes.
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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t=3
y
t=2
t=1
t=0
t = –1
0
x
t = –2
t = –3
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Parametric equations of curves
This diagram shows a sketch of the curve defined by:
t3
x = +2
3
y
y = t2  9
A
Find the coordinates of the
points A and B where the
curve meets the x-axis.
B
x
This point is
called a cusp.
The curve meets the x-axis when y = 0, that is when:
t2  9 = 0
t2 = 9
t = 3
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Parametric equations of curves
When t = 3:
33
x = + 2 = 11
3
When t = –3:
( 3)3
x=
+ 2 = 7
3
So the coordinates of A are (–7, 0) and the coordinates of B
are (11, 0).
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Parametric equations of curves
This diagram shows a sketch of the curve defined by:
x = t2 – 1
y
y = t3 – 4t
x=1
A
The curve meets the line
x = 1 at points A and B.
Find the exact length of the
line segment AB.
0
x
B
When x = 1:
t 2  1=1
t2 = 2
t= 2
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Parametric equations of curves
When t = 2 :
y = ( 2)3  4 2
=2 2 4 2
= 2 2
When t = – 2 :
y = (  2)3 + 4 2
= 2 2 + 4 2
=2 2
So the coordinates of A are (1, 2 2 ) and the coordinates of B
are (1, –2 2 ).
The length of line segment AB = 2 2  2 2
=4 2
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From parametric to Cartesian form
Contents
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
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Converting from parametric to Cartesian form
The Cartesian form of an equation only contains the two
variables x and y.
In many cases, a curve that has been defined parametrically
can be expressed in Cartesian form by eliminating the
parameter. For example,
Find the Cartesian equation for the following pair of
parametric equations:
x = 3t + 1
y = 5 – 2t
In examples of this type, we make t the subject of one of the
equations and then substitute this expression into the other
equation.
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Converting from parametric to Cartesian form
If x = 3t + 1 then
x 1
t=
3
Substituting this into the second equation gives:
 x  1
y = 5  2

 3 
2 x + 2
y 5=
3
3 y  15 = 2x + 2
3 y =17  2x
This Cartesian equation represents a straight line graph.
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Converting from parametric to Cartesian form
Find the Cartesian equation for the following pair of
parametric equations:
x = 5 – t2
y = 3t2 – 4
If x = 5 – t2 then:
t2 = 5  x
The second equation is written in terms of t2 so we can leave
this as it is.
Substituting this value of t2 into the second equation gives:
y = 3(5  x)  4
y =15  3 x  4
y =11 3 x
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Converting from parametric to Cartesian form
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.
x – 3 = 2 sin θ
y – 1 = 2 cos θ
Squaring and adding these equations gives:
( x  3)2 + ( y  1)2 = 22 sin2  + 22 cos2 
= 4(sin2  + cos2  )
=4
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Converting from parametric to Cartesian form
The Cartesian equation is therefore
( x  3)2 + ( y  1)2 = 4
This is the equation of a circle of radius 2 centred at the point
(3,1).
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:
x2
y = 1
2
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Contents
Parametric equations of standard curves
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
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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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Contents
The area under a curve defined parametrically
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
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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:
b
A =  y dx
a
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:
A=
t2
t1
dx
y dt
dt
where t1 and t2 are the limits x = a and x = b rewritten in terms
of the parameter t.
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The area under a curve defined parametrically
For example, consider the curve defined by the parametric
equations:
x = 2t
y = t2 + 3
Suppose we want to find the area
under this curve between x = –2
and x = 4.
y
x
Since t = 2 , these limits can be
written in terms of t as:
t = –1 and t = 2
dx
Also,
=2
dt
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A
–2
4
x
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The area under a curve defined parametrically
The area, A, is given by:
t2
dx
A =  y dt
t1
dt
dx
= 2 and y = t2 + 3 gives:
Substituting t1 = –1, t2 = 2,
dt
2
A =  2(t 2 + 3) dt
1
2
=  2t 2 + 6 dt
1
=
 32 t 3
+ 6 x 
2
1
2
= ( 16
3 +12)  (  3  6)
= 24
So the required area is 24 units2.
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Contents
Examination-style question
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
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Examination-style question
The diagram shows part of
the curve C, defined by the
parametric equations:
1
x = + 4t
t
1
y =  4t
t
y
y + 3x = 12
0
The line y + 3x = 12 cuts the curve
C at points A and B.
A
C
x
B
a) Find the coordinates of the points A and B.
b) Show that the Cartesian equation of the curve C is
x2 – y2 = 16.
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Examination-style question
1
1
a) Substituting x = + 4t and y =  4t into y + 3x = 12 gives:
t
t
1
1

 4t + 3  + 4t  = 12
t
t

1
3
 4t + +12t = 12
t
t
4
+ 8t = 12
t
1+ 2t 2 = 3t
2t 2  3t +1= 0
(2t  1)(t  1) = 0
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Examination-style question
 The line and the curve intersect when t = 21 and when t = 1.
When t =
1
2
:
x=
1
1
2
+ 4( 21 ) = 4
y=
1
1
2
 4( 21 ) = 0
 A is the point (4, 0).
When t = 1:
1
x = + 4(1) = 5
1
1
y =  4(1) = 3
1
 B is the point (5, –3).
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Examination-style question
b) Squaring the parametric equations of the curve gives:
1

2
x =  + 4t 
t

2
and
1
x = 2 + 8 +16t 2
t
2
1

2
y =   4t 
t

2
1
y = 2  8 +16t 2
t
2
Subtracting y2 from x2 gives:
1
1
2
x  y = 2 + 8 +16t  2 + 8  16t 2 = 16
t
t
2
2
 x 2  y 2 = 16
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