3.4 Parametric equations and Cartesian conversion — Practise with help

Work each one first, closed book. Before you eliminate the parameter, decide which equation is easier to rearrange, and whether a trigonometric identity will do the job instead.

Warm-up: remember

Rearrange $x = 3t + 1$ to make $t$ the subject.
Expand $(2t - 1)^2$ .
Write the equation of the circle with centre $(0, 0)$ and radius $7$ .

Answers

Subtract $1$ , then divide by $3$ : $t = \dfrac{x - 1}{3}$ .
$(2t - 1)^2 = 4t^2 - 4t + 1$ .
$x^2 + y^2 = 49$ .

If question 1 was hard, revisit rearranging a formula before you continue.

Practise with help

1. A curve has parametric equations $x = 3t$ , $y = t^2 + 1$ . Find the point where $t = 2$ .

Hint

Substitute $t = 2$ into each equation separately.

Next step

$x = 3 \times 2 = 6$ and $y = 2^2 + 1 = 5$ .

Full working

x = 3 \times 2 = 6, \qquad y = 2^2 + 1 = 5.

The point is $(6, 5)$ .

2. Find the Cartesian equation of $x = 3t$ , $y = t^2 + 1$ .

Hint

The first equation is the simpler one. Make $t$ its subject, then substitute into the second.

Next step

From $x = 3t$ we get $t = \dfrac{x}{3}$ . Substitute that into $y = t^2 + 1$ .

Full working

y = \left(\frac{x}{3}\right)^2 + 1 = \frac{x^2}{9} + 1.

The curve is a parabola. Check with the point from question 1: when $x = 6$ , $y = \frac{36}{9} + 1 = 5$ , which matches.

3. Find the Cartesian equation of $x = t - 4$ , $y = 2t + 1$ .

Hint

Make $t$ the subject of the first equation. It needs one addition.

Next step

From $x = t - 4$ we get $t = x + 4$ . Substitute into $y = 2t + 1$ .

Full working

y = 2(x + 4) + 1 = 2x + 8 + 1 = 2x + 9.

The curve is a straight line with gradient $2$ .

4. Find the Cartesian equation of $x = 4\cos t$ , $y = 4\sin t$ .

Hint

Make $\cos t$ and $\sin t$ the subjects, square both equations, and add them.

Next step

$\cos t = \dfrac{x}{4}$ and $\sin t = \dfrac{y}{4}$ . Use $\cos^2 t + \sin^2 t = 1$ .

Full working

\left(\frac{x}{4}\right)^2 + \left(\frac{y}{4}\right)^2 = 1.

Multiply by $16$ :

x^2 + y^2 = 16.

The curve is the circle of radius $4$ centred on the origin.

5. Find the Cartesian equation of $x = 1 + 3\cos t$ , $y = -2 + 3\sin t$ .

Hint

Rearrange each equation so that the trigonometric function stands alone before you square.

Next step

$\cos t = \dfrac{x - 1}{3}$ and $\sin t = \dfrac{y + 2}{3}$ . Square and add.

Full working

\left(\frac{x - 1}{3}\right)^2 + \left(\frac{y + 2}{3}\right)^2 = 1,

(x - 1)^2 + (y + 2)^2 = 9.

The curve is the circle with centre $(1, -2)$ and radius $3$ .

6. Write the curve $y = x^2 - 3$ using a parameter.

Hint

Choose the simplest possible parameter. Let the parameter be $x$ itself.

Next step

Set $x = t$ . Then read $y$ from the equation of the curve.

Full working

Let $x = t$ . Then

x = t, \qquad y = t^2 - 3.

This parameterisation covers the whole curve, because $t$ runs over every real number. A curve has many parameterisations, and this is the simplest one.

7. A curve is given by $x = t + 2$ , $y = t^2$ , for $0 \le t \le 2$ . Find its Cartesian equation and the range of $x$ it covers.

Hint

Eliminate $t$ in the usual way. Then use the range of $t$ to find the range of $x$ .

Next step

$t = x - 2$ , so $y = (x - 2)^2$ . Now substitute $t = 0$ and $t = 2$ into $x = t + 2$ .

Full working

y = (x - 2)^2.

The parameter runs from $0$ to $2$ , so $x = t + 2$ runs from $2$ to $4$ . The curve is the part of the parabola with

2 \le x \le 4.

The whole parabola $y = (x - 2)^2$ is longer than this, so the range of $t$ is part of the answer.