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Parametric equations

Points of intersection of parametric curves

Points of intersection of parametric curves

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Summary

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 xxx and yyy axes. It is useful to be able to find the value of ttt 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 ttt 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:

x=t2−t−6y=t+1\begin{aligned}x &=t^2-t-6 \\y&=t+1\end{aligned}xy​=t2−t−6=t+1​​​


Find the points of intersection with the coordinate axes.


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

​y=t+10=t+1t=−1\begin{aligned}y &=t+1 \\0 &= t+1 \\t&= -1\end{aligned}y0t​=t+1=t+1=−1​​​


Substitute into the xxx equation:

​x=t2−6t−6x=(−1)2−6(−1)−6x=−4\begin{aligned}x &= t^2 - 6t - 6 \\x&= (-1)^2 - 6(-1) -6 \\x &=-4\end{aligned}xxx​=t2−6t−6=(−1)2−6(−1)−6=−4​​​


Therefore, the curve crosses the xxx-axis at (−4,0)‾\underline{(-4,0)}(−4,0)​.


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

x=t2−t−60=t2−t−60=(t−3)(t+2)t=3t=−2\begin{aligned}x &= t^2 - t-6 \\0 &= t^2-t-6 \\0&=(t-3)(t+2) \\&t=3 \quad t=-2\end{aligned}x00​=t2−t−6=t2−t−6=(t−3)(t+2)t=3t=−2​​​


Substitute into the yyy equation:

​y=t+1y=3+1=4\begin{aligned}y &=t+1 \\ y&=3+1 =4 \\\end{aligned}yy​=t+1=3+1=4​​​
​y=t+1y=−2+1=−1\begin{aligned}y &=t+1 \\ y&=-2+1 =-1 \\\end{aligned}yy​=t+1=−2+1=−1​​​


Therefore, the curve crosses the yyy-axes at (0,4)‾\underline{(0,4)}(0,4)​ and (0,−1)‾\underline{(0,-1)}(0,−1)​.


Example 2

A parametric curve is defined by the following equations for 0<t<π0 \lt t \lt \pi0<t<π​:

x=t2−2y=2cos⁡t\begin{aligned}x&=t^2 - 2 \\y &= 2 \cos t\end{aligned}xy​=t2−2=2cost​​​


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


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

2cos⁡t=0cos⁡t=0t=π2\begin{aligned}2 \cos t &=0 \\ \cos t &=0 \\ t &= \frac {\pi}{2}\end{aligned}2costcostt​=0=0=2π​​​​


Substitute t=π2t=\dfrac{\pi}{2}t=2π​ to find xxx:

x=t2−2x=(π2)2−2x=π24−2\begin{aligned}x &=t^2-2 \\x&= \Big(\dfrac{\pi}{2} \Big)^2 - 2 \\\\x&= \dfrac{\pi^2}{4} - 2 \\\end{aligned}xxx​=t2−2=(2π​)2−2=4π2​−2​​​


Therefore, the curve crosses the xxx-axis at (π24−2,0)‾\underline{\Big( \dfrac{\pi^2}{4}-2, 0 \Big)}(4π2​−2,0)​.


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

t2−2=0t=2\begin{aligned}t^2-2 &=0 \\ t &= \sqrt2\end{aligned}t2−2t​=0=2​​​​


Substitute t=2t = \sqrt2t=2​ to find yyy:

y=2cos⁡ty=2cos⁡(2)\begin{aligned}y &= 2 \cos t \\y&= 2 \cos (\sqrt2)\end{aligned}yy​=2cost=2cos(2​)​​​


Therefore, the curve crosses the yyy-axis at (0,2cos⁡(2))‾\underline{\Big( 0, 2 \cos (\sqrt2) \Big)}(0,2cos(2​))​​



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FAQs - Frequently Asked Questions

How do you find the points of intersection with the coordinate axes for a parametric curve?

You can find the coordinates of the points of intersection by either making x or y equal to 0 and solving to find t. Once you have found t, substitute into the equations to find the coordinates.

How to do you deal with coordinate geometry problems with parametric curves?

Start by find the points of intersection of the curve with the coordinate axes. This will give some idea of where the curve lies, without having to find the Cartesian equation.

How do you find where a parametric curve crosses the y-axis?

Substitute x=0 into the x equation and solve for t. Then substitute this value of t into the y equation to find the y-intercept.

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