Problem 3 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-12: Find the eccentricity, and classify the conic. Sketch the graph, and label the vertices. $$ r=\frac{12}{2-6 \cos \theta} $$

Step-by-Step Solution

Answer

The eccentricity is 3, classifying the conic as a hyperbola with vertices at (-3, 0) and (1.5, π). Sketch shows hyperbola branches on the polar axis.

Step 1: Identify the Conic Type

The given equation is in the form \( r = \frac{ed}{1 - e \cos \theta} \), which represents a conic section with focus at the pole (origin). In the given exercise, we have \( r = \frac{12}{2 - 6\cos \theta} \). Here, \( ed=12 \), \( e=6 \), and the general denominator format is \( 1 - e \cos \theta \). We'll rewrite this equation in standard form by dividing both numerator and denominator by \( 2 \): \[ r = \frac{\frac{12}{2}}{1 - \frac{6}{2} \cos \theta} = \frac{6}{1 - 3 \cos \theta} \]. This indicates \( e=3 \).

Step 2: Calculate the Eccentricity

The parameter \( e \) is already identified in the process above as 3, which is the eccentricity of the conic section. In conic sections:- \( e = 1 \) corresponds to a parabola,- \( e < 1 \) to an ellipse,- \( e > 1 \) to a hyperbola. Hence, with \( e = 3 > 1 \), the conic is classified as a hyperbola.

Step 3: Identify Characteristics and Find Vertices

The given equation \( r = \frac{6}{1 - 3 \cos \theta} \) indicates the hyperbola is oriented along the polar axis (along \( \cos \theta \), which means horizontal orientation). The vertices occur when \( \theta = 0 \) and \( \theta = \pi \):1. When \( \theta = 0 \), \[ r = \frac{6}{1 - 3 \cos 0} = \frac{6}{1 - 3} = -3 \], which represents the vertex on the polar axis direction backward (on the negative axis).2. When \( \theta = \pi \), \[ r = \frac{6}{1 - 3 \cos \pi} = \frac{6}{1 + 3} = \frac{6}{4} = 1.5 \], which is the vertex on the positive polar axis.

Step 4: Sketch the Graph

To sketch the graph, realize it is a hyperbola centered at the pole with two vertices calculated:- One vertex at \( (r, \theta) = (-3, 0) \) (extends negatively or backward on the polar plot),- The other vertex at \( (r, \theta) = (1.5, \pi) \) (extends positively on the polar plot).Since it's a hyperbola, sketch the two open branches extending away from these vertices, reflecting the hyperbolic nature.

Key Concepts

Eccentricity

When we talk about the eccentricity of a conic section, we refer to a number that describes how much a conic section deviates from being circular. This is a crucial parameter that helps classify conics into different types. It’s like a tool that helps us understand the shape of the curve:

• If the eccentricity \( e = 0 \), the conic is a perfect circle.
• When \( 0 < e < 1 \), it forms an ellipse.
• If \( e = 1 \), the conic is a parabola.
• With \( e > 1 \), you have a hyperbola.

In the problem, we found the eccentricity \( e = 3 \). Since \( 3 > 1 \), the conic section is a hyperbola. Understanding eccentricity is essential as it not only helps in identifying the type of conic but also gives insight into its geometric properties.

Hyperbola

A hyperbola is one of the classic types of conic sections. It differs significantly from ellipses and parabolas because it is made of two separate curves called branches. These branches are mirror images and open in opposite directions. The key features of a hyperbola include:

• Two vertices located on the central axis.
• Two foci that lie further away from the center than the vertices.
• An eccentricity greater than 1.

In our equation, \( r = \frac{6}{1 - 3 \cos \theta} \), we determined that it forms a hyperbola due to the eccentricity being greater than 1. The vertices were found at two specific positions on the polar axis. Specifically, when \( \theta = 0 \) and \( \theta = \pi \), we identified it at \( r = -3 \) and \( r = 1.5 \) respectively. This helps us visualize where the central axis lies, and where each branch of the hyperbola will extend from. Recognizing these aspects lets us better understand the nature and direction of hyperbolas.

Polar Coordinates

Polar coordinates provide an alternative way to describe the position of points. Unlike Cartesian coordinates, which use \( x \) and \( y \), polar coordinates utilize distance from a single point and an angle. This form becomes particularly handy when dealing with problems involving rotation or symmetry, such as those with circular or conic sections.
In polar coordinates, a point is represented as \( (r, \theta) \):

• \( r \) is the radial distance from the origin (or pole).
• \( \theta \) is the angle from the polar axis (usually the positive x-axis in a typical coordinate system).

The given exercise uses polar coordinates to express the hyperbola's equation. This is useful because it inherently connects the geometric properties of the hyperbola to its orientation. For example, the given equation \( r = \frac{6}{1 - 3 \cos \theta} \) clearly shows the dependence on \( \cos \theta \). Thus, the structure of the hyperbola reflects its placement along the polar axis. Understanding polar coordinates allows us to see how the geometric figures behave with respect to different angles, providing a richer understanding of complex shapes such as conic sections.