Mg12–Mg14 - Parametric curves and cartesian equations | OCR A-Level Mathematics B (MEI)

Use Mg12 to understand a shared parameter, Mg13 to convert between equation forms with the correct restrictions, and Mg14 to describe circles by an angle. You will also read an unfamiliar curve definition and reconstruct its Cartesian relation without losing an endpoint condition.

One parameter, two coordinates

Can a point go round a circle without ever reaching one particular point on it? A bounded picture need not have a bounded parameter. We will return to that puzzle after learning how a parameter chooses coordinates.

For a simple start, $x=2-t$ , $y=1+3t$ gives $(2,1)$ at $t=0$ and $(0,7)$ at $t=2$ . Both coordinates use the same value of $t$ .

You already know

Substitution replaces a variable consistently. Solving simultaneous equations means finding values that satisfy both; a restriction is part of the mathematical information, not an optional extra.

The parameter can represent time, but it does not have to. It is also useful simply as an ordered label.

Parametric equations

Parametric equations specify coordinates through a shared parameter, for example $x=f(t)$ , $y=g(t)$ , with a stated domain for $t$ . A generated point uses the same parameter value in both equations.

Pairing coordinates from different parameter values generally invents a point outside the curve.

Worked example 1: a line segment

Question. Describe the path of $x=2-t$ , $y=1+3t$ , $0\le t\le2$ .

Plan. Evaluate the parameter endpoints and follow the coordinate changes.

\begin{aligned} t=0&:\ (2,1) &&\text{first point}\\ t=2&:\ (0,7) &&\text{last point}\\ \Delta x&=-\Delta t,\quad\Delta y=3\Delta t &&\text{constant coordinate changes} \end{aligned}

Why this step? Endpoints locate the segment; the signs of the changes locate its direction.

Answer. The segment is traced left and up from $(2,1)$ to $(0,7)$ , including both ends.

Check. At $t=1$ , the intermediate point is $(1,4)$ , halfway along the segment.

Change only the interval, allowing $-1\le t<1$ .

Worked example 2: shifted and open boundaries

Question. Describe the same equations for $-1\le t<1$ .

Plan. Find the new boundary points and carry over the strict endpoint condition.

\begin{aligned} t=-1&:\ (3,-2) &&\text{included start}\\ t=1&:\ (1,4) &&\text{excluded end} \end{aligned}

Why this step? A finite endpoint can be calculated even when the parameter domain excludes it.

Answer. The segment runs left and up from $(3,-2)$ , included, towards $(1,4)$ , excluded.

Check. $t=0$ still gives $(2,1)$ , between the boundaries.

With all real parameters, the same equations describe the whole straight line.

Watch out

Wrong: use $x(0)=2$ and $y(2)=7$ together as $(2,7)$ . Right: $x=2$ forces $t=0$ , so its matching coordinate is $y=1$ .

Keep the equations but choose another interval.

Elimination with the original restrictions

Elimination gives a relation between the coordinates. It must describe exactly the points the parameter permits, rather than a larger curve that happens to contain them.

The preceding line has $t=2-x$ , so $y=7-3x$ . For $0\le t\le2$ , the full Cartesian description includes $0\le x\le2$ . This conversion method is not in the formula sheet: learn it.

Worked example 3: preserve a branch

Question. Eliminate $t$ from $x=2+t^2$ , $y=t-1$ , $t\ge0$ .

Plan. Recover $t$ from the linear equation and translate its domain.

\begin{aligned} t&=y+1 &&\text{rearrange}\\ x&=2+(y+1)^2 &&\text{substitute}\\ y&\ge-1 &&\text{translate }t\ge0 \end{aligned}

Why this step? The squared equation permits negative $t$ too unless the branch restriction is stated.

Answer. $x=2+(y+1)^2$ , $y\ge-1$ ; equivalently $y=-1+\sqrt{x-2}$ , $x\ge2$ .

Check. Every retained point gives $t=y+1\ge0$ , establishing the reverse direction.

Change only the domain to $-1\le t\le2$ . It now includes points on both sides of the vertex.

Worked example 4: a vertex inside the interval

Question. Give the Cartesian description for the same equations with $-1\le t\le2$ .

Plan. Retain the relation and translate both parameter boundaries using $y=t-1$ .

\begin{aligned} x&=2+(y+1)^2 &&\text{same elimination}\\ -2&\le y\le1 &&\text{new coordinate restriction} \end{aligned}

Why this step? An $x$ -interval cannot distinguish all the required and forbidden points on this sideways parabola.

Answer. $x=2+(y+1)^2$ , $-2\le y\le1$ .

Check. The endpoints are $(3,-2)$ , $(6,1)$ , with vertex $(2,-1)$ at $t=0$ .

Division can cause the opposite problem: it can lose a point. Before cancelling $t$ , test $t=0$ in both given equations. In $x=t$ , $y=t^2$ , the origin is a valid point and must not disappear through cancellation.

Watch out

Wrong: $x=2+(y+1)^2$ with no restriction. Right: keep the translated interval; the whole parabola contains points requiring forbidden parameters.

Move the interval towards the lower branch.

Converting from Cartesian form

When MEI asks you to convert Cartesian equations to parametric form, guidance will be given on the choice of parameter. Use that guidance: another correct parametrisation may describe the curve but fail the specified task.

Worked example 5: a given scaling

Question. For $y=2x^2+1$ , $-1\le x\le2$ , use $x=2t$ to give parametric equations and a domain.

Plan. Substitute the prescribed $x$ and divide the interval by $2$ .

\begin{aligned} x&=2t &&\text{given guidance}\\ y&=2(2t)^2+1=8t^2+1 &&\text{substitute}\\ -\frac12&\le t\le1 &&\text{translate }x\text{-domain} \end{aligned}

Why this step? The square acts on the whole $2t$ , so its coefficient becomes $8$ , not $4$ .

Answer. $x=2t$ , $y=8t^2+1$ , $-1/2\le t\le1$ .

Check. $t=-1/2,1$ gives $(-1,3)$ , $(2,9)$ , the required endpoints.

Change only the guidance to $x=-2u$ . The curve remains the same while traversal reverses.

Worked example 6: a reversed scaling

Question. Give the parametrisation of the same curve using $x=-2u$ .

Plan. Substitute the new guidance and reverse the inequality when dividing by a negative number.

\begin{aligned} y&=2(-2u)^2+1=8u^2+1 &&\text{substitute}\\ -1\le-2u\le2&\ \Longrightarrow\ -1\le u\le\frac12 &&\text{translate the interval} \end{aligned}

Why this step? The minus sign changes the parameter order but vanishes inside the square.

Answer. $x=-2u$ , $y=8u^2+1$ , $-1\le u\le1/2$ .

Check. Increasing $u$ starts at $(2,9)$ and ends at $(-1,3)$ , reversing the first route.

A guided parameter can also be $y$ : for $x+2y=7$ , putting $y=t$ gives $x=7-2t$ . For a vertical line $x=3$ , use $x=3$ , $y=t$ . The conversion procedure is not in the formula sheet.

Watch out

Wrong: preserve $-1\le u\le2$ by copying the $x$ -interval. Right: use the actual relation $x=-2u$ , obtaining $-1\le u\le1/2$ .

Try a different reversed scale, keeping the Cartesian curve.

A circle described by an angle

The horizontal and vertical projections of a rotating radius give a parametric circle. For a unit radius at the origin, its endpoint has coordinates $(\cos\theta,\sin\theta)$ : a leftward projection is negative horizontally and a downward projection is negative vertically. Pythagoras therefore gives $\cos^2\theta+\sin^2\theta=1$ . Multiplying by the radius and adding the centre coordinates produces a shifted circle, with the signs still recording position and traversal.

Circle with an angular parameter

For centre $(a,b)$ , radius $r>0$ , and $\theta$ measured anticlockwise from the positive horizontal direction,

x=a+r\cos\theta,\qquad y=b+r\sin\theta.

Then $(x-a)^2+(y-b)^2=r^2$ , using $\cos^2\theta+\sin^2\theta=1$ . The interval $0\le\theta<2\pi$ traces the circle once. Not in the formula sheet: learn it.

One turn is $2\pi$ radians, or $360^\circ$ . Both endpoints together would repeat the start.

Worked example 7: a guided angular description

Question. Use the stated angular convention to parametrise $(x-2)^2+(y+1)^2=16$ . State the point at $\theta=\pi/2$ .

Plan. Read the centre and radius and use signed projections.

\begin{aligned} (a,b)&=(2,-1),\quad r=4 &&\text{read the Cartesian equation}\\ x&=2+4\cos\theta &&\text{horizontal projection}\\ y&=-1+4\sin\theta &&\text{vertical projection} \end{aligned}

Why this step? The radius is $4$ , not $16$ ; the Cartesian equation contains its square.

Answer. These equations with $0\le\theta<2\pi$ trace the full circle once anticlockwise. At $\theta=\pi/2$ , the point is $(2,3)$ .

Check. $(2-2)^2+(3+1)^2=16$ .

Change only the parameter interval to $0\le\theta\le\pi$ . The equations stay fixed.

Worked example 8: identify the arc

Question. Describe the curve generated by the same equations for $0\le\theta\le\pi$ .

Plan. Find the endpoints and use the sign of sine to choose the semicircle.

\begin{aligned} \theta=0&:\ (6,-1) &&\text{rightmost point}\\ \theta=\pi&:\ (-2,-1) &&\text{leftmost point}\\ y+1&=4\sin\theta\ge0 &&\text{upper half} \end{aligned}

Why this step? Endpoint coordinates alone would not tell you whether the upper or lower arc is traced.

Answer. The upper semicircle is traced anticlockwise from $(6,-1)$ to $(-2,-1)$ , with both endpoints included.

Check. Its midpoint in parameter order, $\theta=\pi/2$ , gives the top point $(2,3)$ .

Replacing the sine term by $-4\sin\theta$ instead reflects the path in the horizontal line through the centre and reverses the full-circle traversal.

Watch out

Wrong: $0\le\theta\le2\pi$ includes every point exactly once. Right: $0$ and $2\pi$ give the same point, so exclude one boundary for once-only tracing.

Keep the upper-half-turn interval and translate only the centre, to $(-1,2)$ .

Sketching the path and its traversal

A parametric sketch should show the generated portion and increasing-parameter order, not merely a familiar unrestricted Cartesian curve.

[DIAGRAM: asset_name: Parametric path and traversal; caption: Parametric path and traversal; asset_slug: h640-parametric-paths; description: Equal-scale axes show x equals t squared minus 1 and y equals 2t for minus 2 to 2, with endpoint and vertex labels and traversal arrows, beside a centre 1 comma minus 2 radius 3 circle traced anticlockwise by an angle]

The parabola panel includes its vertex as well as its two endpoints. The circle panel shows a different centre from the preceding examples, $(1,-2)$ , and radius $3$ ; its quarter-turn points check the general angular rule.

Worked example 9: a curve with a turn

Question. Describe the sketch and traversal of $x=t^2-1$ , $y=2t$ , $-2\le t\le2$ .

Plan. Eliminate the parameter and use endpoints plus the vertex.

\begin{aligned} x&=\frac{y^2}{4}-1,\quad -4\le y\le4 &&\text{relation and restriction}\\ t=-2,0,2&:\ (3,-4),\ (-1,0),\ (3,4) &&\text{ordered features} \end{aligned}

Why this step? The parameter increases throughout, even while the horizontal coordinate changes its direction of motion.

Answer. The right-opening parabola segment is traversed up and left to $(-1,0)$ , then up and right to $(3,4)$ .

Check. $t=-1,1$ gives $(0,-2)$ , $(0,2)$ , matching both branches.

Change only the sign of $y$ to $y=-2t$ . The point set stays fixed but its order reverses.

Worked example 10: reverse the traversal

Question. Describe $x=t^2-1$ , $y=-2t$ , $-2\le t\le2$ .

Plan. Re-evaluate the ordered key points and compare the Cartesian relation.

\begin{aligned} x&=\frac{y^2}{4}-1,\quad -4\le y\le4 &&\text{same curve}\\ t=-2,0,2&:\ (3,4),\ (-1,0),\ (3,-4) &&\text{new order} \end{aligned}

Why this step? Squaring removes the sign, so the Cartesian equation alone cannot reveal the change of traversal.

Answer. The segment is traced down and left to the vertex, then down and right to the lower endpoint.

Check. At $t=1$ , the point is $(0,-2)$ , reached after the vertex, as expected.

A plotting calculator needs the correct shared parameter range and a sufficiently fine step. Arrows show parameter order, not equal speed. These sketching rules are not in the formula sheet: learn the method.

Watch out

Wrong: join a few points by straight segments and call the zigzag a parabola. Right: use the eliminated equation to guide a smooth curve through the ordered features.

Keep the reversed equations and retain only the non-negative parameters.

A rational description of a circle

An unfamiliar definition can produce a familiar curve. Read the domain before eliminating: it can explain a missing point even when the resulting equation looks complete.

Consider this short mathematical passage:

For each real number $u\ge0$ , define a point $R(u)$ by
$x=\frac{1-u^2}{1+u^2},\qquad y=\frac{2u}{1+u^2}.$

> The denominator is always positive. The coordinates satisfy a circle equation, but not every point of that circle is generated. For a generated point, $1+x=2/(1+u^2)>0$ and $u=y/(1+x)$. Unlike the angular formula, this definition contains no sine or cosine. That is the surprising fact behind the opening puzzle: rational expressions alone can trace a semicircle while approaching an endpoint forever without reaching it. The supplied description is not a formula-sheet result and is not required recall; use it as given.

Worked example 11: reconstruct the implicit relation

Question. For the passage's domain $u\ge0$ , determine the Cartesian curve and identify its missing endpoint.

Plan. Add the squared coordinates, then use positivity and the recovery formula to check which points occur.

\begin{aligned} x^2+y^2&=\frac{(1-u^2)^2+4u^2}{(1+u^2)^2} &&\text{substitute}\\ &=\frac{1+2u^2+u^4}{(1+u^2)^2}=1 &&\text{expand and factor}\\ y&\ge0,\quad x>-1 &&\text{use the domain and }1+x>0 \end{aligned}

Why this step? The identity proves containment in the circle; the restrictions and reverse formula establish the actual generated set.

Answer. The upper semicircle $x^2+y^2=1$ , $y\ge0$ , with $(-1,0)$ excluded. The start $(1,0)$ is included. For any other upper-semicircle point, set $u=y/(1+x)\ge0$ . The circle identity gives $u^2=(1-x)/(1+x)$ , hence $1+u^2=2/(1+x)$ . Substituting these into both passage equations recovers $x$ and $y$ , proving that every retained point occurs.

Check. $u=0$ gives $(1,0)$ , $u=1$ gives $(0,1)$ , and $u=3$ gives $(-4/5,3/5)$ . Setting $x=-1$ would imply $1=-1$ , so no finite $u$ reaches it.

Change only the domain, from $u\ge0$ to $0\le u\le1$ . The algebraic identity stays the same, while the retained arc changes.

Worked example 12: translate a new passage restriction

Question. Determine the curve produced by the passage equations when $0\le u\le1$ .

Plan. Retain the circle identity and use the sign of $1-u^2$ to locate the arc.

\begin{aligned} x&\ge0,\quad y\ge0 &&\text{non-negative numerators}\\ u=0&:\ (1,0),\qquad u=1:\ (0,1) &&\text{included boundaries} \end{aligned}

Why this step? The same circle equation can describe many arcs; the new parameter interval selects the first quadrant.

Answer. The first-quadrant quarter-circle, including both endpoints, from $(1,0)$ to $(0,1)$ .

Check. At $u=1/2$ , $(x,y)=(3/5,4/5)$ , lying on the retained arc. Conversely $u=y/(1+x)$ lies in $[0,1]$ on this arc because $0\le y\le1\le1+x$ .

The recovery formula is doing real work: it prevents a one-way algebraic identity from being mistaken for an exact description of the point set. Such reading and reconstruction are useful when an unseen MEI passage introduces its own notation.

Watch out

Wrong: $x^2+y^2=1$ , so the full circle is generated. Right: retain the domain's sign and endpoint information, then test the reverse parameter recovery.

The next question supplies the full unfamiliar definition again and changes only its parameter interval.

Quick Recap

Both coordinates use one shared parameter value.
Eliminate and retain the domain; establish the reverse direction when an exact point set is required.
Follow the given guidance when converting to parametric form.
Circle projections give angular parametrisations, with signs and intervals controlling arcs and traversal.
An unfamiliar definition can have a familiar implicit equation while omitting an endpoint.

Yes: the opening puzzle is answered by the rational circle passage. As $u$ grows, its points approach $(-1,0)$ , but no finite allowed parameter reaches that point.

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Mechanics

Mg12–Mg14 - Parametric curves and cartesian equations

One parameter, two coordinates

Elimination with the original restrictions

Converting from Cartesian form

A circle described by an angle

Sketching the path and its traversal

A rational description of a circle

Quick Recap

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