Parametric Equations

What you'll learn

What parametric equations are and how they describe a curve.
How to eliminate a parameter to get a Cartesian equation.
How to find intercepts, domains, circles, line segments, and intersections.
How to sketch parametric curves accurately without overcomplicating them.

1. What a parameter does

You already know coordinates like $(x, y)$ . Parametric equations give $x$ and $y$ separately, both in terms of a third variable.

Definition

Key language

A parametric equation gives coordinates using another variable, usually $t$ or $θ\theta$ .
The extra variable is called a parameter.
A Cartesian equation connects $x$ and $y$ directly, with no parameter left.
The domain is the set of allowed input values, such as allowed values of $t$ or allowed values of $x$ .

Think of the parameter as a slider. For each value of the parameter, you get one point on the curve.

A parameter slider selecting values of t and generating ordered points on a curve.

Example

Plotting points from a parameter

A curve is given by $x=t+2,\ y=t^2-1$ .

Substitute $t = - 1$ :
$x=1,y=0x=1,\quad y=0$
Substitute $t = 0$ :
$x=2,y=−1x=2,\quad y=-1$
Substitute $t = 2$ :
$x=4,y=3x=4,\quad y=3$
So the curve passes through $(1, 0)$ , $(2, - 1)$ , and $(4, 3)$ . As $t$ increases, the points are traced in that order.

Key Idea

Big picture

Parametric equations describe a curve by saying where a moving point is at each value of the parameter.

2. Eliminating the parameter

To eliminate the parameter means to remove $t$ or $θ\theta$ so that you get an equation involving only $x$ and $y$ .

The usual method is:

rearrange one equation to make the parameter the subject;
substitute into the other equation.

Example

Finding a Cartesian equation and sketching

A curve has parametric equations $x=t+4,\ y=t^2+2$ . Find a Cartesian equation and describe the sketch.

The parametric curve becomes the parabola y=(x−4)²+2 with vertex and axis of symmetry clearly shown.

Rearrange $x = t + 4$ to make $t$ the subject:
$t = x - 4$
Substitute into $y=t^2+2$ :
$y=(x-4)^2+2$
Since $t$ is unrestricted here, $x$ can take any real value.
The graph is a parabola opening upwards, with vertex $(4, 2)$ and axis of symmetry $x = 4$ .

Common Mistake

Sign errors when rearranging

From $x = t + 4$ , the correct rearrangement is $t = x - 4$ , not $t = x + 4$ .

3. Finding where a curve crosses the axes

An x-intercept is where the curve crosses the x-axis, so $y = 0$ .
A y-intercept is where the curve crosses the y-axis, so $x = 0$ .

With parametric equations, you often find intercepts fastest by working with the parameter directly.

Example

Crossing the coordinate axes

A curve is given by $x=\ln(t+2),\ y=t^2-9,\ t>-2$ . Find where it crosses the axes.

For an x-intercept, set $y = 0$ :
$t^2-9=0$
Solve for $t$ and check the restriction $t > - 2$ :
$t=3ort=−3t=3\quad \text{or}\quad t=-3$
Reject $t = - 3$ because it is not allowed. For $t = 3$ :
$x=ln⁡(5)x=\ln(5)$
For a y-intercept, set $x = 0$ :
$ln⁡(t+2)=0\ln(t+2)=0$
Since $ln⁡(1)=0\ln(1)=0$ , we get $t + 2 = 1$ , so $t = - 1$ . Then:
$y=(-1)^2-9=-8$
The curve crosses the axes at:
$(ln⁡5,0),(0,−8)(\ln 5,0),\quad (0,-8)$

Tip

Intercept shortcut

For intercepts, do not eliminate the parameter unless you need to. Set $x = 0$ or $y = 0$ in the parametric equations.

4. Recognising circles

A very common parametric form for a circle is:

x=a+rcos⁡θ,y=b+rsin⁡θx=a+r\cos\theta,\quad y=b+r\sin\theta

This represents a circle with centre $(a, b)$ and radius $r$ .

The standard parametric circle has centre (a,b) and radius r, with a point generated by angle θ.

Example

Finding the centre and radius

A circle is defined by $y=−2+4sin⁡θx=3+4\cos\theta,\ y=-2+4\sin\theta$ . Find its centre and radius.

The circle x=3+4cosθ, y=−2+4sinθ has centre (3,−2) and radius 4.

Compare with the standard form $y=b+rsin⁡θx=a+r\cos\theta,\ y=b+r\sin\theta$ .
The centre is $(3, - 2)$ because those are the horizontal and vertical shifts.
The radius is 4 because both the sine and cosine terms have coefficient 4.
To confirm using a Cartesian equation, rearrange:
$x−34=cos⁡θ,y+24=sin⁡θ\frac{x-3}{4}=\cos\theta,\quad \frac{y+2}{4}=\sin\theta$
Square and add, using $cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1$ :
$(x−34)2+(y+24)2=1\left(\frac{x-3}{4}\right)^2+\left(\frac{y+2}{4}\right)^2=1$
Therefore the Cartesian equation is:
$x-3)^2+(y+2)^2=16$

5. Showing a parametric curve is a straight line segment

Sometimes the parametric equations look complicated, but the curve is actually part of a straight line.

A line segment is the finite part of a straight line between two endpoints. If the parameter has a restricted range, you usually only get a segment, not the whole line.

A restricted parameter interval traces only the segment between the endpoint coordinates, not the whole line.

Example

A rational parametric curve that is a line

A curve is given by

x=1+6tt+1,y=3+tt+1,0≤t≤4x=\frac{1+6t}{t+1},\quad y=\frac{3+t}{t+1},\quad 0\le t\le 4

Show that it is part of a straight line and find the length of the segment.

Look for a linear combination of $x$ and $y$ . Try $2 x + 5 y$ :
$2x+5y=2(1+6t)+5(3+t)t+12x+5y=\frac{2(1+6t)+5(3+t)}{t+1}$
Simplify the numerator:
$2x+5y=17+17tt+1=172x+5y=\frac{17+17t}{t+1}=17$
So the curve lies on the straight line:
$2 x + 5 y = 17$
Find the endpoints by using the endpoints of the parameter interval:
$(5,75)t=0:\ (1,3),\quad t=4:\ \left(5,\frac{7}{5}\right)$
Use the distance formula for the length $L$ :
$L=(5−1)2+(75−3)2=16+6425=4295\begin{aligned} L&=\sqrt{(5-1)^2+\left(\frac{7}{5}-3\right)^2}\\ &=\sqrt{16+\frac{64}{25}}\\ &=\frac{4\sqrt{29}}{5} \end{aligned}$

Common Mistake

Check the denominator

If a parametric equation has a denominator, check that it is not zero in the given parameter interval.

6. Using trigonometric identities and domains

For trig parametric equations, your main job is often to spot the right identity.

Useful identities include:

$sin^2 t+\cos^2 t=1$
$sec^2 t=1+\tan^2 t$
$cos 2t=2\cos^2 t-1$

Example

Using $cos 2 t$ to eliminate the parameter

A curve is given by $0≤t≤2πx=2+\cos t,\ y=1+4\cos 2t,\ 0\le t\le 2\pi$ . Find $y$ in terms of $x$ and state the domain.

From $x=2+cos⁡tx=2+\cos t$ , rearrange to get:
$cos⁡t=x−2\cos t=x-2$
Use $cos 2t=2\cos^2 t-1$ :
$y=1+4(2(x−2)2−1)y=1+4\left(2(x-2)^2-1\right)$
Simplify:
$y=8(x-2)^2-3$
Since $−1≤cos⁡t≤1-1\le \cos t\le 1$ , the possible values of $x$ are:
$1≤x≤31\le x\le 3$
The sketch is the part of the upward parabola $y=8(x-2)^2-3$ between $x = 1$ and $x = 3$ , with vertex $(2, - 3)$ and endpoints $(1, 5)$ and $(3, 5)$ .

Only the restricted arc of the parabola is included because the parameter gives 1≤x≤3.

Common Mistake

Forgetting the domain

After eliminating the parameter, always ask what values of $x$ are actually possible. A full parabola may only be a restricted arc or segment.

7. Intersections with another curve

To find intersections, substitute the parametric expressions for $x$ and $y$ into the other curve’s equation. Then solve for the parameter.

Example

Intersecting an ellipse and a circle

A curve has parametric equations $y=23sin⁡tx=4\cos t,\ y=2\sqrt{3}\sin t$ . It meets the circle $x^2+y^2=13$ . Find the intersection points.

The ellipse and circle intersect at four symmetric points: (±2,±3).

Substitute $x=4cos⁡tx=4\cos t$ and $y=23sin⁡ty=2\sqrt{3}\sin t$ into the circle:
$16\cos^2 t+12\sin^2 t=13$
Use $cos^2 t=1-\sin^2 t$ :
$16(1-\sin^2 t)+12\sin^2 t=13$
Simplify:
$sin⁡2t=34\sin^2 t=\frac{3}{4}$
Then $cos⁡2t=14\cos^2 t=\frac{1}{4}$ , so $sin⁡t=±32\sin t=\pm\frac{\sqrt{3}}{2}$ and $cos⁡t=±12\cos t=\pm\frac{1}{2}$ .
Convert back to coordinates:
$x=±2,y=±3x=\pm 2,\quad y=\pm 3$
All four sign combinations occur, giving the four points:
$(2,3),(−2,3),(−2,−3),(2,−3)(2,3),\quad (-2,3),\quad (-2,-3),\quad (2,-3)$

Exam technique

In the exam

First decide whether to eliminate the parameter directly, use a trig identity, or substitute into another equation.
Always carry parameter restrictions through to the final $x$ -domain before sketching.
For intercepts and intersections, reject any parameter values that are outside the given range.

Self review

Check yourself

If $x = t - 5$ and $y=t^2+1$ , what is the Cartesian equation and where is the vertex?
For $x=2+cos⁡tx=2+\cos t$ , why can $x$ never be 5?
When finding an x-intercept, which coordinate must be zero?

Parametric Equations

What you'll learn

What parametric equations are and how they describe a curve.
How to eliminate a parameter to get a Cartesian equation.
How to find intercepts, domains, circles, line segments, and intersections.
How to sketch parametric curves accurately without overcomplicating them.

1. What a parameter does

You already know coordinates like $(x, y)$ . Parametric equations give $x$ and $y$ separately, both in terms of a third variable.

Definition

Key language

A parametric equation gives coordinates using another variable, usually $t$ or $θ\theta$ .
The extra variable is called a parameter.
A Cartesian equation connects $x$ and $y$ directly, with no parameter left.
The domain is the set of allowed input values, such as allowed values of $t$ or allowed values of $x$ .

Think of the parameter as a slider. For each value of the parameter, you get one point on the curve.

A parameter slider selecting values of t and generating ordered points on a curve.

Example

Plotting points from a parameter

A curve is given by $x=t+2,\ y=t^2-1$ .

Substitute $t = - 1$ :
$x=1,y=0x=1,\quad y=0$
Substitute $t = 0$ :
$x=2,y=−1x=2,\quad y=-1$
Substitute $t = 2$ :
$x=4,y=3x=4,\quad y=3$
So the curve passes through $(1, 0)$ , $(2, - 1)$ , and $(4, 3)$ . As $t$ increases, the points are traced in that order.

Key Idea

Big picture

Parametric equations describe a curve by saying where a moving point is at each value of the parameter.

2. Eliminating the parameter

To eliminate the parameter means to remove $t$ or $θ\theta$ so that you get an equation involving only $x$ and $y$ .

The usual method is:

rearrange one equation to make the parameter the subject;
substitute into the other equation.

Example

Finding a Cartesian equation and sketching

A curve has parametric equations $x=t+4,\ y=t^2+2$ . Find a Cartesian equation and describe the sketch.

The parametric curve becomes the parabola y=(x−4)²+2 with vertex and axis of symmetry clearly shown.

Rearrange $x = t + 4$ to make $t$ the subject:
$t = x - 4$
Substitute into $y=t^2+2$ :
$y=(x-4)^2+2$
Since $t$ is unrestricted here, $x$ can take any real value.
The graph is a parabola opening upwards, with vertex $(4, 2)$ and axis of symmetry $x = 4$ .

Common Mistake

Sign errors when rearranging

From $x = t + 4$ , the correct rearrangement is $t = x - 4$ , not $t = x + 4$ .

3. Finding where a curve crosses the axes

An x-intercept is where the curve crosses the x-axis, so $y = 0$ .
A y-intercept is where the curve crosses the y-axis, so $x = 0$ .

With parametric equations, you often find intercepts fastest by working with the parameter directly.

Example

Crossing the coordinate axes

A curve is given by $x=\ln(t+2),\ y=t^2-9,\ t>-2$ . Find where it crosses the axes.

For an x-intercept, set $y = 0$ :
$t^2-9=0$
Solve for $t$ and check the restriction $t > - 2$ :
$t=3ort=−3t=3\quad \text{or}\quad t=-3$
Reject $t = - 3$ because it is not allowed. For $t = 3$ :
$x=ln⁡(5)x=\ln(5)$
For a y-intercept, set $x = 0$ :
$ln⁡(t+2)=0\ln(t+2)=0$
Since $ln⁡(1)=0\ln(1)=0$ , we get $t + 2 = 1$ , so $t = - 1$ . Then:
$y=(-1)^2-9=-8$
The curve crosses the axes at:
$(ln⁡5,0),(0,−8)(\ln 5,0),\quad (0,-8)$

Tip

Intercept shortcut

For intercepts, do not eliminate the parameter unless you need to. Set $x = 0$ or $y = 0$ in the parametric equations.

4. Recognising circles

A very common parametric form for a circle is:

x=a+rcos⁡θ,y=b+rsin⁡θx=a+r\cos\theta,\quad y=b+r\sin\theta

This represents a circle with centre $(a, b)$ and radius $r$ .

The standard parametric circle has centre (a,b) and radius r, with a point generated by angle θ.

Example

Finding the centre and radius

A circle is defined by $y=−2+4sin⁡θx=3+4\cos\theta,\ y=-2+4\sin\theta$ . Find its centre and radius.

The circle x=3+4cosθ, y=−2+4sinθ has centre (3,−2) and radius 4.

Compare with the standard form $y=b+rsin⁡θx=a+r\cos\theta,\ y=b+r\sin\theta$ .
The centre is $(3, - 2)$ because those are the horizontal and vertical shifts.
The radius is 4 because both the sine and cosine terms have coefficient 4.
To confirm using a Cartesian equation, rearrange:
$x−34=cos⁡θ,y+24=sin⁡θ\frac{x-3}{4}=\cos\theta,\quad \frac{y+2}{4}=\sin\theta$
Square and add, using $cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1$ :
$(x−34)2+(y+24)2=1\left(\frac{x-3}{4}\right)^2+\left(\frac{y+2}{4}\right)^2=1$
Therefore the Cartesian equation is:
$x-3)^2+(y+2)^2=16$

5. Showing a parametric curve is a straight line segment

Sometimes the parametric equations look complicated, but the curve is actually part of a straight line.

A line segment is the finite part of a straight line between two endpoints. If the parameter has a restricted range, you usually only get a segment, not the whole line.

A restricted parameter interval traces only the segment between the endpoint coordinates, not the whole line.

Example

A rational parametric curve that is a line

A curve is given by

x=1+6tt+1,y=3+tt+1,0≤t≤4x=\frac{1+6t}{t+1},\quad y=\frac{3+t}{t+1},\quad 0\le t\le 4

Show that it is part of a straight line and find the length of the segment.

Look for a linear combination of $x$ and $y$ . Try $2 x + 5 y$ :
$2x+5y=2(1+6t)+5(3+t)t+12x+5y=\frac{2(1+6t)+5(3+t)}{t+1}$
Simplify the numerator:
$2x+5y=17+17tt+1=172x+5y=\frac{17+17t}{t+1}=17$
So the curve lies on the straight line:
$2 x + 5 y = 17$
Find the endpoints by using the endpoints of the parameter interval:
$(5,75)t=0:\ (1,3),\quad t=4:\ \left(5,\frac{7}{5}\right)$
Use the distance formula for the length $L$ :
$L=(5−1)2+(75−3)2=16+6425=4295\begin{aligned} L&=\sqrt{(5-1)^2+\left(\frac{7}{5}-3\right)^2}\\ &=\sqrt{16+\frac{64}{25}}\\ &=\frac{4\sqrt{29}}{5} \end{aligned}$

Common Mistake

Check the denominator

If a parametric equation has a denominator, check that it is not zero in the given parameter interval.

6. Using trigonometric identities and domains

For trig parametric equations, your main job is often to spot the right identity.

Useful identities include:

$sin^2 t+\cos^2 t=1$
$sec^2 t=1+\tan^2 t$
$cos 2t=2\cos^2 t-1$

Example

Using $cos 2 t$ to eliminate the parameter

A curve is given by $0≤t≤2πx=2+\cos t,\ y=1+4\cos 2t,\ 0\le t\le 2\pi$ . Find $y$ in terms of $x$ and state the domain.

From $x=2+cos⁡tx=2+\cos t$ , rearrange to get:
$cos⁡t=x−2\cos t=x-2$
Use $cos 2t=2\cos^2 t-1$ :
$y=1+4(2(x−2)2−1)y=1+4\left(2(x-2)^2-1\right)$
Simplify:
$y=8(x-2)^2-3$
Since $−1≤cos⁡t≤1-1\le \cos t\le 1$ , the possible values of $x$ are:
$1≤x≤31\le x\le 3$
The sketch is the part of the upward parabola $y=8(x-2)^2-3$ between $x = 1$ and $x = 3$ , with vertex $(2, - 3)$ and endpoints $(1, 5)$ and $(3, 5)$ .

Only the restricted arc of the parabola is included because the parameter gives 1≤x≤3.

Common Mistake

Forgetting the domain

After eliminating the parameter, always ask what values of $x$ are actually possible. A full parabola may only be a restricted arc or segment.

7. Intersections with another curve

To find intersections, substitute the parametric expressions for $x$ and $y$ into the other curve’s equation. Then solve for the parameter.

Example

Intersecting an ellipse and a circle

A curve has parametric equations $y=23sin⁡tx=4\cos t,\ y=2\sqrt{3}\sin t$ . It meets the circle $x^2+y^2=13$ . Find the intersection points.

The ellipse and circle intersect at four symmetric points: (±2,±3).

Substitute $x=4cos⁡tx=4\cos t$ and $y=23sin⁡ty=2\sqrt{3}\sin t$ into the circle:
$16\cos^2 t+12\sin^2 t=13$
Use $cos^2 t=1-\sin^2 t$ :
$16(1-\sin^2 t)+12\sin^2 t=13$
Simplify:
$sin⁡2t=34\sin^2 t=\frac{3}{4}$
Then $cos⁡2t=14\cos^2 t=\frac{1}{4}$ , so $sin⁡t=±32\sin t=\pm\frac{\sqrt{3}}{2}$ and $cos⁡t=±12\cos t=\pm\frac{1}{2}$ .
Convert back to coordinates:
$x=±2,y=±3x=\pm 2,\quad y=\pm 3$
All four sign combinations occur, giving the four points:
$(2,3),(−2,3),(−2,−3),(2,−3)(2,3),\quad (-2,3),\quad (-2,-3),\quad (2,-3)$

Exam technique

In the exam

First decide whether to eliminate the parameter directly, use a trig identity, or substitute into another equation.
Always carry parameter restrictions through to the final $x$ -domain before sketching.
For intercepts and intersections, reject any parameter values that are outside the given range.

Self review

Check yourself

If $x = t - 5$ and $y=t^2+1$ , what is the Cartesian equation and where is the vertex?
For $x=2+cos⁡tx=2+\cos t$ , why can $x$ never be 5?
When finding an x-intercept, which coordinate must be zero?

Parametric Equations

What you'll learn

1. What a parameter does

2. Eliminating the parameter

3. Finding where a curve crosses the axes

4. Recognising circles

5. Showing a parametric curve is a straight line segment

6. Using trigonometric identities and domains

7. Intersections with another curve

Recap questions

You've reached the end

Parametric Equations

What you'll learn

1. What a parameter does

2. Eliminating the parameter

3. Finding where a curve crosses the axes

4. Recognising circles

5. Showing a parametric curve is a straight line segment

6. Using trigonometric identities and domains

7. Intersections with another curve

Recap questions

You've reached the end

Parametric Equations

What you'll learn

1. What a parameter does

2. Eliminating the parameter

3. Finding where a curve crosses the axes

4. Recognising circles

5. Showing a parametric curve is a straight line segment

6. Using trigonometric identities and domains

7. Intersections with another curve

Recap questions

You've reached the end

Parametric Equations Revision Guide

Parametric Equations

What you'll learn

1. What a parameter does

2. Eliminating the parameter

3. Finding where a curve crosses the axes

4. Recognising circles

5. Showing a parametric curve is a straight line segment

6. Using trigonometric identities and domains

7. Intersections with another curve

Recap questions

You've reached the end

Parametric Equations Revision Guide