✔ 最佳答案
Cannot find more than one solution. Can you show the second solution to check?
y = x^3 – 2x + 1 … (1)
y’ = 3x^2 – 2 … (2)
y = x^2 + 2ax … (3)
y’ = 2x + 2a … (4)
Let x = k be the common point of tangency
From (1) & (3), k^3 – 2k + 1 = k^2 + 2ak … (5)
From (2) & (4), 3k^2 – 2 = 2k + 2a => a = (3k^2 – 2k – 2)/2
Sub into (5) k^3 – 2k + 1 = k^2 + k(3k^2 – 2k – 2)
k^3 – 2k + 1 = k^2 + 3k^3 – 2k^2 – 2k
2k^3 – k^2 – 1 = 0
(k – 1)(2k^2 + k + 1) = 0
k = 1 => y = 0
Sub into (3), 0 = 1 + 2a => a = -1/2The common point of tangency is (1,0) and a = -1/2
2010-01-17 21:07:38 補充：
For (0,1), it is on y = x^3 -2x + 1, but not on y = x^2 - 2ax. You can verify this by putting the respective values into the equations. This is the same for (1/2, 1/8)
So your solutions do not match your question. Please check again you question or your answer.
2010-01-18 21:11:11 補充：
Curve1: y = x^3-2x+1; dy/dx = 3x^2-2
Curve2:y=x^2+2ax+1;dy/dx=2x+2a
Let common point be (h,k)
Same slope =>3h^2-2 = 2h+2a =>a=(3h^2-2h-2)/2…(1)
k = h^3-2h+1=h^2+2ah+1
Use(1), h^3-2h-1=h^2+2h(3h^2-2h-2)/2+1
h^3-2h-1=h^2+3h^3-2h^2-2h+1
2h^3-h^2= 0
h^2(2h-1)= 0
h=0 or h=1/2
2010-01-18 21:11:16 補充：
h=0 => k =1, and a = -1
h=1/2 => k=1/8-1+1=1/8, a=(3/4-1-2)/2=-8/9
