Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Name the curve with the given polar equation. If it is a conic, give its eccentricity. Sketch the graph. $$ r=\frac{4}{1+\cos \theta} $$

Short Answer

Expert verified
The curve is a parabola with eccentricity 1.

Step by step solution

01

Identify the form of the polar equation

The given polar equation is \( r = \frac{4}{1+\cos \theta} \). This follows the general form \( r = \frac{ed}{1 + e\cos \theta} \), where \( e \) is the eccentricity, and \( d \) is a constant. In this case, \( ed = 4 \) and the equation is in the standard form of a conic.
02

Determine the eccentricity

From the polar equation \( r = \frac{4}{1+\cos \theta} \), we note that the denominator is \( 1 + e\cos \theta \). Comparing this with \( 1+\cos \theta \), we find that \( e = 1 \). If \( e = 1 \), the conic is a parabola according to conic section properties.
03

Verify the type of conic

A conic section with eccentricity \( e = 1 \) is always a parabola. This confirms that the curve described by the equation \( r = \frac{4}{1+\cos \theta} \) is indeed a parabola.
04

Sketch the graph of the conic

To sketch the graph of the equation \( r = \frac{4}{1+\cos \theta} \), note that it is symmetric with respect to the polar axis because of the \( \cos \theta \) term. The parabola will have its focus at the origin and directrix perpendicular to the axis of symmetry. Since it's opening towards the left (\( -x \) direction), plot points and draw the curve accordingly.

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 Vaia!

Key Concepts

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

Eccentricity
Eccentricity is a key concept when analyzing conic sections. For any conic that is expressed in polar coordinates, eccentricity helps determine the shape of the conic. It is denoted by the letter \( e \). The value of \( e \) dictates the nature of the conic section:
  • If \( e = 0 \), the conic is a circle.
  • If \( 0 < e < 1 \), it is an ellipse.
  • If \( e = 1 \), the conic is a parabola.
  • If \( e > 1 \), the conic is a hyperbola.
When dealing with the polar equation \( r = \frac{ed}{1 + e\cos\theta} \), the term \( e \) is extracted by comparing the formula with the standard form of given equations. In our exercise, the equation \( r = \frac{4}{1+\cos\theta} \) gave us \( e = 1 \), indicating that this equation describes a parabola.
Conic Sections
Conic sections are curves obtained from the intersection of a plane with a double-napped cone. They include circles, ellipses, parabolas, and hyperbolas. These shapes can be described using polar equations, which are particularly useful in defining how they stretch and rotate in the coordinate system. In the context of polar coordinates, these curves take on a form \( r = \frac{ed}{1 + e\cos\theta} \) or \( r = \frac{ed}{1 + e\sin\theta} \). This formula not only helps in identifying the conic type but also in determining its orientation and dimension. By identifying the conic from a polar equation, important features like
  • symmetric properties,
  • focus points,
  • and directrix location
can be understood, leading to insight into the structure of the curve.
Parabolas
Parabolas are distinctive shapes in conic sections. They have a unique property: each point on a parabola is equidistant from a fixed point, called the focus, and a fixed line, called the directrix.In our specific exercise, the equation \( r = \frac{4}{1 + \cos \theta} \) forms a parabola. This polar equation signifies that the directrix is perpendicular to the major axis, which in this situation, aligns with the x-axis. The major axis of a parabola is the line that passes through the focus and is perpendicular to the directrix.Parabolas
  • are widely used in physics, optics, and engineering,
  • often appear when dealing with projectile motion,
  • and have reflective properties that are useful in designing telescopes and satellite dishes.
Understanding the geometric and reflective properties of parabolas is crucial in understanding their uses and applications in various fields.

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

. Plot the Lissajous figure defined by \(x=\cos 2 t\), \(y=\sin 7 t, 0 \leq t \leq 2 \pi .\) Explain why this is a closed curve even though its graph does not look closed.

. Consider the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1\). (a) Show that its perimeter is $$ P=4 a \int_{0}^{\pi / 2} \sqrt{1-e^{2} \cos ^{2} t} d t $$ where \(e\) is the eccentricity. (b) The integral in part (a) is called an elliptic integral. It has been studied at great length, and it is known that the integrand does not have an elementary antiderivative, so we must turn to approximate methods to evaluate \(P .\) Do so when \(a=1\) and \(e=\frac{1}{4}\) using the Parabolic Rule with \(n=4\). (Your answer should be near \(2 \pi\). Why?) AS (c) Repeat part (b) using \(n=20\).

Plot the following parametric curves. Describe in words how the point moves around the curve in each case. (a) \(x=\cos \left(t^{2}-t\right), y=\sin \left(t^{2}-t\right)\) (b) \(x=\cos \left(2 t^{2}+3 t+1\right), y=\sin \left(2 t^{2}+3 t+1\right)\) (c) \(x=\cos (-2 \ln t), y=\sin (-2 \ln t)\) (d) \(x=\cos (\sin t), y=\sin (\sin t)\)

Let \(r_{1}\) and \(r_{2}\) be the minimum and maximum distances (perihelion and aphelion, respectively) of the ellipse \(r=\) \(e d /\left[1+e \cos \left(\theta-\theta_{0}\right)\right]\) from a focus. Show that (a) \(r_{1}=e d /(1+e), r_{2}=e d /(1-e)\), (b) major diameter \(=2 e d /\left(1-e^{2}\right)\) and minor diameter \(=\) \(2 e d / \sqrt{1-e^{2}}\)

Sketch the graph of the given equation. \(25 x^{2}+9 y^{2}+150 x-18 y+9=0\)

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.

Sign-up for free