# Finding the Cartesian equation for a set of parametric equations

• December 5th 2012, 12:23 PM
tangent
Finding the Cartesian equation for a set of parametric equations
I'm trying to solve this problem, but I am having trouble. Can someone give me some pointers?

A curve is defined by the parametric equations:

x = 2t / (1 + t^2)
y = (1 - t^2) / (1 + t^2)

By considering the expression x^2 + y^2, eliminate the parameter t to obtain the Cartesian (x-y) equation of the curve.

Thanks
• December 5th 2012, 01:53 PM
skeeter
Re: Finding the Cartesian equation for a set of parametric equations
so ... what did you determine for the expression $x^2+y^2$ in terms of t ?
• December 5th 2012, 02:18 PM
tangent
Re: Finding the Cartesian equation for a set of parametric equations
This is the equation I formed: x^2 + y^2 = (4t^2 - t^4 + 1) / (1 + t^4). I don't know how to eliminate t from this point.
• December 5th 2012, 02:25 PM
skeeter
Re: Finding the Cartesian equation for a set of parametric equations
first off, note that ...

$(1+t^2)^2 \ne 1 + t^4$

$x^2 = \frac{4t^2}{(1 + t^2)^2}$

$y^2 = \frac{1 - 2t^2 + t^4}{(1 + t^2)^2}$

try adding $x^2+y^2$ again ...
• December 5th 2012, 02:39 PM
tangent
Re: Finding the Cartesian equation for a set of parametric equations
Oops, stupid error! Ok, got it now. The ratio cancels out and I get x2 + y2 = 1. Many thanks.
• December 5th 2012, 05:34 PM
Soroban
Re: Finding the Cartesian equation for a set of parametric equations
Hello, everyone!

This problem (and solution) involves a paradox.

Paramtric equations: . $\begin{Bmatrix}x &=& \dfrac{2t}{1+t^2} & [1] \\ \\[-4mm] y &=& \dfrac{1-t^2}{1+t^2} & [2] \end{Bmatrix}$

Cartesian equation: . $x^2 + y^2 \:=\:1\;\;[3]$

From [3], we have a circle with center at the Origin and radius 1.
. . It has $x$-intercepts $(\pm1,\,0)$ and $y$-intercepts $(0,\,\pm1)$

Let's determine the intercepts using the parametric equations.

For $x$-intercepts, let $y = 0\!: \;\;\frac{1-t^1}{1+t^2} \:=\:0 \quad\Rightarrow\quad 1-t^2 \:=\:0 \quad\Rightarrow\quad t^2 \:=\:1 \quad\Rightarrow\quad t \:=\:\pm1$

Substitute into [1]: . $x \:=\:\frac{2(\pm1)}{1+(\pm1)^2} \;=\;\frac{\pm2}{2} \;=\;\pm1$

The $x$-intercepts are: . $(1,\,0)\text{ and }(-1,\,0)$

For $y$-intercepts, let $x = 0\!:\;\;\frac{2t}{1+t^2} \:=\:0\quad\Rightarrow\quad 2t \:=\:0 \quad\Rightarrow \quad x \:=\:0$

Substitute into [2]: . $y \:=\:\frac{1-0^2}{1+0^2} \quad\Rightarrow\quad y \:=\:1$

The $y$-intercept is $(0,1)$ . . . There is only one!

Where is the other $y$-intercept?