Chapter 8.2, Problem 58E

(a)

To determine

To graph the below polar equation in the viewing rectangle $\left[-3,3\right]$ by $\left[-1,9\right]$ -

  $r=\tan \theta \sec \theta$

Answer to Problem 58E

The graph in the viewing rectangle $\left[-3,3\right]$ by $\left[-1,9\right]$ is

  

Explanation of Solution

Given: Polar equation: $r=\tan \theta \sec \theta$

Formula Used:

A polar equation is any equation that describes a relation between r and θ , where r represents the distance from the pole (origin) to a point on a curve, and θ represents the counter-clockwise angle made by a point on a curve, the pole, and the positive x-axis.

Also, $x=r\cos \theta$ and $y=r\sin \theta$

  $r=\sqrt{x^2+y^2}$ and $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$

Calculation:

Given Polar equation is $r=\tan \theta \sec \theta$

Drawing the polar-form curve for $r=\tan \theta \sec \theta$ on a graph as per the above definition of polar equation, we have:

Let $\theta =0$ ,

  $\begin{array}{l}r=\tan 0\sec 0\\ r=0\end{array}$

Let $\theta =1$ ,

  $\begin{array}{l}r=\tan 1\sec 1\\ r=1.55\times 1.85\\ r=2.79\end{array}$

Let $\theta =-1$ ,

  $\begin{array}{l}r=\tan \left(-1\right)\sec \left(-1\right)\\ r=-1.55\times 1.85\\ r=-2.79\end{array}$

Let $\theta =3$ ,

  $\begin{array}{l}r=\tan 3\sec 3\\ r=\left(-0.14\right)\times \left(-1.01\right)\\ r=0.14\end{array}$

Let $\theta =-3$ ,

  $\begin{array}{l}r=\tan \left(-3\right)\sec \left(-3\right)\\ r=\left(0.14\right)\times \left(-1.01\right)\\ r=-0.14\end{array}$

Plotting the above points on the graph in the viewing rectangle $\left[-3,3\right]$ by $\left[-1,9\right]$ , we have:

  

Conclusion:

Thus, the graph in the viewing rectangle $\left[-3,3\right]$ by $\left[-1,9\right]$ is

  

(b)

To determine

To confirm the above drawn graph is a parabola using the conversion to rectangular coordinates -

  $r=\tan \theta \sec \theta$

Answer to Problem 58E

The polar equation $r=\tan \theta \sec \theta$ is a parabolic equation with rectangular form as $x^2=y$ .

Explanation of Solution

Given: Polar equation: $r=\tan \theta \sec \theta$

Formula Used:

A polar equation is any equation that describes a relation between r and θ , where r represents the distance from the pole (origin) to a point on a curve, and θ represents the counter-clockwise angle made by a point on a curve, the pole, and the positive x-axis.

Also, $x=r\cos \theta$ and $y=r\sin \theta$

  $r=\sqrt{x^2+y^2}$ and $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$

Parabolic Equation is given by −

The standard form is ${\left(x-h\right)}^2=4p\left(y-k\right)$ , where the focus is $\left(h,k+p\right)$ and the directrix is $y=k-p$ .

Calculation:

Given Polar equation is $r=\tan \theta \sec \theta$

  $\begin{array}{l}r=\left(\frac{\sin \theta }{\cos \theta }\right)\left(\frac{1}{\cos \theta }\right)\\ r=\frac{\sin \theta }{{\cos }^2\theta }\end{array}$

Converting the polar equation to rectangular coordinates, we have:

  $\begin{array}{l}r=\frac{\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}{{\left(\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$r$}\right.\right)}^2}\\ r=\frac{y}{r}\times \frac{r^2}{x^2}\\ x^2=y\end{array}$

The above equation satisfies the parabolic equation.

Hence, the polar equation $r=\tan \theta \sec \theta$ is a parabolic equation.

Conclusion:

Thus, the polar equation $r=\tan \theta \sec \theta$ is a parabolic equation with rectangular form as $x^2=y$