Problem 25 - College Algebra

Question

For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity. $$ r=\frac{5}{5-11 \sin \theta} $$

Step-by-Step Solution

Answer

Hyperbola with eccentricity 11/5 and directrix y = -25/11.

Step 1: Identify the standard form of the conic equation

The given conic equation is \( r = \frac{5}{5 - 11 \sin \theta} \). To identify the conic type, we generalize it using the polar form \( r = \frac{ed}{1 - e \sin \theta} \), where \( e \) is the eccentricity and \( d \) corresponds to the numerator and is called the semi-latus rectum, and a constant shift is applied.

Step 2: Match and identify parameters

The given equation \( r = \frac{5}{5 - 11 \sin \theta} \) confirms that \( ed = 5 \) and \( e = 11/5 \) when compared against the standard form \( r = \frac{ed}{1 - e \sin\theta} \) with a directrix that is horizontal (y-axis related), allowing the identification of \( e = 11/5 > 1 \), indicating a hyperbola.

Step 3: Calculate eccentricity

Eccentricity \( e \) is given directly in the equation as \( 11/5 \). Therefore, the eccentricity is \( e = 11/5 \) which confirms the conic is a hyperbola because \( e > 1 \).

Step 4: Find the directrix

The directrix is found using the formula \( d = \frac{ed}{e} = \frac{5}{11/5} \), which simplifies to \( d = 25/11 \). Thus, when \( r = \frac{ed}{1 - e \sin \theta} \), the line \( y = -\frac{25}{11} \) is validated for the position of the directrix, parallel to and below the focus at the origin.

Key Concepts

Polar Coordinates

Polar coordinates provide a unique way to represent points in the plane using a distance and an angle. Unlike the Cartesian coordinate system, where points are defined by x and y values, polar coordinates use

• the radius \( r \), the distance from a reference point known as the pole (analogous to the origin in Cartesian coordinates), and
• the angle \( \theta \), measured from a reference direction (usually the positive x-axis) in radians or degrees.

For presenting curves in polar coordinates, equations often relate the radius \( r \) to the angle \( \theta \). In the exercise, the given equation \( r = \frac{5}{5 - 11 \sin \theta} \) describes a conic section centered at the pole (origin), with the nature of the conic determined by its structure in polar form. Understanding polar coordinates provides crucial insight into the geometric behavior of these conics.

Eccentricity

Eccentricity \( e \) is a parameter that determines the shape of a conic section. It is calculated differently based on the type of conic section:

• A circle has an eccentricity of \( e = 0 \).
• An ellipse has \( 0 < e < 1 \).
• A parabola has \( e = 1 \).
• A hyperbola, like in this exercise, has \( e > 1 \).

By analyzing the formula \( r = \frac{ed}{1 - e \sin \theta} \), we can find the eccentricity of the conic. For this specific problem, equating terms reveals that \( e = \frac{11}{5} \). This value confirms that our conic is a hyperbola, highlighting how eccentricity provides insight into the nature and type of the conic.

Hyperbola

A hyperbola is one of the fascinating conic sections characterized by its open, double-curved shape. It is defined as the set of all points \((x, y)\) for which the difference of the distances to two fixed points (foci) is a constant.
In polar coordinates, a hyperbola can be described by an equation of the form \( r = \frac{ed}{1 - e \cos \theta} \) or \( r = \frac{ed}{1 - e \sin \theta} \). In the problem, the form used is \( r = \frac{ed}{1 - e \sin \theta} \), indicating that the directrix is horizontally aligned. The given eccentricity \( e = 11/5 \) confirms the hyperbolic nature, verifying its characteristic that \( e > 1 \).
Hyperbolas have an interesting property which makes them ideal in various applications, ranging from radio telescopes to navigation systems, where precise focusing capabilities are essential.

Directrix

The directrix of a conic section is a fixed line used in the formal definition of the curve. It serves as a reference for measuring distances that help define the shape of the resulting conic.
In terms of conic sections in polar coordinates, if you have the equation \( r = \frac{ed}{1 - e \sin \theta} \), the directrix determines how the cone is oriented:

• For hyperbolas and ellipses, the directrix allows one to relate the shape of the conic to its eccentricity.
• In the given solution, the directrix was found to be \( y = -\frac{25}{11} \)

This derives from the relationship \( d = \frac{ed}{e} = \frac{5}{11/5} = \frac{25}{11} \), giving insight into the hyperbola's spread around its focal point. Understanding the directrix helps in visualizing and graphing complex conic forms effectively.