NSWMaths Extension 1Syllabus dot point

How do parametric equations describe a curve, and how do we convert between parametric and Cartesian forms?

Sketch curves given parametrically, eliminate the parameter to obtain Cartesian equations, and use parametric form for circles, parabolas and lines

A focused answer to the HSC Maths Extension 1 dot point on parametric equations. Eliminating the parameter, sketching parametric curves, and standard parametrisations of lines, circles and parabolas, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy7 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to interpret a curve given by parametric equations $x = f(t)$ , $y = g(t)$ , eliminate the parameter to find a Cartesian equation when possible, and sketch the resulting curve, including the range of $t$ values used.

The answer

What parametric equations are

A parametric curve in the plane is given by two equations,

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

over a domain of values of the parameter $t$ . As $t$ varies, the point $(x, y)$ traces out a curve.

The parameter is often time (think projectile motion), an angle (circles, ellipses), or just an algebraic placeholder.

Eliminating the parameter

To get a Cartesian equation, eliminate $t$ from the two equations. Strategies:

Solve one equation for $t$ , substitute into the other.
Use an identity (for example, $\cos^2 t + \sin^2 t = 1$ ).
Form a relation that does not require solving for $t$ explicitly.

The resulting equation in $x$ and $y$ is the Cartesian form. Always check the range: parametric curves can be restricted to a portion of the Cartesian curve, depending on the domain of $t$ .

Standard parametrisations

**Line through $(x_0, y_0)$ with direction $(a, b)$ **:

x = x_0 + a t, \qquad y = y_0 + b t.

Circle of radius $r$ centred at origin:

x = r \cos t, \qquad y = r \sin t, \quad t \in [0, 2\pi).

**Circle of radius $r$ centred at $(h, k)$ **:

x = h + r \cos t, \qquad y = k + r \sin t.

**Parabola $y^2 = 4 a x$ **:

x = a t^2, \qquad y = 2 a t,

where $t$ runs over all real numbers. Each value of $t$ gives a unique point on the parabola.

**Ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ **:

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

When to use parametric form

Parametric form is useful when:

The curve is not a function (a circle, a vertical line, anything failing the vertical-line test).
The motion of a point along a curve matters (projectile motion).
The curve is easier to describe via an angle or external variable than via $y$ as a function of $x$ .

Direction of motion

As $t$ increases, the point $(x(t), y(t))$ moves in a definite direction along the curve. For a circle parametrised by $t \in [0, 2\pi)$ with $x = \cos t$ , $y = \sin t$ , motion is anticlockwise starting at $(1, 0)$ .

Note the direction whenever a problem asks about velocity or orientation.

Worked example

Eliminate the parameter to get a Cartesian equation

The curve is $x = t + 1$ , $y = t^2$ . Find the Cartesian equation.

From the first equation, $t = x - 1$ . Substitute: $y = (x - 1)^2$ .

This is a parabola with vertex at $(1, 0)$ .

Circle from parameter

The curve is $x = 3 \cos t$ , $y = 3 \sin t$ for $t \in [0, 2\pi)$ . Eliminate the parameter.

Use $\cos^2 t + \sin^2 t = 1$ . Square and add: $x^2 + y^2 = 9 \cos^2 t + 9 \sin^2 t = 9$ .

Circle of radius $3$ centred at the origin.

Restricted domain

The curve is $x = t^2$ , $y = t$ for $t \in [0, 2]$ . Find the Cartesian form and describe.

From the second equation, $t = y$ . Substitute: $x = y^2$ .

This is the half of the parabola $x = y^2$ with $y \in [0, 2]$ , that is, the upper half from $(0, 0)$ to $(4, 2)$ .

Parametrise a line

Write a parametrisation for the line through $(2, -1)$ with direction $(3, 4)$ .

$x = 2 + 3 t$ , $y = -1 + 4 t$ , $t \in \mathbb{R}$ .

Standard parabola

The parabola $y^2 = 8 x$ has $4 a = 8$ , so $a = 2$ . Parametrise: $x = 2 t^2$ , $y = 4 t$ .

Check: $y^2 = 16 t^2$ and $8 x = 16 t^2$ , agree.

Common traps

Ignoring the domain of $t$: A parametric curve may trace only a portion of the implied Cartesian curve. Always include the range of $t$ when describing it.
Squaring without care: If you square one parametric equation to eliminate $t$ , you may introduce extraneous parts (the other branch). Check by substituting back.
Wrong direction of traversal: IMATH_3 , $y = \sin t$ goes anticlockwise. $x = \cos t$ , $y = -\sin t$ goes clockwise.
Forgetting that the parameter is not in the final Cartesian equation: The Cartesian equation must involve only $x$ and $y$ , not $t$ .
Mistaking parameter elimination for an identity: IMATH_10 , $y = t^3$ does not give $y^2 = x^3$ cleanly: $t = y^{1/3}$ so $x = y^{2/3}$ , which is the same curve but the algebra has to handle the cube root carefully.