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Chapter 10: Problem 8

Write a polar equation of a conic with the focus at the origin and the given data. Hyperbola, eccentricity \(3, \quad\) directrix \(r=-6 \csc \theta\)

Short Answer

Expert verified
The polar equation is \( r = \frac{18}{1 + 3 \sin \theta} \).

Step by step solution

01

Understanding the Problem

We need to write the polar equation of a hyperbola, given its eccentricity and a vertical directrix. The conic section is centered at the origin in polar coordinates.
02

Recall the Polar Equation of a Conic

The general polar equation of a conic section with the focus at the origin is given by: \[ r = \frac{ed}{1 - e \sin \theta} \] or \[ r = \frac{ed}{1 - e \cos \theta} \]. Since our directrix is vertical (involving \( \csc \theta \)), we'll use the equation with \( \sin \theta \).
03

Interpret Given Directrix

The directrix is given by \( r = -6 \csc \theta \), which implies that the conic's directrix line is \( y = -6 \). The distance from the origin to this directrix is \( d = 6 \).
04

Apply Provided Data

Given the eccentricity \( e = 3 \), and knowing the form of the hyperbola with the directrix in terms of \( \sin \theta \), we substitute into the formula: \[ r = \frac{3 \times 6}{1 + 3 \sin \theta} \].
05

Simplify the Polar Equation

Simplify the equation: \[ r = \frac{18}{1 + 3 \sin \theta} \] is the polar equation of the hyperbola.

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Key Concepts

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

Hyperbola in Conics
A hyperbola is one of the several types of conic sections that can be formed by slicing a plane through a double-napped cone. Unlike the circle or the ellipse, a hyperbola consists of two distinct curves called branches. These branches appear as mirror images of one another and open outward. For a hyperbola, there are key defining features:
  • Two foci: distinct points on the plane where the difference in distances from any point on the hyperbola to these foci is constant.
  • Two asymptotes: lines that each branch of the hyperbola approaches but never actually touches.
Knowing how to express a hyperbola using polar coordinates helps explore its properties. Polar equations enable the examination of conics in a coordinate system centered not on the geometric center but rather at one of the foci, which can be particularly useful in orbital mechanics and other applications.
Eccentricity and Its Role
Eccentricity, denoted as \( e \), measures the deviation of a conic section from being circular. For a hyperbola, the eccentricity is always greater than 1. As \( e \) increases, the branches of the hyperbola open wider. Key points to remember about eccentricity include:
  • Eccentricity provides insight into the shape and proportion of the hyperbola.
  • Higher values of eccentricity mean the conic section is more elongated.
  • The eccentricity for our exercise was given as 3, indicating a relatively stretched shape.
Understanding eccentricity is crucial, as it dictates the deformed nature of the hyperbola, thereby influencing its polar equation. It's this property that differentiates conics from one another in the geometric sense.
Defining Directrix in Conics
A directrix is a line used in the definition of conics that assists in controlling their shape and positioning. For conics, there is a simple relationship between the directrix, the conic itself, and the focus:
  • The ratio of the distance of any point from the focus to its perpendicular distance from the directrix is constant and equals the eccentricity \( e \).
  • For hyperbolas, this helps set the orientation of their branches and specifies direction.
In the problem we have, the equation \( r = -6 \csc \theta \) identified a vertical directrix at \( y = -6 \). This directrix aids in drafting the hyperbola鈥檚 equation, guiding the branches鈥� curvature and width through its interaction with other geometric elements like the focus and the eccentricity.
Using Polar Coordinates
Polar coordinates provide a different way to describe points on a plane using the distance from a reference point (usually the origin) and an angle from a reference direction, commonly the positive x-axis. This can be incredibly useful for conic sections:
  • In polar coordinates, the origin often coincides with a focus, not the center.
  • This system simplifies the mathematical description of conics like hyperbolas, where focus-centered attributes are significant.
  • Using polar coordinates can transform complex cartesian equations into more manageable forms by aligning the focus.
With the polar equation,\[ r = \frac{18}{1 + 3 \sin \theta} \], we describe a hyperbola with the focus located at the origin. This alignments help highlight the conic's intrinsic features, making analysis straightforward and highlighting nuances otherwise lost in cartesian forms.

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Most popular questions from this chapter

The orbit of Halley's comet, last seen in 1986 and due to return in \(2062,\) is an ellipse with eccentricity 0.97 and one focus at the sun. The length of its major axis is 36.18 \(\mathrm{AU}\) . [An astronomical unit (AU) is the mean distance between the earth and the sun, about 93 million miles.] Find a polar equa- tion for the orbit of Halley's comet. What is the maximum distance from the comet to the sun?

Find the vertices, foci, and asymptotes of the hyperbola and sketch its graph. $$9 x^{2}-4 y^{2}=36$$

Graph the conics \(r=e /(1-e \cos \theta)\) with \(e=0.4,0.6\) . \(0.8,\) and 1.0 on a common screen. How does the value of \(e\) affect the shape of the curve?

The point in a lunar orbit nearest the surface of the moon is called perilune and the point farthest from the surface is called apolune. The Apollo 11 spacecraft was placed in an elliptical lunar orbit with perilune altitude 110 \(\mathrm{km}\) and apolune altitude 314 \(\mathrm{km}(\) above the moon). Find an equation of this ellipse if the radius of the moon is 1728 \(\mathrm{km}\) and the center of the moon is at one focus.

\(71-76\) Use a graphing device to graph the polar curve. Choose the parameter interval to make sure that you produce the entire curve. $$r=2-5 \sin (\theta / 6)$$
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