Question

How do you convert $ r = \dfrac{3}{{3 - \cos \theta }} $ to rectangular form?

Answer 1

Hint: In order to convert $ r = \dfrac{3}{{3 - \cos \theta }} $ into rectangular form, we should know first what rectangular form is. A rectangular form of an equation is an equation consisting of variables that can be marked on the regular cartesian plane.

Complete step by step solution:
We are given with $ r = \dfrac{3}{{3 - \cos \theta }} $ .
Writing the reciprocal of the equation and we get:
 $
  r = \dfrac{3}{{3 - \cos \theta }} \\
  \dfrac{1}{r} = \dfrac{1}{{\dfrac{3}{{3 - \cos \theta }}}} \\
  \dfrac{1}{r} = \dfrac{{3 - \cos \theta }}{3} = 1 - \dfrac{1}{3}\cos \theta \;
  $
And, we can see that this is representing the value of an ellipse of eccentricity $ \dfrac{1}{3} $ .
From cartesian and polar form, we know that:
 $ r = \sqrt {{x^2} + {y^2}} $ , $ \cos \theta = \dfrac{x}{r} = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }} $ and $ \sin \theta = \dfrac{y}{r} = \dfrac{y}{{\sqrt {{x^2} + {y^2}} }} $ .
Substituting the values of $ r = \sqrt {{x^2} + {y^2}} $ , $ \cos \theta = \dfrac{x}{r} = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }} $ in $ r = \dfrac{3}{{3 - \cos \theta }} $ , we get:
 $ \sqrt {{x^2} + {y^2}} = \dfrac{3}{{3 - \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}}} $
On further solving, we get:
 $
  \sqrt {{x^2} + {y^2}} \left( {3 - \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}} \right) = 3 \\
  3\sqrt {{x^2} + {y^2}} - x\dfrac{{\sqrt {{x^2} + {y^2}} }}{{\sqrt {{x^2} + {y^2}} }} = 3 \\
  3\sqrt {{x^2} + {y^2}} - x = 3 \\
  3\sqrt {{x^2} + {y^2}} = 3 + x \;
  $
Squaring both the sides:
 $
  {\left( {3\sqrt {{x^2} + {y^2}} } \right)^2} = {\left( {3 + x} \right)^2} \\
  9\left( {{x^2} + {y^2}} \right) = 9 + {x^2} + 6x \\
  9{x^2} + 9{y^2} - 9 - {x^2} - 6x = 0 \\
  8{x^2} + 9{y^2} - 6x - 9 = 0 \;
  $
Since, we already saw that the equation represents an ellipse, so the equation obtained is the rectangular form of the ellipse equation.
Therefore, $ r = \dfrac{3}{{3 - \cos \theta }} $ in rectangular form is $ 8{x^2} + 9{y^2} - 6x - 9 = 0 $ .
So, the correct answer is “ $ 8{x^2} + 9{y^2} - 6x - 9 = 0 $ .”.

Note: Polar form of an equation is represented by $ r(\cos \theta ,\sin \theta ) $ .
Rectangular form is written in the cartesian format, in point form that can be graphed on the cartesian plane.
Relation between Polar and Cartesian values: $ r = \sqrt {{x^2} + {y^2}} $ , $ \cos \theta = \dfrac{x}{r} = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }} $ and $ \sin \theta = \dfrac{y}{r} = \dfrac{y}{{\sqrt {{x^2} + {y^2}} }} $ .