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EXAMPLE 4 Transform the following into Polar form
(i) x2 + y2 = 25 (ii) 2x + y = 0 (iii) x2 + y2 - 2ax = 0 (iv) x2+ y2 -2by = 0 (v) x2 - y2 = 4 (vi) x2 + y2 = 4x (vii) x3 + xy2 + 6x2 - 2y2 = 0

Solution :

  1. Putting x = r cos q and y = r sin Now r2 = 25    \ r ± 5 which represents a circle having centre at (0,0) and radius 5.

  2. 2x + y = 0,    Putting x = r cos q and y = r sin q ,    we get    2(r cos q) + r sin q = 0

  3. x2 + y2 - 2ax = 0, Putting x = r cos q and y = r sin q we get, r2 cos2 q + r2 sin2 q = 0

    \ r2(cos2 q) = r (2a cos q)

    \ r2 = 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.

  4. x2 + y2 - 2by = 0, putting x = r cos q, y = r sin q

    \ r2(cos2 q + sin2 q) - 2b (r sin q) = 0

    \ r2(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.

  5. x2 - y2 = 4, putting x = r cos q, y = r sin q

    \ (r2 cos2 q) - (r2 sin2 q) = 4

    \ r2(cos2 q - sin2 q) = 4

    \ r2 cos2q = 4 which is a rectangular Hyperbola.

  6. x2 + y2 = 4x, putting x = r cos q, y = r sin q

    \ r2 (cos2q + sin2 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)

  7. x3 + xy2 + 6x2 - 2y2 = 0, putting x = r cos q, y = r sin q

    x (x2 + y2) + 6x2 - 2y2 = 0

    \ r cos q (r2) + 6 r2 cos2 q - 2r2 sin2 q = 0

    \ r cos q + 6 cos2 q - 2 sin2 q = 0

    \ r cos q = 2 sin2 q - 6 cos2 q \ r cos q = 2 - 8 cos2 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 Co-ordinates
Supplementary Problems

Chapter 8

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