Chapter 9.4, Problem 11E

To determine

To find: Which set of parametric equation represents the given graph.

Answer to Problem 11E

  $\begin{array}{l}x=2t+1\\ y=t^2\end{array}$

Explanation of Solution

Given information:

Graph of parametric equations.

Concept:

Parametric equations of a Parabola in the form ${\left(x-h\right)}^2=4a\left(y-k\right)$

with Vertex $\left(h,k\right)$ is

  $\begin{array}{l}x=h+2at\\ y=k+a{t}^2\end{array}$

From the given graph

On observing

We can say that the graph is Parabola and in form

  ${\left(x-h\right)}^2=4a\left(y-k\right)$

And Vertex of Parabola is $\left(h,k\right)=\left(1,0\right)$

Therefore,

Equation of parabola is

  $\begin{array}{l}\Rightarrow {\left(x-1\right)}^2=4a\left(y-0\right)\\ \Rightarrow {\left(x-1\right)}^2=4ay\end{array}$

From graph another point on parabola is $\left(9,16\right)$

On substituting this point on parabola equation

We get

  $\begin{array}{l}{\left(x-1\right)}^2=4ay\\ \Rightarrow {\left(9-1\right)}^2=4a\left(16\right)\\ \Rightarrow 8^2=64a\\ \Rightarrow 64=64a\\ \Rightarrow a=1\end{array}$

  $\begin{array}{l}x=h+2at\\ y=k+a{t}^2\end{array}$

On substituting values

We get

  $\begin{array}{l}x=h+2at\\ \Rightarrow x=1+2\left(1\right)t\\ \Rightarrow x=1+2t\\ y=k+a{t}^2\\ \Rightarrow y=0+\left(1\right)t^2\\ \Rightarrow y=t^2\end{array}$

Therefore,

The parametric equation which represents given graph is

  $\begin{array}{l}x=2t+1\\ y=t^2\end{array}$