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Chapter 9: Problem 17

Graph and interpret the conic section. $$r=\frac{4}{2 \cos (\theta-\pi / 6)+1}$$

Short Answer

Expert verified
The given equation represents a hyperbola with an eccentricity of 4, oriented towards the right and shifted along the polar axis by 蟺/6.

Step by step solution

01

Express equation in standard form

The given equation, \( r=\frac{4}{2 \cos (\theta-\pi / 6)+1}\), can be rewritten in the form \\( r=\frac{e}{1+e \cos (胃 - 蠒)}\). Compare it to the standard form of the polar equation for a conic section: \( r=\frac{e}{1+e cos(胃 - 蠒)}\). Here, e represents the eccentricity. So, e for the given equation is 4.
02

Determine the type of conic

The eccentricity e, which is 4, determines the type of the conic section. \If e = 1, the conic is a parabola; if e < 1, it's an ellipse; if e > 1, it's a hyperbola. Hence, the given equation represents a hyperbola because e = 4 > 1.
03

Draw the graph of the conic section

Plot points for various values of 胃 and connect these points to get the graph. The polar axis is shifted by \( 蟺/6 \) and the graph opens to the right.

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Key Concepts

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

Eccentricity
Eccentricity is a key feature in determining the shape of a conic section. A conic section can be any of the following: a circle, ellipse, parabola, or hyperbola. The eccentricity, often denoted as \( e \), is a non-negative number that describes the extent to which the conic deviates from being circular.

Here鈥檚 how eccentricity helps you identify different conic sections:
  • If \( e = 0 \), the conic is a circle.
  • If \( e < 1 \), it is an ellipse.
  • If \( e = 1 \), it's a parabola.
  • If \( e > 1 \), the conic is a hyperbola.
In our example, the equation \( r=\frac{4}{2 \cos (\theta-\pi / 6)+1} \) reveals an eccentricity \( e = 4 \). Since \( e \) is greater than 1, the conic section is a hyperbola. Knowing the eccentricity can provide crucial insights even before you start sketching or plotting the graph.
Polar Coordinates
Polar coordinates offer a unique way of describing the positions of points in a plane, using a distance and an angle. This system is especially useful for graphing conic sections that are rotated or have axes that aren't aligned with the typical Cartesian coordinate frame.

In polar coordinates, each point is determined by:
  • \( r \): the radial distance from the origin (also known as the pole).
  • \( \theta \): the angle measured counterclockwise from the positive x-axis (polar axis).
Understanding polar coordinates aids in plotting the equation \( r=\frac{4}{2 \cos (\theta-\pi / 6)+1} \). The term \( \theta - \pi/6 \) indicates a shift in the polar axis by \( \pi/6 \), affecting how you interpret the conic section's orientation.
Graphing Conic Sections
Graphing a conic section involves plotting points that satisfy its equation and connecting these points smoothly. When the conic section is in polar form, such as \( r=\frac{4}{2 \cos (\theta-\pi / 6)+1} \), the steps are similar, but they require a consideration of the polar coordinate system.

Here鈥檚 a simple process to graph our conic:
  • Choose several values of \( \theta \) and compute corresponding \( r \) values using the equation.
  • Plot each point using the \( (r, \theta) \) coordinates on polar graph paper.
  • Because \( e = 4 \) indicates a hyperbola, expect two distinct curves opening in opposite directions, but primarily to the right.
  • Account for the shift of \( \theta \) by \( -\pi/6 \) in your plotting, which effectively rotates the graph.
As you plot these points and connect them, the plot reveals the overall shape and orientation of the hyperbola. The equation's parameters, including eccentricity, play a vital role in shaping the conic.

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Most popular questions from this chapter

If a hyperbolic mirror is in the shape of the right half of \(\frac{x^{2}}{3}-y^{2}=1,\) to which point will light rays following the path \(y=c(x-2)\) reflect?

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Identify the conic section and find each vertex, focus and directrix. $$\frac{(x-1)^{2}}{4}+\frac{(y-2)^{2}}{9}=1$$

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