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Chapter 6: Problem 21

Identify the conic and sketch its graph. \(r=\frac{6}{2+\sin \theta}\)

Short Answer

Expert verified
The graph represents an ellipse with focus at the pole and directrix at y = -3. The eccentricity is 0.5.

Step by step solution

01

Identify the conic

First look at the equation \(r=\frac{6}{2+\sin \theta}\). It is expressed in the general form of a conic in polar coordinates \(r = \frac{ed}{1+e \sin \theta}\) or \(r = \frac{ed}{1-e \sin \theta}\). Here, e (called the eccentricity) determines the type of the conic. The denominator has +sin, therefore the focus of the curve is at the pole and the directrix is a vertical line (y = -d). The constant e is less than 1, which fits the parameters for an ellipse.
02

Calculate the eccentricity and the directrix

To determine the specifics of the ellipse, the eccentricity (e) and the directrix (d) should be calculated from the equation. Here, e = \(\frac{1}{2}\), and d = \(\frac{6}{2}\) = 3. Therefore the ellipse has an eccentricity of 0.5 and a directrix of y = -3.
03

Sketch the graph

Using the information determined in the previous steps, sketch the ellipse. The focus point is positioned at the pole (or origin), and the ellipse is extended vertically, based on the properties of the ellipse in polar coordinates. The ellipse approaches, but doesn't reach, the line y = -3 (the directrix). Since the eccentricity is less than 1, the shape is an vertically extended ellipse and doesn't open towards the directrix.

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