：2. x^2/a^2 + y^2/b^2 = 1, 两边对 x 求导， 得 2x/a^2 + 2yy'/b^2 = 1, y' = -xb^2/(ya^2) 在 P(a/√2, b/√2) y' = -b/a, 法线斜率 k = a/b， Ox 轴到内法线得转角 t = π+arctan(a/b) z = 1-(x^2/a^2+y^2/b^2), z' = -2x/a^2, z'(P) = -√2/a z' ..
