※ 引述《JULIKEBEN (JU)》之铭言： : the length of the portion , cut off by x-axis and y-axis , of any tangent line to the curve : x^(2/3)+y^(2/3)=4 is : a.2 b.4 c.8 d.none of above : 请问要怎麽下手呢?? : 谢谢 设点P(x_0 , y_0)为位於 x^(2/3) + y^(2/3) = 4 第一象限上的点 则 x_0 > 0 , y_0 > 0 , 且 (x_0)^(2/3) + (y_0)^(2/3) = 4 x^(2/3) + y^(2/3) = 4 , 将方程式等号两边对x微分得 2 -1 2 -1 dy (---)(x^(---)) + (---)(y^(---))(----) = 0 3 3 3 3 dx dy y^(1/3) => ---- = - -------- dx x^(1/3) dy | (y_0)^(1/3) => ----| = - ----------- dx |(x_0 , y_0) (x_0)^(1/3) 所以过P之切线方程式为 (y_0)^(1/3) y - y_0 = - (-----------)(x - x_0) (x_0)^(1/3) 令 y = 0 => ((x_0)^(1/3))(y_0) = ((y_0)^(1/3))(x - x_0) => x - x_0 = ((x_0)^(1/3))((y_0)^(2/3)) => x = ((x_0)^(1/3))((x_0)^(2/3) + (y_0)^(2/3)) = (4)((x_0)^(1/3)) 令 A ((4)((x_0)^(1/3)) , 0) 令 x = 0 => y - y_0 = ((y_0)^(1/3))((x_0)^(2/3)) => y = ((y_0)^(1/3))((x_0)^(2/3) + (y_0)^(2/3)) = (4)((y_0)^(1/3)) 令 B (0 , (4)((y_0)^(1/3))) 线段AB的长度 = √(((4)((x_0)^(1/3)))^2 + ((4)((y_0)^(1/3)))^2) = √((16)((x_0)^(2/3) + (y_0)^(2/3))) = √(16*4) = √64 = 8 故选c --

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