２０分！！！amaths1問

dy/dx=2x
設切點為(a,b)
b=a^2 -7 ---(1)
2a=(b-1)/(a-3)
b=2a^2-6a+1------(2)
2a^2-6a+1=a^2 -7
a^2-6a+8=0
a=2 or a=4
b=-3 or b=9
切點為(2,-3)或(4,9)
(2,-3)時，切綫斜率為4
y+3=4(x-2)
y+3=4x-8
4x-y-11=0 <===呢個先啱.....
(4,9)時，切綫斜率為8
y-9=8(x-4)
y-9=8x-32
8x-y-23=0

設切線方程是 y - 1 = m(x-3), 即 y = mx - 3m + 1.
mx - 3m + 1 = x^2 - 7
x^2 - mx + (3m - 8) = 0
由於是切線關係, 判別式 = 0
(-m)^2 - 4(1)(3m-8) = 0
m^2 - 12m + 32 = 0
(m-4)(m-8) = 0
m = 4 或 8
所以切線方程: y = 4x - 11 或 y = 8x - 23
即 4x - y + 11 = 0 或 8x - y - 23 = 0

2007-11-25 21:21:01 補充：
y = 4x - 11 或 y = 8x - 23即 4x - y - 11 = 0 或 8x - y - 23 = 0