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

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.