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
Hello, vanessa123!
. . $\displaystyle \begin{array}{|c||c|c|} t & x & y \\ \hline$\displaystyle \begin{array}{cccc}x &=& t+2 & [1] \\ y &=& 2t^2 & [2]\end{array}$
(a) Sketch the curve represented by these parametric equations.
\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: .$\displaystyle t \:=\:x-2$
Substitute into [2]: .$\displaystyle y \:=\:2(x-2)^2$
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 $\displaystyle y= 2t^2= 2(-2)^2= 2(4)= 8$. If t= -1, then x= t+ 2= -1+ 2= 1 and $\displaystyle y= 2t^2= 2(-1)^2= 2(1)= 2$. 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 $\displaystyle y= 2t^2$ by that: $\displaystyle y= 2(x- 2)^2$.