What is 3.3: Parametric equations?

Parametric equations express both coordinates in terms of a third variable: x=f(t) and y=g(t).

How many marks are 3.3 questions worth?

3.3 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 3.3 appeared in?

We have not yet indexed a verified past-paper appearance for 3.3 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

3.3

Understand and use the parametric equations of curves and conversion between Cartesian and parametric forms.

Draft — not yet indexed

Parametric equations

Worked answers and methods for 3.3 on Edexcel A-level Maths 9MA0.

Explanation

Parametric equations express both coordinates in terms of a third variable: $x=f(t)$ and $y=g(t)$ .
To obtain a Cartesian equation, eliminate the parameter by making $t$ the subject of one equation and substituting into the other, or by using a suitable identity.
To parametrise a Cartesian curve, choose expressions that satisfy it identically, such as $x=a\cos t$ and $y=b\sin t$ for $x^2/a^2+y^2/b^2=1$ .
Record any parameter interval because it determines which part of the Cartesian curve is traced.
A common error is to eliminate $t$ but lose a domain restriction.

Worked example

The curve $x=2t+1$ , $y=t^2-3$ is defined for $t\ge0$ . Find a Cartesian equation and state the corresponding restriction on $x$ .

1.Rearrange $x=2t+1$ to $t=(x-1)/2$ .
2.Therefore $y=((x-1)/2)^2-3=(x-1)^2/4-3$ .
3.Since $t\ge0$ , $x=2t+1\ge1$ .

Answer: $y=\dfrac{(x-1)^2}{4}-3$ , with $x\ge1$

Common mistakes

Don't square an equation to eliminate the parameter and retain Cartesian points not generated by any allowed parameter.
Don't eliminate the parameter correctly but lose the restriction inherited from its stated domain.

Exam tip

After converting to Cartesian form, translate the parameter interval into the corresponding restriction on the Cartesian variable.

Worked practice

Tier 1 · Easy

For

x=2t-1

and

y=t+4

, eliminate

t

to obtain a Cartesian equation.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$y=\dfrac{x}{2}+\dfrac{9}{2}$	2
Notes
From $x=2t-1$ , $t=(x+1)/2$ . Substitute into $y=t+4$ to get $y=(x+1)/2+4=x/2+9/2$ .

(2 marks)

Tier 2 · Standard

The curve

y=(x-1)^2+2

is parameterised so that

x=2t+1

. Express

y

in terms of

t

and find the parameter values where the curve meets the line

y=6

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$y=4t^2+2$ $t=-1$ or $t=1$	4
Notes
Substitute $x=2t+1$ into the Cartesian equation: $y=((2t+1)-1)^2+2=4t^2+2$ . At an intersection with $y=6$ , $4t^2+2=6$ , so $t^2=1$ and therefore $t=-1$ or $t=1$ .

(4 marks)

Tier 3 · Hard

The curve

x=t+1

y=t^2-2t

meets the line

y=3x-6

at two points. Find the exact coordinates of both points.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$\left(\dfrac{7+\sqrt{13}}{2},\dfrac{9+3\sqrt{13}}{2}\right)$ $\left(\dfrac{7-\sqrt{13}}{2},\dfrac{9-3\sqrt{13}}{2}\right)$	5
Notes
At an intersection, $t^2-2t=3(t+1)-6$ , so $t^2-5t+3=0$ . Hence $t=(5\pm\sqrt{13})/2$ . Using $x=t+1$ gives $x=(7\pm\sqrt{13})/2$ , and then $y=3x-6=(9\pm3\sqrt{13})/2$ , with matching signs.

(5 marks)

Tier 1 · Easy

The curve has parametric equations

x=5\cos t

and

y=2\sin t

, where

0\leq t<2\pi

. Eliminate

t

to find a Cartesian equation of the curve.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\dfrac{x^2}{25}+\dfrac{y^2}{4}=1$	2
Notes
From the parametric equations, $\cos t=x/5$ and $\sin t=y/2$ . Using $\cos^2t+\sin^2t=1$ gives $x^2/25+y^2/4=1$ .

(2 marks)

Tier 2 · Standard

A curve

C

is defined by

x=t^2+2

and

y=3t-1

for

t\geq0

. Obtain a Cartesian equation with its restriction. Hence determine exactly where

C

meets

x=6

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$x=\dfrac{(y+1)^2}{9}+2$ , with $y\geq-1$ $(6,5)$	4
Notes
From $y=3t-1$ , $t=(y+1)/3$ . Substitution into $x=t^2+2$ gives $x=(y+1)^2/9+2$ . Since $t\geq0$ , the restriction is $y\geq-1$ . When $x=6$ , $t^2+2=6$ , so the permitted value is $t=2$ , giving $y=5$ .

(4 marks)

Tier 3 · Hard

The Cartesian curve

C

has equation

y^2=4x+8

. By setting

y=2t

, obtain parametric equations for

C

. The line

y=x+1

meets

C

at two points. Find the corresponding exact values of

t

and the exact coordinates of both points.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$x=t^2-2$ , $y=2t$ $t=1+\sqrt2$ gives $(1+2\sqrt2,2+2\sqrt2)$ $t=1-\sqrt2$ gives $(1-2\sqrt2,2-2\sqrt2)$	5
Notes
Putting $y=2t$ into $y^2=4x+8$ gives $4t^2=4x+8$ , so $x=t^2-2$ . At an intersection with $y=x+1$ , $2t=t^2-1$ , hence $t^2-2t-1=0$ and $t=1\pm\sqrt2$ . Since $x=y-1=2t-1$ , matching the signs gives the two stated points.

(5 marks)

Tier 2 · Standard

The curve

C

has parametric equations

x=3e^t

and

y=2e^{-t}

, where

t

is real. Find a Cartesian equation for

C

in the form

xy=k

and state the restrictions on

x

and

y

. Find the exact point on

C

for which

t=\ln 4

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$xy=6$ , with $x>0$ and $y>0$ $(12,\dfrac12)$	4
Notes
Multiplying the parametric equations gives $xy=(3e^t)(2e^{-t})=6$ . Since exponentials are positive for every real $t$ , the traced branch has $x>0$ and $y>0$ . When $t=\ln4$ , $e^t=4$ and $e^{-t}=1/4$ , so the point is $(12,1/2)$ .

(4 marks)

Tier 3 · Hard

For

-2\leq t\leq2

, a point

P

has coordinates

x=t^2-1

and

y=t^3-t

. Show that its path satisfies

y^2=x^2(x+1)

. State the range of

x

and identify all parameter values that produce the self-intersection

(0,0)

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$y^2=x^2(x+1)$ $-1\leq x\leq3$ $t=-1$ and $t=1$	5
Notes
Since $y=t(t^2-1)=tx$ , squaring gives $y^2=t^2x^2$ . Also $t^2=x+1$ , so $y^2=x^2(x+1)$ . Over $-2\leq t\leq2$ , $0\leq t^2\leq4$ , hence $-1\leq x\leq3$ . The point $(0,0)$ requires $x=t^2-1=0$ , so $t=-1$ or $t=1$ ; both values also give $y=0$ .

(5 marks)

Tier 3 · Hard

A curve is defined for real

t

x=\dfrac{1-t^2}{1+t^2}

and

y=\dfrac{2t}{1+t^2}

. Show that every point generated lies on the circle

x^2+y^2=1

. Identify the point on this circle that is not generated by any real value of

t

. Find all values of

t

for which the generated point lies on

y=x

, and give the corresponding exact coordinates.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$x^2+y^2=1$ $(-1,0)$ is not generated $t=-1+\sqrt2$ gives $\left(\dfrac{\sqrt2}{2},\dfrac{\sqrt2}{2}\right)$ $t=-1-\sqrt2$ gives $\left(-\dfrac{\sqrt2}{2},-\dfrac{\sqrt2}{2}\right)$	6
Notes
Substitution gives $x^2+y^2=[(1-t^2)^2+4t^2]/(1+t^2)^2=(1+t^2)^2/(1+t^2)^2=1$ . The point $(-1,0)$ would require $(1-t^2)/(1+t^2)=-1$ , which leads to $1=-1$ , so no real $t$ generates it. On $y=x$ , $2t=1-t^2$ , hence $t^2+2t-1=0$ and $t=-1\pm\sqrt2$ . The line and circle give $2x^2=1$ ; matching the sign obtained from each parameter gives the two stated points.

(6 marks)

Q10

Tier 3 · Hard

10.

For

t>0

, a point

P

is defined by

x=t+\dfrac1t

and

y=t-\dfrac1t

. Obtain a Cartesian equation for the path of

P

and state the restriction on

x

. Prove that the least possible value of

x

2

and give the point where it occurs. Find also the exact point where the path meets the line

x+y=6

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$x^2-y^2=4$ , with $x\geq2$ Least $x=2$ at $(2,0)$ Intersection $\left(\dfrac{10}{3},\dfrac{8}{3}\right)$	5
Notes
Since $x+y=2t$ and $x-y=2/t$ , multiplication gives $x^2-y^2=4$ . For $t>0$ , $(t-1)^2\geq0$ gives $t+1/t\geq2$ , so $x\geq2$ , with equality only at $t=1$ . This gives the point $(2,0)$ . At the intersection with $x+y=6$ , the relation $x+y=2t$ gives $t=3$ . Therefore $x=3+1/3=10/3$ and $y=3-1/3=8/3$ .

(5 marks)

Verified exam appearances

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Other points in 3 Coordinate geometry in the (x, y) plane

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