Parametric equations: parameter elimination, sketches, and standard curves

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,

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. The key shift in viewpoint is that you no longer think of $y$ as a function of $x$; instead $x$ and $y$ are each functions of a third, hidden variable $t$, and the curve is the trail the moving point leaves behind. This is what lets parametric form describe shapes that fail the vertical-line test, like a full circle, which can never be a single $y = f(x)$.

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 $(a, b)$:

Circle of radius $r$ centred at origin:

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

Parabola $y^2 = 4 a x$:

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

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

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.

Trace a parametric curve, stage by stage

The dependable way to sketch a parametric curve, and to see what eliminating the parameter is doing, is to plot a few points and then join them in order of increasing $t$. Below, the curve $x = 2t$, $y = t^2 - 1$ (the curve from the first exam question above) is built one stage at a time.

Stage 1, plot points from a table of $t$ values. Choose a handful of $t$ values and compute $(x, y) = (2t, t^2 - 1)$ for each: $t = -2 \to (-4, 3)$, $t = -1 \to (-2, 0)$, $t = 0 \to (0, -1)$, $t = 1 \to (2, 0)$, $t = 2 \to (4, 3)$. Plot each point and keep its $t$ label, because the labels record the order the point is visited.

Stage 2, join the points in order of increasing $t$. Connecting the points smoothly from $t = -2$ up to $t = 2$ produces a parabola opening upward with its lowest point at $(0, -1)$. Joining in $t$-order (not, say, left to right) is what makes the trace correct.

Stage 3, mark the direction of travel. Add arrows showing how the point moves as $t$ increases: from $t = -2$ it comes down the left branch, through the vertex at $t = 0$, and up the right branch to $t = 2$. The orientation is part of the curve's description and is needed whenever a question asks about velocity or which way the point is going.

Stage 4, eliminate the parameter for the Cartesian equation. From $x = 2t$ we get $t = \frac{x}{2}$; substituting into $y = t^2 - 1$ gives $y = \left(\frac{x}{2}\right)^2 - 1 = \frac{x^2}{4} - 1$. The hidden parameter is gone, leaving a parabola with vertex $(0, -1)$, the same shape we traced. Here $t$ ranges over all reals, so the whole parabola is covered; with a restricted $t$ only part of it would be.

How exam questions ask about parametric equations

The recurring wordings each map to one of a few tasks:

• "Eliminate the parameter to find the Cartesian equation." Solve one equation for $t$ and substitute, or for trig parametrisations isolate $\cos t$ and $\sin t$ and apply $\cos^2 t + \sin^2 t = 1$. The final equation must contain only $x$ and $y$.
• "Sketch the curve" (with a given $t$ domain). Tabulate a few points, join them in $t$-order, and state the range of $x$ or $y$ covered. If $t$ is restricted, draw only the portion traced and mark the endpoints.
• "Describe the curve / what does this represent?" Eliminate $t$, recognise the standard form (line, circle, parabola, ellipse), and state its key features: centre and radius for a circle, vertex for a parabola.
• "Find the parametric equations of [a line / circle / parabola]." Use the standard parametrisations: a line is point plus $t$ times direction; a circle of radius $r$ about $(h, k)$ is $x = h + r\cos t$, $y = k + r\sin t$; the parabola $y^2 = 4ax$ is $x = at^2$, $y = 2at$.
• "In which direction is the point moving / find the velocity?" Use the orientation as $t$ increases. (In the related rates and motion topics this connects to $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, but the orientation alone answers the "which way" version.)
• "Show the point $P$ lies on the curve" or "$P$ is the point with parameter $t$." Substitute the given parameter value to get the coordinates, or substitute the coordinates into both parametric equations and check the same $t$ works in each.