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Chapter 6: Problem 15

Identify the conic and sketch its graph. \(r=\frac{3}{1-\cos \theta}\)

Short Answer

Expert verified
The conic section represented by the given equation \(r = \frac{3}{1 - \cos \theta}\) is a parabola, and its graph is a parabola that opens to the left from the origin towards a directrix 3 units to the right.

Step by step solution

01

Recall the Standard Forms of Conic Sections

The standard forms for polar equations of conic sections are as follows: \n\n- for a circle: \(r = a\)\n- for an ellipse: \(r = \frac{a(1-e^2)}{1 - e\cos \theta}\)\n- for a parabola: \(r = \frac{ed}{1 - e\cos \theta}\)\n- for a hyperbola: \(r = \frac{a(1-e^2)}{1 + e\cos \theta}\)\n\nWhere \(e\) represents eccentricity, \(d\) is the distance from the pole (origin) to the directrix, and \(a\) is the distance from the origin to the vertex for ellipses and hyperbolas.
02

Identify the Given Equation's Form

Compare the given equation \(r = \frac{3}{1 - \cos \theta}\) to the standard forms. Notice that it matches the form for a parabola: \(r = \frac{ed}{1 - e\cos \theta}\), where \(ed = 3\) and \(e = 1\). Thus, the given equation represents a parabola.
03

Sketch the graph

Sketching conics in polar form involves understanding the geometric definition of each conic section. Parabolas have one defining characteristic in polar form: they have a single focus at the origin and the directrix is perpendicular to the polar axis at a distance d from the origin. For this given equation, set \(r = 0\) and solve for \(\theta\) to find the location of the directrix to get \(\theta = \cos^{-1}(1) = 0\). So the directrix is a vertical line at \(d = \frac{3}{e} = 3\) units to the right of the origin. The parabola opens to the left from the focus at the origin towards the directrix.

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Most popular questions from this chapter

Identify the conic as a circle or an ellipse. Then find the center, radius, vertices, foci, and eccentricity of the conic (if applicable), and sketch its graph. \(16 x^{2}+16 y^{2}-64 x+32 y+55=0\)

Sketch the graph of the ellipse, using latera recta. \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\)

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Identify the conic as a circle or an ellipse. Then find the center, radius, vertices, foci, and eccentricity of the conic (if applicable), and sketch its graph. . \(x^{2}+y^{2}-4 x+6 y-3=0\)
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