Parametric to rectangular equations

**CrazyLond** · May 22nd 2008, 08:21 PM

Is this the correct strategy to convert this parametric to rectangular? Do I need to further simplify the answer or is this good enough?

**TKHunny** · May 22nd 2008, 08:50 PM

A correct strategy has two important features:

1) You understand it and can communicate it.
2) It works and doesn't mess up the Domain.

I think you have it.

**Mathstud28** · June 20th 2008, 09:08 AM

Originally Posted by CrazyLond

Is this the correct strategy to convert this parametric to rectangular? Do I need to further simplify the answer or is this good enough?

Probably a better way would be this

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

So solving for t we get

$t=\pm\sqrt{x+\frac{1}{4}}+\frac{1}{2}$

Now imput that into your y

**Soroban** · June 20th 2008, 09:37 AM

Hello, CrazyLond!

Yet another approach . . .

$\begin{array}{cccc}x &=&t^2+t & [1] \\ y &=&t^2-t & [2] \end{array}$

Subtract [1] - [2]: . $x-y \:=\:2t\quad\Rightarrow\quad y \:=\:\frac{x-y}{2}$

Substitute into [2]: . $y \:=\:\left(\frac{x-y}{2}\right)^2 - \left(\frac{x-y}{2}\right)$

This simplifies to: . $x^2 - 2xy + y^2 - 2x - 2y \:=\:0$

I believe this is the parabola: $y^2 = \sqrt{2}\,x$ rotated 45° CCW.
. . (But don't quote me!)

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