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Chapter 11: Problem 34

(a) Find the eccentricity, and identify the conic. (b) Sketch the conic, and label the vertices. $$r=\frac{5}{2-3 \sin \theta}$$

Short Answer

Expert verified
The conic is a hyperbola with eccentricity 3.

Step by step solution

01

Identify the type of conic

The general form of the polar equation of a conic is \( r = \frac{ed}{1 - e\sin \theta} \) or \( r = \frac{ed}{1 - e\cos \theta} \). Compare the given equation \( r = \frac{5}{2 - 3\sin\theta} \) with the general form, recognizing that \( e = 3 \) and \( ed = 5 \).
02

Find the eccentricity

We identified \( e \) as 3. Eccentricity \( e \) characterizes the conic section: if \( e > 1 \), it is a hyperbola; if \( e = 1 \), a parabola; if \( e < 1 \), an ellipse.
03

Identify the conic

Since \( e = 3 \) and \( e > 1 \), the conic is a hyperbola.
04

Determine the vertices

The vertices of the hyperbola can be found when \( \theta = \frac{\pi}{2} \) and \( \theta = \frac{3\pi}{2} \), as these make \( \sin \theta = \pm 1 \). Substitute these into the equation: \( r = \frac{5}{2-3(\pm 1)} \), which simplifies to either \(-5\) or \(5\), positioning the vertices at these radii.
05

Sketch the hyperbola

Plot the polar grid, marking the vertices at \( r = 5 \) and \( r = -5 \), and sketch the hyperbola opening upward and downward through the vertices. Make sure to indicate the direction of the branches correctly.

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Key Concepts

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

Polar Coordinates
Polar coordinates offer a different way to look at points on a plane using angles and distances. Unlike the Cartesian system that employs x and y axes, polar coordinates use a radial distance and an angle from a reference direction.
Imagine a dartboard: the center or bullseye is your origin (called the pole in polar coordinates). The angle ( \( \theta \) ) tells you which direction from the pole you'll measure the distance ( \( r \) , or radius) to find your point.
  • In polar coordinates, every point is represented as \( (r, \theta) \) .
  • If \( r \) is positive, you move away from the pole in the direction of \( \theta \).
  • If \( r \) is negative, you move in the opposite direction of \( \theta \).
In our problem, \( r = \frac{5}{2-3\sin \theta} \) needs to be graphed using polar coordinates, where the equation depends on the \( \sin \theta \) function to determine the conic section's shape.
Eccentricity
Eccentricity is a key concept in analyzing conic sections. It helps to determine the type and shape of a conic section based on its value. Eccentricity is denoted by \( e \) , a ratio that measures how much a conic section deviates from being circular.
Here鈥檚 how eccentricity works for different conics:
  • If \( e = 0 \) , the conic is a circle.
  • If \( 0 < e < 1 \) , you have an ellipse.
  • If \( e = 1 \) , this indicates a parabola.
  • And if \( e > 1 \) , it is a hyperbola.
In our specific exercise, when we compare \( r = \frac{5}{2-3\sin \theta} \) to the general form, we found that \( e = 3 \) . This means, since \( e > 1 \) , the conic section we are dealing with is a hyperbola, characterized by two distinct branches that open away from each other.
Hyperbolas
A hyperbola is a type of conic section that appears like two mirrored curves opening in opposite directions. These curves are a result of a plane intersecting both halves of a double cone. In polar coordinates, hyperbolas can be expressed using eccentricity greater than 1, as seen in the equation \( r = \frac{5}{2-3\sin \theta} \).
The simplest way to imagine a hyperbola is to think about removing material in the form of two opposing dishes from a solid cone.
  • Hyperbolas have two branches. Each branch is shaped like a curve that approaches but never touches specific lines called asymptotes.
  • In our exercise, we determined vertices by setting \( \theta = \frac{\pi}{2} \) and \( \theta = \frac{3\pi}{2} \) . Substituting these into the equation gives the vertices at \( r=5 \) and \( r=-5 \).
  • When sketching a hyperbola in polar coordinates, the branches will open in the respective direction of the determined vertices.
Emphasizing the role of eccentricity, the polar equation governs the shape and orientation of the hyperbola, in this case, causing the divisions along a line specified by the polar angle.

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