Question

Eliminate the parameter $t$. Find a rectangular equation for the plane curve defined by the parametric equations

$x=3t,y=t+3;-2<t<3$

• A. $y=x^2+1;-2<x<2$
• B. $y=-3x+3;-\infty <x<\infty$
• C. $y=\frac{1}{3}x+3;-6<x<9$
• D. $y=\frac{1}{3}x-3;-\infty <x<\infty$

Answer 1

The correct option is C

$y=\frac{1}{3}x+3;-6<x<9$

Explanation for the correct option.

We have parametric equation,

$x=3t,y=t+3;-2<t<3$

Eliminate $t$,

$$\begin{gathered} x=3t \\ \Rightarrow t=\frac{x}{3}...\left(1\right) \\ y=t+3 \\ \Rightarrow t=y-3...\left(2\right) \end{gathered}$$

From (1) and (2), we get

$$\begin{gathered} y-3=\frac{x}{3} \\ \therefore y=\frac{x}{3}+3 \end{gathered}$$

And the domain is

$$\begin{gathered} -2<t<3 \\ \Rightarrow -2<\frac{x}{3}<3 \\ \therefore -6<x<9 \end{gathered}$$

Thus, a rectangular equation for the plane curve defined by the parametric equations $x=3t,y=t+3;-2<t<3$ is$y=\frac{1}{3}x+3;-6<x<9$.

Hence, the correct option is option (C).