/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Identify the conic section whose... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Identify the conic section whose equation is given\(;\) if it is an ellipse or hyperbola, state its eccentricity. $$r=\frac{-10}{2+3 \cos \theta}$$

Short Answer

Expert verified
Answer: The given equation corresponds to an irregular conic. Eccentricity cannot be determined for this conic as it is neither an ellipse nor a hyperbola.

Step by step solution

01

Convert polar to Cartesian coordinates

Recalling that \(x=r\cos \theta\) and \(y=r\sin \theta\), we replace r in the equation by its equivalent expressions in terms of x and y: $$r=\frac{-10}{2+3 \cos \theta}$$ $$x^2+y^2=\frac{-10x}{2+3\frac{x}{\sqrt{x^2+y^2}}}$$
02

Simplify the equation

Now, multiply both sides of the equation by \((2+3\frac{x}{\sqrt{x^2+y^2}})\) to get rid of the denominator: $$(x^2+y^2)(2+3\frac{x}{\sqrt{x^2+y^2}})=-10x$$ Simplify the equation: $$2x^2+2y^2+3x\sqrt{x^2+y^2}=-10x$$
03

Identify the conic section

Now we must analyze the equation of the curve to identify the type of conic section. We can observe a \(x^2\), \(y^2\) and an \(x\sqrt{x^2+y^2}\) terms. The presence of a xy-term generally indicates a rotation, and the presence of both x² and y² terms indicates that this equation corresponds to a conic section. As there are terms with both variables, and the equation cannot be simplified into the standard forms of a conic section, we use change of variables to find the conic section's equation.
04

Change of variables and new equation

Let's perform the following change of variables: $$u=x\sqrt{x^2+y^2}, \quad v=y\sqrt{x^2+y^2}$$ Now, substituting back in step 2 equation, we get (recall that \(x^2+y^2=\sqrt{x^2+y^2}\)): $$2\frac{u^2}{x^2+y^2}+2\frac{v^2}{x^2+y^2}+3u=-10x$$ But notice that \(\frac{u^2+v^2}{x^2+y^2}=x^2+y^2\), thus the equation becomes: $$2(x^2+y^2)+3u=-10x$$ Now substitute \(u\) again to get the equation solely in terms of x and y: $$2(x^2+y^2)+3x\sqrt{x^2+y^2}=-10x$$
05

Analyze the new equation

This is the equation we got in step 2, so it is clear that this conic cannot be simplified using standard methods. We can conclude that it corresponds to an irregular conic. Eccentricity cannot be determined for this conic, as it is neither an ellipse nor a hyperbola.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Eccentricity
Eccentricity is a fundamental concept used in the study of conic sections, which include ellipses, parabolas, and hyperbolas. Mathematically, it is represented by the letter \( e \). Eccentricity measures the deviation of a conic section from being circular. This helps in identifying whether a curve is an ellipse, a hyperbola, or a parabola.

- For ellipses, the eccentricity is between 0 and 1 (\( 0 < e < 1 \)). A circle, which is a special type of ellipse, has an eccentricity of 0, indicating it is perfectly round.- A parabola has an eccentricity of exactly 1 (\( e = 1 \)). - Hyperbolas have an eccentricity greater than 1 (\( e > 1 \)).

Eccentricity is crucial for defining the shape and nature of the conic. The formula for eccentricity differs based on the type of conic and is often characterized by terms in its equation. For students learning about conic sections, understanding eccentricity is key in analyzing and classifying the shapes and their properties.
Polar to Cartesian Conversion
Polar to Cartesian conversion is a process used to transform coordinates from the polar coordinate system into the Cartesian coordinate system. This conversion is necessary when dealing with equations in conic sections that are expressed in polar form.

The polar system uses a radial distance and angle \( (r, \theta) \), whereas the Cartesian system uses \((x, y)\) coordinates.
To convert from polar to Cartesian, utilize the following equations:
  • \( x = r \cos \theta \)
  • \( y = r \sin \theta \)
These equations allow us to express points defined by polar coordinates in terms of \( x \) and \( y \), making it easier to analyze and work with them in standard algebraic contexts. The concept helps in visualizing equations and simplifying complex calculations through manipulation of familiar Cartesian coordinates.
Ellipse
An ellipse is one of the primary types of conic sections characterized by its specific shape and mathematical properties. It resembles an elongated circle and is defined as the set of all points such that the sum of their distances from two fixed points (foci) is constant.

The standard form of an ellipse's equation in Cartesian coordinates is \[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \] where \((h, k)\) are the coordinates of the center, and \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.
The eccentricity of an ellipse, \( e \), is calculated as \( e = \sqrt{1 - \frac{b^2}{a^2}} \) for \( a > b \).
More elliptical shapes have eccentricities closer to 1, while a perfect circle, as previously noted, has an eccentricity of 0. Understanding ellipses involves studying their foci, axes, and how they graphically appear, which is useful in various applications ranging from physics to astronomy.
Hyperbola
A hyperbola is another significant type of conic section. It consists of two separate curves called branches. Each point on a hyperbola is such that the absolute difference of the distances from two fixed points, known as foci, is constant.

The standard equation of a hyperbola in Cartesian form is \[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \] where \((h, k)\) is the center, and \(a\) and \(b\) represent the distances from the center to the vertices and the direction of the asymptotes respectively.
The eccentricity, \( e \) of a hyperbola, is given by \( e = \sqrt{1 + \frac{b^2}{a^2}} \), and it is always greater than 1, differing from ellipses in this aspect.
Hyperbolas are recognizable by their intersecting asymptotes which guide the shape. They appear in natural phenomena such as orbital paths and in many engineering applications due to their properties of reflection and speed modulation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that the length of the latus rectum of the parabola with equation \(\left.y^{2}=4 p x \text { or } x^{2}=4 p y \text { is } 4|p| . \text { [ Hint: Exercise } 73 .\right]\)

Use a calculator in degree mode and assume that air resistance is negligible. A football kicked from the ground has an initial velocity of 75 feet per second. (a) Set up the parametric equations that describe the ball's path. Experiment graphically with different angles to find the smallest angle (within one degree) needed so that the ball travels at least 150 feet. (b) Use algebra and trigonometry to find the angle needed for the ball to travel exactly 150 feet. \([\text {Hint:}\) The ball lands when \(x=150\) and \(y=0 .\) Use this fact and the

Find the equation of the ellipse that satisfies the given conditions. Center (-5,2)\(;\) endpoints of major and minor axes: (0,2), (-5,17),(-10,2),(-5,-13).

Use a calculator in degree mode and assume that air resistance is negligible. A golf ball is hit off the ground at an angle of \(\theta\) degrees with an initial velocity of 100 feet per second. (a) Graph the path of the ball when \(\theta=30^{\circ}\) and when \(\theta=60^{\circ} .\) In which case does the ball land farthest away? (b) Do part (a) when \(\theta=25^{\circ}\) and \(\theta=65^{\circ}\) (c) Experiment further, and make a conjecture as to the results when the sum of the two angles is \(90^{\circ} .\) (d) Prove your conjecture algebraically. [Hint: Find the value of \(t\) at which a ball hit at angle \(\theta\) hits the ground (which occurs when \(y=0\) ); this value of \(t\) will be an expression involving \(\theta .\) Find the corresponding value of \(x\) (which is the distance of the ball from the starting point). Then do the same for an angle of \(90^{\circ}-\theta\) and use the cofunction identities (in degrees) to show that you get the same value of \(x .]\)

If \(a>0\) and \(b>0,\) then the eccentricity of the hyperbola $$\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 \quad \text { or } \quad \frac{(y-k)^{2}}{a^{2}}-\frac{(x-h)^{2}}{b^{2}}=1$$ is the number \(\frac{\sqrt{a^{2}+b^{2}}}{a} .\) Find the eccentricity of the hyperbola whose equation is given. $$6(y-2)^{2}=18+3(x+2)^{2}$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.