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Question: How do you convert \( r = \dfrac{3}{{3 - \cos \theta }} \) to rectangular form?...

How do you convert r=33−cos⁡θr = \dfrac{3}{{3 - \cos \theta }} to rectangular form?

Explanation

Solution

Hint : In order to convert r=33−cos⁡θr = \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=33−cos⁡θr = \dfrac{3}{{3 - \cos \theta }} .
Writing the reciprocal of the equation and we get:
r=33−cos⁡θ 1r=133−cos⁡θ 1r=3−cos⁡θ3=1−13cos⁡θ   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 13\dfrac{1}{3} .
From cartesian and polar form, we know that:
r=x2+y2r = \sqrt {{x^2} + {y^2}} , cos⁡θ=xr=xx2+y2\cos \theta = \dfrac{x}{r} = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }} and sin⁡θ=yr=yx2+y2\sin \theta = \dfrac{y}{r} = \dfrac{y}{{\sqrt {{x^2} + {y^2}} }} .
Substituting the values of r=x2+y2r = \sqrt {{x^2} + {y^2}} , cos⁡θ=xr=xx2+y2\cos \theta = \dfrac{x}{r} = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }} in r=33−cos⁡θr = \dfrac{3}{{3 - \cos \theta }} , we get:
x2+y2=33−xx2+y2\sqrt {{x^2} + {y^2}} = \dfrac{3}{{3 - \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}}}
On further solving, we get:
x2+y2(3−xx2+y2)=3 3x2+y2−xx2+y2x2+y2=3 3x2+y2−x=3 3x2+y2=3+x   \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:
(3x2+y2)2=(3+x)2 9(x2+y2)=9+x2+6x 9x2+9y2−9−x2−6x=0 8x2+9y2−6x−9=0   {\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=33−cos⁡θr = \dfrac{3}{{3 - \cos \theta }} in rectangular form is 8x2+9y2−6x−9=08{x^2} + 9{y^2} - 6x - 9 = 0 .
So, the correct answer is “ 8x2+9y2−6x−9=08{x^2} + 9{y^2} - 6x - 9 = 0 .”.

Note : Polar form of an equation is represented by r(cos⁡θ,sin⁡θ)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=x2+y2r = \sqrt {{x^2} + {y^2}} , cos⁡θ=xr=xx2+y2\cos \theta = \dfrac{x}{r} = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }} and sin⁡θ=yr=yx2+y2\sin \theta = \dfrac{y}{r} = \dfrac{y}{{\sqrt {{x^2} + {y^2}} }} .