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Chapter 10: Problem 53

Identify the conic and write its equation in rectangular coordinates: \(r=\frac{1}{2-2 \cos \theta}\)

Short Answer

Expert verified
The given polar equation corresponds to the circle with radius 1 and center at (1,0) in rectangular coordinates. Its equation is \((x-1)^2 + y^2 = 1\).

Step by step solution

01

Write the given equation

We start with the given polar equation: \(r=\frac{1}{2-2 \cos \theta}\)
02

Convert to Rectangular Coordinates

We will use the two basic relations between polar and rectangular coordinates: \(x= r \cos \theta\) and \(y= r \sin \theta\). First, multiply both sides of the equation by \(r(2 - 2\cos{\theta})\) to get \(r^2 = 2r - 2r \cos{\theta}\). Replace \(r^2\) and \(r \cos \theta\) by their rectangular equivalents, \(x^2 + y^2\) and \(x\), respectively. We get: \(x^2 + y^2 = 2\sqrt{x^2+y^2} - 2x\).
03

Simplify the equation

Next, to arrive at a standard equation for a conic section, re-arrange and square the entire equation to eliminate the square root. First, move the \(2x\) term to the left side: \(x^2 + y^2 + 2x = 2\sqrt{x^2+y^2}\). Then, square both sides of the equation to yield: \((x^2 + y^2 + 2x)^2 = 4(x^2 + y^2)\). Expanding and simplifying, we find the equation for the conic section: \(x^2 - 2x + y^2 = 0\).
04

Complete the square and identify the conic section

The expression \(x^2 - 2x\) can be written as \((x-1)^2 - 1\). So we rewrite the equation as \((x-1)^2 - 1 + y^2 = 0\), or \((x-1)^2 + y^2 = 1\). This is the equation of a circle with radius 1 and center at (1,0).

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Most popular questions from this chapter

In Exercises \(51-60,\) convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the ellipse and give the location of its foci. $$ 36 x^{2}+9 y^{2}-216 x=0 $$

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