[DSE MATHS 2023] 9年5** + 連續3年全卷滿分心得分享， 好人一生平安
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tg: bar_tender123
Fredieeeeeeeeeee
Unfoundedl
大佬今年有冇P1 answer，做到17b卡左
TG：lonelyguy_5332
thxx
thxx
JessicaLingard
各位唔好留tg，我食埋飯upload
勿自爆
勿自爆
Thanks 16a) x^2 + ax + b = 0 have root p and 5p
Show 5a^2 = 36b
16b)
y = mx meet C at P and Q
Circle equation C: x^2 +y^2 -12x -6y + 20 =0
Sub y = mx in, we have
(m^2+1) x^2 - (6m+12)x + 20 = 0
x^2 - (6m+12)x /(m^2+1) + 20/(m^2+1) = 0 --(1)
Let x coordinate of P and Q be a and b
OP : PQ = 1 : 4 ==> a : b = 1 : 5
Note a and b are roots of (1)
Using (a) we have 5[(6m+12)/(m^2+1)]^2 = 36 (20) / (m^2+1)
Solving m = 4/3 or 0
Show 5a^2 = 36b
16b)
y = mx meet C at P and Q
Circle equation C: x^2 +y^2 -12x -6y + 20 =0
Sub y = mx in, we have
(m^2+1) x^2 - (6m+12)x + 20 = 0
x^2 - (6m+12)x /(m^2+1) + 20/(m^2+1) = 0 --(1)
Let x coordinate of P and Q be a and b
OP : PQ = 1 : 4 ==> a : b = 1 : 5
Note a and b are roots of (1)
Using (a) we have 5[(6m+12)/(m^2+1)]^2 = 36 (20) / (m^2+1)
Solving m = 4/3 or 0
jasonfans0415
Paper 1: 101/105
Paper 2: 29/45
有無5*
Paper 2: 29/45
有無5*
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