EJC_H2_2020_Prelim_P2_solution

Section A: Pure Mathematics [40 marks] 1 Suggested solution Suggested solution a) 2 The curve 1C has equation 2 3 2 3 1 x xy x    . The curve 2C has equation     2 2 2 41 1 yx b    , where 0b . (i) Sketch 1C , stating the equations of any asymptotes and the coordinates of any turning points and points where the curve crosses the axes. [4] y x O y x O  0, 6 (1, 3) (2.5, 0) (0, 1/6)

Page 2 of 16 (ii) Hence, find the range of values of b such that there is no point of intersection between 1C and 2C . Using the maximum value of b found, sketch 2C on the same diagram as part (i). [3] Suggested solution (i) 23 2 3 4 3 11 1 x xy x x x       (ii) 2C is a horizontal hyperbola with center 1,4 . Asymptotes:  1 4y b x   For no point of intersection between 1C and 2C , the asymptote of hyperbola must be as steep or less steep than the oblique asymptote of 1C . Note that the gradient of oblique asymptote of 1C is 3. 0 3 b   For 3b , Asymptotes:  3 1 4 3 1, 3 7y x x x       3 A curve C has parametric equations 1 cos ,x   sin 1,y     for 0 2   . 3 1y x  1x x  0, 3 0.155, 2.93   2.15,10.9 y O  0, 3  0,4  2,4 3 1y x  x y O 1x  2.15,10.9 3 7y x 

2020 JC2 H2 Mathematics Preliminary Examination Page 3 of 16 (i) The point P on C has parameter p. Show that the normal to C at P crosses the y-axis at a point Q with coordinates 0, 1p . [5] (ii) The normal to C at P crosses the x-axis at point R. Given that point S is the midpoint of QR, find a cartesian equation of the curve traced by S as p varies. [3] Suggested Solution (i) d sind x   d cos 1d y    d cos 1 d sin y x    Gradient of normal = sin cos 1    Equation of normal at point P  sinsin 1 1 cos cos 1 sin sin sin 1cos 1 sin 1cos 1 py p p x p p py x p p pp py x p p                    When 0, 1x y p   (Shown) (ii) Since equation of normal is sin 1cos 1 py x p p       when 0 1 1 cos sin y p px p    Midpoint of QR   1 1 cos 2sin p px p   1 2 py  Cartesian equation of the curve traced by S     1 cos 2 1 sin 2 1 y yx y      4 Vectors a and b are such that the modulus of a is 2 and b is a unit vector perpendicular to a. (i) State the value of (2 3 )a a b and explain your answer briefly. [1] (ii) Find the numerical area of the parallelogram with adjacent sides defined by 2a b and 5 7a b . [3] A vector c is such that 2 3  b c a b .

Page 4 of 16 (iii) Show that 6 c b a , where  is a constant. [2] (iv) State the geometrical meaning of b c and find the possible values of  if 5b c . [2] Suggested solution (i) (2 3 ) 0 a a b since 2 3a b is a vector that is perpendicular to 2a , and hence it is also perpendicular to a. (ii) Area of the parallelogram (