Parametric equations appear in every UK 9MA0 Paper 2 and every IAL P4. The sketching part often carries 4-6 marks, and students routinely burn 15 minutes plotting 20 evenly-spaced $t$-values.
There's a better way. Five carefully-chosen points, joined with knowledge of the curve's behaviour, gets you the mark in under 4 minutes.
The five points that matter
Given $x = f(t)$ and $y = g(t)$:
- Endpoints ($t = a$ and $t = b$, if the domain is stated).
- Any $t$ where $x$ turns around: solve $\dfrac{dx}{dt} = 0$.
- Any $t$ where $y$ turns around: solve $\dfrac{dy}{dt} = 0$.
- Any $t$ where $y = 0$ (x-axis intercept).
- Any $t$ where $x = 0$ (y-axis intercept).
Between them, you'll capture every stationary point, every intercept, and every turning behaviour in the curve.
Worked example
Sketch $x = t^2 - 4$, $y = t^3 - 3t$ for $-2 \le t \le 2$.
Step 1 — Endpoints.
- $t = -2$: $(x, y) = (0, -2)$
- $t = 2$: $(x, y) = (0, 2)$
Step 2 — Turning points of $x$.
$\dfrac{dx}{dt} = 2t = 0 \Rightarrow t = 0$, giving $(-4, 0)$.
Step 3 — Turning points of $y$.
$\dfrac{dy}{dt} = 3t^2 - 3 = 0 \Rightarrow t = \pm 1$.
- $t = 1$: $(-3, -2)$
- $t = -1$: $(-3, 2)$
Step 4 — Intercepts.
- $y = 0$: $t^3 - 3t = 0 \Rightarrow t(t^2 - 3) = 0 \Rightarrow t = 0, \pm\sqrt{3}$. Only $t = 0$ is in range → $(-4, 0)$. (We already have this.)
- $x = 0$: $t^2 - 4 = 0 \Rightarrow t = \pm 2$. Both already captured.
Step 5 — Join with knowledge.
You now have six points. Sketch them. Then trace the curve by increasing $t$ from $-2$ to $2$ — the curve MOVES from $(0, -2)$ through $(-3, 2)$, through $(-4, 0)$, through $(-3, -2)$, ending at $(0, 2)$. That's a figure-eight-ish loop.
Total time: about 3 minutes.
Why 20 points fails
Plotting evenly spaced $t$-values misses the turning points unless you're lucky. You'll get the shape roughly right but you won't clearly show the stationary points — and the mark scheme awards a specific mark for identifying them.
The Cartesian shortcut
If you can eliminate $t$ and write the curve as $y = h(x)$ or $F(x, y) = 0$, do it. Many parametric curves become recognisable conics (ellipse, hyperbola) once eliminated.
For $x = 3\cos t$, $y = 2\sin t$: use $\cos^2 t + \sin^2 t = 1$ to get $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ — an ellipse. Sketch that in 30 seconds.
Your practice drill
Take three parametric-sketch questions from past papers. For each, work the five points ONLY — endpoints, dx/dt = 0, dy/dt = 0, intercepts. Time yourself. Aim for under 5 minutes per full sketch.
Do this every day for a week and Paper 2's parametric question stops being a 15-minute drain.
Ready to put this into practice?
Real Edexcel-style questions, dark-themed engine, method marks tracked as you go.
Open the formula guide →