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Chapter 10: Problem 35

Give the eccentricities of conic sections with one focus at the origin along with the directrix corresponding to that focus. Find a polar equation for each conic section. $$e=1 / 5, \quad y=-10$$

Short Answer

Expert verified
The polar equation is \( r = \frac{10}{5 + \sin \theta} \).

Step by step solution

01

Understand the problem

We are given a conic section characterized by an eccentricity \( e = \frac{1}{5} \) and a directrix \( y = -10 \). We need to find the type of conic section, its properties, and the equation in polar coordinates.
02

Identify the Conic Section Type

The eccentricity \( e = \frac{1}{5} \) is less than 1, indicating the conic section is an ellipse. In general, different values of eccentricity correspond to different conic sections: if \( e < 1 \) it is an ellipse, \( e = 1 \) a parabola, and \( e > 1 \) a hyperbola.
03

Polar Equation of Conics

The general formula for a conic section in polar coordinates with one focus at the origin is given by \( r = \frac{ed}{1 + e \sin \theta} \) if the directrix is horizontal. Here, \( e \) is the eccentricity and \( d \) is the distance from the origin to the directrix.
04

Substitute Into Polar Equation

For our ellipse, substitute \( e = \frac{1}{5} \) and \( d = 10 \) into the formula. Thus, the equation becomes: \[ r = \frac{\frac{1}{5} \times 10}{1 + \frac{1}{5} \sin \theta} = \frac{2}{1 + \frac{1}{5} \sin \theta} \]
05

Simplify the Polar Equation

Simplify the equation further by multiplying the numerator and denominator by 5 to eliminate the fraction in the denominator:\[ r = \frac{10}{5 + \sin \theta} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Eccentricity
The eccentricity of a conic section is a measure that describes how much the conic deviates from being a circle. It is denoted by the letter \( e \) and is a positive real number. Different conics have different characteristics based on their eccentricity, allowing us to categorize them easily:
  • If \( e < 1 \), the conic is an ellipse, which looks like a stretched circle.
  • If \( e = 1 \), the conic is a parabola, which means it has a symmetry and opens like a letter 'U' or 'V'.
  • If \( e > 1 \), the conic is a hyperbola, where it consists of two separate curves facing away from each other.
In the given problem, with \( e = \frac{1}{5} \), which is less than 1, we are dealing with an ellipse. This value indicates the shape is quite close to being a circle, but still slightly stretched.
Polar Coordinates
Polar coordinates are a way of representing points in a plane using a radius and an angle. Instead of using the traditional \(x\) and \(y\) coordinates, polar coordinates use \(r\) (the distance from the origin) and \(\theta\) (the angle relative to the positive x-axis).
  • Conics in polar coordinates have their focus at the origin, making it convenient to describe them using this system.
  • The polar equation for a conic is \( r = \frac{ed}{1 + e\sin\theta} \), when dealing with a horizontal directrix, or \( r = \frac{ed}{1 + e\cos\theta} \) for a vertical directrix.
This representation simplifies finding the properties of conics like ellipses, parabolas, or hyperbolas directly related to their eccentricity and directrix. For the problem at hand, the coordinates allow us to replace the standard algebraic form with an equation that highlights these geometric features.
Ellipse
An ellipse is a type of conic section that resembles a flattened circle. This shape is characterized by its eccentricity being less than 1. Some properties unique to ellipses include:
  • Two focal points, with any point on the ellipse having a constant total distance to these foci.
  • A consistent shape described by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) in Cartesian coordinates, where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively.
When translated to polar coordinates, the ellipses have the form \( r = \frac{ed}{1 + e\sin\theta} \), which is directly related to their geometrical structure by involving the directrix and the eccentricity. In this case, using the defined \( e = \frac{1}{5} \) and directrix \( y = -10 \), we derived the equation \( r = \frac{10}{5 + \sin \theta} \), showing the relationship of each point of the ellipse to its single focus at the origin.
Directrix
The directrix of a conic section is a crucial reference line used in the definition and formulation of conics. It is not part of the conic itself but contributes to its shape by working alongside the focus:
  • The distance from any point on the conic to the directrix, compared to the distance to the focus, is always a fixed ratio equal to eccentricity.
  • For an ellipse, this distance ratio is less than 1, aligning with the nature of ellipses having "gentler" curves.
In polar coordinates, the directrix provides the distance \( d \) from the focus (origin) to the directrix line. This element is pivotal in deriving the polar equation. For example, in our topic, with the directrix positioned at \( y = -10 \), the polar equation was derived using this concept to show how the ellipse is structured concerning its focus and associated parameters.

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Most popular questions from this chapter

Graph the lines and conic sections. $$r=1 /(1+2 \cos \theta)$$

Sketch each circle in the coordinate plane and label it with both its Cartesian and polar equations. $$x^{2}+y^{2}-\frac{4}{3} y=0$$

Replace the polar equations with equivalent Cartesian equations. Then describe or identify the graph. $$r \sin \theta=-1$$

Give the eccentricities of conic sections with one focus at the origin along with the directrix corresponding to that focus. Find a polar equation for each conic section. $$e=1, \quad x=2$$

Give the eccentricities and the vertices or foci of hyperbolas centered at the origin of the \(x y\) -plane. In each case, find the hyperbola's standard-form equation in Cartesian coordinates. Eccentricity: 3 Foci: (卤3,0)
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