Question

导数题求解第20题的证明

第20题的证明

Answer 1

20. x^(2/3) + y^(2/3) = a^(2/3), 两边对 x 求导得
(2/3)x^(-1/3) + (2/3)y^(-1/3)y' = 0,
解得 y' = -x^(-1/3)/y^(-1/3) = -(y/x)^(1/3)
在点(√2a/4, √2a/4)处切线斜率 k = -1， 法线斜率 是 1.
切线方程 y = -1(x-√2a/4) + √2a/4, 即 x+y = √2a/2，
法线方程 y = x-√2a/4 + √2a/4, 即 x-y = 0。
在任点(m, n)处切线斜率 k = -(n/m)^(1/3)，且 m^(2/3) + n^(2/3) = a^(2/3),
切线方程 y = -(n/m)^(1/3)(x-m) + n,
在 x ，y 轴的截距分别是 x0 = m + m^(1/3)n^(2/3), y0 = n + n^(1/3)m^(2/3),
切线在坐标轴之间的距离的平方
L^2 = (x0)^2 + (y0)^2
= m^2+2m^(4/3)n^(2/3)+m^(2/3)n^(4/3) + n^2+2n^(4/3)m^(2/3)+n^(2/3)m(4/3)
= m^2+n^2+3m^(2/3)n^(2/3)[m^(2/3)+n^(2/3)]
= [m^(2/3)+n^(2/3)]^3 = [a^(2/3)]^3 = a^2. 即为定值。