Show that the parametrization $x=2t+1,y=4t^2-4$for $t\in [-1,\infty )$has the same graph as the one we plotted point by point in the reading.

Consider the parametric curves$x=2f+1,y=4s^2-4,t\in [-1,\infty ).$

Consider the parametric curves $x=2t+1,y=4t^2-4,t\in [-1,\infty )$.

The objective is to parametrize the curves.

To parameterize the curves eliminate the parameter t from the parametric equations.

Take $x=2t+1$

Add $-1$on both sides of the equation.

$$\begin{gathered} x-1=2t+1-1 \\ x-1=2t+1-1 \\ x-1=2t \end{gathered}$$

Now divide the equation by 2 .

$$\begin{gathered} \frac{x-1}{2}=\frac{2t}{2} \\ \frac{x-1}{2}=\frac{2t}{2} \\ \frac{x-1}{2}=t \end{gathered}$$

Now substitute $t=\frac{x-1}{2}$in $y=4t^2-4$then the equation becomes,

$\begin{array}{rcl}y& =& 4{\left(\frac{x-1}{2}\right)}^2-4y\\ & =& 4\frac{(x-1{)}^2}{4}-4y\\ & =& 4\frac{(x-1{)}^2}{4}-4\end{array}$

On further simplification,

Since the parametrization of the curves gives the same equation that of$x=\sin t+1,y={\sin }^{2}t-4.$

Now check the parametrization of the curves $x=\sin t+1,y={\sin }^{2}t-4$.

To parameterize the curves eliminate the parameter t from the parametric equations.

Take$x=\sin t+1$

Add $-1$on both sides of the equation.

$$\begin{gathered} x-1=\sin t+1-1 \\ x-1=\sin t+1t-1 \\ x-1=\sin t \end{gathered}$$

Now substitute$\sin t=x-1$ in the equation $y={\sin }^{2}t-4$.

Then,

Thus, the parameterization of the curves $x=2t+1,y=4t^2-4$ traces exactly the same curve as that of the equations $x=\sin t+1,y={\sin }^{2}t-4$.

Hence the required explanation.