Question

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

$x=\sec t,y=\tan t,t\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

Short answer

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

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

Step-by-step solution

Given information

$x=\sec t,y=\tan t,t\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

Calculation

Consider the parametric equations $x=\sec t,y=\tan t,t\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right).$

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

Take the equation $x=\sec t.$

Square the equation on both sides.

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

Take the equation $y=\tan t$.

Square the equation on both sides. Then.

$y^2={\tan }^{2}t$

Now subtract the equation $y^2={\tan }^{2}t$from $x^2={\sec }^{2}t$.

Thus.

$$\begin{gathered} x^2-y^2={\sec }^{2}t-{\tan }^{2}t \\ {x}^2-y^2=1\left[\text{since}{\sec }^{2}t-{\tan }^{2}t=1\right] \end{gathered}$$

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

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

Then,

$$\begin{gathered} 1^2-y^2=1 \\ 1-y^2=1 \\ y=0\text{simplify} \\ (x,y)=(-1,0) \end{gathered}$$

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

Then.

$$\begin{gathered} (-1{)}^2-y^2=1 \\ 1-y^2=1 \\ y=0\text{simplify} \\ (x,y)=(-1,0) \end{gathered}$$

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

Then,

$$\begin{gathered} 2^2-y^2=1 \\ 4-y^2=1 \end{gathered}$$

Add $-4$on both sides of the equation.

$$\begin{gathered} 4-y^2-4=1-4 \\ 4-y^2-4--3 \\ -y^2=-3 \\ y=\sqrt{3} \\ (x,y)=(2,-\sqrt{3})(2,\sqrt{3}) \end{gathered}$$

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

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