Problem 17 - Algebra and Trigonometry Real Mathematics, Real People

Question

Identify the type of conic represented by the equation. Use a graphing utility to confirm your result. $$r=\frac{4}{1-5 \cos \theta}$$

Step-by-Step Solution

Answer

The type of conic section represented by the equation \(r = \frac{4}{1 - 5 \cos(\theta)}\) is a Hyperbola.

Step 1: Identify the conic section

The general form for conics in Polar coordinates is \(r = \frac{ed}{1 \pm e \cos(\theta - \theta_0)}\) where e is the eccentricity and d is the distance from the pole to the directrix. Here, we can see that the equation \(r = \frac{4}{1 - 5 \cos(\theta)}\) can be rearranged as \(-5r \cos(\theta) + r = 4\). From the rearranged form, it is clear that eccentricity 'e' is equal to 5. Since e > 1, this suggests that the conic section is a Hyperbola.

Step 2: Graph the equation

By graphing \(r = \frac{4}{1 - 5 \cos(\theta)}\) using a graphing utility, it would be clear that the equation represents a Hyperbola, which would confirm the earlier calculation.

Key Concepts

Understanding Hyperbolas

A hyperbola is a type of conic section that can appear when slicing a cone with a plane in a manner that results in two distinct curves. These curves have unique properties and are defined by their eccentricity, which measures how "stretched" they are. In a hyperbola, the two curves mirror each other.
A hyperbola is characterized by having two foci and two asymptotes, which guide the shape of the curves. The standard Cartesian equation for a hyperbola has the form:

• Horizontal hyperbola: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
• Vertical hyperbola: \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)

In polar coordinates, a hyperbola's equation is given as \( r = \frac{ed}{1 \pm e \cos(\theta - \theta_0)} \), where the eccentricity \( e \) is greater than 1. With this parameter, we can determine its classification as hyperbolic. The equation from our exercise, \( r = \frac{4}{1 - 5 \cos(\theta)} \), exemplifies a hyperbola because its eccentricity \( e = 5 \), which is distinctly greater than 1.

Exploring Eccentricity in Conics

Eccentricity is a crucial concept in understanding conic sections, as it defines their shape. It is denoted by \( e \) and varies for different conics. Here's how it affects each type:

• For a circle, \( e = 0 \), meaning the conic is perfectly round.
• An ellipse has \( 0 < e < 1 \). It looks like a stretched circle.
• A parabola has \( e = 1 \), resulting in a U-shaped curve.
• For hyperbolas, \( e > 1 \), leading to two separate curves.

Eccentricity determines not just the shape, but also how the conic will behave around its points, like vertices and foci. In polar coordinates, the eccentricity also tells us whether our conic is a hyperbola if it's greater than 1. In the given exercise, since the eccentricity \( e = 5 \), we see a high degree of stretch, confirming the shape as a hyperbola.

Using Graphing Utilities

Graphing utilities are incredibly useful when it comes to visualizing equations and confirming analytical results. They can quickly plot complex equations involving polar coordinates, like those of a hyperbola.
A graphing utility, such as a graphing calculator or software like Desmos, helps to:

• Plot the shape of the conic section in real-time.
• Confirm the type of conic by visual appearance.
• Illustrate the symmetry and orientation of the hyperbola.

Using these tools, we can input the polar equation from our exercise: \( r = \frac{4}{1 - 5 \cos(\theta)} \). The graphing utility will display the hyperbolic curves, confirming the eccentricity and type deduced earlier. Visualizing the graph alongside your analytical solution provides a complete understanding of the conic's properties and behavior.