EXAMPLE 4 Transform the following into Polar form
(i) x^{2} + y^{2} = 25 (ii) 2x + y = 0 (iii) x^{2} + y^{2}  2ax = 0 (iv) x^{2}+ y^{2} 2by = 0
(v) x^{2}  y^{2} = 4 (vi) x^{2} + y^{2} = 4x (vii) x^{3} + xy^{2} + 6x^{2}  2y^{2} = 0
Solution :
Putting x = r cos
q and y = r sin Now r^{2} = 25 \ r ± 5 which represents a circle having centre at (0,0) and
radius 5.
2x + y = 0, Putting x = r cos q and y = r sin q , we get 2(r cos q) + r sin q = 0
x^{2} + y^{2}  2ax = 0, Putting x = r cos
q and y = r sin q we get,
r^{2} cos^{2} q + r^{2} sin^{2} q = 0
\ r^{2}(cos^{2} q) = r (2a cos q)
\ r^{2} = r (2a cos q)
\ r = 2a cos q which is a circle of radius "2a", passing through (0,0) and centre (2a, 0) on the polar axis.
x^{2} + y^{2}  2by = 0, putting x = r cos q, y = r sin q
\ r^{2}(cos^{2} q + sin^{2}
q)  2b (r sin q) = 0
\ r^{2}(1) = r (2b sin q) : \ r = 2b sin q
which is a circle with radius '2b' and centre
at (0, 2b), lies on the ^ar line through pole (0, 0) to the polar axis.
x^{2}  y^{2} = 4, putting x = r cos q, y = r sin q
\ (r^{2} cos^{2} q)  (r^{2}
sin2 q) = 4
\ r^{2}(cos^{2} q  sin^{2} q) = 4
\ r^{2} cos^{2}q = 4 which is a rectangular Hyperbola.
x^{2} + y^{2} = 4x, putting x = r cos q, y = r sin q
\ r^{2} (cos^{2}q + sin^{2} q) = 4 (r cos
q)
\ r = 4 cos q which is a circle having centre (4, 0) on the polar axis which passes through (0, 0)
x^{3} + xy^{2} + 6x^{2}  2y^{2} = 0, putting x = r cos
q, y = r sin q
x (x^{2} + y^{2}) + 6x^{2}  2y^{2} = 0
\ r cos q (r^{2}) + 6 r^{2} cos^{2} q  2r^{2} sin^{2} q = 0
\ r cos q + 6
cos^{2} q  2 sin^{2} q = 0
\ r cos q = 2
sin^{2} q  6 cos^{2} q \ r cos q = 2  8 cos^{2} q
\ r = 2 sec q  8 cos q
\ r = 2 (sec q  4 cos q)

Index
7.1 Scalers & Vectors 7.2 Algebra of Vectors 7.3 Representation of a vector in a plane 7.4 Dotor Scalar product 7.5 Polar Coordinates Supplementary Problems
Chapter 8
