You CAN Ace Calculus  

17calculus > conics > polar form  


Topics You Need To Understand For This Page
Calculus Main Topics
Single Variable Calculus 

MultiVariable Calculus 
Tools
math tools 

general learning tools 
additional tools 
Related Topics and Links
external links you may find helpful 

Conics in Polar Coordinates 

on this page: ► parabola ► ellipse ► hyperbola ► determining the type of conic section from the equation 
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. 
Search 17Calculus 
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.
MIP4U  Graphing Conic Sections Using Polar Equations  Part 1  
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 

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

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

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

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

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

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

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

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

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

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

\(\displaystyle{r = \frac{1}{10.75\cos(\theta)} }\) 
Hyperbola 

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

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

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

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

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

Find the polar equation of the hyperbola with eccentricity = \(1.5\) and directrix \(y=2\). 
solution 
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.
MIP4U  Ex: Determine the Type of Conic Section Given a Polar Equation  
Practice 3 

Identify the conic, find the eccentricity and directrix and sketch the conic with equation \(\displaystyle{\frac{9}{6+2\cos\theta}}\). 
solution 
Practice 4 

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

Write the polar equation of the conic with directrix \(x=3\) and eccentricity = \(2/3\). 
solution 
Practice 6 

Find the polar equation of the ellipse with eccentricity = \(1/2\) and directrix \(r=\sec\theta\). 
solution 
Practice 7 

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

Graph \(\displaystyle{r=\frac{8}{4+2\sin\theta}}\) and label all key components. 
solution 
Practice 9 

Graph \(\displaystyle{r=\frac{8}{24\sin\theta}}\) and label all key components. 
solution 
Practice 10 

Find the intercepts and foci of \(\displaystyle{ r=\frac{4}{42\cos\theta} }\). 
solution 
Practice 11 

Find the intercepts and foci of \(\displaystyle{ r=\frac{6}{33\sin\theta} }\) 
solution 
Practice 12 

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