Converting Between Parametric and Cartesian Forms

Being able to convert between parametric equations and Cartesian equations is a key skill in the Edexcel 9MA0 specification. The conversion reveals the shape of the curve and allows you to apply techniques from other parts of the syllabus.

Parametric to Cartesian: Eliminating the Parameter

The general strategy is to express the parameter t (or θ) in terms of x from one equation, then substitute into the other to eliminate the parameter.

Method 1: Direct Substitution

Example: Convert x = 2t + 1, y = 3t − 2 to Cartesian form.

From the first equation: t = (x − 1)/2.

Substitute into the second: y = 3((x − 1)/2) − 2 = (3x − 3)/2 − 2 = (3x − 7)/2.

So: 2y = 3x − 7, or 3x − 2y − 7 = 0. This is a straight line.

Method 2: Using Trigonometric Identities

When the parametric equations involve sin and cos, use the identity sin²θ + cos²θ = 1.

Example: Convert x = 3cos(θ), y = 3sin(θ) to Cartesian form.

cos(θ) = x/3, sin(θ) = y/3.

Substitute into sin²θ + cos²θ = 1: (y/3)² + (x/3)² = 1 x² + y² = 9.

This is a circle with centre (0, 0) and radius 3.

Example: Convert x = 2 + 4cos(θ), y = −1 + 4sin(θ) to Cartesian form.

cos(θ) = (x − 2)/4, sin(θ) = (y + 1)/4.

((x − 2)/4)² + ((y + 1)/4)² = 1 (x − 2)² + (y + 1)² = 16.

Circle with centre (2, −1) and radius 4.

Method 3: Substitution with Powers

Example: Convert x = t², y = 2t to Cartesian form.

From y = 2t: t = y/2. Substitute: x = (y/2)² = y²/4. So y² = 4x.

This is a parabola with vertex at the origin opening to the right.

More Complex Conversions

Using sec²θ = 1 + tan²θ

Example: Convert x = tan(θ), y = sec(θ) to Cartesian form.

Since sec²θ = 1 + tan²θ: y² = 1 + x², so y² − x² = 1.

This is a hyperbola.

Using Double Angle Identities

Example: Convert x = cos(2t), y = sin(t) to Cartesian form.

Using cos(2t) = 1 − 2sin²(t): x = 1 − 2y², so y² = (1 − x)/2.

Cartesian to Parametric

There are often many valid parametrisations for a given curve.

Example: Write y = x² + 1 in parametric form.

Simplest choice: let x = t, then y = t² + 1.

Or: let x = 2t, then y = 4t² + 1.

Example: Write (x − 1)² + (y + 2)² = 9 in parametric form.

x = 1 + 3cos(θ), y = −2 + 3sin(θ).

Restrictions and Domains

When converting, consider the domain of the parameter and how it affects the Cartesian curve.

Example: x = t², y = t for all real t.

Cartesian form: x = y² (since t = y, x = t² = y²).

But since x = t² ≥ 0 for all real t, the Cartesian curve x = y² is only valid for x ≥ 0. The parametric form traces only the right-hand part of the parabola — which is in fact all of x = y² anyway.

However: if x = t², y = t for t ≥ 0 only, then x = y² with y ≥ 0 — only the upper branch.

Exam Tip: When converting parametric to Cartesian, always check whether the domain of the parameter restricts the Cartesian curve. State any restrictions explicitly — examiners award marks for this.

Summary Table

Parametric Form	Cartesian Form	Method
x = at + b, y = ct + d	Linear (eliminate t directly)	Solve for t
x = a cos θ, y = a sin θ	x² + y² = a² (circle)	sin²θ + cos²θ = 1
x = at², y = 2at	y² = 4ax (parabola)	Substitute t = y/(2a)
x = a cos θ, y = b sin θ	x²/a² + y²/b² = 1 (ellipse)	sin²θ + cos²θ = 1
x = a sec θ, y = b tan θ	x²/a² − y²/b² = 1 (hyperbola)	sec²θ − tan²θ = 1

Key Terms

Term	Definition
Eliminate the parameter	The process of removing t or θ to obtain a Cartesian equation
Cartesian equation	An equation relating x and y directly
Trigonometric identity	An equation involving trig functions that is true for all valid inputs
Parametrisation	The process of expressing a Cartesian curve using a parameter

Summary

To convert parametric to Cartesian, eliminate the parameter using substitution or trigonometric identities.
The identity sin²θ + cos²θ = 1 is the most commonly used tool for trigonometric parametrisations.
To convert Cartesian to parametric, choose a suitable parameter (often letting x = t or using trig functions).
Always state any domain restrictions that arise from the original parametric domain.

A-Level Deep Dive: Converting Between Parametric and Cartesian Forms

Spec mapping

Edexcel 9MA0 Year 2 Pure section 8 — Parametric equations covers parametric equations in modelling in a variety of contexts; convert between Cartesian and parametric forms of a curve, including the use of trigonometric identities to eliminate parameters (refer to the official specification document for exact wording). Although this lesson focuses on the conversion mechanic, the skill is unavoidably synoptic. It feeds into section 8 (Parametric differentiation, where $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ requires a confident parametric set-up), section 5 (Trigonometry, supplying the identities $\cos^2 t + \sin^2 t = 1$ and $1 + \tan^2 t = \sec^2 t$ that drive elimination), section 7 (Coordinate geometry, where ellipses and hyperbolas are identified by their Cartesian forms after elimination), and 9MA0-03 Mechanics, where projectile trajectories $y = x \tan\alpha - \dfrac{gx^2}{2u^2 \cos^2 \alpha}$ are obtained by eliminating $t$ from the displacement components $x = u t \cos\alpha$ and $y = u t \sin\alpha - \tfrac{1}{2}g t^2$ . The Edexcel formula booklet does not list elimination strategies — these must be drilled.

Worked example with full mark scheme

Question (8 marks):

A curve $C$ has parametric equations $x = 3 \cos t, \quad y = 2 \sin t, \quad 0 \leq t < 2\pi.$

(a) Show that the Cartesian equation of $C$ is $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ . (4)

(b) State the range of $x$ and the range of $y$ on $C$ , and identify the curve. (2)

Solution with mark scheme:

(a) Step 1 — isolate $\cos t$ and $\sin t$ .

From $x = 3 \cos t$ , $\cos t = \dfrac{x}{3}$ . From $y = 2 \sin t$ , $\sin t = \dfrac{y}{2}$ .

M1 — both trigonometric functions correctly isolated. Common error: dividing the wrong way (writing $\cos t = \tfrac{3}{x}$ ). That single slip propagates through and forfeits every subsequent mark.

Step 2 — invoke the Pythagorean identity.

The identity $\cos^2 t + \sin^2 t = 1$ holds for all real $t$ , so:

$\left(\dfrac{x}{3}\right)^2 + \left(\dfrac{y}{2}\right)^2 = 1$

M1 — substituting the isolated expressions into the correct identity. Examiners look for an explicit citation of $\cos^2 t + \sin^2 t = 1$ ; an unannounced jump from the parametric pair to the Cartesian form loses the method mark even if the algebra is right.

Step 3 — simplify to the printed form.

$\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$

A1 — correct algebraic form.

A1 — answer presented exactly as printed (numerators $x^2$ , $y^2$ ; denominators $9$ , $4$ ; right-hand side $1$ ). Edexcel "show that" conventions demand the printed expression appear verbatim on the candidate's working.

(b) Step 1 — extract ranges from the parametric definitions.

Since $\cos t \in [-1, 1]$ , $x = 3 \cos t \in [-3, 3]$ . Since $\sin t \in [-1, 1]$ , $y = 2 \sin t \in [-2, 2]$ .

B1 — both ranges correct, given as closed intervals (the endpoints are attained because $0 \leq t < 2\pi$ covers a full period).

Step 2 — identify the curve.

The Cartesian form $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ is the standard form of an ellipse with semi-major axis $3$ along the $x$ -axis and semi-minor axis $2$ along the $y$ -axis, centred at the origin.

B1 — curve correctly named and the semi-axes identified.

(c) Substitute $x = \tfrac{3\sqrt{2}}{2}$ : $\cos t = \tfrac{1}{2} \cdot \tfrac{\sqrt{2}}{1} \cdot \tfrac{1}{1} = \tfrac{\sqrt{2}}{2}$ . Substitute $y = \sqrt{2}$ : $\sin t = \tfrac{\sqrt{2}}{2}$ .

M1 — both trig values evaluated.

Both $\sin t$ and $\cos t$ positive places $t$ in the first quadrant. The standard angle is $t = \dfrac{\pi}{4}$ .

A1 — exact answer in radians.

Total: 8 marks.

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): A curve $C$ has parametric equations $x = 2 \sec t$ , $y = 3 \tan t$ , for $-\tfrac{\pi}{2} < t < \tfrac{\pi}{2}$ .

(a) Find a Cartesian equation for $C$ in the form $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ , stating the values of $a$ and $b$ . (4)

(b) State, with reason, which branch of the curve is traced as $t$ varies in the given range. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — isolating: $\sec t = \tfrac{x}{2}$ and $\tan t = \tfrac{y}{3}$ .
M1 (AO1.1b) — applying the identity $\sec^2 t - \tan^2 t = 1$ (equivalently $1 + \tan^2 t = \sec^2 t$ ), substituting to get $\left(\tfrac{x}{2}\right)^2 - \left(\tfrac{y}{3}\right)^2 = 1$ .
A1 (AO1.1b) — Cartesian equation $\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1$ .
A1 (AO2.5) — correctly stated $a = 2$ , $b = 3$ (matching the printed form).

(b)

M1 (AO2.4) — observing that $\sec t > 0$ on $-\tfrac{\pi}{2} < t < \tfrac{\pi}{2}$ (since $\cos t > 0$ throughout), so $x = 2 \sec t > 0$ .
A1 (AO2.4) — concluding that only the right-hand branch of the hyperbola (where $x \geq 2$ ) is traced, with reason given.

Total: 6 marks split AO1 = 3, AO2 = 3. The split is balanced because Edexcel uses parametric questions to test both procedural fluency (knowing which identity to use) and reasoning (linking domain restrictions on $t$ to which part of the Cartesian curve is realised).

Synoptic links

Connects to:

Section 5 — Trigonometric identities: the entire conversion machinery in this lesson is just a redeployment of $\cos^2 t + \sin^2 t = 1$ , $1 + \tan^2 t = \sec^2 t$ and $1 + \cot^2 t = \csc^2 t$ . Recognising which identity to reach for is a pattern-matching skill: pairs $(\cos t, \sin t)$ trigger Pythagoras; pairs $(\sec t, \tan t)$ trigger the secant identity.
Section 7 — Coordinate geometry of conics: ellipses $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ and hyperbolas $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ are introduced via their Cartesian forms but parametrise naturally as $x = a \cos t,\ y = b \sin t$ and $x = a \sec t,\ y = b \tan t$ respectively. Elimination is the bridge between the two viewpoints.
Section 8 — Parametric differentiation: once a curve is given parametrically, $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ avoids the need to convert at all — but if a tangent or normal is requested at a specific Cartesian point, conversion is unavoidable.
Section 8 — Integration with a substitution: computing the area enclosed by a parametric curve uses $\int y \, dx = \int y(t) \, x'(t) \, dt$ . Recognising a parametric circle or ellipse via elimination is the first step before deciding whether to integrate parametrically or via a Cartesian substitution.
9MA0-03 Mechanics — projectiles: eliminating $t$ from $x = u t \cos\alpha$ and $y = u t \sin\alpha - \tfrac{1}{2}g t^2$ gives the trajectory equation $y = x \tan\alpha - \dfrac{g x^2}{2 u^2 \cos^2 \alpha}$ . This is the canonical applied example of parametric-to-Cartesian conversion and frequently appears as a "show that" in Mechanics papers.

Mark-scheme literacy

Parametric-elimination questions on 9MA0 split AO marks more evenly than algebra-and-functions questions:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Isolating the parameter, citing the correct identity, performing the algebraic substitution, simplifying to the requested form
AO2 (reasoning / interpretation)	30–40%	Identifying the curve (circle/ellipse/hyperbola), explaining domain restrictions on the Cartesian form arising from the parameter range, justifying which branch is traced
AO3 (problem-solving)	0–15%	Mechanics trajectories, modelling problems where the parametric form encodes physical meaning

Examiner-rewarded phrasing: "using the identity $\cos^2 t + \sin^2 t = 1$ "; "since $\cos t \in [-1, 1]$ , $x \in [-a, a]$ "; "the parametric range $0 \leq t < 2\pi$ traces the curve exactly once"; "this is an ellipse with semi-major axis $a$ and semi-minor axis $b$ "; "the right-hand branch of the hyperbola because $\sec t > 0$ on this interval". Phrases that lose marks: leaving the answer with $t$ still present ("not eliminated"); stating "this is a curve" without naming the conic; failing to give the range of $x$ or $y$ when the parameter range is restricted.

A specific Edexcel pattern: when the question gives a restricted parameter range (e.g. $0 \leq t \leq \pi$ rather than $0 \leq t < 2\pi$ ), a B1 mark is reserved for noting which arc of the Cartesian curve is traced. Skipping this remark forfeits the mark even if the Cartesian equation is impeccable.

Grade-band model answers

3-mark question

Question: A curve has parametric equations $x = 2t - 1$ , $y = t^2 + 3$ . Find a Cartesian equation for the curve.

Grade C response (~210 words):

From the first equation, $t = \dfrac{x + 1}{2}$ .

Substitute into the second:

$y = \left(\dfrac{x + 1}{2}\right)^2 + 3 = \dfrac{(x + 1)^2}{4} + 3$ .

Multiply through by 4: $4y = (x + 1)^2 + 12$ , so $(x + 1)^2 = 4y - 12$ , i.e. $y = \dfrac{(x + 1)^2}{4} + 3$ .

Converting Between Parametric and Cartesian Forms

Converting Between Parametric and Cartesian Forms

Parametric to Cartesian: Eliminating the Parameter

Method 1: Direct Substitution

Method 2: Using Trigonometric Identities

Method 3: Substitution with Powers

More Complex Conversions

Using sec²θ = 1 + tan²θ

Using Double Angle Identities

Cartesian to Parametric

Restrictions and Domains

Summary Table

Key Terms

Summary

A-Level Deep Dive: Converting Between Parametric and Cartesian Forms

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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