Chapter 9, Problem 8CT

To determine

The symmetry of the polar equation $r^2\cos \theta =5$ with respect to pole, the polar axis, and the line $\theta =\frac{\pi }{2}$ .

Answer to Problem 8CT

Solution:

The polar equation $r^2\cos \theta =5$ is symmetric with respect to pole, polar axis, and the line $\theta =\frac{\pi }{2}$ .

Explanation of Solution

Given information:

The polar equation, $r^2\cos \theta =5$

Use the test of symmetry with respect to polar axis, pole and line $\theta =\frac{\pi }{2}$ .

Polar axis: Replace $\theta$ with $-\theta$ .

  $r^2\cos \left(-\theta \right)=5$ .

Using $\cos \left(-\theta \right)=\cos \theta$ ,

  $\Rightarrow r^2\cos \left(\theta \right)=5$

The resulting equation is same as original, so the graph is symmetric with respect to polar axis.

The line $\theta =\frac{\pi }{2}$ : Replace $\theta$ with $\pi -\theta$ .

  $r^2\cos \left(\pi -\theta \right)=5$ .

  $\Rightarrow r^2\left[\cos \pi \cos \theta +\sin \pi \sin \theta \right]=5$

  $\Rightarrow r^2\left[\left(-1\right)\cos \theta +\left(0\right)\sin \theta \right]=5$

  $\Rightarrow -r^2\cos \left(\theta \right)=5$ .

Since the resulting equation is not same as original, the graph may or may not be symmetrical about the line $\theta =\frac{\pi }{2}$ .

The pole: Replace $r$ with $-r$ .

  ${\left(-r\right)}^2\cos \theta =5$

  $\Rightarrow r^2\cos \theta =5$ .

The resulting equation is same as original, so the graph is symmetric with respect to pole.

Since the graph is symmetric about pole and polar axis, check the symmetry the line $\theta =\frac{\pi }{2}$ by replacing $\theta$ with $-\theta$ and $r$ with $-r$

  ${\left(-r\right)}^2\cos \left(-\theta \right)=5$

  $\Rightarrow r^2\cos \theta =5$

The resulting equation is same as original, so the graph is symmetric with respect to line $\theta =\frac{\pi }{2}$ .

Therefore the polar equation $r^2\cos \theta =5$ is symmetric with respect to pole, polar axis, and the line $\theta =\frac{\pi }{2}$ .