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Chapter 6: Problem 63

Convert the polar equation to rectangular form. $$r=\frac{6}{2-3 \sin \theta}$$

Short Answer

Expert verified
The rectangular form of the given polar equation is: \(4x^4 + 8x^2y^2 -12x^2y^3 -36x^4y^2 + 4y^4 -12y^5+ 9y^6 +36x^2y^3 -108y^3 + 36y^4 +144x^2+144y^2 - 144y^4 + 216y^3 -216y^2 - 72 = 0\).

Step by step solution

01

Substitution of \(r\) into Polar Equation

First, the given relation \(r = \frac{6}{2-3\sin(\theta)}\) will be multiplied through by \(r\) to give \(r^2 = \frac{6r}{2-3\sin(\theta)}\). Then the known relations for rectangular coordinates will be substituted in, namely, \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\), giving us \(x^2 + y^2 = \frac{6 \sqrt{x^2 + y^2}}{2-3y}\).
02

Simplify the Equation

Next, multiply the equation throughout by \((2-3y)\) to isolate the denominator, resulting in \((2-3y)(x^2 +y^2) = 6\sqrt{x^2 + y^2}\). This can be written as \(2x^2 + 2y^2 - 3y^3 - 3yx^2 = 6\sqrt{x^2 + y^2}\). Then, to isolate the radical term, the equation can be simplified further to \(2x^2 +2y^2 - 3y^3 -3yx^2 - 6\sqrt{x^2 + y^2} = 0\).
03

Square the Resulting Equation to Eliminate the Radical

Finally, square \(2x^2 +2y^2 - 3y^3 -3yx^2 - 6\sqrt{x^2 + y^2}\) which removes the square root to yield \(4x^4 + 8x^2y^2 -12x^2y^3 -36x^4y^2 + 4y^4 -12y^5+ 9y^6 +36x^2y^3 -108y^3 + 36y^4 +144x^2+144y^2 - 144y^4 + 216y^3 -216y^2 - 72 = 0\), the final equation in rectangular coordinates.

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Most popular questions from this chapter

\(r=4+3 \cos \theta\)

In Exercises 33-48, find a polar equation of the conic with its focus at the pole. $$ \begin{array}{ll} {\text { Conic }} & \text { Vertex or Vertices } \\ \text { Ellipse } & (2, \pi / 2),(4,3 \pi / 2) \\ \end{array} $$

\(r=3 \sin \left(\frac{5 \theta}{2}\right)\)

\(r=8 \sin \theta \cos ^{2} \theta\)

\(r=5 \csc \theta\)
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