What are parametric curves?

A parametric description gives x and y each as a function of a parameter t: x = f(t), y = g(t). As t varies, the point (x, y) traces a curve.

Name the curve x^216 - y^29 = 1 and state its asymptotes. [2 marks]

SEAB Original-style practice: A curve is defined parametrically by x = t^2, y = 2t for t R. Find the Cartesian equation and describe the curve.

From y = 2t we get t = y2. Substitute into x = t^2: x = (y2)^2 = y^24, so y^2 = 4x. This is a parabola opening to the right with vertex at the origin and axis along the positive x-axis. As t ranges over all reals, x = t^2 ≥ 0 and y = 2t takes all real values. Markers reward eliminating the parameter, the Cartesian form y^2 = 4x, and identifying the rightward-opening parabola.

SingaporeMathsSyllabus dot point

How do we recognise standard conic graphs and work with curves defined parametrically?

Recognise and sketch the standard conics (circle, ellipse, parabola, hyperbola) from their equations, and sketch and analyse curves defined parametrically

A focused answer to the H2 Mathematics outcome on conics and parametric curves. Recognising circles, ellipses, parabolas and hyperbolas from their equations, and sketching and converting curves given parametrically.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to recognise the standard conic sections from their equations and sketch them, and to work with parametric curves: sketching them, eliminating the parameter to obtain a Cartesian equation, and describing the curve traced out.

The answer

The standard conics

Circle: $(x - a)^2 + (y - b)^2 = r^2$ , centre $(a, b)$ , radius $r$ .
Ellipse: $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ , centred at the origin, semi-axes $a$ (horizontal) and $b$ (vertical).
Parabola: $y^2 = 4ax$ opens right; $x^2 = 4ay$ opens up. A vertex sits at the origin in the standard forms.
Hyperbola: $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ opens left-right with asymptotes $y = \pm\dfrac{b}{a}x$ .

Recognising which form you have comes from the signs and whether the squared terms are added (circle or ellipse) or subtracted (hyperbola).

Parametric curves

A parametric description gives $x$ and $y$ each as a function of a parameter $t$ : $x = \mathrm{f}(t)$ , $y = \mathrm{g}(t)$ . As $t$ varies, the point $(x, y)$ traces a curve.

To find the Cartesian equation, eliminate $t$ : solve one equation for $t$ and substitute, or use an identity (such as $\cos^2 t + \sin^2 t = 1$ ) when trigonometric functions appear.

Sketching a parametric curve

Plot a few values of $t$ , note the range of $x$ and $y$ , watch the direction of travel as $t$ increases, and identify any symmetry. The range of the parameter restricts which part of the full Cartesian curve is actually traced.

Worked example

A curve is given parametrically by $x = 3\cos t$ , $y = 2\sin t$ for $0 \leq t < 2\pi$ . Find the Cartesian equation and describe the curve.

Step 1: Isolate the trigonometric ratios

\cos t = \frac{x}{3}, \qquad \sin t = \frac{y}{2}

Step 2: Use the Pythagorean identity

Since $\cos^2 t + \sin^2 t = 1$ :

\left(\frac{x}{3}\right)^2 + \left(\frac{y}{2}\right)^2 = 1 \implies \frac{x^2}{9} + \frac{y^2}{4} = 1

Step 3: Identify the curve

This is an ellipse centred at the origin with semi-axes $3$ (horizontal) and $2$ (vertical).

Step 4: Confirm the range traced

As $t$ runs from $0$ to $2\pi$ , the point sweeps the entire ellipse once, starting at $(3, 0)$ and travelling anticlockwise.

Examples in context

Example 1. Projectile path. A projectile has $x = 20t$ , $y = 15t - 5t^2$ . Eliminating $t = \dfrac{x}{20}$ gives $y = 15\left(\dfrac{x}{20}\right) - 5\left(\dfrac{x}{20}\right)^2$ , a downward parabola - the trajectory - showing parametric form is natural for motion.

Example 2. An ellipse from circular motion. Scaling a circle unequally, $x = 5\cos t$ , $y = 3\sin t$ , produces the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ . This is how a tilted circular orbit appears in projection.

Try this

Q1. Name the curve $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$ and state its asymptotes. [2 marks]

Cue. A hyperbola with asymptotes $y = \pm\dfrac{3}{4}x$ .

Q2. A curve has $x = t + 1$ , $y = t^2$ . Find its Cartesian equation. [2 marks]

Cue. $t = x - 1$ , so $y = (x - 1)^2$ , an upward parabola with vertex $(1, 0)$ .

Q3. State the centre and radius of $(x - 2)^2 + (y + 3)^2 = 25$ . [2 marks]

Cue. Centre $(2, -3)$ , radius $5$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksThe curve has equation

\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1

. Name the curve, state its intercepts with the axes, and sketch it.

Show worked answer →

This is an ellipse centred at the origin.

Setting $y = 0$ : $\dfrac{x^2}{9} = 1$ , so $x = \pm 3$ (the $x$ -intercepts). Setting $x = 0$ : $\dfrac{y^2}{4} = 1$ , so $y = \pm 2$ (the $y$ -intercepts).

The semi-axes are $3$ along the $x$ -axis and $2$ along the $y$ -axis, so the ellipse is wider than it is tall, passing through $(\pm 3, 0)$ and $(0, \pm 2)$ .

Markers reward naming the ellipse, both pairs of intercepts, and a correctly proportioned sketch.

Original5 marksA curve is defined parametrically by

x = t^2

y = 2t

for

t \in \mathbb{R}

. Find the Cartesian equation and describe the curve.

Show worked answer →

From $y = 2t$ we get $t = \dfrac{y}{2}$ .

Substitute into $x = t^2$ : $x = \left(\dfrac{y}{2}\right)^2 = \dfrac{y^2}{4}$ , so $y^2 = 4x$ .

This is a parabola opening to the right with vertex at the origin and axis along the positive $x$ -axis. As $t$ ranges over all reals, $x = t^2 \geq 0$ and $y = 2t$ takes all real values.

Markers reward eliminating the parameter, the Cartesian form $y^2 = 4x$ , and identifying the rightward-opening parabola.