✔ 最佳答案
(1) C1 : y = x^3 – x
C2 : y = (x – a)^3 – (x – a)
Solving C1 and C2 for intersecting points,
x^3 – x = (x – a)^3 – (x – a)
x^3 – x = x^3 – 3ax^2 + 3a^2x – a^3 – x + a
0 = – 3ax^2 + 3a^2x – a^3 + a
3ax^2 – 3a^2x + (a^3 – a) = 0
For at least one common point, the above quadratic equation should have at least one real root = > Discriminate >= 0
(3a^2)^2 – 4(3a)(a^3 – a) >= 0
9a^4 – 12a^4 + 12a^2 >= 0
-3a^2(a^2 – 4) >= 0
(a + 2)(a – 2) <= 0
-2 <= a <= 2
(2) y = x^2
dy/dx = 2x
At x = p, dy/dx = 2p and the equation of the tangent is (y – p^2)/(x – p) = 2p
y – p^2 = 2px – 2p^2
y = 2px – p^2
Sub this line into y = (x – 2)^2 + 2
2px – p^2 = x^2 – 4x + 4 + 2
x^2 – (4 + 2p)x + p^2 + 6 = 0
Tangency => discriminant = 0
(4 + 2p)^2 – 4(p^2 + 6) = 0
4p^2 + 16p + 16 – 4p^2 – 24 = 0
16p = 8
p =1/2
The equation of the tangent is y = 2(1/2)x – (1/2)^2
Or y = x – 1/4 (4y = 4x – 1)
2010-01-26 17:38:16 補充：
可否告訴我那部份錯了?
