Points of intersection of parametric curves

In a nutshell

You need to know how to deal with coordinate geometry problems involving parametric curves. In particular, you should know how to find the coordinates of the points where the curve crosses the $x$ and $y$ axes. It is useful to be able to find the value of $t$ at the points of intersection.

Points of intersection

It is possible to have some idea of what a parametric curve looks like, without having to find the Cartesian equation. In particular, finding the value of $t$ for a particular point can help to find the points of intersection with the coordinate axes.

Example 1

A parametric curve is defined by the following equations:

$\begin{aligned}x &=t^2-t-6 \\y&=t+1\end{aligned}$ 

Find the points of intersection with the coordinate axes.

To find where the curve crosses the $x$ -axis, substitute $y=0$ :

$\begin{aligned}y &=t+1 \\0 &= t+1 \\t&= -1\end{aligned}$

Substitute into the $x$ equation:

$\begin{aligned}x &= t^2 - 6t - 6 \\x&= (-1)^2 - 6(-1) -6 \\x &=-4\end{aligned}$

Therefore, the curve crosses the $x$ -axis at $\underline{(-4,0)}$ .

To find where the curve crosses the $y$ -axis, substitute $x=0$ :

$\begin{aligned}x &= t^2 - t-6 \\0 &= t^2-t-6 \\0&=(t-3)(t+2) \\&t=3 \quad t=-2\end{aligned}$ 

Substitute into the $y$ equation:

\begin{aligned}y &=t+1 \\ y&=3+1 =4 \\\end{aligned}

\begin{aligned}y &=t+1 \\ y&=-2+1 =-1 \\\end{aligned}

Therefore, the curve crosses the $y$ -axes at $\underline{(0,4)}$ and $\underline{(0,-1)}$ .

Example 2

A parametric curve is defined by the following equations for $0 \lt t \lt \pi$ :

$\begin{aligned}x&=t^2 - 2 \\y &= 2 \cos t\end{aligned}$ 

Find the exact coordinates of the points of intersection of the curve with the coordinate axes.

To find where the curve crosses the $x$ -axis, substitute $y=0$ :

$\begin{aligned}2 \cos t &=0 \\ \cos t &=0 \\ t &= \frac {\pi}{2}\end{aligned}$ 

Substitute $t=\dfrac{\pi}{2}$ to find $x$ :

$\begin{aligned}x &=t^2-2 \\x&= \Big(\dfrac{\pi}{2} \Big)^2 - 2 \\\\x&= \dfrac{\pi^2}{4} - 2 \\\end{aligned}$ 

Therefore, the curve crosses the $x$ -axis at $\underline{\Big( \dfrac{\pi^2}{4}-2, 0 \Big)}$ .

To find where the curve crosses the $y$ -axis, substitute $x=0$ :

$\begin{aligned}t^2-2 &=0 \\ t &= \sqrt2\end{aligned}$ 

Substitute $t = \sqrt2$ to find $y$ :

$\begin{aligned}y &= 2 \cos t \\y&= 2 \cos (\sqrt2)\end{aligned}$ 

Therefore, the curve crosses the $y$ -axis at $\underline{\Big( 0, 2 \cos (\sqrt2) \Big)}$ 