Problem 10 - Algebra and Trigonometry

Question

Write a polar equation of a conic that has its focus at the origin and satisfies the given conditions. Ellipse, eccentricity 0.4, vertex at \((2,0)\)

Step-by-Step Solution

Answer

The polar equation is \( r = \frac{0.672}{1 + 0.4\cos\theta} \).

Step 1: Recall the Polar Equation of an Ellipse

The general polar equation for a conic section with the focus at the origin is given by \( r = \frac{ed}{1 + e\cos\theta} \), where \( e \) is the eccentricity and \( d \) is the semi-latus rectum. Since this is an ellipse, \( e < 1 \).

Step 2: Identify the Given Values

The eccentricity \( e = 0.4 \) and the vertex of the ellipse is at \((2,0)\). In a polar ellipse, the distance to the vertex is the semi-major axis \( a \). So, \( a = 2 \).

Step 3: Determine the Semi-Latus Rectum

The semi-latus rectum \( p \) can be found using the relationship \( a = \frac{p}{1 - e^2} \). Plug in the given values: \( 2 = \frac{p}{1 - 0.4^2} \). Solve for \( p \): \( p = 2(1 - 0.16) = 2(0.84) = 1.68 \).

Step 4: Write the Polar Equation

Substitute the values of eccentricity \( e = 0.4 \) and semi-latus rectum \( p = 1.68 \) into the polar equation \( r = \frac{ed}{1 + e\cos\theta} \). The polar equation becomes \( r = \frac{0.4 \times 1.68}{1 + 0.4\cos\theta} \). Simplify the expression: \( r = \frac{0.672}{1 + 0.4\cos\theta} \).

Key Concepts

Eccentricity

Eccentricity is a key concept in understanding conic sections. It is a measure of how "stretched" a conic section is. For ellipses, the eccentricity, denoted as \( e \), ranges from 0 to just under 1. An eccentricity of 0 would mean a perfect circle, while an eccentricity closer to 1 appears more stretched or elongated.

In the context of the original exercise, the ellipse has an eccentricity of 0.4. This value tells us that the shape is not a perfect circle but is not very elongated either. It's moderately stretched, maintaining a well-rounded form.

Understanding eccentricity helps us visualize the shape of conic sections without necessarily having to draw them. It determines how the conic section deviates from a circle, influencing other properties like its directrix and its latus rectum.

Semi-Latus Rectum

The semi-latus rectum is an important concept in the geometry of conic sections. It represents the distance from the focus to the curve along a line perpendicular to the major axis or directrix. This distance defines the width of the conic at its most curved part.

For the ellipse discussed, the exercise calculates the semi-latus rectum, denoted as \( p \), using the formula \( a = \frac{p}{1 - e^2} \), where \( a \) is the semi-major axis and \( e \) is the eccentricity. Substituting \( a = 2 \) and \( e = 0.4 \), we found \( p = 1.68 \).

The semi-latus rectum helps in determining the precise nature of the conic's curve, especially in how it expands out from the focus. It's a vital part of converting between different forms of conic section equations, particularly polar and Cartesian.

Conic Sections

Conic sections are the curves obtained by intersecting a plane with a cone. They include ellipses, parabolas, hyperbolas, and circles, each defined by distinct properties and equations.

Ellipses, with their closed-loop, are characterized by an eccentricity less than 1, like our example with \( e = 0.4 \). They have two axes—major and minor—determining their shape, with very specific geometric properties relating to their foci and directrices.

A unique aspect of conics in polar form is they can elegantly capture these shapes with an equation like \( r = \frac{ed}{1 + e\cos\theta} \). This polar equation shifts focus from one directrix, allowing for easy graphing when the focus is at the origin.

• Ellipses have eccentricities between 0 and 1.
• Parabolas have an eccentricity of exactly 1.
• Hyperbolas have an eccentricity greater than 1.

Understanding these geometrical characteristics of conic sections gives insight into their relationships with each other and their mathematical significance.