# Parametric Equations

> Describe a curve with a third variable, the parameter. Convert lines, parabolas and circles between parametric and Cartesian form.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/parametric-equations/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The drone's flight log
A delivery drone does not record a curve. Every second it logs two numbers: how far east it is and how far north it is. At time $t = 1$ it is at $(2, 3)$, at $t = 2$ it is at $(4, 5)$, and so on.

Each coordinate is its own little function of time. Writing a path this way, with $x$ and $y$ both given in terms of a third variable, is called **parametric form**. It is how video games, robots and GPS trackers describe motion.

**In short:** Parametric equations give $x$ and $y$ each in terms of a third variable, the parameter $t$. Eliminate $t$ to get the Cartesian equation: for example, $x = h + r\cos t$, $y = k + r\sin t$ is the circle with centre $(h, k)$ and radius $r$.

## Starting point: what you already know

- A **Cartesian equation** links $x$ and $y$ directly, such as $y = 2x + 1$ or $x^2 + y^2 = 9$.
- The **circle** centred at the origin with radius $r$ is $x^2 + y^2 = r^2$.
- **Substitution:** if $x = 3$ and $y = x + 2$, then $y = 5$.
- The identity $\sin^2 t + \cos^2 t = 1$ is true for every angle $t$.

## What parametric form means

In **parametric form** a curve is given by two equations:

$$x = f(t), \qquad y = g(t)$$

The letter $t$ is the **parameter**. Each value of $t$ gives one point $(x, y)$. As $t$ changes, the point moves and traces the curve.

### Worked example: Plot a few points
$x = t + 1$, $y = t^2$.

| $t$ | $-2$ | $-1$ | 0 | 1 | 2 |
|---|---|---|---|---|---|
| $x$ | $-1$ | 0 | 1 | 2 | 3 |
| $y$ | 4 | 1 | 0 | 1 | 4 |

The points lie on a parabola with its vertex at $(1, 0)$.

## Getting back to the Cartesian equation

To find the Cartesian equation, **eliminate the parameter**: get rid of $t$ so only $x$ and $y$ are left.

**Method 1: make $t$ the subject of one equation and substitute.**

### Worked example: A parabola
$x = t + 1$, $y = t^2$. From the first, $t = x - 1$. Substitute: $y = (x - 1)^2$.

### Worked example: A straight line
$x = 2t$, $y = 3 - t$. From the first, $t = \dfrac{x}{2}$, so $y = 3 - \dfrac{x}{2}$, a straight line with gradient $-\dfrac{1}{2}$.

**Method 2: use an identity.** This is the trick for circles.

### Worked example: A circle
$x = 2\cos t$, $y = 2\sin t$. Then $\cos t = \dfrac{x}{2}$ and $\sin t = \dfrac{y}{2}$. Since $\cos^2 t + \sin^2 t = 1$:
$$\frac{x^2}{4} + \frac{y^2}{4} = 1 \quad\Rightarrow\quad x^2 + y^2 = 4$$
A circle with centre $(0, 0)$ and radius 2.

*Diagram: A circle of radius 2 centred at the origin. A point P on it at angle t has x equals 2 cos t and y equals 2 sin t.*
*As t goes from 0 to 2π, the point P(2 cos t, 2 sin t) goes once around the circle.*

## Writing a curve in parametric form

You can go the other way too. For the parabola $y = x^2$, let $x = t$. Then $y = t^2$. A curve has many possible parametric forms; any choice that traces the same points works.

### Exam tip: Circle anywhere
The circle with centre $(h, k)$ and radius $r$ is
$$x = h + r\cos t, \qquad y = k + r\sin t, \qquad 0 \le t < 2\pi$$

**Mind map**

```text
Parametric equations
  Form
    x = f(t), y = g(t)
    t is the parameter
  To Cartesian
    Make t the subject, substitute
    Use sin² t + cos² t = 1
  Shapes
    Line: x and y linear in t
    Parabola: x = t, y = t²
    Circle: x = r cos t, y = r sin t
  Watch
    Limits on t limit the curve
```

### Year 11: The parameter can limit the curve
$x = t^2$, $y = t^2 + 1$ gives $y = x + 1$. But $x = t^2$ can never be negative, so the curve is only the **half line** $y = x + 1$ for $x \ge 0$.

Always check what values $x$ and $y$ can actually take.

### Year 11: A shifted circle
$x = 3 + 5\cos t$, $y = -1 + 5\sin t$. Then $\cos t = \dfrac{x - 3}{5}$ and $\sin t = \dfrac{y + 1}{5}$, so
$$(x - 3)^2 + (y + 1)^2 = 25$$
Centre $(3, -1)$, radius 5.

**Other identities.** Pick the identity that links the two trig ratios.
- Double angles: $x = \cos t$, $y = \cos 2t$ gives $y = 2\cos^2 t - 1 = 2x^2 - 1$, only for $-1 \leq x \leq 1$. And $x = \sin t$, $y = \sin 2t = 2\sin t\cos t$: square to remove the sign problem, $y^2 = 4x^2(1 - x^2)$.
- Square and add: $x = \cos t + \sin t$, $y = \cos t - \sin t$ give $x^2 + y^2 = 2$, because the $2\sin t\cos t$ parts cancel.
- $\sec^2 t - \tan^2 t = 1$: $x = 2\sec t$, $y = 3\tan t$ gives $\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1$, a **hyperbola**.
- Exponentials: $x = e^t$, $y = e^{2t} = (e^t)^2$ gives $y = x^2$, but only for $x > 0$.

### Year 12: Parametric equations of motion
In Year 12 projectile motion, a ball's position is $x = ut\cos\theta$ and $y = ut\sin\theta - \tfrac{1}{2}gt^2$ with time $t$ as the parameter. Eliminating $t$ shows the path is a parabola. See the chapter on vectors and projectile motion.

**Gradients from parametric equations.** By the chain rule,
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$
For $x = t^2$, $y = t^3 - 3t$: $\dfrac{dy}{dx} = \dfrac{3t^2 - 3}{2t}$. At $t = 2$ the gradient is $\dfrac{9}{4}$, at the point $(4, 2)$, so the tangent is $y - 2 = \frac{9}{4}(x - 4)$.

**Horizontal tangents** are where $\dfrac{dy}{dt} = 0$ (and $\dfrac{dx}{dt} \neq 0$); here $t = \pm 1$, at $(1, -2)$ and $(1, 2)$. **Vertical tangents** are where $\dfrac{dx}{dt} = 0$.

## Quick summary

- Parametric form gives $x$ and $y$ each in terms of a parameter $t$.
- To get the Cartesian equation, eliminate $t$: make it the subject and substitute, or use $\sin^2 t + \cos^2 t = 1$.
- $x = h + r\cos t$, $y = k + r\sin t$ is a circle with centre $(h, k)$ and radius $r$.
- Check the possible values of $x$ and $y$, because the parameter may restrict the curve.

## Quiz

Try these questions from the chapter quiz.

1. For $x = 2t$, $y = t + 1$, which point is at $t = 3$?
   - A. $(3, 4)$
   - B. $(6, 4)$
   - C. $(6, 3)$
   - D. $(5, 4)$
   - Answer: B. $(6, 4)$. $x = 6$, $y = 4$.

2. In parametric equations, the letter $t$ is called the:
   - A. gradient
   - B. parameter
   - C. intercept
   - D. radius
   - Answer: B. parameter. It is the third variable that both $x$ and $y$ depend on.

3. $x = t$, $y = t^2 - 3$. The Cartesian equation is:
   - A. $y = x^2 - 3$
   - B. $y = x - 3$
   - C. $x = y^2 - 3$
   - D. $y = (x - 3)^2$
   - Answer: A. $y = x^2 - 3$. Replace $t$ with $x$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/parametric-equations/#quiz