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EXAMPLE 5 Transform the following into rectangular form:
(i) r = -3 (ii) r = a cos q (iii) r = b sin q (iv) q = a (v) b cosec q (vi) r = a sec q (vii) r = a (1 + cos q) (viii) r = a (1 + sin q) (ix) r =

Solution :

  1. r = -3,   squaring   both   sides   we  get r2 = 9      then r2(cos2 q + sin2 q) = 9

    \ r2 cos2 q + r2 sin2 q = 9

    \ (r cos q)2 + (r sin q)2 = 9

    put r cos q = x and r sin q = y we get x2 + y2 = 9 which is a circle of radius '3' and centre (0, 0).

  2. r = a cos q

    \ r2 = ra cos q

    \   r2(cos2 q + sin2 q) = a (r cos q)

    \   r2 cos2 q + r2 sin2 q = a (r cos q)

    \   (r cos q)2 + (r sin q)2 = a (r cos q)

    \   (x)2 + (y)2 = a (x) ... x = r cos q and y = r sin q

    \ x2 + y2 - ax = 0

  3. r = b sin q \ r2 = b r sin q

    \ x2 + y2 = b (y)

    since r2 = x2 + y2 and r sin q = y

    \ x2 + y2 - by = 0

  4. q = a which is a straight line through origin.

  5. r = b cosec q

    \ r =

    \ r sin q = b

    \ y = b which is a straight line parallel to x - axis at a distance 'b' unit from the origin.

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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