1.03cPureStretch

Parametric Equations

Some curves cannot be written as $y = f (x)$ at all — a full circle fails immediately, because most $x$ values have two $y$ values. Parametric equations solve this by defining $x$ and $y$ separately in terms of a third variable, the $parameter$ , so the curve is described by where a point is at each value of that parameter.

35 min Coordinate Geometry

The big picture

The change of viewpoint is worth dwelling on. Cartesian form describes a curve as a relationship between coordinates; parametric form describes it as a $journey$ — first this point, then that one. That is why parametric equations are the natural language for anything moving: projectile motion in Mechanics is exactly $x$ and $y$ given separately in terms of time. It is also why they unlock curves that fail the vertical line test: a parameter can revisit the same $x$ with a different $y$ because it has moved on. Later you will differentiate parametrically to find gradients on such curves without ever eliminating the parameter.

What you'll be able to do

Interpret a pair of parametric equations and plot points from them
Convert parametric to Cartesian form by eliminating the parameter
Use trigonometric identities to eliminate a trigonometric parameter
Find intersections with the axes and with other curves in parametric form

Two equations, one curve

A parametric definition gives $x = f (t)$ and $y = g (t)$ . Each value of $t$ produces one point, and as $t$ runs through its range the point traces the curve.

Take $x = t^{2}$ , $y = 2 t$ . At $t = - 2$ the point is $(4, - 4)$ ; at $t = 0$ it is $(0, 0)$ ; at $t = 2$ it is $(4, 4)$ . Note that $t = \pm 2$ give the same $x$ but different $y$ — impossible for a function $y = f (x)$ , and completely unremarkable parametrically.

The parameter is usually $t$ or $θ$ . It often has no physical meaning at all; it is simply the variable doing the bookkeeping.

Why this works

The parameter carries information the Cartesian equation throws away — the order in which points are visited, and how fast. Two different parametrisations can trace the same curve in opposite directions.

Eliminating the parameter

To convert to Cartesian form, make $t$ the subject of the simpler equation and substitute into the other. Choose the equation where $t$ is easiest to isolate — usually the linear one.

The result is a single relationship between $x$ and $y$ . Sometimes it is a familiar curve in disguise, which is exactly the point of the exercise.

1The first equation is linear in

t

t = x - 1

2Substitute into the second:

y = (x - 1)^{2} - 3

Answer

y = (x - 1)^{2} - 3

, a parabola with vertex

(1, - 3)

Tip — State any restriction the parametrisation imposes. For $x = t^{2}$ the Cartesian curve only exists for $x \geq 0$ , because a square is never negative.

Trigonometric parameters and the Pythagorean identity

When the parameter appears inside sine and cosine, isolating it means using inverse functions and that gets messy fast. The better route is to rearrange so that $cos θ$ and $sin θ$ stand alone, then use $sin^{2} θ + cos^{2} θ = 1$ to eliminate $θ$ in one move.

This is why the circle is the standard example. $x = r cos θ$ , $y = r sin θ$ gives $(\frac{x}{r})^{2} + (\frac{y}{r})^{2} = 1$ , which rearranges to $x^{2} + y^{2} = r^{2}$ — the circle equation from the previous lesson, arrived at from a completely different direction.

sin^{2} θ + cos^{2} θ = 1

The identity that eliminates a trig parameter.

x = a cos θ, y = a sin θ \Rightarrow x^{2} + y^{2} = a^{2}

The parametric circle.

1Isolate the trig functions:

cos θ = \frac{x}{3}

and

sin θ = \frac{y}{5}

2Square both and add, so the identity applies:

\frac{x ^{2}}{9} + \frac{y ^{2}}{25} = cos^{2} θ + sin^{2} θ

3The right-hand side is 1.

Answer

\frac{x ^{2}}{9} + \frac{y ^{2}}{25} = 1

— an ellipse.

Why this works

Squaring and adding is the whole technique whenever both $sin$ and $cos$ of the same angle appear. Look for it as soon as you see a trigonometric parameter — it converts the problem to algebra immediately.

Working directly in parametric form

You do not always need the Cartesian equation. To find where a curve crosses the $x$ -axis, set $y = 0$ , solve for the parameter, and substitute back into $x$ . Crossing the $y$ -axis works the same way with $x = 0$ .

For an intersection with a line, substitute both parametric expressions into the line equation and solve for the parameter. Each solution gives one intersection point.

1Set

y = 0

t^{3} - 4 t = 0

, so

t (t^{2} - 4) = 0

2Roots

t = 0

t = 2

t = - 2

3Substitute into

x

t = 0

gives

x = - 4

;

t = \pm 2

both give

x = 0

AnswerIt crosses at

(- 4, 0)

and

(0, 0)

— the latter reached at both

t = 2

and

t = - 2

Tip — When two parameter values give the same point, the curve crosses itself there. That self-intersection is invisible in the Cartesian equation and is one of the things parametric form is genuinely better at showing.

Think like an examiner

When converting, state any restriction on $x$ or $y$ that the parametrisation forces — it is frequently the final mark.
For trigonometric parameters, show the squaring-and-adding step explicitly and name the identity used.
If asked for the Cartesian equation "in the form $y = f (x)$ ", finish the rearrangement rather than leaving an implicit relation.
Parametric questions often end by asking for a specific point — remember to substitute the parameter value back rather than stopping at $t$ .

Common misconceptions

Trying to eliminate

θ

from

x = cos θ

y = sin θ

using

θ = arccos x

Square and add instead:

x^{2} + y^{2} = cos^{2} θ + sin^{2} θ = 1

. Far cleaner and always available.

Giving

y = x^{2}

for

x = t^{2}

y = t^{4}

with no restriction.

Correct as an equation, but

x = t^{2} \geq 0

, so the curve is only the right-hand half.

Substituting a parameter value into the wrong equation.

Solve for

t

using one equation, then substitute that

t

into the other to get the matching coordinate.

Assuming each point corresponds to exactly one parameter value.

A curve can revisit a point —

t = \pm 2

may give the same

(x, y)

. That is a self-intersection, not an error.

Parametric essentials

x = f (t), y = g (t)

Both coordinates from one parameter.

sin^{2} θ + cos^{2} θ = 1

Eliminates a trig parameter.

x = a cos θ, y = a sin θ

Circle of radius

a

x = a cos θ, y = b sin θ \Rightarrow \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

Ellipse.

Stretch yourself

A curve is given by $x = t^{2} - 2 t$ , $y = t + 1$ . Find its Cartesian equation, and determine the coordinates of the point where the curve is furthest to the left.

Hint — The second equation is linear — make $t$ the subject there. For the leftmost point, think about what value minimises $x$ .

Questions students ask

Why use parametric equations at all?

They describe curves that fail the vertical line test, they encode direction and speed along a curve, and they are the natural form for motion problems where both coordinates depend on time.

Does the parameter always mean time?

No. In mechanics it usually does, but in pure coordinate geometry

t

θ

is often just a convenient variable with no physical interpretation.

Is the Cartesian equation always better?

No. It loses the direction of travel and can hide self-intersections, and for some curves eliminating the parameter produces something far messier than the parametric pair.

How do I find a gradient without converting?

Differentiate each equation with respect to the parameter and use

\frac{d y}{d x} = \frac{d y / d t}{d x / d t}

. That is parametric differentiation, covered in the calculus section.

Key takeaways

Parametric form defines $x$ and $y$ separately through a parameter, describing the curve as a journey.
It can represent curves that are not functions of $x$ , including circles and self-intersecting curves.
Eliminate the parameter by making it the subject of the simpler equation and substituting.
For trigonometric parameters, isolate $sin$ and $cos$ , then square and add.
State any restriction on $x$ or $y$ that the parametrisation imposes.

How this fits the course

Build on

Trigonometric Identities (1.05)soon

Leads to

Parametric differentiation (1.07)soon

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Parametric Equations

Two equations, one curve

Eliminating the parameter

Trigonometric parameters and the Pythagorean identity

Working directly in parametric form

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Related topics

Parametric Equations

Two equations, one curve

Eliminating the parameter

Trigonometric parameters and the Pythagorean identity

Working directly in parametric form

Practise this topic

Related topics