Chapter 10.6, Problem 46E

To determine

To graph the curve represented by the parametric equations.

Explanation of Solution

Given information:

Given parametric equation is

. $\begin{array}{l}x=\sec \theta \\ y=\tan \theta \end{array}$

Calculation:

Since both $\sec \theta$ and $\tan \theta$ are not defined for $\theta =\frac{\pi }{2}$ and $\theta =\frac{3\pi }{2}$ , the parametric equations are defined for $0\le \theta <\frac{\pi }{2},\frac{\pi }{2}<\theta <\frac{3\pi }{2}$ and $\frac{3\pi }{2}<\theta \le 2\pi$ , so the domain for $\theta$ is $\left[0,\frac{\pi }{2}\right)\cup \left(\frac{\pi }{2},\frac{3\pi }{2}\right)\cup \left(\frac{3\pi }{2},2\pi \right]$ . Hence both $x$ and $y$ take on both positive and located in all the four quadrants.

Let us find the rectangular equation by eliminating the parameter $\theta$ from the two given equation.

From the first of the parametric equations, $\theta$ in terms of $x$ is given by

  $\sec \theta =x$

Similarly from the second equation $\theta$ in terms of $y$ is given by

  $\tan \theta =y$

Use the trigonometric identity ${\sec }^2\theta -{\tan }^2\theta =1$ to get the rectangular equation as

  $x^2-y^2=1$

This is the equation of an hyperbola. It is known that equation of an hyperbola having centre at $\left(h,k\right)$ ,vertices at $\left(\pm a,0\right)$ and horizontal transverse axis is given by

  $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=\mathrm{1...........}(3)$

Comparing equation (2) with the standard equation of an hyperbola given by equation (3), it is seen that the hyperbola represented by equation (2) has centreat $\left(0,0\right)$ , vertices at $\left(\pm 1,0\right)$ and transverse axis along $x$ axis.

Using graphing utility, the graph of the hyperbola is drawn in the viewing window $\left[-4,4\right]\times \left[-4,4\right]$ and given below. The hyperbola, with $a=b$ is a special hyperbola known as rectangular hyperbola..