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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.

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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

\(e=3/4\)

\(d=4/3\)

\(\displaystyle{r = \frac{1}{1+0.75\sin(\theta)} }\)

\(e=3/4\)

\(d=4/3\)

\(\displaystyle{r = \frac{1}{1-0.75\sin(\theta)} }\)

\(e=3/4\)

\(d=4/3\)

\(\displaystyle{r = \frac{1}{1+0.75\cos(\theta)} }\)

\(e=3/4\)

\(d=4/3\)

\(\displaystyle{r = \frac{1}{1-0.75\cos(\theta)} }\)

Hyperbola

\(e=2\)

\(d=1\)

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

\(e=2\)

\(d=1\)

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

\(e=2\)

\(d=1\)

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

\(e=2\)

\(d=1\)

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

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 \(\displaystyle{r=\frac{5}{10-15\sin \theta}}\), 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 \(\displaystyle{r=\frac{8}{2-2\cos\theta}}\) 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 \(\displaystyle{r=\frac{8}{2-4\sin\theta}}\) and label all key components.

solution

Practice 10

Find the intercepts and foci of \(\displaystyle{ r=\frac{4}{4-2\cos\theta} }\).

solution

Practice 11

Find the intercepts and foci of \(\displaystyle{ r=\frac{6}{3-3\sin\theta} }\)

solution

Practice 12

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

solution

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