Parametric Equations — Converting to Cartesian Form, and Domain/Range

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

What a parametric curve is, and the one reliable way to eliminate t

Spec 3.1 names the whole of this topic in one line: "Parametric equations of curves and conversion between cartesian and parametric forms." A parametric curve is a pair of equations, $x=f(t)$ and $y=g(t)$ , where a single third variable $t$ (the parameter) drives both $x$ and $y$ at once — every value of $t$ produces one point $(f(t), g(t))$ on the curve, and letting $t$ range over its own traces the whole curve out. "Eliminating the parameter" means finding a single equation relating $x$ and $y$ directly, with no $t$ anywhere in it, that every one of those points satisfies.

There is exactly one reliable method, and it is pure algebra: rearrange whichever of the two given equations is simpler to solve for $t$ (or, for a trig pair, for $\cos t$ or $\sin t$ ), then substitute that expression into the OTHER equation everywhere $t$ appears. This works because both $x$ and $y$ are tied to the same shared $t$ at every point on the curve — solving one equation for $t$ expresses that shared value in terms of $x$ , and substituting removes the need to know $t$ at all, leaving a relationship that holds between $x$ and $y$ directly.

Two specific instincts are worth naming and setting aside before this gets any practice, because both are real, documented, and score nothing. First: guessing the shape the final answer ought to take, then working to make it fit, rather than deriving it — a real examiner report records exactly this on a genuine conversion question: "a small minority assumed the general form of the answer, substituted for x in terms of t and compared this to the expression for y but this strategy was often unsuccessful" (Jan 2021, Q4). Second: reaching for differentiation, on the assumption that a question involving two variables and a parameter must want calculus — a different real report is direct about this: "A few candidates thought it appropriate to use calculus and scored no marks" (Oct 2022, Q1). $\frac{dy}{dx}$ describes a gradient, a completely different object from the direct relationship between $x$ and $y$ this topic asks for.

What "eliminating the parameter" is actually doing

In plain terms

Imagine a fitness tracker that, once every second, logs two numbers at once: how far you've walked, and your current heart rate. Every second produces one paired reading — distance and heart rate, read off the same clock tick. If you write down every pair of numbers for a walk and then throw away the actual clock times (never write down which second each pair came from), you're left with something new: a direct trace of heart rate against distance, with the shared clock gone completely. That's what eliminating a parameter is — $x$ and $y$ were each secretly driven by the same hidden $t$ , and eliminating $t$ finds the direct relationship between $x$ and $y$ that survives once the shared driver is thrown away.

In this lesson's own vocabulary: $t$ is the parameter, the hidden shared driver; $x=f(t)$ and $y=g(t)$ are the parametric equations; and the single equation left once $t$ is removed — found by solving one equation for $t$ and substituting into the other — is the cartesian equation of the same curve. The two forms describe exactly the same set of points; only the description changes.

Formally

Given $x=f(t)$ , $y=g(t)$ for $t$ in some domain $D$ , eliminating the parameter means finding a single relation $F(x,y)=0$ (or, where solvable, $y=h(x)$ ) satisfied by every point $(f(t),g(t))$ for $t\in D$ . The relation itself is found algebraically — by solving $x=f(t)$ for $t$ (or for $\cos t$ / $\sin t$ , in a trig pair) and substituting into $y=g(t)$ — but $F(x,y)=0$ alone, considered as bare algebra, may hold for a wider set of points than the curve $D$ actually produces; which $x$ -values (and which $y$ -values) genuinely belong to the curve is a separate question, addressed later in this lesson, that depends on exactly how $t$ was restricted.

Worked, in full

Eliminating t from a polynomial pair — x = 2t + 1, y = t² − 3t (VERIDIAN-original)

01
Pick the simpler equation to solve for $t$ — here, $x=2t+1$ is linear in $t$ , so rearrange it directly: $t = \dfrac{x-1}{2}$ .
Earns: M1 — rearranges the linear equation for t. This is the necessary first move, and the one step every version of the wrong methods this lesson has already named — guessing the answer's shape, or reaching for calculus — skips entirely.
02
Substitute this expression for $t$ into the OTHER equation, $y=t^2-3t$ , at EVERY occurrence of $t$ : $y = \left(\dfrac{x-1}{2}\right)^2 - 3\left(\dfrac{x-1}{2}\right)$ .
Earns: M1 — substitutes into both occurrences of t, dependent on stage 1. A partial substitution — replacing t in one term and leaving it in the other — is a real, documented error on this exact question type: "A disappointing number of responses ended up with y in terms of both x and t as they only partially rearranged" (Oct 2022, Q1).
03
Expand each part separately, before combining anything: $\left(\dfrac{x-1}{2}\right)^2 = \dfrac{x^2-2x+1}{4}$ , and $3\left(\dfrac{x-1}{2}\right) = \dfrac{3x-3}{2} = \dfrac{6x-6}{4}$ .
Earns: A1 — correct expansion of the squared term and correct common-denominator form of the linear term, kept separate before combining. This is the step where a dropped sign most often hides.
04
Combine over the common denominator, with both minus signs surviving: $y = \dfrac{x^2-2x+1}{4} - \dfrac{6x-6}{4} = \dfrac{x^2-2x+1-6x+6}{4} = \dfrac{x^2-8x+7}{4}$ .
Earns: A1 — correctly combines the two fractions, keeping the $-6x$ and the $+6$ (from $-(-6)$ ) both intact.
05
State the cartesian equation, with $t$ genuinely absent from it: $y = \dfrac{x^2-8x+7}{4}$ , equivalently $4y = x^2-8x+7$ .
Earns: B1 — states the final equation with t completely eliminated. This is the one thing every wrong path this lesson has named — the guessed-form strategy, the calculus detour, a partial substitution — never actually manages to do cleanly.

Complete it yourself

Complete the chain — eliminate t from x = t + 3, y = t² + 2t − 1 (VERIDIAN-original)

01
Solve the simpler equation, $x=t+3$ , for $t$ : $t = x-3$ .
02
Substitute into $y=t^2+2t-1$ , at every occurrence of $t$ : $y = (x-3)^2 + 2(x-3) - 1$ .

Trig-parameter pairs: isolate cos t and sin t first, then square and add

When $x$ and $y$ are each written in terms of $\cos t$ and $\sin t$ — the second common shape this spec point tests — solving for $t$ itself is not the clean move it was above; $t$ sits inside a trig function, not standing alone. The $\sin^2\theta + \cos^2\theta \equiv 1$ (met on this course in the wma12-trigonometric-equations prerequisite, spec 6.1) is the tool that replaces direct substitution here: it lets $\cos t$ and $\sin t$ be eliminated together, in one combined step, rather than $t$ itself being isolated first.

The method, in order: isolate $\cos t$ and $\sin t$ SEPARATELY, each equal to exactly 1 times itself — subtract any constant, then divide by any coefficient, for BOTH equations, before doing anything else. Only once both are fully isolated, square each one and add the two results together. The left side becomes $\cos^2 t + \sin^2 t$ , which the identity replaces with $1$ in a single line; the right side is left as a relationship purely in $x$ and $y$ , with $t$ gone.

The order matters, and skipping it is the trig-pair version of the same partial-rearrangement trap already named above: squaring before fully isolating — for instance squaring $x=5+3\cos t$ directly, with the $5$ and the $3$ still attached — produces an expanded mess that never simplifies down to the identity's clean $1$ . The identity only fires once $\cos t$ and $\sin t$ each stand alone, with coefficient exactly $1$ , on their own side of their own equation.

Worked, in full

Eliminating θ from a trig pair — x = 5 + 3cos θ, y = −2 + 4sin θ (VERIDIAN-original)

01
Isolate $\cos\theta$ and $\sin\theta$ separately, fully — subtract the constant, then divide by the coefficient, for EACH equation, before doing anything else: from $x=5+3\cos\theta$ , $\cos\theta = \dfrac{x-5}{3}$ ; from $y=-2+4\sin\theta$ , $\sin\theta = \dfrac{y+2}{4}$ .
Earns: M1 — fully isolates cos θ and sin θ, each with coefficient exactly 1. This is the necessary state before the identity can be used at all; a $\cos\theta$ left multiplied by 3 cannot be squared and added directly to get the identity's clean 1.
02
Square both fully-isolated expressions: $\cos^2\theta = \left(\dfrac{x-5}{3}\right)^2$ and $\sin^2\theta = \left(\dfrac{y+2}{4}\right)^2$ .
Earns: M1 — squares both, dependent on stage 1. Squaring before isolating (e.g. $(3\cos\theta)^2$ instead of $\cos^2\theta$ ) does not lead anywhere the identity can reach.
03
Add the two squared expressions and invoke $\sin^2\theta + \cos^2\theta \equiv 1$ : $\left(\dfrac{x-5}{3}\right)^2 + \left(\dfrac{y+2}{4}\right)^2 = \cos^2\theta + \sin^2\theta = 1$ .
Earns: A1 — correctly invokes the identity to replace the left side with 1, eliminating θ entirely. This line is the one place the whole trig method actually earns its keep — everything before it was setup.
04
State the cartesian equation: $\dfrac{(x-5)^2}{9} + \dfrac{(y+2)^2}{16} = 1$ — an ellipse, centre $(5,-2)$ .
Earns: A1 — correct final equation, with the denominators written as $3^2=9$ and $4^2=16$ .

Mechanism

Why the parameter's domain becomes the cartesian function's domain — and why its range needs more than two numbers

When $t$ is restricted to a domain $D$ — a CLOSED, bounded interval, say $a \leq t \leq b$ , is the simplest case, covered first below, but $D$ can equally be OPEN and/or UNBOUNDED (such as $t>0$ , which both excludes its own lower end and has no upper one), a genuinely different case covered later in this same block — the SET of $x$ -values genuinely reached — $\{f(t) : t \in D\}$ — is exactly what defines the valid domain of the cartesian equation. The bare equation, as pure algebra with no reference to $t$ , might make sense for a wider range of $x$ ; only the $x$ -values a real point of the curve actually reaches belong to it. When $f$ is strictly monotonic over $D$ (a linear $x=f(t)$ with nonzero coefficient always is), the two endpoints of $D$ map straight onto the two endpoints of the $x$ -domain, with every value in between covered too — so for a monotonic $f$ , checking the two endpoints of $t$ genuinely is enough to state the whole $x$ -domain, correctly. The of $y$ is a structurally different question, and this is exactly where it stops being safe to apply the same shortcut. $y=g(t)$ is very often NOT monotonic over $D$ — a quadratic in $t$ has exactly one — and if that turning point sits at some $t^\ast$ INTERIOR to $D$ , the values $g$ takes near $t^\ast$ are not bounded between $g(a)$ and $g(b)$ at all: they can dip below both, or rise above both, at that one interior point alone. So the very shortcut that correctly finds the $x$ -domain (because $x$ happens to be monotonic here) fails outright for the $y$ -range whenever $y$ is not — precisely the failure a real examiner report records directly: "Common incorrect solutions followed attempts to substitute either end of the domain in the parametric equation for y. This usually resulted in only one of the two marks being scored, with the minimum value missing" (Oct 2021, Q5(c)). The fix reuses exactly the completing-the-square reading this lesson's own prerequisite lesson already establishes for a directly-given quadratic: rewrite $y$ as a function of $x$ , find any interior turning point (by , or by setting $\frac{dy}{dx}=0$ , equivalently $\frac{dy}{dt}=0$ ), CHECK that the $t$ -value at that turning point genuinely lies inside the stated domain — a turning point outside $D$ belongs to a part of the curve that was never drawn — and only then combine that interior value with the two endpoint values to state the complete range. This is also the precise reason domain and range are never interchangeable answers to each other: domain is a statement about which INPUTS are used, fixed entirely by where $t$ is allowed to range; range is a statement about which OUTPUTS actually occur, which depends on the full behaviour of $g$ across that domain, turning points included — a structurally different question, and "gave answers in terms of x rather than y" (Jan 2021, Q4) is exactly what happens when a range question is answered with a domain-shaped answer instead. Two further cases complete the picture, both turning on whether $D$ is closed and bounded or not. FIRST, domain-finding when $D$ is open and/or unbounded: there is no literal endpoint value of $t$ to substitute at all — $t>0$ excludes its own lower end and has no upper one — so the closed-bounded shortcut above does not apply directly; each excluded or infinite end has to be reasoned about as a LIMIT instead, asking what $x$ approaches as $t$ tends to the excluded value, and what $x$ approaches (or does) as $t\to\infty$ , then combining the two limiting values into an open and/or unbounded $x$ -domain (e.g. $x>k$ , never $x\geq k$ , whenever the limiting value $k$ itself is never actually reached by any real value of $t$ ). SECOND, range-finding when $g$ , rewritten as a function of $x$ , is monotonic and has NO interior turning point at all: the completing-the-square method above assumes a turning point exists to find, and searching for one that does not leaves no method whatsoever for the boundary values a monotonic function's range still genuinely needs. The fix here checks two different kinds of boundary directly instead: evaluate the cartesian function AT its own excluded endpoint — valid whenever the function is continuous there, even though that exact $x$ -value is itself excluded from the stated domain, because the value is still the genuine limit the range approaches — and separately find the value the function approaches as $x\to\infty$ (for a rational function $\frac{ax+b}{cx+d}$ , this second boundary is the ratio of leading coefficients, $\frac{a}{c}$ , since the constants $b$ and $d$ become negligible once $x$ is large enough). Both boundaries can equally be reached by working with $t$ directly instead of $x$ — taking the very same two limits of $t$ that gave the open/unbounded $x$ -domain above, and substituting them into $y(t)$ rather than into $g(x)$ .

Worked, in full

Eliminating t from a reciprocal pair, with an open/unbounded domain — real WMA14 anchor: x = 1/t + 2, y = (1−2t)/(3+t), t > 0 (WMA14/01, January 2021, Q4(a))

01
Rearrange $x=\dfrac1t+2$ for $t$ — this is a RECIPROCAL relationship, not a linear one: subtract 2 from both sides, then take the reciprocal of both sides: $x-2=\dfrac1t$ , so $t=\dfrac{1}{x-2}$ .
Earns: M1 — "Attempts to find t in terms of x and substitutes into y..." (WMA14/01, January 2021, Q4(a) mark scheme). Genuinely different from every earlier worked example in this lesson: every $x=f(t)$ elimination so far has been linear in $t$ , with nothing to invert — here $t$ sits inside a reciprocal, and has to be found by inverting the whole relationship, not by simple rearrangement.
02
Substitute $t=\dfrac{1}{x-2}$ into the DENOMINATOR of $y=\dfrac{1-2t}{3+t}$ first, and clear the compound (fraction-within-a-fraction) result by multiplying every term by $(x-2)$ : $3+\dfrac{1}{x-2} = \dfrac{3(x-2)+1}{x-2} = \dfrac{3x-5}{x-2}$ .
Earns: A1 — "Correct numerator or denominator with fraction removed (allow unsimplified)" (same mark scheme), credited here for the denominator. Clearing a compound fraction — a fraction nested inside another fraction — by multiplying every term through by the same factor is the one algebraic mechanic no earlier example in this lesson has needed: every previous pair combined polynomial terms over a single shared denominator, never one fraction sitting inside another.
03
Do the same for the NUMERATOR: $1-2t = 1-\dfrac{2}{x-2} = \dfrac{(x-2)-2}{x-2} = \dfrac{x-4}{x-2}$ .
Earns: A1 — the same real criterion covers whichever of numerator or denominator is reached; both are needed before they can be combined, and this is the second of the two, cleared the same way as the first.
04
Combine: $y=\dfrac{(x-4)/(x-2)}{(3x-5)/(x-2)}$ — the shared $(x-2)$ cancels top and bottom, leaving $y=\dfrac{x-4}{3x-5}$ , matching the target form $g(x)=\dfrac{ax+b}{cx+d}$ with $a=1,\,b=-4,\,c=3,\,d=-5$ , all integers.
Earns: A1 — final accuracy mark for the fully cleared and simplified rational function, matching the printed target form exactly.
05
State the domain of $x$ — but $t>0$ is OPEN (excludes $t=0$ ) and UNBOUNDED (no upper limit), so neither end of $t$ can simply be substituted the way a closed interval's two endpoints can. Reason about each end as a LIMIT instead: $x=\dfrac1t+2$ is strictly DECREASING as $t$ increases ( $\dfrac{dx}{dt}=-\dfrac{1}{t^2}<0$ for every $t>0$ ), so as $t\to\infty$ , $x\to0+2=2$ (approached, never reached — no value of $t$ is actually infinite), and as $t\to0^+$ , $\dfrac1t\to+\infty$ , so $x\to\infty$ (also never reached, since $t=0$ itself is excluded). Combine the two limits: $x>2$ .
Earns: B1 — "k = 2 or x > 2" (same mark scheme). Genuinely different reasoning from every earlier domain statement in this lesson, which mapped a CLOSED, bounded $t$ -domain onto $x$ by substituting its two literal endpoint values directly — here neither end of $t$ can be substituted at all, because $t=0$ is excluded and " $t=\infty$ " is not a number; only reasoning about what $x$ approaches at each end supplies the domain.

Same question, every valid method

Hence, or otherwise, state the range of $g$ — real WMA14 anchor, continuing directly from the worked chain above: $g(x)=\dfrac{x-4}{3x-5}$ , $x>2$ (WMA14/01, January 2021, Q4(b), 2 marks).

2 valid methods · every one reaches $-2 < g(x) < \dfrac13$ . $g'(x)=\dfrac{7}{(3x-5)^2}>0$ throughout $x>2$ (independently checked), so $g$ is strictly increasing with no interior turning point at all — the completing-the-square method taught elsewhere in this lesson has nothing to find here, because there genuinely is nothing to find. · 2 marks available

01
The domain is $x>2$ , so $x=2$ itself is excluded — but $g$ is continuous there, so evaluate $g(2)$ anyway: $g(2)=\dfrac{2-4}{3(2)-5}=\dfrac{-2}{1}=-2$ .
Method mark for obtaining one of the two boundaries — the real mark scheme explicitly credits "attempting g(2) for their g" for this mark, even though $x=2$ lies outside the stated domain: because $g$ is continuous at $x=2$ , the value $g(2)$ is still the genuine limit the range approaches as $x\to2^+$ .
M1
02
Find the other boundary as $x\to\infty$ : for a rational function $\dfrac{ax+b}{cx+d}$ , the constants $b$ and $d$ become negligible next to $ax$ and $cx$ once $x$ is large, so $g(x)\to\dfrac{a}{c}=\dfrac13$ (confirmed: $g(100)\approx0.325$ , $g(1000)\approx0.333$ , closing in on $\frac13$ from below). Combine both, keeping the inequalities strict since neither boundary is actually reached: $-2<g(x)<\dfrac13$ .
Accuracy mark for the complete range, both boundaries correct and strict. The real mark scheme credits this route's second boundary directly: "or for attempting their a/their c."
A1

Whenever the excluded endpoint of the x-domain is one specific finite number and g is continuous there — evaluating g directly is usually the fastest route to that one boundary, and reading off a/c is faster still than re-deriving a limit from scratch.

Diagram — The endpoints trap, drawn: two equal endpoints hiding a lower dip

x-axis: x · y-axis: y

y = x² − 4x, 0 ≤ x ≤ 4: The cartesian curve recovered from x = t − 1, y = t² − 6t + 5 for 1 ≤ t ≤ 5 (this lesson's own marked-solution question). An upward-opening parabola arc, restricted to the domain the parameter actually produces.

Left endpoint (0, 0): t = 1 — one edge of the valid t-domain, mapped through x = t − 1.
Right endpoint (4, 0): t = 5 — the other edge. Both endpoints happen to give the identical y-value, 0 — exactly what makes the trap below so easy to fall into.
Interior minimum (2, −4): t = 3, found by completing the square. Not visible from the two endpoints alone, and strictly lower than either of them — this is the value a real script most often loses.

Common error: Substituting only t = 1 and t = 5 into y(t), seeing both give y = 0, and concluding the range is just the single value y = 0 — or, more generally, stopping at the two endpoint values without checking whether anything different happens between them.

Correct: The curve dips to y = −4 at the interior point x = 2 (t = 3) before rising back to 0 at the right endpoint. The true range is −4 ≤ y ≤ 0, and only the interior minimum — found by completing the square, or by differentiating — reveals the lower bound; the two endpoints alone supply only the upper one.

examiner-report · Oct 2021 · Q5(c)

Marked, line by line

A curve has parametric equations $x = t - 1$ , $y = t^2 - 6t + 5$ , for $1 \leq t \leq 5$ . (a) Show that the cartesian equation of the curve can be written as $y = x^2 - 4x$ , stating the domain of $x$ for which this equation represents the curve. (4) (b) Hence find the range of the function found in part (a). (2) — VERIDIAN-original question, built specifically to demonstrate the real, documented domain/range trap (Oct 2021, Q5(c)): substituting only the two ends of the parameter's domain into y(t) finds the endpoints' own y-values but misses an interior turning point that can sit lower (or higher) than either. Not a reproduction of any past-paper question; the per-line mark allocations are modelled on WMA14's own verified mark-scheme conventions (§4) rather than transcribed from a real scheme, which for an original question does not exist.

6 marks available

(a) — 4 marks

“Show that” — the answer is already printed above

Matching the printed result isn’t the same as deriving it — real examiner reports describe scripts that adjust flawed working just to still land on it. Write your own full working below before checking it against the mark scheme.

(b) — 2 marks

101
Complete the square: $y = x^2-4x = (x-2)^2 - 4$ , so the minimum value of $y$ is $-4$ , at $x=2$ . Check this lies in the valid domain: $x=2$ corresponds to $t=x+1=3$ , and $1 \leq 3 \leq 5$ , so this interior turning point genuinely belongs to the curve.
Method mark for finding the interior turning point (by completing the square, or equivalently setting dy/dx = 0) AND checking it lies within the stated t-domain — both parts required, since a turning point outside the valid domain would be irrelevant to this particular curve.
M1
102
Evaluate the two endpoints: $x=0$ gives $y=0$ ; $x=4$ gives $y=16-16=0$ — both ends of the domain give the same value, which is therefore the maximum on this interval (the parabola opens upward, so nothing beyond the two ends and the one interior minimum needs checking). State the range: $-4 \leq y \leq 0$ .
Accuracy mark for combining the interior minimum with the endpoint values to state the complete range — both halves are required, and finding only one of them is exactly the real documented shortfall on this question type: candidates who substitute only the endpoints typically score only one of the two available marks, "with the minimum value missing" (Oct 2021, Q5(c)).
A1

In your own words

In one sentence: why can checking only the two endpoint values of a restricted parameter be enough to find a cartesian function's domain, but not enough, on its own, to find its range?

Named traps

guessed-target-form-instead-of-deriving-itfudged-reverse-fit: Confirmed directly: "A small minority assumed the general form of the answer, substituted for x in terms of t and compared this to the expression for y but this strategy was often unsuccessful" (Jan 2021, Q4). The trap is specifically INFORMAL, ungoverned guessing and pattern-matching — trying values, or eyeballing a match, without ever setting up a genuine algebraic identity and solving it for the unknowns. It is NOT the same as assuming a general form outright: this very question's own real mark scheme credits a fully valid, full-marks "Alternative 1" built on exactly that starting move — assume $g(x)=\frac{ax+b}{cx+d}$ (the form the question itself gives), substitute $x=\frac1t+2$ into it, and derive $a,b,c,d$ by equating the result to $y=\frac{1-2t}{3+t}$ as an identity in $t$ (matching coefficients of the powers of $t$ on each side), verbatim: "M1: Assume g(x)=(ax+b)/(cx+d) and substitute in x=1/t+2 … A1: g(x)=(a+(b+2a)t)/(c+(d+2c)t) … A1(M1 on EPEN): Correct numerator or denominator … A1: y=(x-4)/(3x-5)" (WMA14/01, January 2021, Q4(a) mark scheme — see this lesson's own real worked chain above, which reaches the same $g(x)=(x-4)/(3x-5)$ by the other valid route). The distinction that actually matters: deriving the unknowns by a genuine coefficient-matching identity is a legitimate forward derivation; guessing them and checking whether they happen to fit a couple of points is not.
endpoint-substitution-attempted-on-an-open-or-unbounded-domain: The lesson's closed-bounded domain method (substitute the parameter's two literal endpoint values directly) has no endpoint to substitute at all when the parameter's domain is open and/or unbounded — e.g. $t>0$ , which excludes $t=0$ and has no upper bound. Confirmed on the real anchor this lesson's worked chain above is built on: the real mark scheme states the domain result plainly — "k = 2 or x > 2" (WMA14/01, January 2021, Q4(a) mark scheme) — reached only by reasoning about limits at each excluded/unbounded end of t, never by substituting a literal value of t that does not exist. Confusing the two domain shapes — reaching for direct substitution on an open or unbounded interval, the way the closed-bounded case correctly allows — leaves no method for the domain at all.
domain-confused-with-range: Confirmed on the same question: "Candidates find the concept of domain and range difficult and it was clear that some confused range with domain and gave answers in terms of x rather than y" (Jan 2021, Q4). Domain is a statement about which x-values are used, fixed entirely by the parameter's own restriction; range is a statement about which y-values actually occur, and depends on the full behaviour of y over that domain. Asked for one, giving the other is not a smaller version of a correct answer — it answers a different question.
partial-rearrangement-leaves-t-in-the-equation: Confirmed directly: "A disappointing number of responses ended up with y in terms of both x and t as they only partially rearranged" (Oct 2022, Q1). Eliminating the parameter is only finished once t no longer appears ANYWHERE in the final equation — substituting into one occurrence of t while leaving another untouched produces an expression that looks like progress but is not yet a cartesian equation at all.
calculus-reached-for-on-an-elimination-question: Confirmed on the same question: "A few candidates thought it appropriate to use calculus and scored no marks" (Oct 2022, Q1). Converting between parametric and cartesian form is pure algebraic substitution; differentiating produces a gradient function that still depends on t, or that is simply not the relationship between x and y the question asked for — a plausible-looking wrong tool for this specific question type, not a shortcut to the right one.
endpoints-substituted-for-the-whole-range: Confirmed directly: "Common incorrect solutions followed attempts to substitute either end of the domain in the parametric equation for y. This usually resulted in only one of the two marks being scored, with the minimum value missing" (Oct 2021, Q5(c)). Two endpoint values only describe the two ends of the parameter's domain — they say nothing about an interior turning point, which is exactly where the true minimum (or maximum) of a non-monotonic y(t) can sit, unannounced, between them.

Retrieval — with feedback on every choice

Question 1

3 marks

A curve has parametric equations $x=t-2$ , $y=t^2+4t$ . Find the cartesian equation.

Question 2

3 marks

Eliminate $\theta$ from $x=4\cos\theta - 1$ , $y = 2\sin\theta+3$ .

Question 3

2 marks

A curve has parametric equation $x=3t-2$ for $-1 \leq t \leq 4$ . What is the domain of $x$ for the resulting cartesian function?

Question 4

3 marks

A curve has parametric equations $x=t-3$ , $y=t^2-2t-3$ , for $0 \leq t \leq 4$ . Find the range of the resulting cartesian function.

Question 5

2 marks

A real examiner report on a parametric-conversion question states that some candidates "assumed the general form of the answer, substituted for x in terms of t and compared this to the expression for y" — and that this strategy was "often unsuccessful." What is this approach actually doing wrong?

Question 6

2 marks

A real examiner report on a different parametric-conversion question states: "A few candidates thought it appropriate to use calculus and scored no marks." Why does differentiating x=f(t) and y=g(t) and forming dy/dx not answer a "convert to cartesian form" question?

Reference — not a study method, a lookup

Eliminate t: solve ONE equation for t (or for cos t/sin t), substitute into the OTHER — or, if a target general form is given, substitute x(t) into it and derive every coefficient by matching it to y(t) as a genuine identity in t. Don't just guess coefficients to fit the target without deriving them.
Trig pair: isolate cos t and sin t FULLY (subtract, then divide) — THEN square both and add, using sin²t + cos²t ≡ 1.
A parametric elimination question is pure algebra — reaching for differentiation on it scores nothing (Oct 2022, Q1).
Domain (of x) comes from mapping the parameter's own bounds through x(t). Range (of y) is a separate, y-shaped question — never answer one with the other.
Closed, bounded t-domain (e.g. 1 ≤ t ≤ 5): map the domain by substituting the two literal endpoint values directly.
Open and/or unbounded t-domain (e.g. t > 0): there is no literal endpoint to substitute — reason about what x approaches as a LIMIT at each excluded or infinite end instead, then combine into an open/unbounded x-domain.
To find the range: check the two endpoint t-values AND look for an interior turning point (complete the square, or differentiate) — endpoints alone can miss an interior min/max.
Before using an interior turning point, check its t-value actually lies inside the stated domain — one outside it belongs to a part of the curve that was never drawn.
If g(x) is monotonic (no turning point exists at all — check g'(x) never changes sign), stop looking for one. Find the range's two boundaries instead: evaluate g at its own excluded endpoint directly (valid if g is continuous there), and find the limiting value as x → ∞ (for a rational g(x) = (ax+b)/(cx+d), this is a/c).

Not affiliated with or endorsed by Pearson Edexcel. Every quotation attributed to an examiner report in this lesson was independently checked against WMA14-verified-facts.md §5.3, itself checked against the actual Pearson PDFs (§6). UPDATED (mark-scheme-bullet coverage audit, see WMA14-verified-facts.md's own Lesson-audit log): §5.3 originally held examiner-report PROSE ONLY for this topic, with no real numeric scenario attached — that gap has since been closed for one question. The real, verbatim per-line mark scheme for WMA14/01 January 2021 Q4 — x = 1/t + 2, y = (1−2t)/(3+t), t > 0, both parts (a) and (b) — was re-extracted directly from the primary mark-scheme PDF and is now used, fully and accurately, in one worked chain and one method-comparison block above (real M1/A1/B1 codes, real quoted criteria, real numbers throughout, including the real credited "Alternative 1" assume-and-match route). Every OTHER equation, number, domain, range, worked chain, chain-drill and MCQ in this lesson remains VERIDIAN-original — computed and independently checked by hand (sympy) before being written in, including every deliberately wrong distractor, each produced by genuinely applying a documented wrong method (a sign slip solving for t, substituting x in place of t without eliminating it, swapping which coefficient's square belongs under which variable) rather than chosen to merely look plausible. Because the marked-solution question (the closed-bounded, interior-turning-point case) is original, its per-line M/A/B mark allocations are modelled on WMA14's own verified mark-scheme conventions (§4 of the facts bank) rather than transcribed from a real scheme; the two Q4-anchored blocks above are the one exception, transcribed from a real scheme that does exist. The evidence base for this specific topic remains genuinely thinner than this paper's other two lessons: five quotable examiner-report findings, from three different series/questions (Jan 2021 Q4, Oct 2021 Q5(c), Oct 2022 Q1) — Jan 2021 Q4 and Oct 2022 Q1 each yielding two distinct findings, Oct 2021 Q5(c) one — with every finding documenting a DIFFERENT specific trap, not one trap independently confirmed three times over, the shape the vectors lesson's own "three-series-confirmed" claim has; that count is unchanged by this audit, which added a real mark-scheme SKELETON for an anchor already-quoted in prose, not a new trap. No claim anywhere in this lesson escalates any single named trap beyond the one series and question that actually documents it.

Question 13 marks

A curve has parametric equations $x=t-2$ , $y=t^2+4t$ . Find the cartesian equation.

A $y = 2x+8$ , from $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t+4}{1} = 2t+4$ , then substituting $t=x+2$
This computes a gradient function, not the cartesian equation — differentiating is the real documented wrong tool for this question type: "A few candidates thought it appropriate to use calculus and scored no marks" (Oct 2022, Q1).
B $y = x^2+4x$ , using $x$ directly in place of $t$ throughout, without first solving $x=t-2$ for $t$
This never actually eliminates the parameter — it substitutes $x$ where $t$ should have been rearranged first, silently ignoring the $-2$ shift between them entirely. $x$ and $t$ are related but not equal; the relationship $t=x+2$ has to be used, not skipped.
C $y = x^2-8x+12$ , from a sign slip rearranging $x=t-2$ as $t=2-x$ instead of $t=x+2$
$x=t-2$ rearranges to $t=x+2$ (add 2 to both sides), not $t=2-x$ (which would solve $-x=t-2$ , a different equation). The sign slip changes every subsequent term.
$y = x^2+8x+12$ , from $t=x+2$ substituted into $y=t^2+4t$
Correct. $t = x+2$ rearranges $x=t-2$ ; substituting gives $y=(x+2)^2+4(x+2) = x^2+4x+4+4x+8 = x^2+8x+12$ .

Traps tested: Calculus reached for on an elimination question · Parameter not actually eliminated · Sign error rearranging for t

Question 23 marks

Eliminate $\theta$ from $x=4\cos\theta - 1$ , $y = 2\sin\theta+3$ .

A $\dfrac{(x-1)^2}{16}+\dfrac{(y-3)^2}{4}=1$
A sign slip isolating $\cos\theta$ : $x=4\cos\theta-1$ rearranges to $\cos\theta=\frac{x+1}{4}$ (add 1, then divide), not $\frac{x-1}{4}$ .
$\dfrac{(x+1)^2}{16}+\dfrac{(y-3)^2}{4}=1$
Correct. Isolating gives $\cos\theta=\frac{x+1}{4}$ and $\sin\theta=\frac{y-3}{2}$ ; squaring both and adding invokes $\cos^2\theta+\sin^2\theta=1$ .
C $\dfrac{(x+1)^2}{4}+\dfrac{(y-3)^2}{16}=1$
The two denominators are swapped: $x$ 's own coefficient is 4 (so its squared denominator should be $4^2=16$ ), and $y$ 's coefficient is 2 (denominator $2^2=4$ ) — this answer has attached each squared coefficient to the wrong variable.
D $\dfrac{x+1}{4}+\dfrac{y-3}{2}=1$
This adds the two isolated expressions without squaring either — $\cos\theta+\sin\theta$ has no fixed value, so this line uses no genuine identity at all. The identity needs the SQUARES, $\cos^2\theta+\sin^2\theta$ , not the bare terms.

Traps tested: Sign error isolating before squaring · Denominators swapped between x and y · Identity applied without squaring

Question 32 marks

A curve has parametric equation $x=3t-2$ for $-1 \leq t \leq 4$ . What is the domain of $x$ for the resulting cartesian function?

A $-1 \leq x \leq 4$
This copies the parameter's OWN bounds, $-1 \leq t \leq 4$ , directly onto $x$ without ever applying $x=3t-2$ to transform them. $x$ and $t$ are different variables related by that equation — the bounds have to be mapped through it, not copied across.
BThe domain cannot be determined without also knowing $y$
The domain of $x$ depends only on the equation $x=3t-2$ and the given bounds on $t$ — it is found the same way regardless of what $y$ happens to be.
$-5 \leq x \leq 10$
Correct. $x=3t-2$ increases throughout as $t$ increases (coefficient $+3$ ), so the two endpoints of $t$ map directly to the two endpoints of $x$ : $t=-1$ gives $x=-5$ ; $t=4$ gives $x=10$ .
D $-10 \leq x \leq 5$
This is what a sign slip produces — solving as though $x=2-3t$ instead of $x=3t-2$ : $t=-1$ would give $5$ , $t=4$ would give $-10$ . The original equation has $-2$ , not $+2$ , and $3t$ , not $-3t$ .

Traps tested: Parameter bounds copied without mapping through x · Domain assumed to require y · Sign error when mapping domain

Question 43 marks

A curve has parametric equations $x=t-3$ , $y=t^2-2t-3$ , for $0 \leq t \leq 4$ . Find the range of the resulting cartesian function.

$-4 \leq y \leq 5$ — found by completing the square on $y=x^2+4x$ to locate an interior minimum of $-4$ at $x=-2$ ( $t=1$ , within the domain), together with checking both endpoints ( $y=-3$ at $x=-3$ , $y=5$ at $x=1$ )
Correct. $x=t-3$ , $y=t^2-2t-3$ eliminates to $y=x^2+4x=(x+2)^2-4$ , minimum $-4$ at $x=-2$ ( $t=1$ , inside $[0,4]$ ). Domain of $x$ : $t=0 \to x=-3$ ; $t=4 \to x=1$ . Endpoints: $y(-3)=-3$ , $y(1)=5$ . The range runs from the interior minimum up to the larger of the two endpoint values.
B $-3 \leq y \leq 5$ — found by substituting $t=0$ and $t=4$ into $y=t^2-2t-3$
This checks only the two endpoints and misses the interior minimum entirely: the true minimum is $-4$ at $t=1$ (interior to $[0,4]$ ), lower than either endpoint value of $-3$ or $5$ .
C $-3 \leq y \leq 1$ — the domain of $x$
This states the $x$ -domain ( $-3 \leq x \leq 1$ ) as if it answered a question about the range of $y$ — a direct instance of the documented confusion between the two.
D $y \geq -4$ — the minimum value found by completing the square
This finds the correct interior minimum but never checks the endpoints for an upper bound. On a bounded domain (here $0 \leq t \leq 4$ ), the range has a genuine finite ceiling, found only by also checking the ends: $y=5$ at $x=1$ .

Traps tested: Endpoints substituted for the whole range · Domain confused with range · Interior extremum treated as the whole range

Question 52 marks

A real examiner report on a parametric-conversion question states that some candidates "assumed the general form of the answer, substituted for x in terms of t and compared this to the expression for y" — and that this strategy was "often unsuccessful." What is this approach actually doing wrong?

ANothing — comparing two expressions for the same variable is a legitimate way to eliminate a parameter
The real report is explicit that this strategy "was often unsuccessful" — comparing an assumed general form against the actual expression for y is a fundamentally weaker method than deriving one equation from the other by direct substitution.
BIt fails only because most candidates made an arithmetic mistake while comparing coefficients
The report attributes the failure to the STRATEGY itself, not to arithmetic slips within it — guessing the target form and working to match it is unreliable even when every individual step of arithmetic is executed correctly.
CIt fails only for trigonometric parametric equations, not polynomial ones
The real quote makes no mention of which form of parametric equations was involved, and the underlying problem — working backward from an assumed shape instead of deriving forward — applies equally to a polynomial pair or a trig pair.
It compares a guessed shape for the answer against the expression for y without ever completing a genuine coefficient-matching derivation — so unless every unknown is actually solved for by an identity, an apparent match is not yet a true consequence of the two given equations
Correct. Deriving forward — solve one equation for t and substitute into the other, OR assume a general form and derive its coefficients by matching it against the other equation as a genuine identity — guarantees every line is a true consequence of what was actually given. Guessing an answer's shape and merely comparing to it, without completing that derivation, does not carry that guarantee — but completing it DOES: this very question's own real mark scheme credits exactly that assume-and-derive route as a full, valid "Alternative 1" (see this lesson's trap-taxonomy entry above). The failure named here is the guessing, not the general form.

Traps tested: Guessing treated as legitimate · Guessing failure blamed on arithmetic not method · Trap wrongly scoped to one form

Question 62 marks

A real examiner report on a different parametric-conversion question states: "A few candidates thought it appropriate to use calculus and scored no marks." Why does differentiating x=f(t) and y=g(t) and forming dy/dx not answer a "convert to cartesian form" question?

AIt fails because the chain rule cannot be applied to parametric equations
This is false — the chain rule is exactly how parametric differentiation works (spec 5.1, a later WMA14 topic), and it is applied correctly here. The failure is not in the calculus itself; it is that differentiating answers a different question from the one asked.
$\frac{dy}{dx}$ is a gradient function — it describes how steeply the curve rises at each point, not the relationship between $x$ and $y$ itself, and forming it does nothing to eliminate $t$ from the system
Correct. A cartesian equation states which pairs $(x,y)$ lie on the curve; a gradient function states how $y$ changes relative to $x$ at a point, a genuinely different object. Neither the process of forming $\frac{dy}{dx}$ nor its result removes $t$ from the description of the curve.
CIt does answer the question, but only when $t$ appears linearly in both equations
The real report states plainly that this approach "scored no marks" — it is not a partially-valid shortcut that works in a special case; forming a gradient function is simply a different task from eliminating the parameter, regardless of how t appears.
DIt fails only because $\frac{dy}{dx}$ needs to be integrated afterward, which candidates forgot to do
Integrating $\frac{dy}{dx}$ would not recover the cartesian equation either — it is not a missing final step on an otherwise correct method, but a different operation from the direct substitution this question type actually requires.

Traps tested: Chain rule inapplicability invented · Wrong tool partially excused · Wrong tool treated as nearly right

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Oct 2021 · Q5(c) — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA14.

Parametric Equations — Converting to Cartesian Form, and Domain/Range

What a parametric curve is, and the one reliable way to eliminate t

What "eliminating the parameter" is actually doing

Eliminating t from a polynomial pair — x = 2t + 1, y = t² − 3t (VERIDIAN-original)

Complete the chain — eliminate t from x = t + 3, y = t² + 2t − 1 (VERIDIAN-original)

Trig-parameter pairs: isolate cos t and sin t first, then square and add

Eliminating θ from a trig pair — x = 5 + 3cos θ, y = −2 + 4sin θ (VERIDIAN-original)

Why the parameter's domain becomes the cartesian function's domain — and why its range needs more than two numbers

Eliminating t from a reciprocal pair, with an open/unbounded domain — real WMA14 anchor: x = 1/t + 2, y = (1−2t)/(3+t), t > 0 (WMA14/01, January 2021, Q4(a))

Deep dive

Practice this for real