椭圆方程X^2/4+y^2/3=1,试确定t的取值范围,使得椭圆上有2个不同的点关于直线y=4x+t对称

设这两点为M(x1,y1) N(x2,y2)MN中点(x0,y0)(x1)^2/4+(y1)^2/3=1 (x2)^2/4+(y2)^2/3=1 两式相减得(x1+x2)/(y1+y2)=-4/3*(y2-y1)/(x2-x1)注意到(x1+x2)/(y1+y2)=x0/y0(y2-y1)/(x2-x1)=-1/4所以x0/y0=1/3又y0=4x0+t得x0...