1.由题知得2a=4,a=2,e=c/a=√2/2,c=√2,b=√(a^2-c^2)=√2椭圆方程是x^2/4+y^2/2=1.
2.设动点P坐标为(x,y)则由动点P关于直线y=2x的对称点为P1(x1,y1)得(y+y1)/2=2(x+x1)/2,(y-y1)/(x-x1)=-1/2化简两式子得,2x-y=y1-2x1,2y+x=2y1+x1,前一式子×2加上后一式子消去y得3x1-4y1=-5x,而点P在x^2/4+y^2/2=1上,所以0