F.5 maths (20marks!!)

1.
Let L2 : y = mx + c
Q(0,2) is on L2, so 2 = m(0) + c, c = 2
so L2 is y = mx + 2
Sub. this into equation of circle, we get
x^2 + (mx + 2)^2 - 6x - 6(mx + 2) + 10 = 0
x^2 + m^2x^2 + 4 + 4mx - 6x - 6mx - 12 + 10 = 0
(1 + m^2)x^2 - x(6 + 2m) + 2 = 0.......(1)
Since L2 touches the circle, so delta of the equation = 0, that is
(6 + 2m)^2 - 4(1 + m^2)(2) = 0
36 + 4m^2 + 24m - 8 - 8m^2 = 0
- 4m^2 + 24m + 28 = 0
m^2 - 6m - 7 = 0
(m - 7)(m + 1) = 0
m = 7 or - 1.
m = - 1 is for L1(negative slope), m = 7 is for L2 (positive slope).
so L2 is y = 7x + 2.
Sub m = 7 into (1) above, 50x^2 - 20x + 2 = 0
25x^2 - 10x + 1 = 0
(5x - 1)^2 = 0
x = 1/5 which is the x - coordinate of B
From L2, y = 7(1/5) + 2 = 17/5 which is the y coordinate of B, so B is (1/5, 17/5).
d.
Since PB is always perpendicular to PA, so slope of PB x slope of PA = - 1
so (y - 1)/(x - 1) *(y - 17/5)/(x - 1/5) = - 1
(y - 1)/(x - 1) *(5y - 17)/(5x - 1) = - 1
equation of locus is
(x - 1)(5x - 1) + (y - 1)(5y - 17) = 0
5x^2 - 6x + 1 + 5y^2 - 22y + 17 = 0
5x^2 + 5y^2 - 6x - 22y + 18 = 0


2011-08-13 18:42:11 補充：
19b. By cosine rule, AC^2 = AE^2 + EC^2 - 2(AE)(EC) cos (angle AEC). So you can find angle AEC. Angle BEC = 360 - 2 x angle AEC. = 92.1 degree.

2011-08-13 18:48:17 補充：
19c. By cosine rule, CB^2 = EC^2 + EB^2 - 2(EC)(EB) cos ( angle BEC). EC = EB, so CB can be found. Again by cosine rule, CB^2 = NC^2 + NB^2 - 2(NC)(NB) cos (angle CNB), so angle between plane = angle CNB can be found.

2011-08-13 18:52:18 補充：
19d. Let M be the point on AB such that NM is perpendicular to AB. Angle NBA = 60 degree, so NM = NB sin 60 = 5 sin 60. So angle between plane = arcsin ( NE/NM).