1602 parametric equations

41: Parametric Equations41: Parametric Equations
© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Vol. 2: A2 Core ModulesVol. 2: A2 Core Modules

Parametric Equations
Module C4
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θθ sin3,cos3 == yx
The Cartesian equation of a curve in a plane is an
equation linking x and y.
Some of these equations can be written in a way
that is easier to differentiate by using 2 equations,
one giving x and one giving y, both in terms of a 3rd
variable, the parameter.
Letters commonly used for parameters are s, t and
θ. ( θ is often used if the parameter is an angle. )
e.gs. tytx 4,2 2
==

Converting between Cartesian and Parametric forms
We use parametric equations because they are
simpler, so we only convert to Cartesian if asked to
do so !
e.g. 1 Change the following to a Cartesian equation
and sketch its graph:
tytx 4,2 2
==
Solution: We need to eliminate the parameter t.
We substitute for t from the easier equation:
⇒= ty 4
4
y
t =
Subst. in :2 2
tx =
2
4
2 





=⇒
y
x
8
2
y
x =⇒

The Cartesian equation is
8
2
y
x =
We usually write this as xy 82
=
Either, we can sketch using a graphical calculator
with
xy 8±=
and entering the graph in 2 parts.
Or, we can notice that the equation is quadratic with
x and y swapped over from the more usual form.

The sketch is
The curve is called a parabola.
xy 82
=
tytx 4,2 2
==Also, the parametric equations
show that as t increases, x increases faster than y.

e.g. 2 Change the following to a Cartesian equation:
Solution: We need to eliminate the parameter θ.
BUT θ appears in 2 forms: as and
so, we need a link between these 2 forms.
θcos θsin
Which trig identity links and ?θcos θsin
ANS: 1sincos 22
≡+ θθ
To eliminate θ we substitute into this expression.

θcos
3
=⇒
x
9
sin
2
2 y
=θ
9
cos
2
2 x
=⇒ θ
θsin
3
=
y
1sincos 22
≡+ θθ 1
99
22
=+
yx
922
=+ yxMultiply by 9:
becomes
θθ sin3,cos3 == yxSo,
N.B. = not ≡
We have a circle, centre (0, 0), radius 3.

Since we recognise the circle in Cartesian form,
it’s easy to sketch.
However, if we couldn’t eliminate the parameter or
didn’t recognise the curve having done it, we can
sketch from the parametric form.
If you are taking Edexcel you may want to skip this
as you won’t be asked to do it.
SKIP

Solution:
Let’s see how to do it without eliminating the
parameter.
We can easily spot the min and max values of x and y:
22 ≤≤− x and 33 ≤≤− y
( It doesn’t matter that we don’t know which angle
θ is measuring. )
For both and the min is −1 and the
max is +1, so
θcos θsin
e.g. Sketch the curve with equations
It’s also easy to get the other coordinate at each
of these 4 key values e.g. 002 =⇒=⇒= yx 
θ

⇒== θθ sin3,cos2 yx 22 ≤≤− x and 33 ≤≤− y
We could draw up a
table of values finding
x and y for values of
θ but this is usually
very inefficient. Try
to just pick out
significant features.
x
x
x

90=θ
x
0=θ

⇒== θθ sin3,cos2 yx 22 ≤≤− x and
x
33 ≤≤− y
This tells us what happens to x and y.

90
Think what happens to and as θ increases
from 0 to .
θcos θsin
We could draw up a
table of values finding
x and y for values of
θ but this is usually
very inefficient. Try
to just pick out
significant features.
x
x
x

90=θ
x
0=θ

⇒== θθ sin3,cos2 yx 22 ≤≤− x and
x
Symmetry now
completes the diagram.
33 ≤≤− y
This tells us what happens to x and y.

90
Think what happens to and as θ increases
from 0 to .
θcos θsin
x
x
x

90=θ
x
0=θ

⇒== θθ sin3,cos2 yx 22 ≤≤− x and 33 ≤≤− y
Symmetry now
completes the diagram.
x
x
x

90=θ
x
0=θ

So, we have the parametric equations of an ellipse
( which we met in Cartesian form in Transformations ).
The origin is at the
centre of the ellipse.
x
x
x
x
Ox

You can use a graphical calculator to sketch
curves given in parametric form. However, you
will have to use the setup menu before you enter
the equations.
You will also have to be careful about the range
of values of the parameter and of x and y. If
you don’t get the right scales you may not see
the whole graph or the graph can be distorted
and, for example, a circle can look like an
ellipse.
By the time you’ve fiddled around it may have
been better to sketch without the calculator!

The following equations give curves you need to
recognise:
θθ sin,cos ryrx ==
atyatx 2,2
==
)(sin,cos babyax ≠== θθ
a circle, radius r, centre
the origin.
a parabola, passing through
the origin, with the x-axis as
an axis of symmetry.
an ellipse with centre at the
origin, passing through the
points (a, 0), (−a, 0), (0, b), (0, −b).

To write the ellipse in Cartesian form we use the
same trig identity as we used for the circle.
)(sin,cos babyax ≠== θθSo, for
use 1sincos 22
≡+ θθ
12
2
2
2
=+
b
y
a
x
⇒
1
22
=





+





b
y
a
x
⇒
The equation is usually left in this form.

There are other parametric equations you might be
asked to convert to Cartesian equations. For example,
those like the ones in the following exercise.
Exercise
θθ tan2,sec4 == yx
t
ytx
3
,3 ==
( Use a trig identity )
1.
2.
Sketch both curves using either parametric or Cartesian
equations. ( Use a graphical calculator if you like ).

Solution:
θθ tan2,sec4 == yx1.
Use θθ 22
sectan1 ≡+
22
42
1 





=





+⇒
xy
164
1
22
xy
=+⇒
We usually write this in a form similar to the
ellipse:
1
416
22
=−
yx
Notice the minus sign. The curve is a hyperbola.

θθ tan2,sec4 == yxSketch:
1
416
22
=−
yx
or
A hyperbola
Asymptotes

t
ytx
3
,3 ==
( Eliminate t by substitution. )
2.
Solution:
3
3
x
ttx =⇒=
⇒=
t
y
3
Subs. in
x
y
9
=⇒
9=⇒ xy
3
3
x
y =
The curve is a rectangular hyperbola.
x
x 3
3
33 ×÷= =

t
ytx
3
,3 == 9=xySketch: or
A rectangular hyperbola.
Asymptotes

The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.

Converting between Cartesian and Parametric forms
We use parametric equations because they are
simpler, so we only convert to Cartesian if asked to
do so !
e.g. 1 Change the following to a Cartesian equation
and sketch its graph:
tytx 4,2 2
==
Solution: We need to eliminate the parameter t.
Substitution is the easiest way.
⇒= ty 4
4
y
t =
Subst. in :2 2
tx =
2
4
2 





=⇒
y
x
8
2
y
x =⇒

e.g. 2 Change the following to a Cartesian equation:
Solution: We need to eliminate the parameter θ.
BUT θ appears in 2 forms: as and
so, we need a link between these 2 forms.
θcos θsin
To eliminate θ we substitute into the expression.
1sincos 22
≡+ θθ

The following equations give curves you need to
recognise:
θθ sin,cos ryrx ==
atyatx 2,2
==
)(sin,cos babyax ≠== θθ
a circle, radius r, centre
the origin.
a parabola, passing through
the origin, with the x-axis an
axis of symmetry.
an ellipse with centre at the
origin, passing through the
points (a, 0), (−a, 0), (0, b), (0, −b).

1602 parametric equations

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