求通過點(3,1),切於曲線y=x^2-7的各切線方程(答案:4x+y-11=0 , 8x-y-23=0)
我要式！
２０分！！！amaths1問
2007-11-26 4:40 am
回答 (2)
2007-11-26 5:09 am
✔ 最佳答案
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
2007-11-26 5:09 am
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
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
收錄日期: 2021-04-13 18:11:58
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