Introduction to Parametric Equations
A satellite traces an ellipse around Earth. A projectile follows a parabolic arc through the air. A wheel leaves a cycloid path on the road. These curves are almost impossible to describe efficiently with a single equation in $x$ and $y$ — but with a parameter, they become elegant. In this lesson you'll learn to think in three variables at once, eliminate the parameter, and unlock a powerful new language for curves.
The unit circle can be described by $x = \cos t$, $y = \sin t$. Without using a formula — what value of $t$ gives the point $(1, 0)$? What about $(0, 1)$? Write your thoughts before reading on.
There are only two core skills in this lesson. Learn to write parametric equations $x = f(t),\; y = g(t)$, and then eliminate the parameter to recover the Cartesian form. Everything else builds on these two moves.
Every parametric problem lives on one of two roads: express both $x$ and $y$ via a parameter $t$ to describe a curve, or eliminate $t$ algebraically to find the relationship between $x$ and $y$ directly.
Key facts
- Parametric equations are of the form $x = f(t)$, $y = g(t)$
- The parameter $t$ controls both coordinates simultaneously
- $\cos^2 t + \sin^2 t = 1$ is the key trig identity for eliminating circular parameters
Concepts
- How the parameter traces a curve as it varies
- Why parametric equations may only trace part of the Cartesian curve
- The difference between parametric and Cartesian representations
Skills
- Eliminate the parameter to find the Cartesian equation
- Convert between parametric and Cartesian forms for lines, parabolas, and circles
- Identify any restrictions imposed by the parameter range
Instead of writing $y$ directly in terms of $x$, we can write both $x$ and $y$ in terms of a third variable called a parameter:
As $t$ varies, the point $(x, y)$ traces out a curve in the plane. The parameter $t$ often represents time, angle, or some other continuous quantity.
- Line: $x = 1 + 2t$, $y = 3 - t$ — as $t$ varies, the point moves along the line.
- Unit circle: $x = \cos t$, $y = \sin t$ for $0 \le t \le 2\pi$ — traces the full circle.
- Parabola: $x = t$, $y = t^2$ — gives $y = x^2$, the standard parabola.
Parametric equations: x = f(t), y = g(t) where t is the parameter; As t varies, the point (x, y) traces a curve in the Cartesian plane
Pause — copy the parametric definition into your book: $x = f(t)$, $y = g(t)$; as $t$ varies, the point $(x,y)$ traces a curve; $t$ is often time but may represent any quantity.
Quick check: For the parametric equations $x = t$, $y = t^2$, what is the value of $y$ when $t = 3$?
We just saw that parametric equations $x = f(t)$, $y = g(t)$ represent a curve in the plane with $t$ as a link variable. That raises a question: how do we convert this to a single Cartesian equation $y = h(x)$ (or an implicit form) by removing $t$? This card answers it → solve one equation for $t$ and substitute into the other; for trig parametrics, isolate $cos t$ and $sin t$ then use $cos^2 t + sin^2 t = 1$.
To convert parametric equations to Cartesian form, we eliminate the parameter:
- Solve one equation for the parameter $t$.
- Substitute that expression into the other equation.
- Simplify to get a direct relationship between $x$ and $y$.
Important: If the parameter is restricted to a range (e.g. $0 \le t \le \pi$), the parametric form may only trace part of the Cartesian curve. Always check.
The three-step method for eliminating the parameter to find the Cartesian equation.
To eliminate the parameter: (1) Solve one equation for t. (2) Substitute into the other. (3) Simplify.; For trig parametrics: use ^2 t + ^2 t = 1 — rearrange each equation to get t and t, then apply the identity.
Pause — copy both elimination methods into your book: algebraic — solve one equation for $t$, substitute into the other; trigonometric — isolate $\cos t$ and $\sin t$ from each equation, then apply $\cos^2 t + \sin^2 t = 1$.
Did you get this? True or false: the parametric equations $x = t^2$, $y = t$ (for all real $t$) give the full parabola $x = y^2$.
Worked examples · 3 in a row, reveal as you go
Eliminate the parameter: $x = t + 2$, $y = t^2 - 1$.
Eliminate the parameter: $x = \cos t$, $y = \sin t$ for $0 \le t \le \pi$.
Eliminate the parameter: $x = 3\cos t$, $y = 2\sin t$ for $0 \le t \le 2\pi$.
Fill the gap: For $x = 2t$ and $y = t^2 + 1$, eliminating the parameter gives $y = $ . (Hint: express $t$ in terms of $x$.)
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the parametric equations $x = \cos t$, $y = \sin t$ for $0 \le t \le 2\pi$ describe the entire unit circle.
Activities · practice with the ideas
Eliminate the parameter from $x = 2t$, $y = t^2 + 1$.
For $x = 1 + \cos t$, $y = \sin t$, find the Cartesian equation and identify the curve.
For $x = t^2$, $y = t^3$ with $t \ge 0$, eliminate the parameter and describe the resulting curve.
Explain why $x = \cos t$, $y = \sin t$ for $0 \le t \le \pi$ does NOT describe the full unit circle.
Write parametric equations for the line passing through $(1, 2)$ with slope $3$.
Odd one out: Which of these parametric equations does NOT produce a circle (or part of one)?
Earlier you were asked: what value of $t$ gives $(1, 0)$ and $(0, 1)$ on $x = \cos t$, $y = \sin t$?
For $(1, 0)$: $\cos t = 1$ and $\sin t = 0$, so $t = 0$. For $(0, 1)$: $\cos t = 0$ and $\sin t = 1$, so $t = \dfrac{\pi}{2}$. The parameter $t$ represents the angle in radians — and this is why parametric equations are so powerful for describing circular motion and other curves that are awkward to express as $y = f(x)$.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Eliminate the parameter from $x = 2t$, $y = t^2 + 1$, and state the Cartesian equation. (2 marks)
Q2. Find the Cartesian equation for $x = 1 + \cos t$, $y = \sin t$ and identify the curve. (2 marks)
Q3. For $x = t^2$, $y = t^3$ with $t \ge 0$, eliminate the parameter and describe the curve, including any restrictions. (2 marks)
Comprehensive answers (click to reveal)
Activity 1: 1. $t = x/2$, so $y = (x/2)^2 + 1 = x^2/4 + 1$ · 2. $\cos t = x - 1$, $\sin t = y$; using $\cos^2 t + \sin^2 t = 1$: $(x-1)^2 + y^2 = 1$ — a circle centred at $(1, 0)$ radius $1$ · 3. $t = x^{1/2}$, $y = x^{3/2}$; curve is the right half of $y^2 = x^3$ ($x \ge 0$, $y \ge 0$) · 4. For $0 \le t \le \pi$, $\sin t \ge 0$ so $y \ge 0$ — only the upper semicircle is traced · 5. $x = 1 + t$, $y = 2 + 3t$ (many valid answers)
Q1 (2 marks): $t = x/2$ [1]. $y = (x/2)^2 + 1 = \dfrac{x^2}{4} + 1$ [1].
Q2 (2 marks): $\cos t = x - 1$, $\sin t = y$; applying the identity: $(x-1)^2 + y^2 = 1$ [1]. This is a circle with centre $(1, 0)$ and radius $1$ [1].
Q3 (2 marks): From $x = t^2$ with $t \ge 0$: $t = \sqrt{x}$, so $y = (\sqrt{x})^3 = x^{3/2}$ [1]. This is the upper half of the curve $y^2 = x^3$, restricted to $x \ge 0$, $y \ge 0$ [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering parametric equations questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.