Question

Graph each polar equation. $$ r=\frac{1}{3-2 \cos \theta} \quad(\text {ellipse}) $$

Short answer

Identify conic, calculate key points, sketch ellipse.

Step-by-step solution

- Understand the Polar Equation

Given the polar equation is \( r=\frac{1}{3-2 \cos \theta} \), identify that it represents a conic section (ellipse) with a focus at the pole and a directrix.

- Determine the Conic Type

To confirm the conic type, check the denominator of the polar equation. Analyze the term \( 3-2\cos\theta \). Since the coefficient of \(\cos\theta\) (which is -2) leads to an eccentricity \(e\) less than 1, this indicates an ellipse.

- Identify Key Features

For the ellipse, command \(r = \frac{ed}{1 + e \cos \theta}\). Comparing this with the given equation, determine the eccentricity \(e = \frac{2}{3}\). The directrix can be found using the form of the equation.

- Evaluate Critical Points

Find values of \(r\) at critical points \( \theta = 0, \pi/2, \pi, 3\pi/2 \). Example: For \( \theta = 0 \), \( r = \frac{1}{3-2*1}=1 \) etc.

- Sketch the Graph

Plot the calculated points. Sketch the polar plot with these coordinates ensuring to illustrate an ellipse centered around the pole and passing through the identified critical points.

Key concepts

ellipse in polar coordinates

Understanding how an ellipse can be represented in polar coordinates is essential for mastering polar equations. An ellipse in polar coordinates has a distinct form that usually involves a directrix and an eccentricity. The given polar equation is \( r = \frac{1}{3-2 \cos \theta} \). This equation mirrors the standard form for conic sections in polar coordinates, which is \( r = \frac{ed}{1 + e \, \cos \, \theta} \) where \(e\) is the eccentricity and \(d\) is the distance from the focus to the directrix. In this context:
\- \(r\): Radius
\- \(\theta\): Angle
\- \(\cos \theta \): Determines the horizontal component of the radius

By comparing our given equation with the standard form, we can see that the denominator of our given equation helps us identify the ellipse by investigating the coefficient of \( \cos \theta \). Here, the coefficient is 2, which plays a key role in figuring out the eccentricity.

eccentricity of conic sections

Eccentricity is a measure of how much a conic section deviates from being circular. For ellipses, eccentricity values are always less than 1. This value helps to determine the shape and nature of the ellipse. In our given equation \(r = \frac{1}{3-2 \cos \theta}\), we can determine the eccentricity by comparing it with the general form \( r=\frac{ed}{1+e \cos \theta} \). From the standard form:
\- The numerator contains the product of the eccentricity \(e\) and the directrix distance \(d\).
\- The denominator's constant term indicates the distance between the center and the directrix.

Given the equation \(r = \frac{1}{3-2 \cos \theta}\), we can identify the coefficient of \( \cos \theta \) as \(-2 \). Therefore, \(e = \frac{2}{3}\), revealing that our ellipse has an eccentricity of \( \frac{2}{3}\). This value implies that the ellipse is moderately elongated and not too narrow or too wide.

polar graph plotting

Plotting a polar equation requires identifying key points and understanding the shape. To graph the equation \( r = \frac{1}{3-2 \cos \theta} \), we need to compute \( r \) for critical points of the angle \( \theta \). Typical values include \( \theta = 0, \pi/2, \pi, 3\pi/2 \):

\- When \( \theta = 0 \): \( r = \frac{1}{3-2 \times 1} = \frac{1}{1} = 1\)
\- When \( \theta = \pi/2 \): \( r = \frac{1}{3-2 \times 0} = \frac{1}{3} \)
\- When \( \theta = \pi \): \( r = \frac{1}{3-2 \times (-1)} = \frac{1}{5} \)
\- When \( \theta = 3\pi/2 \): \( r = \frac{1}{3-2 \times 0} = \frac{1}{3} \)

Plotting these points on a polar graph involves converting them into Cartesian coordinates if necessary. For instance, the point \( (r, \theta) = (1, 0) \) translates to (1,0) in Cartesian. After plotting these key points, connect them smoothly to reveal the ellipse. This representation helps to understand how the ellipse behaves in polar coordinates.