Parametric equationsEdexcel International A Level Maths: Revision notes

A curve can be described by giving $x$ and $y$ each as a function of a third variable, the parameter, usually $t$ or $\theta$: $x=f(t)$, $y=g(t)$. As $t$ varies, the point $(x,y)$ traces out the curve. The allowed values of $t$ (the domain) matter: for $x=3t-1$, $y=t^2+2$ with $t\in\mathbb{R}$ the whole parabola is drawn, but if $0\leq t\leq2$ only part of it is. The parameter can stand for time, as when a ball has $x=20t$ and $y=15t-5t^2$, or for an angle.

To find the point for a given $t$, substitute into both equations: at $t=-1$ on $x=3t-1$, $y=t^2+2$ the point is $(-4,3)$. To find where the curve meets a line, put the line's condition into the right equation and solve for $t$. For $y=11$: $t^2+2=11$, so $t=\pm3$, then $x=8$ or $x=-10$. For the $y$-axis use $x=0$, and for the $x$-axis use $y=0$. Always substitute back into both equations to get the coordinates. A line such as $y=x-4$ meeting the curve gives an equation in the parameter.

To remove the parameter, make $t$ the subject of the simpler equation and substitute into the other. From $x=3t-1$: $t=\frac{x+1}{3}$, so $y=\left(\frac{x+1}{3}\right)^2+2=\frac{(x+1)^2}{9}+2$. For $x=2t$, $y=\frac2t$: $t=\frac x2$, so $y=\frac4x$. Look for a way to combine the equations: if $x=t^2-1$ and $y=2t$ then $t=\frac y2$ and $x=\frac{y^2}{4}-1$, so $y^2=4(x+1)$. State any restriction, such as $x\neq0$ for $y=\frac4x$.

When the equations use $\sin$ and $\cos$ of the same angle, use $\sin^2\theta+\cos^2\theta=1$. For $x=3\sin t$, $y=2\cos t$: $\sin t=\frac x3$ and $\cos t=\frac y2$, so $\frac{x^2}{9}+\frac{y^2}{4}=1$, an ellipse. For $x=2+3\sin\theta$, $y=1+3\cos\theta$: $\sin\theta=\frac{x-2}{3}$, $\cos\theta=\frac{y-1}{3}$, so $(x-2)^2+(y-1)^2=9$, a circle with centre $(2,1)$ and radius 3. Other identities may help, such as $\cos 2t=1-2\sin^2t$.

To write a Cartesian curve parametrically, choose a simple parameter. For $y=x^2+1$ take $x=t$, $y=t^2+1$. For $xy=4$ take $x=2t$, $y=\frac2t$ (or $x=t$, $y=\frac4t$). For the circle $(x-1)^2+(y+2)^2=16$ take $x=1+4\cos\theta$, $y=-2+4\sin\theta$. The answer is not unique, but check that it reproduces the Cartesian equation.

Parametric equations suit motion because $x$ and $y$ depend on time. For the ball $x=20t$, $y=15t-5t^2$: it lands when $y=0$, so $5t(3-t)=0$ and $t=3$, giving $x=60$ m. The greatest height occurs midway at $t=1.5$: $y=11.25$ m. Eliminating $t$ gives the path $y=\frac{3x}{4}-\frac{x^2}{80}$. Always give answers in the units used, and remember that the domain of $t$ may restrict the part of the curve drawn.