91影视

The parametric curves,$x=\sinh t,y={\cosh }^{2}t,t鈭�\mathrm{鈩�}$

The goal is to draw the parametric curve.

Assume $t=-2,-1,0,1,2$when drawing the graph for the parametric equations.

Find the values $x,y$by substituting different $t$values in the parametric equations.

The point $(x,y)$When $t=-2$is,

$(x,y)=\left(\sinh t,{\cosh }^{2}t\right)$

$(x,y)=\left(\sinh (-2),{\cosh }^2(-2)\right)[$since by substituting $t=-2]$

$(x,y)=(-3.62,14.13)$simplify

The point $(x,y)$When $t=-1$is,

$$\begin{gathered} (x,y)=\left(\sinh t,{\cosh }^{2}t\right) \\ (x,y)=\left(\sinh (-1),{\cosh }^2(-1)\right)[\text{since by substituting}t=-1] \\ (x,y)=\left(-\sinh (-1),{\cosh }^2\left(1\right)\right) \\ (x,y)=(-1.17,2.37) \end{gathered}$$

The point $(x,y)$When $t=0$is,

$$\begin{gathered} (x,y)=\left(\sinh t,{\cosh }^{2}t\right) \\ (x,y)=\left(\sinh \left(0\right),{\cosh }^2\left(0\right)\right)[\text{since by substituting}t=0] \\ (x,y)=(0,1)\text{simplify} \end{gathered}$$

The point $(x,y)$ When $t=1$ is,

The point $(x,y)$ When $t=2$ is,

The points are represented in tabular form as follows:

Below is a graphical illustration of the situation.

Therefore, the solution is the required graph.