Chapter 9, Problem 14CT

a.

To determine

To find: Set of parametric equations.

Answer to Problem 14CT

  $\begin{array}{l}x=t\\ y=\frac{4-t^2}{2}\end{array}$

Explanation of Solution

Given information:

  $x^2+2y=4$

  $t=x$

  $x^2+2y=4$

Substitute $x=t$ in above equation

We get

  $\begin{array}{l}{x}^2+2y=4\\ \Rightarrow t^2+2y=4\\ \Rightarrow 2y=4-t^2\\ \Rightarrow y=\frac{4-t^2}{2}\end{array}$

The set $x=t$ and $y=\frac{4-t^2}{2}$ represents the set of parametric equations of given the Rectangular equation $x^2+2y=4$

b.

To determine

To find: Set of parametric equations.

Answer to Problem 14CT

  $\begin{array}{l}x=2-t\\ y=\frac{4-{\left(2-t\right)}^2}{2}\end{array}$

Explanation of Solution

Given information:

  $x^2+2y=4$

  $t=2-x$

  $x^2+2y=4$

  $\begin{array}{l}t=2-x\\ \Rightarrow x=2-t\end{array}$

Substitute $x=2-t$ in the equation $x^2+2y=4$

We get

  $\begin{array}{l}{x}^2+2y=4\\ \Rightarrow {\left(2-t\right)}^2+2y=4\\ \Rightarrow 2y=4-{\left(2-t\right)}^2\\ \Rightarrow y=\frac{4-{\left(2-t\right)}^2}{2}\end{array}$

Therefore, the set $x=2-t$ and $y=\frac{4-{\left(2-t\right)}^2}{2}$ also represents the set of parametric equations of the given Rectangular equation $x^2+2y=4$