DHS Polar Coordinates (9649) (Revision Solutions)

Polar Coordinates H2 Double Math Polar Coordinates (Solutions) [Solution] ,0 21c o s ar   12 12 ,1c o s 1c o s aarr  12 8 3 arr 12 8 1c o s 1c o s 3 aa a  Since POQ is a straight line, 10,  then 21  11 8 1c o s 1c o s 3 aa a    11 11 8 1c o s 1c o s 3 2 1 14 1c o s 3  1 3sin 2  (Since 110, s i n 0   ) 2 or 33   Coordinates of    21 2 33 3, o r 2 ,Pa a   .

Polar Coordinates 2.

Polar Coordinates [Solution] Since ,b   and 0r when 3 4   , least possible 4b (i)  2 22 2 2 2 2 2 2 2 2 2d sin cos 1 cos cosd rra b a b b a b a b b           22 211 c o sab b   Arc length of one loop 44 2 22 00 d d1 1 5 c o s 4 dd rra       44 00 11 5 d 4 daa a     (ii) At A, ,8 ra  . Cartesian coordinates of A cos , sin88aa   At B, 35 ,48 8 ra     . Cartesian coordinates of B 55cos , sin88aa   Gradient of line AB 53 2sin sin 2cos sin 388 8 8 cot53 2 8cos cos 2sin sin88 8 8 aa aa             Thus AB : 3sin cot cos88 8ya xa      33cot cot cos sin88 8 8yx a a      [Note answer is not simplified] Alternative Note that triangle AOB is right-angled at O, and is also isosceles. Thus 8   . AB : sin tan cos88 8ya xa      tan 2 sin88yx a   

Polar Coordinates [Solution] (i) (a) For 10 , 0 sin 1. Thus 0 sin 1 and 1. sin r           (b) Since sin sin ,   11 sinsin    . C is symmetrical about the line 2   . (a) 0,  1 cos sin x     , sin sin 0 sin y     Thus 0y is an asymptote to C and thus 0  is an asymptote. (iii) 11 1 sin sin 222 y    Thus 5,66   . Area 2 6 11 1 3 12 d 22s i n 2 2 2            2 6 3ln cos cot 2 3ln 2 3 2 ec       3.

Polar Coordinates 4 The curve C1 has polar equation cos 2ra a  for 02  , where a is a positive constant. The curve C2 is obtained by rotating C1 through an angle of 2  anticlockwise about the pole. (i) Find a polar equation for C2 in terms of cos 2 . [1] (ii) On the same diagram, sketch the curves C1 and C2, indicating the polar coordinates of the points of intersections of the curves. [2] (iii) Find the exact area of the regions that lie within both C1 and C2. [4] (iv) A design for a necklace is made by taking the combined graphs of C1 and C2, but with the perimeter of the regions described in part (iii) removed. Find the perimeter of the n e c k l a c e . [ 2 ] [CJC/FM/2017/Promo/Q4] [Solution] 4 [CJC\Promo\P\Q4] (i) A polar equat