Parametric and Cartesian Forms – A Level Maths Year 2

1 What Parametric Equations Are

A Cartesian equation links $x$ and $y$ directly, like $y = x^2$. Parametric equations give $x$ and $y$ separately in terms of a third variable, the parameter (usually $t$ or $\theta$): $\;x = \mathrm{f}(t),\; y = \mathrm{g}(t)$. Each value of $t$ gives one point $(x, y)$.

Worked Example 1 — Plotting from a table

Sketch the curve $x = t^2$, $y = 2t$ for $-3 \leqslant t \leqslant 3$.

①Make a table of values.

$t$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$x = t^2$	$9$	$4$	$1$	$0$	$1$	$4$	$9$
$y = 2t$	$-6$	$-4$	$-2$	$0$	$2$	$4$	$6$

②Plot the points $(x, y)$ and join them in order of $t$.

③The curve is a parabola lying on its side. As $t$ increases, the point moves from bottom right, through the origin, to top right.

The parameter carries extra information. The Cartesian equation tells you the shape; the parametric form also tells you where on the curve you are for each $t$, and in which direction the point moves. That is why it is the natural language of motion.

2 Converting to Cartesian Form

Eliminate the parameter

Make $t$ the subject of the simpler equation, then substitute into the other

Worked Example 2 — Substitution

Find a Cartesian equation for $x = 2t + 1$, $y = t^2 - 3$.

①From the first: $t = \dfrac{x - 1}{2}$.

②Substitute: $y = \left(\dfrac{x - 1}{2}\right)^2 - 3 = \dfrac{(x - 1)^2}{4} - 3$.

③A parabola with vertex at $(1, -3)$ (where $t = 0$).

Worked Example 3 — A reciprocal parameter

Find a Cartesian equation for $x = \dfrac{1}{t}$, $y = t + 2$, where $t \neq 0$.

①$t = \dfrac{1}{x}$, so $y = \dfrac{1}{x} + 2$.

②Since $x = \tfrac1t$ can never be $0$, the curve exists for $x \neq 0$ — exactly as the Cartesian equation requires.

When $x$ and $y$ involve trigonometric functions, you usually cannot make $\theta$ the subject neatly. Instead, isolate the trig functions and use an identity.

Identities that eliminate $\theta$

$\sin^2\theta + \cos^2\theta \equiv 1$ · $1 + \tan^2\theta \equiv \sec^2\theta$ · $\cos 2\theta \equiv 1 - 2\sin^2\theta \equiv 2\cos^2\theta - 1$

Worked Example 4 — A circle

Show that $x = 2 + 3\cos\theta$, $y = -1 + 3\sin\theta$, $0 \leqslant \theta < 2\pi$, is a circle, and state its centre and radius.

①Isolate: $\cos\theta = \dfrac{x - 2}{3}$ and $\sin\theta = \dfrac{y + 1}{3}$.

②Use $\cos^2\theta + \sin^2\theta = 1$: $\dfrac{(x - 2)^2}{9} + \dfrac{(y + 1)^2}{9} = 1$.

③$(x - 2)^2 + (y + 1)^2 = 9$: centre $(2, -1)$, radius $3$.

In general $x = a + r\cos\theta$, $y = b + r\sin\theta$ is the circle with centre $(a, b)$ and radius $r$, traced anticlockwise once as $\theta$ goes from $0$ to $2\pi$.

Worked Example 5 — An ellipse

Find a Cartesian equation for $x = 4\cos\theta$, $y = 2\sin\theta$.

①$\cos\theta = \dfrac{x}{4}$, $\;\sin\theta = \dfrac{y}{2}$.

②$\dfrac{x^2}{16} + \dfrac{y^2}{4} = 1$ — an ellipse ("stretched circle") crossing the axes at $(\pm 4, 0)$ and $(0, \pm 2)$.

Worked Example 6 — Using a double-angle identity

Find a Cartesian equation for $x = \sin\theta$, $y = \cos 2\theta$, and state the possible values of $x$.

①$y = \cos 2\theta = 1 - 2\sin^2\theta$.

②Substitute $\sin\theta = x$: $y = 1 - 2x^2$.

③$x = \sin\theta$ can only take values $-1 \leqslant x \leqslant 1$, so the curve is only the part of the parabola $y = 1 - 2x^2$ with $-1 \leqslant x \leqslant 1$.

The Cartesian equation on its own describes the whole parabola. The parametric curve is only a piece of it. Always state the restriction.

Worked Example 7 — Using $1 + \tan^2\theta \equiv \sec^2\theta$

Find a Cartesian equation for $x = \sec\theta$, $y = \tan\theta$.

①$\sec^2\theta - \tan^2\theta = 1$.

②So $x^2 - y^2 = 1$.

3 Domain and Range of a Parametric Curve

The values of $t$ allowed determine which $x$ values occur (the domain of the Cartesian function) and which $y$ values occur (the range). Find them from the parametric equations, not from the Cartesian equation.

Worked Example 8 — Restricted parameter

A curve has equations $x = 3 + 2t$, $y = t^2$, $-1 \leqslant t \leqslant 2$. Find its Cartesian equation, and the domain and range.

①$t = \dfrac{x - 3}{2}$, so $y = \dfrac{(x - 3)^2}{4}$.

②$x = 3 + 2t$ increases with $t$: from $x = 1$ to $x = 7$. Domain $1 \leqslant x \leqslant 7$.

③$y = t^2$ for $-1 \leqslant t \leqslant 2$: smallest value $0$ (at $t = 0$), largest $4$ (at $t = 2$). Range $0 \leqslant y \leqslant 4$.

Substituting only the end values $t = -1$ and $t = 2$ would give $1 \leqslant y \leqslant 4$, which is wrong — $t = 0$ is inside the interval and gives $y = 0$.

Worked Example 9 — Half an ellipse

Describe the curve $x = 2\cos t$, $y = \sin t$, $0 \leqslant t \leqslant \pi$.

①$\dfrac{x^2}{4} + y^2 = 1$, an ellipse.

②For $0 \leqslant t \leqslant \pi$, $\sin t \geqslant 0$, so only the upper half: $-2 \leqslant x \leqslant 2$, $0 \leqslant y \leqslant 1$.

③It starts at $(2, 0)$ when $t = 0$, passes through $(0, 1)$ and ends at $(-2, 0)$ when $t = \pi$.

4 Intersections with the Axes and with Lines

$x$-axis: solve $y(t) = 0$ for $t$, then find $x$.
$y$-axis: solve $x(t) = 0$ for $t$, then find $y$.
A line or curve with a Cartesian equation: substitute $x(t)$ and $y(t)$ into it, solve for $t$, then find the points.

Worked Example 10 — A curve that crosses itself

A curve has equations $x = t^2 - 4$, $y = t^3 - t$. Find where it meets the coordinate axes.

①$x$-axis: $t^3 - t = t(t - 1)(t + 1) = 0$, so $t = 0, 1, -1$.

②$t = 0$ gives $(-4, 0)$. Both $t = 1$ and $t = -1$ give $(-3, 0)$ — the curve passes through this point twice, so it crosses itself there.

③$y$-axis: $t^2 - 4 = 0$, so $t = \pm 2$, giving $y = 8 - 2 = 6$ and $y = -8 + 2 = -6$: the points $(0, 6)$ and $(0, -6)$.

Worked Example 11 — Meeting a line

Find where the curve $x = t^2$, $y = 2t$ meets the line $y = x - 3$.

①Substitute: $2t = t^2 - 3 \Rightarrow t^2 - 2t - 3 = 0 \Rightarrow (t - 3)(t + 1) = 0$.

②$t = 3$: $(9, 6)$. $t = -1$: $(1, -2)$.

③Check: $6 = 9 - 3$ ✓ and $-2 = 1 - 3$ ✓.

A repeated root for $t$ means the line is a tangent to the curve. Try it with $y = x + 1$: $2t = t^2 + 1$ gives $(t - 1)^2 = 0$.

5 Quick Reference

Parametric

$x = \mathrm{f}(t)$, $y = \mathrm{g}(t)$.

Eliminate $t$

Make $t$ the subject, substitute.

Trig parameter

Isolate $\sin$ and $\cos$, use $\sin^2 + \cos^2 = 1$.

Circle

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

Ellipse

$x = p\cos\theta$, $y = q\sin\theta$: $\dfrac{x^2}{p^2} + \dfrac{y^2}{q^2} = 1$.

Domain / range

From the $t$ values; watch for turning values inside.

Axes

$y(t) = 0$ or $x(t) = 0$, solve for $t$.

Line

Substitute $x(t)$, $y(t)$ into it.

Self-crossing

Two values of $t$ give the same point.

6 Practice Questions

Question 1

A curve has equations $x = t + 1$, $y = t^2$. Find the points with $t = -2, 0, 2$, and a Cartesian equation.

▶ Show solution

$t = -2$: $(-1, 4)$. $t = 0$: $(1, 0)$. $t = 2$: $(3, 4)$.

$t = x - 1$, so $y = (x - 1)^2$.

Question 2

Find a Cartesian equation for $x = 3t$, $y = \dfrac{6}{t}$, $t \neq 0$.

▶ Show solution

$t = \dfrac{x}{3}$, so $y = \dfrac{6}{x/3} = \dfrac{18}{x}$, i.e. $xy = 18$.

(Quicker: multiply the two equations, $xy = 3t \times \tfrac6t = 18$.)

Question 3

Find a Cartesian equation for $x = 5\cos\theta$, $y = 5\sin\theta$, and describe the curve.

▶ Show solution

$x^2 + y^2 = 25\cos^2\theta + 25\sin^2\theta = 25$.

A circle, centre the origin, radius $5$.

Question 4

Show that $x = 1 + 2\cos\theta$, $y = 3 + 2\sin\theta$ is a circle and find its centre and radius.

▶ Show solution

$\cos\theta = \dfrac{x - 1}{2}$, $\sin\theta = \dfrac{y - 3}{2}$.

$\left(\dfrac{x - 1}{2}\right)^2 + \left(\dfrac{y - 3}{2}\right)^2 = 1 \Rightarrow (x - 1)^2 + (y - 3)^2 = 4$.

Centre $(1, 3)$, radius $2$.

Question 5

Find a Cartesian equation for $x = 3\cos\theta$, $y = 2\sin\theta$, and state where the curve crosses the axes.

▶ Show solution

$\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$, an ellipse.

Crosses the $x$-axis at $(\pm 3, 0)$ and the $y$-axis at $(0, \pm 2)$.

Question 6

A curve has equations $x = \cos\theta$, $y = \cos 2\theta$, $0 \leqslant \theta \leqslant \pi$. Find a Cartesian equation and state the domain and range.

▶ Show solution

$\cos 2\theta = 2\cos^2\theta - 1$, so $y = 2x^2 - 1$.

For $0 \leqslant \theta \leqslant \pi$, $\cos\theta$ takes every value from $1$ down to $-1$: domain $-1 \leqslant x \leqslant 1$.

Then $y = 2x^2 - 1$ ranges from $-1$ (at $x = 0$) to $1$ (at $x = \pm 1$): range $-1 \leqslant y \leqslant 1$.

Question 7

A curve has equations $x = t^2$, $y = t^3 - 3t$. Find the points where it meets the $x$-axis, and show that it crosses itself.

▶ Show solution

$y = 0$: $t(t^2 - 3) = 0$, so $t = 0$ or $t = \pm\sqrt{3}$.

$t = 0$ gives $(0, 0)$. Both $t = \sqrt3$ and $t = -\sqrt3$ give $x = 3$, $y = 0$.

Two different values of $t$ give the same point $(3, 0)$, so the curve crosses itself there.

Question 8

The curve $x = 2t$, $y = t^2 + 1$ meets the line $y = 2x - 2$. (a) Find the points of intersection. (b) Show that the line $y = x$ is a tangent to the curve, and find the point of contact.

▶ Show solution

(a) $t^2 + 1 = 4t - 2 \Rightarrow t^2 - 4t + 3 = 0 \Rightarrow (t - 1)(t - 3) = 0$.

$t = 1$: $(2, 2)$. $t = 3$: $(6, 10)$.

(b) $t^2 + 1 = 2t \Rightarrow (t - 1)^2 = 0$: a repeated root, so the line touches the curve. Point of contact $t = 1$: $(2, 2)$.

Question 9

A curve has equations $x = 1 + \tan\theta$, $y = \sec^2\theta$, $-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}$. Find a Cartesian equation in the form $y = \mathrm{f}(x)$ and state the range of $\mathrm{f}$.

▶ Show solution

$\tan\theta = x - 1$ and $\sec^2\theta = 1 + \tan^2\theta$.

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

On this interval $\tan\theta$ takes every real value, so $x$ can be any real number. The range is $y \geqslant 1$ (minimum when $\theta = 0$, at $(1, 1)$).

Question 10

A figure-of-eight curve has equations $x = 4\sin t$, $y = 2\sin 2t$, $0 \leqslant t < 2\pi$.

(a) Find the points where the curve meets the $x$-axis, and the values of $t$ for each. (b) Show that a Cartesian equation is $16y^2 = x^2(16 - x^2)$. (c) Find the greatest value of $y$ and the corresponding exact value of $x$. (d) State the possible values of $x$.

▶ Show solution

(a) $\sin 2t = 0 \Rightarrow 2t = 0, \pi, 2\pi, 3\pi \Rightarrow t = 0, \tfrac{\pi}{2}, \pi, \tfrac{3\pi}{2}$.

These give $x = 0, 4, 0, -4$. So the points are $(0, 0)$ (when $t = 0$ and $t = \pi$ — the crossing point), $(4, 0)$ and $(-4, 0)$.

(b) $y = 2\sin 2t = 4\sin t\cos t = x\cos t$, so $y^2 = x^2\cos^2 t = x^2(1 - \sin^2 t)$.

$\sin t = \dfrac{x}{4}$, so $y^2 = x^2\left(1 - \dfrac{x^2}{16}\right)$, and multiplying by $16$: $16y^2 = x^2(16 - x^2)$. $\blacksquare$

(c) $y = 2\sin 2t$ has greatest value $2$, when $2t = \tfrac{\pi}{2}$, i.e. $t = \tfrac{\pi}{4}$. Then $x = 4\sin\tfrac{\pi}{4} = 2\sqrt2$.

(Check in (b): $16(4) = 8(16 - 8) = 64$ ✓.)

(d) $x = 4\sin t$, so $-4 \leqslant x \leqslant 4$.

Parametric and Cartesian Forms · A Level Maths Year 2 · Created with MathJax