Question

How do you convert $ r = 2\cos 2\theta $ into rectangular form?

Answer 1

Hint: In this problem, we have given an equation in the polar form. Here we are asked to convert the given polar form equation into the rectangular form. To convert the given equation into the equation into the rectangular form we need to use some identities and by using that identity we can get a required solution.

Complete step-by-step solution:
The polar form equation is $ r = 2\cos 2\theta $ .
Now, the identities that we are going to use are,
 $ x = r\cos \theta $
 $ y = r\sin \theta $
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 the terms, so take $ {r^2} $ as common.
 $ \Rightarrow {x^2} + {y^2} = {r^2}({\cos ^2}\theta + {\sin ^2}\theta ) - - - - - (1) $
Now we know that, $ {\cos ^2}\theta + {\sin ^2}\theta = 1 $
Substitute the value of $ {\cos ^2}\theta + {\sin ^2}\theta $ in equation (1), we get
 $ \Rightarrow {x^2} + {y^2} = {r^2} $
Also we can write $ \cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta $ , so we can substitute this value in the given polar equation. Then we get,
 $ r = 2\cos 2\theta = 2({\cos ^2}\theta - {\sin ^2}\theta ) - - - - - (*) $
Now, further simplification using the identities…
 $ {x^2} - {y^2} = {r^2}{\cos ^2}\theta - {r^2}{\sin ^2}\theta $ , now take the $ {r^2} $ common outside.
 $ \Rightarrow {x^2} - {y^2} = {r^2}({\cos ^2}\theta - {\sin ^2}\theta ) $
On rewriting we get
 $ \Rightarrow \dfrac{{{x^2} - {y^2}}}{{{r^2}}} = {\cos ^2}\theta - {\sin ^2}\theta - - - - - (2) $
Now use (*) in equation (2), we get
 $ r = \dfrac{2}{{{r^2}}}({x^2} - {y^2}) $
Now, substitute the value of $ {r^2} $ , we get
 $ \Rightarrow r = \dfrac{{2({x^2} - {y^2})}}{{({x^2} + {y^2})}} $
Also, we know that $ {x^2} + {y^2} = {r^2} \Rightarrow \sqrt {{x^2} + {y^2}} = r $
 $ \Rightarrow \sqrt {{x^2} + {y^2}} = \dfrac{{2({x^2} - {y^2})}}{{({x^2} + {y^2})}} $
Taking cross multiply and we get
 $ \Rightarrow {\left( {{x^2} + {y^2}} \right)^{\dfrac{3}{2}}} = 2\left( {{x^2} - {y^2}} \right) $
On rewriting we get
 $ \Rightarrow {\left( {{x^2} + {y^2}} \right)^{\dfrac{3}{2}}} - 2\left( {{x^2} - {y^2}} \right) = 0 $ .

This is the rectangular form of the given polar equation.
 $ {\left( {{x^2} + {y^2}} \right)^{\dfrac{3}{2}}} - 2\left( {{x^2} - {y^2}} \right) = 0 $

Additional Information: In two dimensions, the Cartesian coordinate (x, y) specifies the location of a point P in the plane. Another two- dimensional coordinate system is polar coordinate system is polar coordinates. As r ranges from $ 0 $ to infinity and $ \theta $ ranges from $ 0{\text{ to 2}}\pi $ , the point P specified by the polar coordinate $ (r,\theta ) $ covers every point in the plane.

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. In that first one we need to write $ x = r\cos \theta $ and $ y = r\sin \theta $ . Next we need to evaluate $ \cos \theta $ and $ \sin \theta $ . Then we have to multiply $ \cos \theta $ by r to find the x-coordinate of the rectangular form and multiply $ \sin \theta $ by r to find the y-coordinate of the rectangular form.