y'=4kx(x^2-3)y''=4k(x^2-3)+8kx^2=12k(x^2-1)=0,x=1或-1,此时y'=-8kxx=1时y'=-8k,法线斜率=1/8k,方程为y-4k=(x-1)/8k过原点所以k=±√2/8x=-1时y'=8k,法线方程y-4k=-(x+1)/8k过原点,解得k=±√2/8综上k=±√2/8...