Short Answer

Use polar coordinates to graph the conics in Exercises 44–51.
$r = \frac{2}{1 + 3 \cos θ}$

The graph is

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

$r = \frac{2}{1 + 3 \cos θ} .$ The given equation is

Step 2. Comparison.

On comparing the given equation with the standard form,

$r = \frac{e u}{1 + e \cos θ} The eccentricity e = 3 Here e u = 2 3 \cdot u = 2 \Rightarrow u = \frac{2}{3}$

Step 3. Graph of the equation.

For $r = \frac{2}{1 + 3 \cos θ}$ the eccentricity is $e = 2$ , so the graph is a hyperbola. The directrix is perpendicular to the polar axis at a distance $u = \frac{2}{3}$ units to the right of the pole. The directrix will be the vertical line $x = \frac{2}{3} .$

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Short Answer

Use polar coordinates to graph the conics in Exercises 44–51.
$r = \frac{2}{1 + 3 \cos θ}$

Step by Step Solution

Step 1. Given information.

Step 2. Comparison.

Step 3. Graph of the equation.

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Use polar coordinates to graph the conics in Exercises 44–51. r=21+3cosθ

Step by Step Solution

Step 1. Given information.

Step 2. Comparison.

Step 3. Graph of the equation.

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Use polar coordinates to graph the conics in Exercises 44–51.
$r = \frac{2}{1 + 3 \cos θ}$