As you learned on the main conics page, there is a standard equation for conics, i.e. \(Ax^2+Bxy+Cy^2+\) \( Dx+Ey+F=0\). Conics are particularly nice in polar coordinates and the equations are, in many ways, easier to represent and use.
The polar equation for a conic will be in one of these four forms.
\(\displaystyle{ r = \frac{ed}{1\pm e \sin\theta}}\) 
\(\displaystyle{ r = \frac{ed}{1\pm e \cos\theta}}\)  

e is the eccentricity  
\(\abs{d}\) is the distance between the focus at the pole and its directrix 
In order to write these equations in this form, we require that the focus (or one of the foci) be located at the origin. When we have a conic in this form, we can use the eccentricity to classify the equation, as follows.
type 
eccentricity  

ellipse 
\(0 < e < 1\)  
parabola 
\(e=1\)  
hyperbola 
\(e > 1\) 
This video clip gives a nice overview of conic sections in polar coordinates and the presenter uses an example of a parabola to explain the equations.
video by MIP4U 

To get a better understanding of these equations, we will look at examples of each of the three types of conics (parabolas, ellipses and hyperbolas). To help you understand these equations, get out a piece of paper and a pencil and do some calculations to convince yourself why these graphs look like they do.
Parabola 

\(e=1\) 
\(d=2\) 

\(\displaystyle{r = \frac{2}{1+\sin(\theta)} }\) 
\(e=1\) 
\(d=2\) 

\(\displaystyle{r = \frac{2}{1\sin(\theta)} }\) 
\(e=1\) 
\(d=2\) 

\(\displaystyle{r = \frac{2}{1+\cos(\theta)} }\) 
\(e=1\)  \(d=2\) 

\(\displaystyle{r = \frac{2}{1\cos(\theta)} }\) 
Problem Statement 

Find the polar equation of the parabola with vertex \( (4, 3\pi/2) \).
Solution 

video by Krista King Math 

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Ellipse 

\(e=3/4\) 
\(d=4/3\) 

\(\displaystyle{r = \frac{1}{1+0.75\sin(\theta)} }\) 
\(e=3/4\) 
\(d=4/3\) 

\(\displaystyle{r = \frac{1}{10.75\sin(\theta)} }\) 
\(e=3/4\) 
\(d=4/3\) 

\(\displaystyle{r = \frac{1}{1+0.75\cos(\theta)} }\) 
\(e=3/4\) 
\(d=4/3\) 

\(\displaystyle{r = \frac{1}{10.75\cos(\theta)} }\) 
Here is a video of an application of an ellipse in polar coordinates. At this point, you should be able to understand the equations in this video. We hope you find it interesting to see an application of these equations.
video by TU Delft Online Learning 

Hyperbola 

\(e=2\) 
\(d=1\) 

\(\displaystyle{r = \frac{2}{1+2\sin(\theta)} }\) 
\(e=2\) 
\(d=1\) 

\(\displaystyle{r = \frac{2}{12\sin(\theta)} }\) 
\(e=2\) 
\(d=1\) 

\(\displaystyle{r = \frac{2}{1+2\cos(\theta)} }\) 
\(e=2\) 
\(d=1\) 

\(\displaystyle{r = \frac{2}{12\cos(\theta)} }\) 
Problem Statement 

Find the polar equation of the hyperbola with eccentricity = \(1.5\) and directrix \(y=2\).
Solution 

video by Krista King Math 

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Determining the Type of Conic Section From the Equation 

After studying the previous sets of graphs, you should have started to get a handle on how the graphs and equations are related. You will probably be asked to determine the type of conic from the equation. You already know that the eccentricity will help you a lot to determine the general type.
This video contains several examples, showing details on what to look for.
video by MIP4U 

Problem Statement 

Identify the conic, find the eccentricity and directrix and sketch the conic with equation \(\displaystyle{ \frac{9}{6+2\cos\theta} }\).
Solution 

video by Krista King Math 

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Problem Statement 

Identify the conic given by the polar equation \(\displaystyle{ r = \frac{5}{1015\sin \theta} }\), then determine the directrix and eccentricity.
Solution 

video by PatrickJMT 

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Problem Statement 

Write the polar equation of the conic with directrix \(x=3\) and eccentricity = \(2/3\).
Solution 

video by PatrickJMT 

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Problem Statement 

Find the polar equation of the ellipse with eccentricity = \(1/2\) and directrix \( r = \sec \theta \).
Solution 

video by Krista King Math 

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Problem Statement 

Graph \(\displaystyle{ r = \frac{8}{22\cos\theta} }\) and label all key components.
Solution 

video by MIP4U 

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Problem Statement 

Graph \(\displaystyle{ r = \frac{8}{4+2\sin\theta} }\) and label all key components.
Solution 

video by MIP4U 

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Problem Statement 

Graph \(\displaystyle{ r = \frac{8}{24\sin\theta} }\) and label all key components.
Solution 

video by MIP4U 

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Problem Statement 

Find the intercepts and foci of \(\displaystyle{ r = \frac{4}{42\cos\theta} }\).
Solution 

video by MIP4U 

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Problem Statement 

Find the intercepts and foci of \(\displaystyle{ r = \frac{6}{33\sin\theta} }\).
Solution 

video by MIP4U 

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Problem Statement 

Find the intercepts and the foci of \(\displaystyle{ r = \frac{12}{2+6\cos\theta} }\).
Solution 

video by MIP4U 

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