Question

A Parabola in Polar Coordinates

(a) Graph the polar equation$r=\tan \theta \sec \theta$ in the viewing rectangle$[-3,3]$ by.$[-1,9]$

(b) Note that your graph in part (a) looks like a parabola. Confirm this by converting the equation to rectangular coordinate

Short answer

1. Graph of given equation $r=\tan \theta \sec \theta$is-

By conversion into rectangular coordinates we get $x^2=y$

Step-by-step solution

Step 1. Given information

An equation is given as $r=\tan \theta \sec \theta$

Step 2. Concept used

A polar equation is an equation that describes a relation between r and$\theta$ , where r represents the distance from the pole (origin) to a point on a curve, and$\theta$ represents the counterclockwise angle made by a point on a curve, the pole, and the positive -axis.

To convert polar coordinates into rectangular coordinates substitute:

$\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \\ r^2=x^2+y^2\end{array}$

Step 3. Calculation

First we will find values,

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,$\theta =3$

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

Let,$\theta =-1$

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

Let,$\theta =-3$

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

Now plot the graph-

Step 1. Given information

An equation is given as $r=\tan \theta \sec \theta$

Step 2. Concept used

A polar equation is an equation that describes a relation between r and$\theta$ , where r represents the distance from the pole (origin) to a point on a curve, and$\theta$ represents the counterclockwise angle made by a point on a curve, the pole, and the positive x-axis.

To convert polar coordinates into rectangular coordinates substitute:

$\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \\ r^2=x^2+y^2\end{array}$

Step 3. Calculation

Changing given equation into the rectangular coordinates we get,

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

Further substituting with the rectangular coordinates we get,

$\begin{array}{l}r=\frac{\frac{y}{r}}{{\left(\frac{x}{r}\right)}^2}\\ r=\frac{y}{r}\times \frac{r^2}{x^2}\\ x^2=y\end{array}$