Chapter 1.4, Problem 43E

(a)

To determine

To draw: Thegraph of equations $x=a\sec t$ and $y=b\tan t$ for $a=1,2,\text{or}3$ and $b=1,2,\text{or}3$ in the interval $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ . Also, describe the role of a andb .

Explanation of Solution

Given information:The given parametric equations are $x=a\sec t$ and $y=b\tan t$ with the conditions $a=1,2,\text{or}3$ and $b=1,2,\text{or}3$ in the interval $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ .

Graph:

Consider $a=1$ and $b=1,2,\text{or}3$ to draw the graph.

The parametric equation for $a=1$ and $b=1$ are:

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

The parametric equation for $a=1$ and $b=2$ are:

  $\begin{array}{l}x=\sec t\\ y=2\tan t\end{array}$

The parametric equation for $a=1$ and $b=3$ are:

  $\begin{array}{l}x=\sec t\\ y=3\tan t\end{array}$

Now, draw the graph of parametric equations in the same viewing window in the interval $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ as shown below.

  

Figure (1)

Interpretation: From the graph it can be observed that the graph of parametric equation for different values of a andb is the right half of a hyperbola that lies in Quadrant I and IV. The value of a gives x -intercept, b describes the shape of hyperbola.

As the value of b increases, the curve of hyperbola goes away from the x -axis and becomes shallower.

(b)

To determine

To draw: The graph of equations $x=a\sec t$ and $y=b\tan t$ for $a=2$ and $b=3$ in the interval $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$ .

Explanation of Solution

Given information:The given parametric equations are $x=a\sec t$ and $y=b\tan t$ .The values are $a=2$ and $b=3$ in the interval $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$ .

Graph:

Substitute 2 for a in the equation $x=a\sec t$ .

  $x=2\sec t$

Substitute 3 for b in the equation $y=b\tan t$ .

  $y=3\tan t$

Now, draw the graph of parametric equations in the interval $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$ as shown below.

  

Figure (2)

Interpretation: From the graph it can be observed that the graph of parametric equations is a left half of the hyperbola in second quadrant and third quadrant.

(c)

To determine

To draw: The graph of equations $x=a\sec t$ and $y=b\tan t$ for $a=2$ and $b=3$ in the interval $\left(-\frac{\pi }{2},\frac{3\pi }{2}\right)$ .

Explanation of Solution

Given information:The given parametric equations are $x=a\sec t$ and $y=b\tan t$ .The values are $a=2$ and $b=3$ in the interval $\left(-\frac{\pi }{2},\frac{3\pi }{2}\right)$ .

Graph:

Substitute 2 for a in the equation $x=a\sec t$ .

  $x=2\sec t$

Substitute 3 for b in the equation $y=b\tan t$ .

  $y=3\tan t$

Now, draw the graph of parametric equations in the interval $\left(-\frac{\pi }{2},\frac{3\pi }{2}\right)$ as shown below.

  

Figure (3)

Interpretation: From the graph it can be observed that the graph of parametric equations is a hyperbola. A student must be careful while drawing the graph in an interval that includes $\pm \frac{\pi }{2}$ as $\sec t$ and $\tan t$ are not continuous at this point. So, the graph does not include the asymptotes of hyperbola.

(d)

To determine

To prove: The equation ${\left(\frac{x}{a}\right)}^2-{\left(\frac{y}{b}\right)}^2=1$ by algebra.

Explanation of Solution

Given information:The given parametric equations are $x=a\sec t$ and $y=b\tan t$ .

Proof:

It is known that the standard trigonometric identity for $\sec t$ and $\tan t$ is:

  ${\sec }^{2}t-{\tan }^{2}t=1$

Substitute $\frac{x}{a}$ for $\sec t$ and $\frac{y}{b}$ for $\tan t$ in the above identity.

  $\begin{array}{c}{\sec }^{2}t-{\tan }^{2}t=1\\ {\left(\frac{x}{a}\right)}^2-{\left(\frac{y}{b}\right)}^2=1\end{array}$

Hence, it is proved that ${\left(\frac{x}{a}\right)}^2-{\left(\frac{y}{b}\right)}^2=1$ .

(e)

To determine

To evaluate: Repeatpart (a), (b), (c) and (d) for parametric equations $x=a\tan t$ and $y=b\sec t$ .

Explanation of Solution

Given information:The given parametric equations are $x=a\tan t$ and $y=b\sec t$ .

Part (a): Consider $a=1,2,\text{or}3$ and $b=1$ to draw the graph of given parametric equations in the interval $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ .

The parametric equation for $a=1$ and $b=1$ are:

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

The parametric equation for $a=2$ and $b=1$ are:

  $\begin{array}{l}x=2\tan t\\ y=\sec t\end{array}$

The parametric equation for $a=3$ and $b=1$ are:

  $\begin{array}{l}x=3\tan t\\ y=\sec t\end{array}$

Now, draw the graph of parametric equations in the same viewing window in the interval $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ as shown below.

  

Figure (4)

From the graph it can be observed that the graph of parametric equations is the top half of a hyperbola that lies in first and second Quadrant. The value of b gives y -intercepts and the values ofadescribe the shape of hyperbola.

As the value of a increases, the curve of hyperbola goes away from the y -axis and becomes shallower.

Part (b): Consider $a=2$ and $b=3$ to draw the graph of given parametric equations in the interval $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ .

Substitute 2 for a in the equation $x=a\tan t$ .

  $x=2\tan t$

Substitute 3 for b in the equation $y=b\sec t$ .

  $y=3\sec t$

Now, draw the graph of parametric equations in the interval $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$ as shown below.

  

Figure (5)

From the graph it can be observed that the graph of parametric equations is the bottom half of the hyperbola that lies in third and fourth quadrant.

Part (d): It is known that the standard trigonometric identity for $\sec t$ and $\tan t$ is:

  ${\sec }^{2}t-{\tan }^{2}t=1$

Substitute $\frac{x}{a}$ for $\tan t$ and $\frac{y}{b}$ for $\sec t$ in the above identity.

  $\begin{array}{c}{\sec }^{2}t-{\tan }^{2}t=1\\ {\left(\frac{y}{b}\right)}^2-{\left(\frac{x}{a}\right)}^2=1\end{array}$