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Question: How do you convert \(r = \dfrac{6}{{2 - 3\sin (\theta )}}\) into Cartesian form?...

How do you convert r=62−3sin⁡(θ)r = \dfrac{6}{{2 - 3\sin (\theta )}} into Cartesian form?

Explanation

Solution

Here we have to convert the given term into Cartesian form. We will use the Cartesian format structure and substitute the values. After doing some calculation we get the required answer.

Complete step-by-step solution:
It is given the r=62−3sin⁡(θ)r = \dfrac{6}{{2 - 3\sin (\theta )}}
Now we use the Cartesian form of the equation can be expressed in the format:
r(cos⁡θ,sin⁡θ)=(x,y)r(\cos \theta ,\sin \theta ) = (x,y)
Also, we use that: r=x2+y2r = \sqrt {{x^2} + {y^2}} and sin⁡θ=yr\sin \theta = \dfrac{y}{r}
Therefore, on substituting the values from above in the equation we get:
⇒r=62−3(yr)\Rightarrow r = \dfrac{{6}}{{2 - 3\left( {\dfrac{y}{r}} \right)}}
On simplifying the denominator, we get:
⇒r=62−3yr\Rightarrow r = \dfrac{{6}}{{2 - \dfrac{{3y}}{r}}}
Now on taking the L.C.M in the denominator of the right-hand side we get:
⇒r=62r−3yr\Rightarrow r = \dfrac{{6}}{{\dfrac{{2r - 3y}}{r}}}
On rearranging the equation, we get:
⇒r=6r×r2r−3y\Rightarrow r = \dfrac{{6r \times r}}{{2r - 3y}}
Now since rr is present on both the left-hand side and the right-hand side in the numerator, we can cancel it and write the equation as:
⇒r=6r2r−3y\Rightarrow r = \dfrac{{6r}}{{2r - 3y}}
On rearranging the equation, we get:
⇒r=3−32y\Rightarrow r = 3 - \dfrac{3}{2}y
Now on squaring both sides:
⇒r2=(3−32y)2\Rightarrow {r^2} = {\left( {3 - \dfrac{3}{2}y} \right)^2}
Now since we know that the value of r=x2+y2r = \sqrt {{x^2} + {y^2}} , on substituting and squaring in the equation we get:
⇒x2+y2=(3−32y)2\Rightarrow {x^2} + {y^2} = {\left( {3 - \dfrac{3}{2}y} \right)^2}
Now on squaring the terms on the right-hand side, we get:
⇒x2+y2=(32−2×3×32y+(32y)2)2\Rightarrow {x^2} + {y^2} = {\left( {{3^2} - 2 \times 3 \times \dfrac{3}{2}y + {{\left( {\dfrac{3}{2}y} \right)}^2}} \right)^2}
On squaring the equation, we get:
⇒x2+y2=(9−2×3×32y+94y2)2\Rightarrow {x^2} + {y^2} = {\left( {9 - 2 \times 3 \times \dfrac{3}{2}y + \dfrac{9}{4}{y^2}} \right)^2}
On cancel the term and we get
⇒x2+y2=9−9y+94y2\Rightarrow {x^2} + {y^2} = 9 - 9y + \dfrac{9}{4}{y^2}
On taking the L.C.M on the right-hand side and simplifying the equation we get:
⇒4x2−5y2+36y−36=0\Rightarrow 4{x^2} - 5{y^2} + 36y - 36 = 0, which is the required answer.

The given expression in cartesian form is 4x2−5y2+36y−36=04{x^2} - 5{y^2} + 36y - 36 = 0.

Note: Another word for the Cartesian format is the rectangular format, rectangular or Cartesian coordinates are written in the format (x,y)(x,y) while on the other hand, polar coordinates are written in the form (r,θ)(r,\theta ).
It is to be remembered that there exists a relationship between the polar and Cartesian format which is
x=rcos⁡θx = r\cos \theta and y=rsin⁡θy = r\sin \theta .
In the above question we have used the formula of (a−b)2{(a - b)^2} which is a2−2ab+b2{a^2} - 2ab + {b^2}.
It is to be remembered that whenever we square a term in the square root, it becomes the original number for example (x)2=x{\left( {\sqrt x } \right)^2} = x.