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Graph each conic section. If the conic is a parabola, specify (using rectangular coordinates) the vertex and the directrix. If the conic is an ellipse, specify the center, the eccentricity, and the lengths of the major and minor axes. If the conic is a hyperbola, specify the center, the eccentricity, and the lengths of the transverse and conjugate axes. $$r=\frac{24}{2-3 \cos \theta}$$

Short Answer

Expert verified
The conic is a hyperbola with eccentricity \( \frac{3}{2} \).

Step by step solution

01

Identify the Type of Conic

Recognize that the given equation \( r = \frac{24}{2 - 3 \cos \theta} \) is in the form of a polar equation. It has the form \( r = \frac{ed}{1 - e \cos \theta} \) or \( r = \frac{ed}{1 - e \sin \theta} \), which represents a conic section with an eccentricity \( e \). Since \( e = 3/2 > 1 \), the conic section is a hyperbola.
02

Determine the Parameters of the Hyperbola

For a hyperbola in polar form, the eccentricity \( e \) is given by the coefficient of \( \cos \theta \), which is 3/2 in this case. The directrix \( d \) is found using the equation \( r = \frac{ed}{1 - e \cos \theta} = \frac{24}{2 - 3 \cos \theta} \), so \( ed = 24 \) and \( d = 16 \).
03

Convert to Cartesian Coordinates

To better understand the hyperbola, convert the polar equation to rectangular form: \( r = \frac{24}{2 - 3 \cos \theta} \) implies \( r(2 - 3 \cos \theta) = 24 \). Replace \( r \) with its Cartesian form \( r = \sqrt{x^2 + y^2} \) and \( \cos \theta = \frac{x}{\sqrt{x^2 + y^2}} \) to get \( \sqrt{x^2 + y^2} (2 - 3 \frac{x}{\sqrt{x^2 + y^2}}) = 24 \). This simplifies to \( 2\sqrt{x^2 + y^2} = 24 + 3x \).
04

Center, Eccentricity and Axis

The center of the hyperbole is the point around which the hyperbola is symmetric, commonly at the origin in polar coordinates. The eccentricity \( e \) is 3/2, as determined earlier. The lengths of the transverse and conjugate axes can be calculated, but are primarily represented in polar form by the relation involving \( e \).
05

Graph the Hyperbola

To graph the hyperbola, plot points by calculating values of \( r \) for different angles \( \theta \) from 0 to 2\( \pi \). Plot the corresponding Cartesian coordinates. The hyperbola will open along the x-axis, centered at the pole.

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Key Concepts

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

Hyperbola
A hyperbola is a type of conic section formed when a double cone is cut by a plane such that the angle between the plane and the base of the cone is smaller than the angle between the side of the cone and its axis. Unlike ellipses and parabolas, hyperbolas consist of two separate curves called branches.

Key characteristics include:
  • Two distinct curves or branches
  • An eccentricity greater than 1 (e > 1)
  • A center, which is the midpoint between the foci of the hyperbola
  • Axes of symmetry, usually intersecting at the center
In the case of the polar equation \( r = \frac{24}{2 - 3 \cos \theta} \), we've identified it as a hyperbola due to the eccentricity \( e = \frac{3}{2} \), which is greater than 1.

Thus, this hyperbola opens outwards along the x-axis and has two branches that extend infinitely outward. The parameter 'e', or eccentricity, indicates how stretched the hyperbola is compared to a perfect circle.
Polar Coordinates
Polar coordinates offer an alternative method to Cartesian coordinates for representing points in a plane. Instead of using x and y, polar coordinates describe a point with the distance from the origin (r) and the angle (\(\theta\)) from a reference direction (usually the positive x-axis).

The key components of polar coordinates are:
  • The radial coordinate \( r \), representing the distance from the origin
  • The angular coordinate \( \theta \), usually in radians, which specifies the direction from the reference axis
The conversion between polar and Cartesian is performed using the formulas:
  • \( x = r \cos \theta \)
  • \( y = r \sin \theta \)
  • \( r = \sqrt{x^2 + y^2} \)
  • \( \theta = \tan^{-1}(y/x) \)
In our example, \( r = \frac{24}{2 - 3 \cos \theta} \) uses polar coordinates, which help to graph complex shapes like the hyperbola by understanding how points relate to a central origin or focus.

This system simplifies the expression of conic sections and other curves where patterns related to angles and distances are more apparent.
Graphing Conic Sections
Graphing conic sections involves understanding the shape, position, and orientation of curves like ellipses, parabolas, and hyperbolas, and plotting these in a coordinate system. Each conic section has its unique attributes that aid in graphing:

- **Vertices and Foci**: For hyperbolas and ellipses, plotting these points helps in determining the shape of the graph.- **Axes**: Hyperbolas have transverse and conjugate axes, fundamental in defining its extent and orientation in the plane.
For the given polar equation \( r = \frac{24}{2 - 3 \cos \theta} \), converting to Cartesian coordinates can help graph the hyperbola by understanding its behavior in the x-y plane.

To graph:
  • Calculate \( r \) for various \( \theta \), especially those that cause the expression \((2 - 3 \cos \theta)\) to create maximum deviation in \( r \).
  • Use these \( r \) and \( \theta \) values to find corresponding Cartesian points.
  • Plot these points to visualize the hyperbola's branches.
This method simplifies the interaction between polar and Cartesian systems and helps in understanding complex shapes that are regular in polar form.

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Most popular questions from this chapter

In designing an arch, architects and engineers sometimes use a parabolic arch rather than a semicircular arch. (One reason for this is that, in general, the parabolic arch can support more weight at the top than can the semicircular arch.) In the following figure, the blue arch is a semicircle of radius \(1,\) centered at the origin. The red arch is a portion of a parabola. As is indicated in the figure, the two arches have the same base and the same height. Assume that the unit of distance for each axis is the meter. GRAPH CANT COPY (a) Find the equation of the parabola in the figure. (b) Using calculus, it can be shown that the area under this parabolic arch is \(\frac{4}{3} \mathrm{m}^{2} .\) Assuming this fact, show that the area beneath the parabolic arch is approximately \(85 \%\) of the area beneath the semicircular arch. (c) Using calculus, it can be shown that the length of this parabolic arch is \(\sqrt{5}+\frac{1}{2} \ln (2+\sqrt{5})\) meters. Assuming this fact, show that the length of the parabolic arch is approximately \(94 \%\) of the length of the semicircular arch.

Find the coordinates of a point \(P\) in the first quadrant on the ellipse \(9 x^{2}+25 y^{2}=225\) such that \(\angle F_{2} P F_{1}\) is a right angle.

As is the case for most asteroids in our solar system, the orbit of the asteroid Gaspra is located between the orbits of Mars and Jupiter. Given that its distances from the Sun at perihelion and aphelion are \(2.132 \mathrm{AU}\) and \(2.288 \mathrm{AU}\) respectively, compute the length of the semimajor axis of the orbit and the eccentricity. Round both answers to two decimal places. Remark: Within the past decade the Galileo spacecraft (on a mission to Jupiter) took images of Gaspra, which can be viewed on the World Wide Web. For instance, at the time of this writing, one NASA website containing these images is httpi//www.jpl.nasa.gov/galileo. From this web page, use the search feature there to find "Gaspra."

Determine the equation of the hyperbola satisfying the given conditions. Write each answer in the form \(A x^{2}-B y^{2}=\) Cor in the form \(A y^{2}-B x^{2}=C\). Asymptotes \(y=\pm \frac{1}{2} x ;\) vertices (±2,0)

At the point \(P\) on the hyperbola \(x^{2} / a^{2}-y^{2} / b^{2}=1\) a tangent line is drawn that meets the lines \(x=a\) and \(x=-a\) at \(S\) and \(T,\) respectively. Show that the circle with \(\overline{S T}\) as a diameter passes through the two foci of the hyperbola.

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