Parametric equation

**vanessa123** · December 8th 2009, 06:06 PM

Sketch the curve represented by this parametric equation, and then eliminate the parameter and obtain the general form of the rectangular equation.

x=t+2
y=2t^2

**acc100jt** · December 8th 2009, 08:03 PM

I eliminated the parameter and i got the following,

$x=t+2 \Rightarrow t=x-2$
Substitute $t=x-2$ into $y=2t^{2}$
$y=2(x-2)^2$

It is a quadratic curve with turning point at (2, 0)

**Soroban** · December 8th 2009, 08:18 PM

Hello, vanessa123!

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

(a) Sketch the curve represented by these parametric equations.

. . $\begin{array}{|c||c|c|} t & x & y \\ \hline<br /> \text{-}2 & 0 & 8 \\ \text{-}1 & 1 & 2 \\ 0 & 2 & 0 \\ 1 & 3 & 2 \\ 2 & 4 & 8 \end{array}$

Code:

        |
        *           *
        |
        |
        |
        |  *     *
    - - + - - * - - - - -
        |     2

(b) Eliminate the parameter and obtain the rectangular equation.

From [1], we have: . $t \:=\:x-2$

Substitute into [2]: . $y \:=\:2(x-2)^2$

**vanessa123** · December 8th 2009, 08:33 PM

Originally Posted by Soroban

Hello, vanessa123!

. . $\begin{array}{|c||c|c|} t & x & y \\ \hline<br /> \text{-}2 & 0 & 8 \\ \text{-}1 & 1 & 2 \\ 0 & 2 & 0 \\ 1 & 3 & 2 \\ 2 & 4 & 8 \end{array}$

Code:

        |
        *           *
        |
        |
        |
        |  *     *
    - - + - - * - - - - -
        |     2

From [1], we have: . $t \:=\:x-2$

Substitute into [2]: . $y \:=\:2(x-2)^2$

hi soroban, i don't understand how you got that table? can you please help me. and how did u eliminate the parameter.

**HallsofIvy** · December 9th 2009, 04:41 AM

For the table, Soroban just used the equations given. The first row has t= -2. He chose that as a reasonable starting place. If t= -2, then x= t+ 2= -2+ 2= 0 and $y= 2t^2= 2(-2)^2= 2(4)= 8$ . If t= -1, then x= t+ 2= -1+ 2= 1 and $y= 2t^2= 2(-1)^2= 2(1)= 2$ . If t= 0, then x= t+ 2= 0+ 2= 2 and [tex]y= 2t^2= 2(0)^2= 0, etc.

As for how he eliminated the parameter, acc10jt showed that nicely: from x= t+ 2, x- 2= t. Now replace "t" in $y= 2t^2$ by that: $y= 2(x- 2)^2$ .

**mr fantastic** · December 9th 2009, 04:51 AM

Originally Posted by vanessa123

Sketch the curve represented by this parametric equation, and then eliminate the parameter and obtain the general form of the rectangular equation.

x=t+2
y=2t^2

Is there a restriction on the values of t? If so, the required graph will not be of all of the curve corresponding to the equation.

**vanessa123** · December 9th 2009, 07:14 AM

Originally Posted by mr fantastic

Is there a restriction on the values of t? If so, the required graph will not be of all of the curve corresponding to the equation.

no restrictions. thank you for all your help.

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