## 17Calculus - Conics in Polar Coordinates

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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.

$$\displaystyle{ r = \frac{ed}{1\pm e \sin\theta}}$$ 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.

eccentricity type ellipse parabola hyperbola

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 [4min-7secs]

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 $$e=1$$

$$d=2$$

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

$$d=2$$

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

$$d=2$$

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

$$d=2$$

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

Practice

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

Problem Statement

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

Hint

You are not given all the information about this parabola in order to determine an exact parabola. So your final answer will have a variable of $$\theta$$ in it.

Problem Statement

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

Hint

You are not given all the information about this parabola in order to determine an exact parabola. So your final answer will have a variable of $$\theta$$ in it.

Solution

### 1610 video

video by Krista King Math

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Ellipse $$e=3/4$$

$$d=4/3$$

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

$$d=4/3$$

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

$$d=4/3$$

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

$$d=4/3$$

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

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.

### TU Delft Online Learning - The Trajectory Equation [14min-31secs]

Practice

Find the polar equation of the ellipse with eccentricity = $$1/2$$, directrix $$r=4 \sec \theta$$ and focus $$(0,0)$$.

Problem Statement

Find the polar equation of the ellipse with eccentricity = $$1/2$$, directrix $$r=4 \sec \theta$$ and focus $$(0,0)$$.

Solution

### 2635 video

video by Krista King Math

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A conic section is given by the polar equation $$\displaystyle{ r = \frac{10}{3-2\cos\theta} }$$. Find the eccentricity, identify the conic, locate the directrix and sketch the conic.

Problem Statement

A conic section is given by the polar equation $$\displaystyle{ r = \frac{10}{3-2\cos\theta} }$$. Find the eccentricity, identify the conic, locate the directrix and sketch the conic.

Solution

### 2636 video

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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

Find the polar equation of the hyperbola with eccentricity = $$1.5$$ and directrix $$y=2$$.

Problem Statement

Find the polar equation of the hyperbola with eccentricity = $$1.5$$ and directrix $$y=2$$.

Solution

### 1611 video

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.

### MIP4U - Ex: Determine the Type of Conic Section Given a Polar Equation [4min-15secs]

video by MIP4U

Practice

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

Problem Statement

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

Solution

### 1612 video

video by Krista King Math

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Identify the conic given by the polar equation $$\displaystyle{ r = \frac{5}{10-15\sin \theta} }$$, then determine the directrix and eccentricity.

Problem Statement

Identify the conic given by the polar equation $$\displaystyle{ r = \frac{5}{10-15\sin \theta} }$$, then determine the directrix and eccentricity.

Solution

### 1613 video

video by PatrickJMT

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Write the polar equation of the conic with directrix $$x=3$$ and eccentricity = $$2/3$$.

Problem Statement

Write the polar equation of the conic with directrix $$x=3$$ and eccentricity = $$2/3$$.

Solution

### 1614 video

video by PatrickJMT

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

Problem Statement

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

Solution

### 1615 video

video by Krista King Math

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Graph $$\displaystyle{ r = \frac{8}{2-2\cos\theta} }$$ and label all key components.

Problem Statement

Graph $$\displaystyle{ r = \frac{8}{2-2\cos\theta} }$$ and label all key components.

Solution

### 1616 video

video by MIP4U

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

Problem Statement

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

Solution

### 1617 video

video by MIP4U

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Graph $$\displaystyle{ r = \frac{8}{2-4\sin\theta} }$$ and label all key components.

Problem Statement

Graph $$\displaystyle{ r = \frac{8}{2-4\sin\theta} }$$ and label all key components.

Solution

### 1618 video

video by MIP4U

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

Problem Statement

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

Solution

### 1619 video

video by MIP4U

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

Problem Statement

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

Solution

### 1620 video

video by MIP4U

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

Problem Statement

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

Solution

### 1621 video

video by MIP4U

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### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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