difficult differentiation

2007-10-21 6:01 am

Given a curve C : y = (x^3) - 3(x^2) +2x -1 . A(1 , -1) and B(a , b) are two points on the curve and O is the origin.
(a) Prove that the tangent to the curve at A passes through O.
(b) If the tangent to the curve at B passes through O , show that 2(a^3) - 3(a^2) +1 = 0
(c) Hence , find the equations of the tangents drawn from O to the curve C.

回答 (1)

用戶195274

2007-10-21 6:34 am

✔ 最佳答案

For part (c) only,

From (b), we get 2(a^3) - 3(a^2) +1 = 0
Note that a = 1 is a root of the equation.
2a^3 - 3a^2 + 1 = 0
(a-1)(2a^2 - a - 1) = 0
(a-1)(2a+1)(a-1) = 0
a = 1 or -1/2
Since the x-coordinate of A is 1, the x-coordinate of B is -1/2
y-coordinates of B is (-1/2)^3-3(-1/2)^2 +2(-1/2)-1 = -23/8.

The equations of two tangents drawn from O to C are the tangents to the curve at A and at B (since both tangents passes through O as well)
Equation of AO: y-0 = [(-1-0)/(1-0)] (x - 0), i.e. x+y=0
Equation of BO: y-0 = [(-23/8-0)/(-1/2-0)] (x-0), i.e. 23x-4y= 0

Required equations of tangents are x+y=0 and 23x-4y=0

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