Define a curve parametrically by $x = f(t)$ and $y = g(t)$, eliminate the parameter $t$ by algebra or by a Pythagorean identity to obtain the Cartesian equation, parametrise standard curves (lines, parabolas, circles, ellipses and the rectangular hyperbola), and determine the direction of travel (orientation) and any excluded points

What this dot point is asking

Up to now every curve you have met has been written as a single equation linking $x$ and $y$, such as $y = x^2$ or $x^2 + y^2 = 9$. NESA now wants you to describe a curve a second way: give $x$ and $y$ each as a separate function of a third variable, called the parameter and usually written $t$ (or $\theta$ for an angle). The pair

are the parametric equations of the curve. As $t$ runs through its values, the point $(f(t), g(t))$ moves and traces out the curve.

There are three skills bundled here:

• Read a parametric definition and understand that one value of $t$ gives one point, so the parameter acts like a clock that drives the point along the curve.
• Eliminate the parameter to recover the ordinary Cartesian equation in $x$ and $y$, either by algebra (solve one equation for $t$ and substitute) or, for trigonometric pairs, by a Pythagorean identity.
• Describe the motion: the direction of travel (orientation) as $t$ increases, and any excluded points the curve never reaches because some value of $t$ is forbidden.

The letter $t$ is used because it often stands for time, which is the clearest way to picture the idea: a projectile's horizontal and vertical positions are each a function of time, and eliminating time leaves the shape of its path. This Year 11 page is the first-contact treatment; the Year 12 page parametric equations builds on it for the parabola $x^2 = 4ay$ and its chords, tangents and loci. The "act on the picture, feature by feature" habit you used for inverse relations and adding ordinates carries over, with $t$ now driving the point.

The answer

What parametric equations are

A parametric curve is given by two equations, one for each coordinate, in terms of a parameter:

Substituting a value of $t$ produces the coordinates of one point $(f(t), g(t))$ on the curve. Letting $t$ vary sweeps the point along the curve, so the parameter behaves like time on a clock: at each instant $t$ the moving point is somewhere definite, and the trail it leaves is the curve.

The standard worked illustration is a projectile. Suppose a ball is launched so that after $t$ seconds its horizontal distance is $x = 16t$ and its height is $y = -5t^2 + 25t$ (in metres, with $t$ from $0$ to $5$). Each value of $t$ gives the ball's position at that instant. The table of positions and the path are below.

The same two facts that make parametric form useful are visible here: $x$ and $y$ are each a clean function of $t$ even though the path itself is a curve, and the parameter records when as well as where, so the arrow of increasing $t$ gives the direction of travel.

Eliminating the parameter by algebra

To get the ordinary Cartesian equation of the curve, eliminate the parameter: remove $t$ so that only $x$ and $y$ remain. The reliable method when one equation is easy to invert is:

1. Solve the simpler equation for $t$ (usually the one that is linear in $t$).
2. Substitute that expression for $t$ into the other equation.
3. Simplify to a single equation in $x$ and $y$.

For the projectile, the first equation $x = 16t$ is linear, so $t = \dfrac{x}{16}$. Substituting into $y = -5t^2 + 25t$:

The path is a concave-down parabola through $x = 0$ and $x = 80$, peaking at $(40, 31.25)$. Eliminating $t$ has thrown away the timing information (we no longer know when the ball is at a given point) but kept the shape.

A second, even simpler example is the parabola $x = 2t$, $y = t^2$. From $x = 2t$ we get $t = \dfrac{x}{2}$, so $y = \left(\dfrac{x}{2}\right)^2 = \dfrac{x^2}{4}$, that is $x^2 = 4y$. The variable point $(2t, t^2)$ runs along this parabola as $t$ sweeps the whole number line, like a bent and stretched copy of it.

Eliminating the parameter with a Pythagorean identity

When $x$ and $y$ involve $\cos\theta$ and $\sin\theta$ you usually cannot solve neatly for $\theta$. Instead, isolate the trigonometric functions and use the identity

The circle $x = r\cos\theta$, $y = r\sin\theta$ is the model case. Take $r = 3$, so $x = 3\cos\theta$ and $y = 3\sin\theta$. Then $\cos\theta = \dfrac{x}{3}$ and $\sin\theta = \dfrac{y}{3}$, and squaring and adding:

so $x^2 + y^2 = 9$, the circle of radius $3$ centred at the origin. The parameter $\theta$ is the angle the radius makes with the positive $x$-axis, and as $\theta$ increases the point moves anticlockwise around the circle. Notice that here the map from parameter to point is many-to-one: every $\theta$ gives one point, but each point comes from infinitely many $\theta$ differing by multiples of $2\pi$.

If the two amplitudes differ, the same identity gives an ellipse. For $x = 4\cos\theta$, $y = 2\sin\theta$, we have $\cos\theta = \dfrac{x}{4}$ and $\sin\theta = \dfrac{y}{2}$, so

This is the great advantage of the trig parametrisation: a circle or ellipse is a relation (it fails the vertical line test), but parametrically it is described by a pair of functions, which are far easier to handle.

There is a companion identity for the other pair. Dividing $\cos^2\theta + \sin^2\theta = 1$ by $\cos^2\theta$ gives

which eliminates $\theta$ from a $\sec\theta$, $\tan\theta$ pair. For $x = 2\sec\theta$, $y = 2\tan\theta$ we get $\sec\theta = \dfrac{x}{2}$ and $\tan\theta = \dfrac{y}{2}$, so $\left(\dfrac{x}{2}\right)^2 - \left(\dfrac{y}{2}\right)^2 = 1$, that is $x^2 - y^2 = 4$, a hyperbola. Because $|\sec\theta| \ge 1$ always, this forces $|x| \ge 2$, so the whole strip $-2 < x < 2$ is empty: a worked reminder that the parametrisation can quietly exclude part of the plane.

Reading the orientation and the excluded points

Two features of a parametric curve are invisible in the bare Cartesian equation, so you must read them from the parametric form.

The orientation (direction of travel) is the way the point moves as $t$ increases. Find it by computing the point at two or three increasing values of $t$ and seeing which way it goes. For the projectile, $t = 0$ gives $(0, 0)$, $t = 2.5$ gives the peak $(40, 31.25)$ and $t = 5$ gives $(80, 0)$, so the motion is left to right. For the circle, $\theta = 0$ gives $(3, 0)$ and $\theta = \tfrac{\pi}{2}$ gives $(0, 3)$, so the motion is anticlockwise.

An excluded point is a point the curve never reaches because the value of $t$ that would produce it is not allowed (or the curve only uses part of the parameter's range). The model case is the rectangular hyperbola $x = t$, $y = \dfrac{1}{t}$. Eliminating $t$ gives $xy = 1$, but $\dfrac{1}{t}$ is undefined at $t = 0$, so $t = 0$ is excluded. There is no parameter value giving $x = 0$ and none giving $y = 0$, so the curve approaches but never reaches the origin: both axes are asymptotes.

A restricted range of the parameter excludes points too. If the projectile is only defined for $0 \le t \le 5$, the Cartesian parabola $y = \dfrac{5x(80 - x)}{256}$ is drawn only for $0 \le x \le 80$; the parts of the full parabola beyond the launch and landing points are not part of the curve. Always carry the parameter's restriction across to the Cartesian curve.

Eliminating the parameter, stage by stage

The full routine, plot then eliminate, is best seen as a short sequence. Take the parabola $x = 2t$, $y = t^2$ and build it step by step.

Stage 1, plot the table points

Make a table for a few values of $t$: $t = -2$ gives $(-4, 4)$, $t = -1$ gives $(-2, 1)$, $t = 0$ gives $(0, 0)$, $t = 1$ gives $(2, 1)$ and $t = 2$ gives $(4, 4)$. Plot these points.

Stage 2, join into a curve

Joining the points in order of increasing $t$ gives a smooth parabola opening upward with its vertex at the origin.

Stage 3, mark the orientation

As $t$ increases the point moves down the left arm to the vertex (reached at $t = 0$) and then up the right arm. Draw an arrow showing this left-to-right sweep.

Stage 4, eliminate $t$

From $x = 2t$, $t = \dfrac{x}{2}$, so $y = \left(\dfrac{x}{2}\right)^2 = \dfrac{x^2}{4}$, that is $x^2 = 4y$, the Cartesian equation of the finished curve.

How exam questions ask about parametric curves

The wording tells you which skill to use.

• "Eliminate the parameter / find the Cartesian equation." Solve the linear equation for $t$ and substitute, or isolate $\cos\theta$ and $\sin\theta$ and use $\cos^2\theta + \sin^2\theta = 1$ (or $\sec^2\theta - \tan^2\theta = 1$).
• "Describe the curve." Name it (line, parabola, circle, ellipse, rectangular hyperbola) and give its centre, radius, intercepts or vertex.
• "State the direction of travel / orientation." Evaluate the point at two or three increasing values of $t$ and say which way it moves.
• "State any restriction / excluded value." Identify a forbidden $t$ (a zero denominator, or $\cos\theta = 0$ for $\sec\theta$) and name the missing point or strip; if $t$ is restricted, restrict $x$ to match.
• "Parametrise this curve." Reverse the standard pairs: line $x = at + b$, $y = ct + d$; parabola $x^2 = 4ay$ as $x = 2at$, $y = at^2$; circle $x^2 + y^2 = r^2$ as $x = r\cos\theta$, $y = r\sin\theta$.
• "This models a projectile / motion." Treat $t$ as time, eliminate it for the path, and use the factored form to read the peak and the range.