伐头岭
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'<x> = -2x/a^2, z'<x>(P) = -√2/az'<y> = -2y/b^2, z'<y>(P) = -√2/b∂z/∂L = cost z'<x> + sint z'<y>= (-√2/a)[-a/√(a^2+b^)] + (-√2/b)[-b√(a^2+b^)]= 2√2/√(a^2+b^)