答案是对的,但这样的题计算量太大了,而且怎么做都很麻烦：焦点(2,0),设直线：y=k(x-2)令A(x1,y1),B(x2,y2),则：y1=kx1-2k,y2=kx2-2kMA=(x1+2,y1-2),MB=(x2+2,y2-2)MA·MB=0,即：(x1+2,y1-2)·(x2+2,y2-2)=(x1+2)(x2+...(x1+2,y1-2)·(x2+2,y2-2)=(x1+2)(x2+2)+(kx1-2k-2)(kx2-2k-2)=x1x2+2(x1+x2)+4+k^2x1x2-2k(k+1)(x1+x2)+4(k+1)^2+4 这个式子 后面的4怎么算出来的。为啥 我算出来没4。求解！不好意思，我有个地方写的不对：(x1+2,y1-2)·(x2+2,y2-2)=(x1+2)(x2+2)+(kx1-2(k+1))(kx2-2(k+1))=x1x2+2(x1+x2)+4+k^2x1x2-2k(k+1)(x1+x2)+4(k+1)^2-------------这里没有4=(k^2+1)x1x2+(2-2k^2-2k)(x1+x2)+4(k+1)^2+4=0后面的没问题了