1.6.3 Parametric equations of curves (A-level only)

What you'll learn

How a parameter can define both coordinates of a point on a curve.
How to plot, interpret and use parametric equations, including the direction of travel.
How to eliminate a parameter to obtain a Cartesian equation.
How to convert suitable Cartesian equations into parametric form.

Before You Start: Cartesian Equations

You are already familiar with Cartesian equations, which connect the coordinates $x$ and $y$ directly. For example,

y=x^2+3

describes a parabola.

To find a point on this curve, you choose an $x$ -coordinate and calculate the corresponding $y$ -coordinate. Parametric equations describe points differently: both coordinates are calculated from a third variable.

What Are Parametric Equations?

Definition

Parameter

A parameter is an extra variable, often written as $t$ , which determines the coordinates of a point on a curve.

In a pair of parametric equations, $x$ and $y$ are each given in terms of the same parameter:

\qquad y=g(t).

As $t$ changes, the point $(x,y)\left(x,y\right)$ moves and traces out a parametric curve.

Key Idea

One parameter controls both coordinates

For each permitted value of $t$ , calculate $x$ and $y$ using the same value of $t$ . This produces one point $(x,y)\left(x,y\right)$ on the curve.

For example, consider

x=2t,y=t2.x=2t,\qquad y=t^2.

Example

Generating points on a parametric curve

Choose some values of the parameter, such as $t = - 2, - 1, 0, 1, 2$ .
Substitute each value into both equations:
$t=−2:x=−4,y=4,t=−1:x=−2,y=1,t=0:x=0,y=0,t=1:x=2,y=1,t=2:x=4,y=4.\begin{aligned} t=-2 &: \quad x=-4,\quad y=4,\\ t=-1 &: \quad x=-2,\quad y=1,\\ t=0 &: \quad x=0,\quad y=0,\\ t=1 &: \quad x=2,\quad y=1,\\ t=2 &: \quad x=4,\quad y=4. \end{aligned}$
Plot the resulting points. They lie on a parabola with vertex at the origin.
Follow the points in order of increasing $t$ . The point moves down the left branch towards the origin, then rises along the right branch.

Graph of x equals 2t and y equals t squared, with labelled parameter values and arrows showing increasing t

Direction and Restrictions

Direction of Travel

Unlike a Cartesian equation by itself, parametric equations can indicate the direction in which a curve is traced. You find this by considering what happens as the parameter increases.

For $x = 2 t$ and $y=t^2$ , increasing $t$ moves the point from the upper-left branch, through the origin, and then up the right branch.

Tip

Showing direction on a sketch

Calculate points for several increasing values of $t$ , plot them in order, and add arrowheads to the curve.

Restrictions on the Parameter

A question may restrict the parameter, for example:

−1≤t≤2.-1\le t\le 2.

This means that only the part of the curve generated by those values is included. The restriction may produce a complete curve, an arc, a line segment or just part of a larger curve.

Example

Finding the endpoints of a restricted curve

The curve is given by

x=t+1,y=t2,−1≤t≤2.x=t+1,\qquad y=t^2,\qquad -1\le t\le 2.

At the lower endpoint $t = - 1$ ,
$x=0,y=1,x=0,\qquad y=1,$
so one endpoint is $(0,1)\left(0,1\right)$ .
At the upper endpoint $t = 2$ ,
$x=3,y=4,x=3,\qquad y=4,$
so the other endpoint is $(3,4)\left(3,4\right)$ .
Since $x = t + 1$ increases with $t$ , the curve is traced from $(0,1)\left(0,1\right)$ to $(3,4)\left(3,4\right)$ .

Common Mistake

Ignoring the parameter interval

After finding the Cartesian equation, do not automatically sketch the whole Cartesian curve. The permitted values of $t$ may generate only part of it.

Converting Parametric Equations to Cartesian Form

Definition

Eliminating the parameter

To eliminate the parameter means to combine the parametric equations so that the final equation contains only $x$ and $y.

The usual method is:

Rearrange one equation to make the parameter the subject.
Substitute that expression into the other equation.
Simplify and include any resulting restriction.

When the Parameter Is Easy to Isolate

Example

Eliminating a linear parameter

Convert

x=3t−1,y=2t2+5x=3t-1,\qquad y=2t^2+5

to Cartesian form.

The equation for $x$ is easiest to rearrange:
$\quad\Rightarrow\quad t=\frac{x+1}{3}.$
Substitute this into the equation for $y$ :
$y=2(x+13)2+5.y=2\left(\frac{x+1}{3}\right)^2+5.$
Simplify:
$y=29(x+1)2+5.y=\frac{2}{9}(x+1)^2+5.$
This is a parabola with vertex $(−1,5)\left(-1,5\right)$ .

Common Mistake

Losing brackets when substituting

If $t=x+13t=\frac{x+1}{3}$ and the equation contains $t^2$ , substitute $(x+13)2\left(\frac{x+1}{3}\right)^2$ . The whole expression must be squared.

When Squaring or Using an Identity Helps

Sometimes you do not need to make $t$ the subject directly. Instead, identify expressions that combine conveniently.

For example, if

x=acos⁡t,y=asin⁡t,x=a\cos t,\qquad y=a\sin t,

then

xa=cos⁡t,ya=sin⁡t.\frac{x}{a}=\cos t,\qquad \frac{y}{a}=\sin t.

Using the identity $sin^2t+\cos^2t=1$ gives

x2a2+y2a2=1,\frac{x^2}{a^2}+\frac{y^2}{a^2}=1,

x^2+y^2=a^2.

This is a circle with centre at the origin and radius $a$ , provided the parameter covers a full revolution.

Example

Converting a trigonometric parametrisation

Convert

x=4cos⁡t,y=3sin⁡tx=4\cos t,\qquad y=3\sin t

to Cartesian form.

Rearrange each equation:
$cos⁡t=x4,sin⁡t=y3.\cos t=\frac{x}{4},\qquad \sin t=\frac{y}{3}.$
Square both equations and use $sin^2t+\cos^2t=1$ :
$x216+y29=1.\frac{x^2}{16}+\frac{y^2}{9}=1.$
This is an ellipse centred at the origin, with horizontal semi-axis 4 and vertical semi-axis 3.

Common Mistake

Squaring can lose directional information

A Cartesian equation obtained by squaring may describe the correct set of points, but it does not show the direction in which the parametric curve is traced.

Using a Cartesian Equation to Find Coordinates

You do not always need to eliminate $t$ completely. If a point satisfies an extra condition, use whichever parametric equation makes that condition easiest to apply.

Example

Finding points with a given coordinate

The curve is

x=t2−1,y=2t+3.x=t^2-1,\qquad y=2t+3.

Find the points where $x = 3$ .

Apply the condition to the equation for $x$ :
$t2−1=3⇒t2=4.t^2-1=3 \quad\Rightarrow\quad t^2=4.$
Solve for all possible parameter values:
$t=±2.t=\pm2.$
When $t = 2$ , $y = 7$ . When $t = - 2$ , $y = - 1$ . Therefore the points are
$(3,7)and(3,−1).\left(3,7\right)\quad\text{and}\quad\left(3,-1\right).$

Common Mistake

Missing a parameter value

An equation such as $t^2=4$ has two solutions. Check every permitted value of $t$ , because different parameter values may produce different points.

Converting Cartesian Equations to Parametric Form

To parametrise a Cartesian curve means to choose a parameter and express both $x$ and $y$ in terms of it.

There is usually more than one correct parametrisation.

A Simple Choice

For a graph written as $y = f (x)$ , you can often choose

x = t

and then replace $x$ by $t$ in the equation for $y$ .

Example

Parametrising a parabola

Write a parametric form of

y=x^2-4x+1.

Choose $x = t$ .
Substitute $x = t$ into the Cartesian equation:
$y=t^2-4t+1.$
One valid parametrisation is therefore
$x=t,y=t2−4t+1.x=t,\qquad y=t^2-4t+1.$

Choosing a Useful Parametrisation

Sometimes a structured choice is more useful. For the circle

x^2+y^2=a^2,

the standard parametrisation is

x=acos⁡t,y=asin⁡t.x=a\cos t,\qquad y=a\sin t.

For an ellipse

x2a2+y2b2=1,\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,

a standard parametrisation is

x=acos⁡t,y=bsin⁡t.x=a\cos t,\qquad y=b\sin t.

Key Idea

Parametrisations are not unique

Different pairs of parametric equations can trace the same Cartesian curve. They may trace it at different rates, in different directions or over different parameter intervals.

Exam technique

In the exam

To eliminate $t$ , first identify which equation is easiest to rearrange or which identity links the two equations.
Keep any parameter restriction and translate it into endpoints or restrictions on $x$ and $y$ where possible.
For coordinate conditions, solve for every valid value of $t$ before calculating the corresponding points.
On a sketch, mark key points and use increasing values of $t$ to show the direction of travel.

Self review

Check yourself

Can you eliminate $t$ from $x = 2 t + 1$ and $y=t^2-3$ ?
What Cartesian curve is described by $x=5cos⁡tx=5\cos t$ and $y=2sin⁡ty=2\sin t$ ?
How would you find the endpoints and direction when a parameter interval is given?

What you'll learn

How a parameter can define both coordinates of a point on a curve.
How to plot, interpret and use parametric equations, including the direction of travel.
How to eliminate a parameter to obtain a Cartesian equation.
How to convert suitable Cartesian equations into parametric form.

Before You Start: Cartesian Equations

You are already familiar with Cartesian equations, which connect the coordinates $x$ and $y$ directly. For example,

y=x^2+3

describes a parabola.

What Are Parametric Equations?

Definition

Parameter

A parameter is an extra variable, often written as $t$ , which determines the coordinates of a point on a curve.

In a pair of parametric equations, $x$ and $y$ are each given in terms of the same parameter:

\qquad y=g(t).

As $t$ changes, the point $(x,y)\left(x,y\right)$ moves and traces out a parametric curve.

Key Idea

One parameter controls both coordinates

For each permitted value of $t$ , calculate $x$ and $y$ using the same value of $t$ . This produces one point $(x,y)\left(x,y\right)$ on the curve.

For example, consider

x=2t,y=t2.x=2t,\qquad y=t^2.

Example

Generating points on a parametric curve

Choose some values of the parameter, such as $t = - 2, - 1, 0, 1, 2$ .
Substitute each value into both equations:
$t=−2:x=−4,y=4,t=−1:x=−2,y=1,t=0:x=0,y=0,t=1:x=2,y=1,t=2:x=4,y=4.\begin{aligned} t=-2 &: \quad x=-4,\quad y=4,\\ t=-1 &: \quad x=-2,\quad y=1,\\ t=0 &: \quad x=0,\quad y=0,\\ t=1 &: \quad x=2,\quad y=1,\\ t=2 &: \quad x=4,\quad y=4. \end{aligned}$
Plot the resulting points. They lie on a parabola with vertex at the origin.
Follow the points in order of increasing $t$ . The point moves down the left branch towards the origin, then rises along the right branch.

Graph of x equals 2t and y equals t squared, with labelled parameter values and arrows showing increasing t

Direction and Restrictions

Direction of Travel

Unlike a Cartesian equation by itself, parametric equations can indicate the direction in which a curve is traced. You find this by considering what happens as the parameter increases.

For $x = 2 t$ and $y=t^2$ , increasing $t$ moves the point from the upper-left branch, through the origin, and then up the right branch.

Tip

Showing direction on a sketch

Calculate points for several increasing values of $t$ , plot them in order, and add arrowheads to the curve.

Restrictions on the Parameter

A question may restrict the parameter, for example:

−1≤t≤2.-1\le t\le 2.

This means that only the part of the curve generated by those values is included. The restriction may produce a complete curve, an arc, a line segment or just part of a larger curve.

Example

Finding the endpoints of a restricted curve

The curve is given by

x=t+1,y=t2,−1≤t≤2.x=t+1,\qquad y=t^2,\qquad -1\le t\le 2.

At the lower endpoint $t = - 1$ ,
$x=0,y=1,x=0,\qquad y=1,$
so one endpoint is $(0,1)\left(0,1\right)$ .
At the upper endpoint $t = 2$ ,
$x=3,y=4,x=3,\qquad y=4,$
so the other endpoint is $(3,4)\left(3,4\right)$ .
Since $x = t + 1$ increases with $t$ , the curve is traced from $(0,1)\left(0,1\right)$ to $(3,4)\left(3,4\right)$ .

Common Mistake

Ignoring the parameter interval

After finding the Cartesian equation, do not automatically sketch the whole Cartesian curve. The permitted values of $t$ may generate only part of it.

Converting Parametric Equations to Cartesian Form

Definition

Eliminating the parameter

To eliminate the parameter means to combine the parametric equations so that the final equation contains only $x$ and $y.

The usual method is:

Rearrange one equation to make the parameter the subject.
Substitute that expression into the other equation.
Simplify and include any resulting restriction.

When the Parameter Is Easy to Isolate

Example

Eliminating a linear parameter

Convert

x=3t−1,y=2t2+5x=3t-1,\qquad y=2t^2+5

to Cartesian form.

The equation for $x$ is easiest to rearrange:
$\quad\Rightarrow\quad t=\frac{x+1}{3}.$
Substitute this into the equation for $y$ :
$y=2(x+13)2+5.y=2\left(\frac{x+1}{3}\right)^2+5.$
Simplify:
$y=29(x+1)2+5.y=\frac{2}{9}(x+1)^2+5.$
This is a parabola with vertex $(−1,5)\left(-1,5\right)$ .

Common Mistake

Losing brackets when substituting

If $t=x+13t=\frac{x+1}{3}$ and the equation contains $t^2$ , substitute $(x+13)2\left(\frac{x+1}{3}\right)^2$ . The whole expression must be squared.

When Squaring or Using an Identity Helps

Sometimes you do not need to make $t$ the subject directly. Instead, identify expressions that combine conveniently.

For example, if

x=acos⁡t,y=asin⁡t,x=a\cos t,\qquad y=a\sin t,

then

xa=cos⁡t,ya=sin⁡t.\frac{x}{a}=\cos t,\qquad \frac{y}{a}=\sin t.

Using the identity $sin^2t+\cos^2t=1$ gives

x2a2+y2a2=1,\frac{x^2}{a^2}+\frac{y^2}{a^2}=1,

x^2+y^2=a^2.

This is a circle with centre at the origin and radius $a$ , provided the parameter covers a full revolution.

Example

Converting a trigonometric parametrisation

Convert

x=4cos⁡t,y=3sin⁡tx=4\cos t,\qquad y=3\sin t

to Cartesian form.

Rearrange each equation:
$cos⁡t=x4,sin⁡t=y3.\cos t=\frac{x}{4},\qquad \sin t=\frac{y}{3}.$
Square both equations and use $sin^2t+\cos^2t=1$ :
$x216+y29=1.\frac{x^2}{16}+\frac{y^2}{9}=1.$
This is an ellipse centred at the origin, with horizontal semi-axis 4 and vertical semi-axis 3.

Common Mistake

Squaring can lose directional information

A Cartesian equation obtained by squaring may describe the correct set of points, but it does not show the direction in which the parametric curve is traced.

Using a Cartesian Equation to Find Coordinates

You do not always need to eliminate $t$ completely. If a point satisfies an extra condition, use whichever parametric equation makes that condition easiest to apply.

Example

Finding points with a given coordinate

The curve is

x=t2−1,y=2t+3.x=t^2-1,\qquad y=2t+3.

Find the points where $x = 3$ .

Apply the condition to the equation for $x$ :
$t2−1=3⇒t2=4.t^2-1=3 \quad\Rightarrow\quad t^2=4.$
Solve for all possible parameter values:
$t=±2.t=\pm2.$
When $t = 2$ , $y = 7$ . When $t = - 2$ , $y = - 1$ . Therefore the points are
$(3,7)and(3,−1).\left(3,7\right)\quad\text{and}\quad\left(3,-1\right).$

Common Mistake

Missing a parameter value

An equation such as $t^2=4$ has two solutions. Check every permitted value of $t$ , because different parameter values may produce different points.

Converting Cartesian Equations to Parametric Form

To parametrise a Cartesian curve means to choose a parameter and express both $x$ and $y$ in terms of it.

There is usually more than one correct parametrisation.

A Simple Choice

For a graph written as $y = f (x)$ , you can often choose

x = t

and then replace $x$ by $t$ in the equation for $y$ .

Example

Parametrising a parabola

Write a parametric form of

y=x^2-4x+1.

Choose $x = t$ .
Substitute $x = t$ into the Cartesian equation:
$y=t^2-4t+1.$
One valid parametrisation is therefore
$x=t,y=t2−4t+1.x=t,\qquad y=t^2-4t+1.$

Choosing a Useful Parametrisation

Sometimes a structured choice is more useful. For the circle

x^2+y^2=a^2,

the standard parametrisation is

x=acos⁡t,y=asin⁡t.x=a\cos t,\qquad y=a\sin t.

For an ellipse

x2a2+y2b2=1,\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,

a standard parametrisation is

x=acos⁡t,y=bsin⁡t.x=a\cos t,\qquad y=b\sin t.

Key Idea

Parametrisations are not unique

Different pairs of parametric equations can trace the same Cartesian curve. They may trace it at different rates, in different directions or over different parameter intervals.

Exam technique

In the exam

To eliminate $t$ , first identify which equation is easiest to rearrange or which identity links the two equations.
Keep any parameter restriction and translate it into endpoints or restrictions on $x$ and $y$ where possible.
For coordinate conditions, solve for every valid value of $t$ before calculating the corresponding points.
On a sketch, mark key points and use increasing values of $t$ to show the direction of travel.

Self review

Check yourself

Can you eliminate $t$ from $x = 2 t + 1$ and $y=t^2-3$ ?
What Cartesian curve is described by $x=5cos⁡tx=5\cos t$ and $y=2sin⁡ty=2\sin t$ ?
How would you find the endpoints and direction when a parameter interval is given?

What you'll learn

Before You Start: Cartesian Equations

What Are Parametric Equations?

Direction and Restrictions

Direction of Travel

Restrictions on the Parameter

Converting Parametric Equations to Cartesian Form

When the Parameter Is Easy to Isolate

When Squaring or Using an Identity Helps

Using a Cartesian Equation to Find Coordinates

Converting Cartesian Equations to Parametric Form

A Simple Choice

Choosing a Useful Parametrisation

Review 1.6.3 Parametric equations of curves (A-level only) by teaching Genie

1.6.3 Parametric equations of curves (A-level only)

What you'll learn

Before You Start: Cartesian Equations

What Are Parametric Equations?

Direction and Restrictions

Direction of Travel

Restrictions on the Parameter

Converting Parametric Equations to Cartesian Form

When the Parameter Is Easy to Isolate

When Squaring or Using an Identity Helps

Using a Cartesian Equation to Find Coordinates

Converting Cartesian Equations to Parametric Form

A Simple Choice

Choosing a Useful Parametrisation

Review 1.6.3 Parametric equations of curves (A-level only) by teaching Genie

1.6.3 Parametric equations of curves (A-level only)

What you'll learn

Before You Start: Cartesian Equations

What Are Parametric Equations?

Direction and Restrictions

Direction of Travel

Restrictions on the Parameter

Converting Parametric Equations to Cartesian Form

When the Parameter Is Easy to Isolate

When Squaring or Using an Identity Helps

Using a Cartesian Equation to Find Coordinates

Converting Cartesian Equations to Parametric Form

A Simple Choice

Choosing a Useful Parametrisation

Review 1.6.3 Parametric equations of curves (A-level only) by teaching Genie

1.6.3 Parametric equations of curves (A-level only) Revision Guide

1.6.3 Parametric equations of curves (A-level only)

What you'll learn

Before You Start: Cartesian Equations

What Are Parametric Equations?

Direction and Restrictions

Direction of Travel

Restrictions on the Parameter

Converting Parametric Equations to Cartesian Form

When the Parameter Is Easy to Isolate

When Squaring or Using an Identity Helps

Using a Cartesian Equation to Find Coordinates

Converting Cartesian Equations to Parametric Form

A Simple Choice

Choosing a Useful Parametrisation

Review 1.6.3 Parametric equations of curves (A-level only) by teaching Genie

1.6.3 Parametric equations of curves (A-level only) Revision Guide