Pure · Coordinate geometry

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Converting to Cartesian form

Eliminating the parameter to find the Cartesian equation of a parametric curve, by substitution or rearrangement, and carrying the domain restriction from the parameter through to the Cartesian equation.

Pure · Coordinate geometry

Converting to Cartesian form

Why it works

A parametric pair

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

and a single Cartesian equation in

x

and

y

can describe the same curve. Converting from one to the other means getting rid of the parameter — eliminating $t$ — so that only

x

and

y

are left.

The standard method: make $t$ the subject, then substitute. Rearrange the simpler of the two equations to get

t = \ldots

in terms of

x

(or

y

), and put that into the other equation. If

x = t + 1, \qquad y = t^2,

then

t = x - 1

, and substituting into

y = t^2

gives

y = (x-1)^2

. Done — a Cartesian equation with no

t

in sight.

When $t$ is awkward to isolate, look for a combination that cancels it. For

x = 2/t,\ y = t + 1

you can still write

t = 2/x

and substitute to get

y = \tfrac{2}{x} + 1

. Sometimes adding, subtracting or multiplying the two equations is cleaner than isolating

t

— and for trigonometric parametrics you use an identity instead of algebra (next concept).

Carry the restriction across — this is where marks are lost. The Cartesian equation on its own may describe more of a curve than the parametric form does, because the domain of

t

limits which points are actually reached. You must state the resulting restriction on

x

(or

y

$x = \sqrt{t},\ y = t - 4$ with $t \ge 0$ : here $t = x^2$ , so $y = x^2 - 4$ — but

x = \sqrt{t} \ge 0

, so the curve is only the part with

x \ge 0

, half of the parabola.

$x = t + 1,\ y = \tfrac1t$ (with $t \ne 0$ ): $y = \dfrac{1}{x - 1}$ , valid for

x \ne 1

(since

t = 0

is excluded,

x

never equals

1

).

Always finish by asking: as $t$ ranges over its domain, what values can $x$ take? That set is the domain of the Cartesian equation, and quoting it is part of the answer.