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

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

Explanation

Solution

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 –
r2=x2+y2 ⇒r=x2+y2   {r^2} = {x^2} + {y^2} \\\ \Rightarrow r = \sqrt {{x^2} + {y^2}} \;
And cos⁡θ=xx2+y2\cos \theta = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}
Using the above two values in the given polar equation, we get –
x2+y2=3(xx2+y2) ⇒x2+y2=3x ⇒x2+y2−3x=0   \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 94\dfrac{9}{4} on both sides of the above equation, we get –
x2−3x+94+y2=94 x2−2×32x+(32)2+y2=94 ⇒(x−32)2+y2=94   {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−32)2+y2=94{(x - \dfrac{3}{2})^2} + {y^2} = \dfrac{9}{4} .
So, the correct answer is “ (x−32)2+y2=94{(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)(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,θ)(r,\theta ) where r is the distance of the point from the origin and θ\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 - x2+y2=r2{x^2} + {y^2} = {r^2} and by trigonometry we have - cos⁡θ=basehypotenuse=xx2+y2\cos \theta = \dfrac{{base}}{{hypotenuse}} = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }} .