Question

In Exercises 24-34 sketch the parametric curve by eliminating the parameter.

\(x=\cosh t, y=\sinh t, t \in \mathbb{R}\)

Short answer

The graphical representation by using the points $(1,0)(2,-\sqrt{3})(2, \sqrt{3})(3,-2 \sqrt{2})(3,2 \sqrt{2})(4,-\sqrt{15})(4, \sqrt{15})$ is as follows,

Therefore, the equation after elimination of the parameter is $x^{2}-y^{2}=1$ or $y^{2}=x^{2}-1$.

Step-by-step solution

Given information

$x=\cosh t, y=\sinh t, t \in \mathbb{R}$

Calculation

Consider the parametric equations $x=\cosh t_{v} y=\sinh t, t \in \mathbb{R}$.

The objective is to sketch the parametric curve by eliminating the parameter.

Take the equation $x=\cosh t$.

Square the equation on both sides.

$x^{2}=\cosh ^{2} t$

Take the equation $y=\sinh t$.

Square the equation on both sides. then

$$

y^{2}=\sinh ^{2} t

$$

Now subtract the equations $y^{2}=\sinh ^{2} t$ from $x^{2}=\cosh ^{2} t$.

Thus,

$$

\begin{aligned}

&x^{2}-y^{2}=\cosh ^{2} t-\sinh ^{2} t \\

&x^{2}-y^{2}=1 \quad\left[\operatorname{since} \cosh ^{2} t-\sinh ^{2} t=1\right] \\

&x^{2}-1=y^{2} \\

&y^{2}=x^{2}-1

\end{aligned}

$$

In order to draw the graph of the equation assume $x=1,2,3,4$.

Substitute $x=1$ in the equation $y^{2}=x^{2}-1$.

Then,

$$

\begin{aligned}

&y^{2}=1^{2}-1 \\

&y=0 \\

&(x, y)=(1,0)

\end{aligned}

$$

Substitute $x=2$ in the equation $y^{2}=x^{2}-1$.

Then.

$$

\begin{aligned}

&y^{2}=(2)^{2}-1 \\

&y^{2}=4-1 \\

&y^{2}=3 \\

&y=\pm \sqrt{3} \\

&(x, y)=(2,-\sqrt{3})(2, \sqrt{3})

\end{aligned}

$$

Substitute $x=3$ in the equation $y^{2}=x^{2}-1$.

Then,

$$

\begin{aligned}

y^{2} &=(3)^{2}-1 \\

y^{2} &=9-1 \\

y &=\sqrt{8}-\pm 2 \sqrt{2} \\

(x, y) &=(3,-2 \sqrt{2})(3,2 \sqrt{2})

\end{aligned}

$$

Substitute $x=4$ in the equation $y^{2}=x^{2}-1$.

Then,

$$

\begin{aligned}

&y^{2}=(4)^{2}-1 \\

&y^{2}=16-1 \\

&y=\pm \sqrt{15} \quad\left[\text { since } y^{2}=15 \Rightarrow y=\pm \sqrt{15}\right] \\

&(x, y)=(4,-\sqrt{15})(4, \sqrt{15})

\end{aligned}

$$

The graphical representation by using the points $(1,0)(2,-\sqrt{3})(2, \sqrt{3})(3,-2 \sqrt{2})(3,2 \sqrt{2})(4,-\sqrt{15})(4, \sqrt{15})$ is as follows,

Therefore, the equation after elimination of the parameter is $x^{2}-y^{2}=1$ or $y^{2}=x^{2}-1$.