曲线:x³-3x²+2x直线y=kx,且直线与曲线相切与点(x0,y0)(x0≠0),求直线的方程及切点坐标
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曲线:x³-3x²+2x直线y=kx,且直线与曲线相切与点(x0,y0)(x0≠0),求直线的方程及切点坐标
曲线:x³-3x²+2x直线y=kx,且直线与曲线相切与点(x0,y0)(x0≠0),求直线的方程及切点坐标
曲线:x³-3x²+2x直线y=kx,且直线与曲线相切与点(x0,y0)(x0≠0),求直线的方程及切点坐标
为说话方便,设曲线方程为 f(x).
∵过点(x0,y0) ∴ y0 = k * x0 = f(x0),化简得 k = x0^2 - 3 x0 + 2
∵相切 ∴f'(x0) = 3 x0^2 - 6 x0 + 2 = k
∴ x0^2 - 3 x0 + 2 = 3 x0^2 - 6 x0 + 2
解得 x0 = 0 (舍去)或 x0 = 3/2
代入 k = x0^2 - 3 x0 + 2 = -1/4
∴直线方程为 y = -1/4 ,切点坐标为 (3/2,-3/8)
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