Graph the polar equation. $$ \mathrm{r}=\frac{18}{3-6 \cos \theta} $$

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To graph the polar equation

$$
r = \frac{18}{3 - 6 \cos \theta},
$$

we can start by rewriting it in a more recognizable form. This equation resembles the standard form of a conic section in polar coordinates, which is given by

$$
r = \frac{ed}{1 - e \cos \theta}
$$

where $$ e $$ is the eccentricity and $$ d $$ is the distance from the pole to the directrix.

### Step 1: Identify parameters

We can rewrite the given equation as:

$$
r = \frac{18}{3 - 6 \cos \theta} = \frac{18/3}{1 - (6/3) \cos \theta} = \frac{6}{1 - 2 \cos \theta}.
$$

From this, we can identify:

- $$ e = 2 $$ (the coefficient of $$ \cos \theta $$)
- $$ d = 6 $$

### Step 2: Determine the type of conic

Since $$ e > 1 $$, this indicates that the conic section is a hyperbola.

### Step 3: Find the directrix

The directrix for a polar equation of the form $$ r = \frac{ed}{1 - e \cos \theta} $$ is located at a distance $$ d $$ from the pole, along the direction of $$ \theta = 0 $$ (the positive x-axis). In this case, the directrix is at $$ x = -d = -6 $$.

### Step 4: Analyze the graph

1. **Asymptotes**: For hyperbolas in polar coordinates, the asymptotes can be found by setting $$ r $$ to infinity, which occurs when the denominator approaches zero. This happens when $$ 3 - 6 \cos \theta = 0 $$, or $$ \cos \theta = \frac{1}{2} $$. This corresponds to $$ \theta = \frac{\pi}{3} $$ and $$ \theta = -\frac{\pi}{3} $$ (or $$ \theta = \frac{5\pi}{3} $$).

2. **Plotting points**: To get a better understanding of the shape, we can calculate a few values of $$ r $$ for specific angles $$ \theta $$:

- For $$ \theta = 0 $$:
     $$
     r = \frac{18}{3 - 6 \cdot 1} = \frac{18}{-3} = -6 \quad (\text{point at } (6, 0))
     $$
   - For $$ \theta = \frac{\pi}{3} $$:
     $$
     r = \frac{18}{3 - 6 \cdot \frac{1}{2}} = \frac{18}{0} \quad (\text{asymptote})
     $$
   - For $$ \theta = \frac{2\pi}{3} $$:
     $$
     r = \frac{18}{3 - 6 \cdot (-\frac{1}{2})} = \frac{18}{6} = 3 \quad (\text{point at } (3, \frac{2\pi}{3}))
     $$
   - For $$ \theta = \pi $$:
     $$
     r = \frac{18}{3 - 6 \cdot (-1)} = \frac{18}{9} = 2 \quad (\text{point at } (2, \pi))
     $$

### Step 5: Sketch the graph

1. The graph will have branches extending outwards from the directrix at $$ x = -6 $$.
2. The branches will approach the asymptotes at angles $$ \frac{\pi}{3} $$ and $$ -\frac{\pi}{3} $$.

### Conclusion

The graph of the polar equation $$ r = \frac{18}{3 - 6 \cos \theta} $$ is a hyperbola with branches that open to the right, asymptotic to the lines defined by $$ \theta = \frac{\pi}{3} $$ and $$ \theta = -\frac{\pi}{3} $$. The directrix is located at $$ x = -6 $$.

You can plot this using polar graphing software or graphing calculators to visualize the hyperbola accurately.
The graph of $$ r = \frac{18}{3 - 6 \cos \theta} $$ is a hyperbola with branches that open to the right, asymptotic to the lines at $$ \theta = \frac{\pi}{3} $$ and $$ \theta = -\frac{\pi}{3} $$, and a directrix at $$ x = -6 $$.

Graph the polar equation. \[ \mathrm{r}=\frac{18}{3-6 \cos \theta} \]

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