设焦点F2坐标为(c,0),M坐标为(x1,y1),N坐标(x2,y2)由于F2是△BMN的重心,其横坐标c=(x1+x2+0)/3,纵坐标0=(y1+y2+b)/3∴x1+x2=3c,y1+y2=-b,将x12/a2 +y12/b2=1,x22/a2 +y22/b2=1两式相减,移项用平方差公式展开得:(x1-x2)(x1+x2)/a2=-(y1-y2)(y1+y2)/b2 (y1-y2)/(x1-x2)=-(b2/a2)(x1+x2)/(y1+y2)=(-b2/a2)[3c/(-b)]=3b...
|