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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
"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with
permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"
Parametric Equations
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
==
Parametric Equations
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 =⇒
Parametric Equations
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.
Parametric Equations
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.
Parametric Equations
e.g. 2 Change the following to a Cartesian equation:
θθ sin3,cos3 == yx
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.
Parametric Equations
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.
Parametric Equations
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
Parametric Equations
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
θθ sin3,cos2 == yx
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
Parametric Equations
⇒== 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
Parametric Equations
⇒== 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
Parametric Equations
⇒== 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
Parametric Equations
⇒== θθ sin3,cos2 yx 22 ≤≤− x and 33 ≤≤− y
Symmetry now
completes the diagram.
x
x
x

90=θ
x
0=θ
Parametric Equations
⇒== θθ sin3,cos2 yx 22 ≤≤− x and 33 ≤≤− y
Symmetry now
completes the diagram.
x
x
x

90=θ
x
0=θ
Parametric Equations
θθ sin3,cos2 == yx
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
Parametric Equations
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!
Parametric Equations
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).
Parametric Equations
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.
Parametric Equations
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 ).
Parametric Equations
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.
Parametric Equations
θθ tan2,sec4 == yxSketch:
1
416
22
=−
yx
or
A hyperbola
Asymptotes
Parametric Equations
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 ×÷= =
Parametric Equations
t
ytx
3
,3 == 9=xySketch: or
A rectangular hyperbola.
Asymptotes
Parametric Equations
Parametric Equations
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.
Parametric Equations
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
==
Parametric Equations
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 =⇒
Parametric Equations
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.
Parametric Equations
e.g. 2 Change the following to a Cartesian equation:
θθ sin3,cos3 == yx
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
Parametric Equations
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.
Parametric Equations
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).

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1602 parametric equations

  • 1. 41: Parametric Equations41: Parametric Equations © Christine Crisp ““Teach A Level Maths”Teach A Level Maths” Vol. 2: A2 Core ModulesVol. 2: A2 Core Modules
  • 2. Parametric Equations Module C4 "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"
  • 3. Parametric Equations 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 ==
  • 4. Parametric Equations 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 =⇒
  • 5. Parametric Equations 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.
  • 6. Parametric Equations 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.
  • 7. Parametric Equations e.g. 2 Change the following to a Cartesian equation: θθ sin3,cos3 == yx 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.
  • 8. Parametric Equations 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.
  • 9. Parametric Equations 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
  • 10. Parametric Equations 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 θθ sin3,cos2 == yx 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
  • 11. Parametric Equations ⇒== 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
  • 12. Parametric Equations ⇒== 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
  • 13. Parametric Equations ⇒== 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
  • 14. Parametric Equations ⇒== θθ sin3,cos2 yx 22 ≤≤− x and 33 ≤≤− y Symmetry now completes the diagram. x x x  90=θ x 0=θ
  • 15. Parametric Equations ⇒== θθ sin3,cos2 yx 22 ≤≤− x and 33 ≤≤− y Symmetry now completes the diagram. x x x  90=θ x 0=θ
  • 16. Parametric Equations θθ sin3,cos2 == yx 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
  • 17. Parametric Equations 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!
  • 18. Parametric Equations 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).
  • 19. Parametric Equations 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.
  • 20. Parametric Equations 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 ).
  • 21. Parametric Equations 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.
  • 22. Parametric Equations θθ tan2,sec4 == yxSketch: 1 416 22 =− yx or A hyperbola Asymptotes
  • 23. Parametric Equations 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 ×÷= =
  • 24. Parametric Equations t ytx 3 ,3 == 9=xySketch: or A rectangular hyperbola. Asymptotes
  • 26. Parametric Equations 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.
  • 27. Parametric Equations 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 ==
  • 28. Parametric Equations 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 =⇒
  • 29. Parametric Equations 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.
  • 30. Parametric Equations e.g. 2 Change the following to a Cartesian equation: θθ sin3,cos3 == yx 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
  • 31. Parametric Equations 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.
  • 32. Parametric Equations 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).