Chapter 6.3, Problem 15E

Solution

To eliminate: The parameter and identify the graph. Use the graph to support your answer.

The general equation is $x={\left(y-1\right)}^2$ and the graph is shown in figure $\left(1\right)$ .

Given information:

The equations are $x=t^2$ and $y=t+1$ .

Calculation:

Simplify the equation $y=t+1$ for $t$ .

  $\begin{array}{c}y=t+1\\ y-1=t\end{array}$

Substitute $y-1$ for $t$ to the equation $x=t^2$ .

  $x={\left(y-1\right)}^2$

If the $x$ and $y$ are interchanged, $y={\left(x-1\right)}^2$ the equation is a parabola opening upward with vertex at $\left(1,0\right)$ from the vertex form $y=a{\left(x-h\right)}^2+k$ . Hence, the graph of the equation is its inverse which is a parabola opening to the right with vertex at $\left(0,1\right)$ . This is validated by the graph obtained from the parametric equations.

  

Figure (1)

Therefore, the general equation is $x={\left(y-1\right)}^2$ and the graph is shown in figure $\left(1\right)$ .