试确定实数m的取值范围,使椭圆x^2/4+y^2/3=1上存在关于直线y=2x+m对称的点

设对称点为(x1,y1)(x2,y2)代入x^2/4+y^2/3=1得x1^2/4+y1^2/3=1x1^2/4+y1^2/3=1两式相减(x1+x2)(x1-x2)/4+(y1+y2)(y1-y2)/3=0(y1-y2)/(x1-x2)=-1/2x1+x2=2x0y1+y2=2y0所以3x0=2y0而（x0,y0）在直线y=2x+m上所以解得x0...