Question

How do you convert the polar equation $r = \dfrac{2}{{2 - \cos \theta }}$ into rectangular form?

Answer 1

Hint: In this problem, we have to convert the given polar form into a rectangular form. We should know about the rectangular coordinates, the polar coordinates and coordinate conversion equations to convert from a polar to a rectangular form. We change the polar form to rectangular form step by step, to get a rectangular form.

Formula used:
${\cos ^2}\theta + {\sin ^2}\theta = 1$
Perfect square trinomial rule: $\left( {{a^2} + {b^2}} \right) = {a^2} + 2ab + {b^2}$

Complete step by step solution:
We know that the rectangular coordinates are $\left( {x,y} \right)$ and the polar coordinates are $\left( {r,\theta } \right)$.
We also know that the coordinate conversion equations are,
$x = r\cos \theta $…(i)
$y = r\sin \theta $…(ii)
Now just squaring and adding the $x$, $y$ values, we get
${x^2} = {r^2}{\cos ^2}\theta $ and ${y^2} = {r^2}{\sin ^2}\theta $
Then, ${x^2} + {y^2} = {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta $
Also, ${r^2}$ is common in both terms, so take ${r^2}$ as common.
$ \Rightarrow {x^2} + {y^2} = {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)$…(iii)
As we know that, ${\cos ^2}\theta + {\sin ^2}\theta = 1$.
Substitute the value of ${\cos ^2}\theta + {\sin ^2}\theta $ in equation (iii), we get
$ \Rightarrow {x^2} + {y^2} = {r^2}$
Now on taking square root on both sides, we get
$ \Rightarrow r = \sqrt {{x^2} + {y^2}} $…(ii)
We know that the given polar form is,
$r = \dfrac{2}{{2 - \cos \theta }}$
We can multiply $2 - \cos \theta $ on both sides, we get
$ \Rightarrow r\left( {2 - \cos \theta } \right) = \dfrac{{2\left( {2 - \cos \theta } \right)}}{{\left( {2 - \cos \theta } \right)}}$
Now we can cancel the similar terms, we get
$ \Rightarrow 2r - r\cos \theta = 2$
Now we can substitute the equation (i) and (iii) in above step, we get
$ \Rightarrow 2\sqrt {{x^2} + {y^2}} - r \times \dfrac{x}{r} = 2$
$ \Rightarrow 2\sqrt {{x^2} + {y^2}} - x = 2$
We can now add $x$ on both sides, we get
$ \Rightarrow 2\sqrt {{x^2} + {y^2}} = 2 + x$
We can now square on both sides to get,
$ \Rightarrow 4\left( {{x^2} + {y^2}} \right) = {\left( {2 + x} \right)^2}$
Use algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$.
$ \Rightarrow 4{x^2} + 4{y^2} = {x^2} + 4x + 4$
Move all terms to one side of the equation.
$ \Rightarrow 3{x^2} + 4{y^2} - 4x - 4 = 0$

Final solution: Hence, the rectangular form of the given polar equation is $3{x^2} + 4{y^2} - 4x - 4 = 0$.

Note: In this problem our aim is to convert the given polar coordinate equation which is in the form $\left( {r,\theta } \right)$ into rectangular coordinates, for this we followed some important steps.
Relationship between Polar and Rectangular Coordinates:
Frequently, it will be useful to superimpose a rectangular $xy$-coordinate system on top of a polar coordinate system, making the positive $x$-axis coincide with the polar axis. If this is done, then every point P will have both rectangular coordinates $\left( {x,y} \right)$ and polar coordinates $\left( {r,\theta } \right)$. These coordinates are related by the equations
$x = r\cos \theta $, $y = r\sin \theta $…(1)
These equations are well suited for finding $x$ and $y$ when $r$ and $\theta $ are known. However, to find $r$ and $\theta $ when $x$ and $y$ are known, it is preferable to use the identities ${\sin ^2}\theta + {\cos ^2}\theta = 1$ and $\tan \theta = \sin \theta /\cos \theta $ to rewrite (1) as
${r^2} = {x^2} + {y^2}$, $\tan \theta = \dfrac{y}{x}$…(2)