# Parametric Equations

## Parametric Equations Revision

**Parametric Equations**

So far we have seen graphs from **Cartesian equations** – this is where a **single equation** that links x and y defines a graph.

Sometimes, for **graphs that are more complicated**, it is easier to have **two equations**, one for x and one for y, that are linked by a shared parameter. This is a **parametric equation**.

There are **six skills** you need to know for **parametric equations**.

Make sure you are happy with the following topics before continuing.

**Skill 1: Parametric Equations**

You need to understand what **parametric equations** mean and how to **plot graphs from them**.

**Example: **Sketch the graph of x=t-1 and y=t^{2}.

Create a table of points:

Now plot these points on a graph.

You should also be able to answer questions such as “what is the value of y when t=4” and “what value of t corresponds to y=1”. All questions of this form can be **solved by substituting in the right values** into the **parametric equation**.

**Example: **For the graph x=t-1 and y=t^{2}, find the value of y when x=5.

x=t-1

5=t-1

t=6

\begin{aligned}y&=t^{2}\\[1.2em]&=6^{2}\\[1.2em]&=36\end{aligned}

**Skill 2: Parametric Equations of Circles**

While t is the **most common parameter**, some **parametric equations** use **trigonometric functions**, so use the parameter \theta.

The **parametric equation** **of a circle** with radius r and centre (a,b) is:

x=r\cos(\theta)+a

y=r\sin(\theta)+b

**Example: **The picture on the right shows a **circle** with centre (3,4) and radius 5.

It has **cartesian equation** (x-3)^{2}+(y-4)^{2}=25

It has **parametric equation** x=5\cos(\theta)+3 and y=5\sin(\theta)+4.

**Note: **If the radius were different in the x equation and the y equation then we would have an **ellipse**.

**Skill 3: Intersections of Graphs and Parametric Equations**

To find the intersection of a **parametric equation** and **Cartesian equation** on a graph, **substitute in the values** for t for x and y into the **Cartesian equation**, then solve for t, then find the x and y values corresponding to that value of t.

**Example: **Find where the graph x=3t+4 and y=2t^{2} intersects the line y=x+1

**Substitute in** expressions in t

2t^{2}=3t+4+1

2t^{2}=3t+5

2t^{2}-3t-5=0

(2t-5)(t+1)=0

t=\dfrac{5}{2} or t=-1

x=3t+4x=3\times\dfrac{5}{2}+4 or x=3\times(-1)+4

x=\dfrac{15}{2}+4 or x=-3+4

x=\dfrac{23}{2} or x=1

y=x+1When x = \dfrac{23}{2}, y=\dfrac{23}{2}+1 = \dfrac{25}{2}

When x = 1, y=1+1 = 2

**Skill 4: Finding the Cartesian Equation**

It is possible to get back to the **cartesian equation** of a graph from its **parametric form**. The way to do this is to **rearrange the equation** of one of x or y into the format t=f(x) or t=f(y), then **substitute in** this expression for t into the other equation.

**Example: **Find the **Cartesian equation** of y=t^{2}, x=t+5

x=t+5

t=x-5

\begin{aligned}y&=t^{2}\\[1.2em]&=(x-5)^{2}\end{aligned}

**Skill 5: Using Trig Identities to find the Cartesian Equation**

**Note: **For more about **trig identities**, go to the **Trigonometric Identities** page.

For **parametric equations** involving \theta and **trigonometric functions**, it is often necessary to use** trig identities** to recover the **cartesian form**.

**Example: **Find the **Cartesian equation** of x=1+cos(\theta), y=cos(2\theta)

x=1+cos(\theta)

cos(\theta)=x-1

\begin{aligned}y&=cos(2\theta)\\[1.2em]&=2cos^{2}(\theta)-1\\[1.2em]&=2(x-1)^{2}-1\\[1.2em]&=2(x^{2}-2x+1)-1\\[1.2em]&=2x^{2}-4x+2-1\\[1.2em]&=2x^{2}-4x+1\end{aligned}

**Skill 6: Modelling with Parametric Equations**

**Parametric equations** can be used to model **real life situations**. However, this can place **restrictions** on the parameter because **some scenarios are not possible in real life**.

- The equations could model for a
**parameter**where only some x and y values**make sense**, e.g. positive values. - Restrictions could exist to
**avoid repeating values**of x and y, such as limiting \theta to 0<\theta<2\pi. **Avoiding dividing by zero**is another reason for**restrictions**, for example if we had x=\dfrac{1}{t-1} we must say t\neq 1.

The values of the **parameter** that are allowed are collectively called the **domain** of the **parameter**.

**Example: **A tennis ball is thrown into the air. Its horizontal distance from the point of throwing is x=25t, and its vertical distance from the point of throwing is y=9t-t^{2}. Find its vertical distance when it has travelled 50m horizontally.

We can approach this problem **in the same way** we would approach a problem that is not modelling a real thing.

\begin{aligned}x&=25t\\[1.2em]&=50\end{aligned}

25t=50

t=2

\begin{aligned}y&=9t-t^{2}\\[1.2em]&=9\times2-2^{2}\\[1.2em]&=18-4\\[1.2em]&=14\text{ m}\end{aligned}

## Parametric Equations Example Questions

**Question 1: **Consider the parametric equation x=2t+1, y=t^{2}+3t

a) Find x and y when t=3.

b) What is the value of t when x=4?

c) What is the value of y when x=5?

**[4 marks]**

a)

\begin{aligned}x&=2t+1\\[1.2em]&=2\times3+1\\[1.2em]&=6+1\\[1.2em]&=7\end{aligned}

\begin{aligned}y&=t^{2}+3t\\[1.2em]&=3^{2}+3\times3\\[1.2em]&=9+9\\[1.2em]&=18\end{aligned}

b) x=2t+1

4=2t+1

2t=3

t=\dfrac{3}{2}

c) x=2t+1

5=2t+1

2t=4

t=2

\begin{aligned}y&=t^{2}+3t\\[1.2em]&=2^{2}+3\times2\\[1.2em]&=4+6\\[1.2em]&=10\end{aligned}

**Question 2: **State the centre and radius of the circles from these parametric equations:

a) x=2\cos(\theta)+1

y=2\sin(\theta)-1

b) x=5\cos(\theta)+2

y=5\sin(\theta)+5

c) x=16\cos(\theta)-12

y=16\sin(\theta)-13**[3 marks]**

a) Centre: (1,-1)

Radius: 2

b) Centre: (2,5)

Radius: 5

c) Centre: (-12,-13)

Radius: 16

**Question 3: **Where does the line y=8x+3 intersect the graph x=3t+1 and y=t^{2}+17t+3?

**[4 marks]**

y=8x+3

x=3t+1

y=t^{2}+17t+3

\begin{aligned}t^{2}+17t+3&=8(3t+1)+3\\[1.2em]&=24t+8+3\\[1.2em]&=24t+11\end{aligned}

t^{2}-7t-8=0

(t-8)(t+1)=0

t=8 or t=-1

x=3t+1

x=3\times8+1 or x=3\times(-1)+1

x=24+1 or x=-3+1

x=25 or x=-2

y=8x+3

When x = 25, y=8\times25+3 = 200+3 = 203

When x=-2, y=8\times(-2)+3 = -16 + 3 = -13

**Question 4: **Find the Cartesian equation of the parametric equation:

y=t^{3}+3t^{2}+6t+9

x=\dfrac{1}{2}t-3

**[3 marks]**

x=\dfrac{1}{2}t-3

2x=t-6

t=2x+6

\begin{aligned}y&=t^{3}+3t^{2}+6t+9\\[1.2em]&=(2x+6)^{3}+3(2x+6)^{2}+6(2x+6)+9\\[1.2em]&=8x^{3}+72x^{2}+216x+216+3(4x^{2}+24x+36)\\[1.2em]&+12x+36+9\\[1.2em]&=8x^{3}+72x^{2}+216x+216+12x^{2}+72x+108\\[1.2em]&+12x+36+9\\[1.2em]&=8x^{3}+84x^{2}+300x+369\end{aligned}

**Question 5: **Find the Cartesian equation of the parametric equation:

x=\sin(2\theta)

y=\sin^{2}(\theta)\cos^{2}(\theta)

**[2 marks]**

\dfrac{x}{2}=\sin(\theta)\cos(\theta)

\begin{aligned}y&=\sin^{2}(\theta)\cos^{2}(\theta)\\[1.2em]&=(\sin(\theta)\cos(\theta))^{2}\\[1.2em]&=(\dfrac{x}{2})^{2}\\[1.2em]&=\dfrac{x^{2}}{4}\end{aligned}

**Question 6: **A kite flies through the air according to the parametric equation:

y=8t-t^{2}

where x is horizontal distance and y is vertical distance.

How far has the kite travelled horizontally when it hits the ground.

**[3 marks]**

Hitting ground is y=0.

y=8t-t^{2}

8t-t^{2}=0

t(8-t)=0

t=0 or t=8

t=0 is the start, so the solution we want is t=8.

\begin{aligned}x&=25t+4\\[1.2em]&=25\times8+4\\[1.2em]&=200+4\\[1.2em]&=204\end{aligned}

## Parametric Equations Worksheet and Example Questions

### Parametric Equations 1

A Level### Parametric Equations 2

A Level## You May Also Like...

### MME Learning Portal

Online exams, practice questions and revision videos for every GCSE level 9-1 topic! No fees, no trial period, just totally free access to the UK’s best GCSE maths revision platform.