91Ó°ÊÓ

To test for symmetry, one can examine if the equation is symmetrical with respect to the line \( \theta = \pi / 2 \) (the vertical axis), and the pole (the origin, r=0). \n 1. Symmetry with respect to the vertical line: Replace \( \theta \) with \( -\theta \), if the equation remains same then it is symmetrical with respect to the vertical line. \n 2. Symmetry with respect to the pole: Replace \( r \) with \( -r \), if the equation remains same, then it is symmetrical with respect to the pole. \n After replacing \( \theta \) with \( -\theta \) in the given equation \( r = \frac{1}{1 - \cos \theta} \), we get \( r = \frac{1}{1 - \cos(-\theta)} = \frac{1}{1 - \cos \theta} \). Therefore, the equation is symmetrical with respect to the vertical line. But, when we replace \( r \) with \( -r \), equation becomes \( -r = \frac{1}{1 - \cos \theta} \). This isn't the same as the original polar equation, so it is not symmetrical about the pole.

Because the equation is symmetric with respect to the vertical line, only the graph for \( 0 \leq \theta \leq \pi \) needs to be plotted. For other values of \( \theta \), the graph will reflect about the vertical line. In the domain \( 0 \leq \theta \leq \pi \), calculate the value of r for various values of \( \theta \). For better results, select values for \( \theta \) that makes \( \cos \theta \) equal to 0,1, and -1. Plot the corresponding points (r, \( \theta \)) in the polar coordinate system and then draw a smooth curve through these points. Finally, draw a mirror image of the plotted graph about the vertical line to complete the entire graph because the function is symmetric about this line.