Question

How do you write the polar equation \[r = 3\cos \theta \] in rectangular form?

Answer 1

Hint: In this question, we have to convert a polar equation into rectangular form. For converting the given equation into the rectangular for, we will express both the left-hand side and the right-hand side of the given equation in terms of x and y so that we get the all the quantities in terms of x and y and by further solving the equation, we get the rectangular form of the given polar equation.

Complete step-by-step answer:
We know that –
 $
  {r^2} = {x^2} + {y^2} \\
   \Rightarrow r = \sqrt {{x^2} + {y^2}} \;
  $
And $ \cos \theta = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }} $
Using the above two values in the given polar equation, we get –
 $
  \sqrt {{x^2} + {y^2}} = 3(\dfrac{x}{{\sqrt {{x^2} + {y^2}} }}) \\
   \Rightarrow {x^2} + {y^2} = 3x \\
   \Rightarrow {x^2} + {y^2} - 3x = 0 \;
  $
Adding $ \dfrac{9}{4} $ on both sides of the above equation, we get –
 $
  {x^2} - 3x + \dfrac{9}{4} + {y^2} = \dfrac{9}{4} \\
  {x^2} - 2 \times \dfrac{3}{2}x + {(\dfrac{3}{2})^2} + {y^2} = \dfrac{9}{4} \\
   \Rightarrow {(x - \dfrac{3}{2})^2} + {y^2} = \dfrac{9}{4} \;
  $
Hence the given polar equation is written in rectangular form as $ {(x - \dfrac{3}{2})^2} + {y^2} = \dfrac{9}{4} $ .
So, the correct answer is “ $ {(x - \dfrac{3}{2})^2} + {y^2} = \dfrac{9}{4} $ ”.

Note: There are two types of coordinates for plotting a point on the graph paper namely rectangular coordinate system and polar coordinate system. Rectangular coordinate system is the most commonly used coordinate system and is of the form $ (x,y) $ where x is the distance of this point from the y-axis and y is the distance of the point from the x-axis. The polar coordinate system is of the form $ (r,\theta ) $ where r is the distance of the point from the origin and $ \theta $ is the counterclockwise angle between the line joining the point and the origin, and the x-axis.
From these definitions, we see that a right-angled triangle is formed by x, y and r, where r is the hypotenuse, x is the base and y is the height of the triangle, so by Pythagoras theorem, we have - $ {x^2} + {y^2} = {r^2} $ and by trigonometry we have - $ \cos \theta = \dfrac{{base}}{{hypotenuse}} = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }} $ .