Question

由方程X^3+2Y^3-6aXY=0确定的隐方程y=y(x),满足条件y(2a)=2a,试证：x=2a为驻点,

Answer 1

证明：

方程X^3+2Y^3-6aXY=0,X^3+2Y^3=6aXY

两边分别对X求导,并化简得X^2+2Y^2*Y'=2a(Y+XY'),

X=2a时,Y(2a)=2a,代入得4a^2+8a^2*Y'(2a)=2a[2a+2a*Y'(2a)],

当a不等于0时,1+2Y'(2a)=1+Y'(2a),Y'(2a)=0,即X=2a为驻点；

当a=0时,原方程为方程X^3+2Y^3=0,易知X=0时,Y(0)=0,Y'(0)即X=0为驻点,满足X=2a为驻点；

综上,对任意实数a都有,X=2a为驻点.

由上X^2+2Y^2*Y'=2a(Y+XY'),两边再次对X求导,得X+2Y*Y'+Y^2*Y''=a(2Y'+XY''),

将X=2a,Y(2a)=2a,Y'(2a)=0代入,得2a+4a^2*Y''(2a)=2a^2*Y''(2a),

当a不等于0时,1+2aY''(2a)=0,Y''(2a)=-1/(2a),

当a>0时,Y''(2a)