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 equation for C2 is   cos 2 cos 2 cos 22ra a a a a a        (ii) (iii) Area =  224 0 181 c o s 2 d2 a    by symmetry  22 4 0 4 1 2cos 2 cos 2 da     2 4 0 3c o s 442 c o s 2 d22a     42 0 3s i n 44s i n 228a    23 42 a units2

Polar Coordinates (iv)  d1c o s 2 2 s i n 2d rra a     2 24 4 dPerimeter = 4 d d rr       224 4 41 c o s 2 2 s i n 2 da         13.710a 13.7a units

Polar Coordinates 5 Show that the gradient of tangen t of the polar curve with equation 1 1c o sr   is cot .2  [5] [EJC/FM/2017/Promo/Q8b] [Solution] 5  2 1s i n 1c o s 1c o s drr d       Note that   2 2 sin 1d sin cossin cos 1c o s1c o sd d dd sin 1cos sin cos sind 1c o s1c o s r ry rx r                     22 2cossin cos 1 cos 1c o s 2 cotsin cos sin 1 cos sin 2 2sin cos22                  Alternatively:       22 sin 1 cos sin coscos sincos 1c o s 1c o s 1c o s dxxr d                     22 cos 1 cos sin sinsin 1 cossin 1c o s 1c o s 1c o s dyyr d               2dy 2cosd1 c o sd 2 cotdxds i n 2 2sin cosd2 2 y x          

Polar Coordinates 6 The curve C has equation 222 2 150xx y y  . (a) Show that a polar equation for C can be expressed in the form 2 sin 2 Pr QR   , where P, Q and R are integers to be found and   . [3] (b) Hence, find the polar coordinates of the points on C which are the furthest from the pole O. [3] [JJC/FM/2017/Promo/Q2] [Solution] (a) 222 2 150xx y y      22 2 2 150 2 cos sin 150 xy x y rr r       2 2 2 sin cos 150 150 sin 22 2 r r      300 4s i n 2  (Shown) 300, 4, 1PQ R    (b) 2 300 4s i n 2r   For r to be maximum, sin 2 1  32, 22 3,44       Max. 300 1041r   Polar coordinates are 310, and 10,44       .

Polar Coordinates 7 The straight line with polar equation 1 1 sin cosr ab    , is a tangent to the circle with the polar equation, 2 2c o src  , where a, b and c are real numbers, 22 0ab and 0c . By first finding the Cartesian equations of the respective polar equations, find the possible value(s) of c in terms of a and/or b. [8] [MJC/FM/2017/Promo/Q2] [Solution] (i) 1 1 sin cosr ab    11 sin cos 1 1 1 , when 0 ra rb ay bx bxya a      2 2c o src   2 22 22 2 22 2c o s 2 rc r xy c x x cy c     Case 1 (when a is non-zero) Substituting 1 bxy a  into  2 22x cy c gives:     2 2 2 22 2 2 2 2 2 22 2 2 1 21 2 0 21 0 - ( * ) bxxc c a ax c x c b x b x a c ab x a c b x               Since the line is a tangent to the circle, there is only one solution. Thus the discriminant to (*) is zero.       222 2 222 2 42 2 2 2 2 42 2 2 24 1 0 0 20 20 Da c b a b ac b a b ac ab c b a b ac ab c a                 

Polar Coordinates Since a is non-zero, 22 21 0ac b c   2 2 22 22 22 4 2 bb a bb ac aa     Case 2 (when a is zero) The Cartesian Equation of the line reduces to 101yb x x b   This implies that the line is tang ential to the circle at its non-zero x-intercept  2, 0c , since equation is  2 22x cy c . Thus, 11 2 2ccbb 