Here we will learn to find the relation between Cartesian and Polar CoOrdinates.
Let XOX’ and YOY’ be a set of rectangular Cartesian axes of polar Coordinates through the origin O. now, consider a polar Coordinates system whose pole and initial line coincide respectively with the origin O and the positive xaxis of the Cartesian system. Let P be any point on the plane whose Cartesian and polar Coordinates are (x, y) and (r, θ) respectively. Draw PM perpendicular to OX. Then we have,
OM = x, PM = y, OP = r and < MOP = θ
Now, from the rightangled triangle MOP we get,
x/r = cos θ or, x = r cos θ …… (1)
and
y/r = sin θ or, y = r sin …… (2)
Using (1) and (2) we can find Cartesian Coordinates (x, y) of the point whose polar Coordinates (r, θ) are given.
Again, from the right angled triangle OPM we get,
r² = x² + y²
or, r = √(x² + y²) …… (3)
and tan θ = y/x or, θ = tan\(^{1}\) y/x ……… (4)
Using (3) and (4) we can find the polar Coordinates (r, θ) of the points whose Cartesian Coordinates (x, y) are given.
Note:
If the Cartesian Coordinates (x, y) of a point are given then to find the value of the vectorial angle θ by the transformation equation θ = tan\(^{1}\) y/x we should note the quadrant in which the point (x, y) lies.
Examples on the relation between Cartesian and Polar CoOrdinates.
1. The cartesian coordinates of a point are ( 1, √3); find its polar coordinates.
Solution:
If the pole and initial line of the polar system coincide with the origin and positive xaxis respectively of the cartesian system and the cartesian and polar coordinates of a point are ( x, y ) and ( r, θ ) respectively, then
x = r cos θ and y= r sin θ.
In the given problem, x = 1 and y = √3
Therefore, r cos θ = 1 and r sin θ = √3
Therefore, r² Cos² θ + r² sin² = ( 1)² + (√3)²
And tan θ = (r sin θ)/(r cos θ) = (√3)/(1) = √3 = tan π/3
Or, tan θ =tan(π+ π/3) [Since, the point ( 1,  √3) lise in the third quadrant]
Or, tan θ = tan 4π/3
Therefore, θ = 4π/3
Therefore, the polar coordinates of the point ( 1,  √3) are (2, 4π/3).
2. Find the cartesian coordinates of the point whose polar coordinates are (3,  π/3).
Solution:
Let (x, y) be the cartesian coordinates of the point whose polar coordinates are (3,  π/3). Then,
x= r cos θ = 3 cos ( π/3) = 3 cos π/3 = 3 ∙ 1/2 = 3/2
and y = r sin θ = 3 sin ( π/3) = 3 sin π/3 = (3√3)/2.
Therefore, the required cartesian coordinates of the point (3, π/3) are (3/2, (3√3)/2)
3. Transfer, the cartesian form of equation of the curve x²  y² = 2ax to its polar form.
Solution:
Let OX and OY be the rectangular cartesian axes and the pole and the initial line of the polar system coincide with O and OX respectively. If (x, y) be the cartesian coordinates of the point whose polar coordinates are (r, θ), then we have,
x = r cos θ and y = r sin θ.
Now, x²  y² = 2ax
or, r² cos² θ  r² sin² θ = 2a.r cos θ
or, r² (cos² θ  sin² θ) = 2ar cos θ
or, r cos 2 θ = 2a cos θ (Since, r ≠0)
which is the required polar form of the given cartesian equation.
4. Transform the polar form of equation \(r^{\frac{1}{2}}\) = \(a^{\frac{1}{2}}\)
cos θ/2 to its cartesian form.
Solution:
Let OX and OY be the rectangular cartesian axes and the pole and the initial line of the polar system coincide with O and OX respectively. If (x, y) be the cartesian coordinates of the point whose polar coordinates are (r, θ), then we have,
x = r cos θ and y = r sin θ.
Clearly, x² + y²
= r² cos² θ + r² sin² θ
= r²
Now, \(r^{\frac{1}{2}}\) = \(a^{\frac{1}{2}}\) cos θ/2
or, r = a cos² θ/2 (squaring both sides)
or, 2r = a ∙ 2 cos² θ/2
or, 2r = = a(1 + cosθ); [Since, cos² θ/2 = 1 + cosθ]
or, 2r² = a(r + r cosθ) [multiplying by r (since, r ≠0)]
or, 2(x² + y ²) = ar + ax [r² = x² + y² and r cos θ = x]
or, 2x² + 2y²  ax = ar
or, (2x² + 2y²  ax)² = a²r² [Squaring both sides]
or, (2x² + 2y²  ax)² = a² (x² + y²),
which is the required cartesian form of the given polar form of equation.
● Coordinate Geometry
11 and 12 Grade Math
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