To determine the symmetries of the polar graph of the function [tex]\( r = \frac{1}{1 - 5 \sin \theta} \)[/tex], we need to consider different types of symmetries: x-axis symmetry, y-axis symmetry, and symmetry about the origin. Let's analyze each symmetry:
1. X-axis symmetry:
- A function [tex]\( r = f(\theta) \)[/tex] is symmetric about the x-axis if [tex]\( r(-\theta) = r(\theta) \)[/tex].
- For the given function, let’s replace [tex]\(\theta\)[/tex] with [tex]\(-\theta\)[/tex]:
[tex]\[
r(-\theta) = \frac{1}{1 - 5 \sin(-\theta)} = \frac{1}{1 + 5 \sin(\theta)}
\][/tex]
- Since this expression is not equal to [tex]\( \frac{1}{1 - 5 \sin(\theta)} \)[/tex], it indicates that the function is not symmetric about the x-axis.
2. Y-axis symmetry (or symmetry about the line [tex]\(\theta = \frac{\pi}{2}\)[/tex]):
- A function [tex]\( r = f(\theta) \)[/tex] is symmetric about the y-axis if [tex]\( r(\pi - \theta) = r(\theta) \)[/tex].
- For the given function, let’s replace [tex]\(\theta\)[/tex] with [tex]\(\pi - \theta\)[/tex]:
[tex]\[
r(\pi - \theta) = \frac{1}{1 - 5 \sin(\pi - \theta)} = \frac{1}{1 - 5 (\sin \pi \cos \theta - \cos \pi \sin \theta)} = \frac{1}{1 - 5 (0 \cdot \cos \theta - (-1) \sin \theta)} = \frac{1}{1 + 5 \sin(\theta)}
\][/tex]
- Again, this expression is not equal to [tex]\( \frac{1}{1 - 5 \sin(\theta)} \)[/tex], suggesting that the function is not symmetric about the y-axis.
3. Symmetry about the origin:
- A function [tex]\( r = f(\theta) \)[/tex] is symmetric about the origin if [tex]\( r(\theta + \pi) = - r(\theta) \)[/tex].
- For the given function, let’s replace [tex]\(\theta\)[/tex] with [tex]\(\theta + \pi\)[/tex]:
[tex]\[
r(\theta + \pi) = \frac{1}{1 - 5 \sin(\theta + \pi)} = \frac{1}{1 - 5 (\sin \theta \cos \pi + \cos \theta \sin \pi)} = \frac{1}{1 - 5 (\sin \theta \cdot (-1) + \cos \theta \cdot 0)} = \frac{1}{1 + 5 \sin(\theta)}
\][/tex]
- This expression is not [tex]\( - \frac{1}{1 - 5 \sin(\theta)} \)[/tex], which indicates that the function is not symmetric about the origin.
Given this detailed analysis, we find that:
- The graph of [tex]\( r = \frac{1}{1 - 5 \sin \theta} \)[/tex] exhibits no symmetry about the x-axis, y-axis, or the origin.
Therefore, the best answer among the choices provided is:
[tex]\[
\boxed{\text{D. No symmetry}}
\][/tex]
