等边三角形ABC中,AB=a,O为中心,过O的直线交AB于M,交AC于N,求1/(OM^2)+1/(ON^2)的最大最小
3个回答

取O为原点.OA为y轴.AC方程3x+√3y=1,AB方程-3x+√3y=1.设MN方程y=kx.

算得M(1/(-3+√3k),k/(-3+√3k)),

N(1/(3+√3k),k/(3+√3k)),

(1/OM²)+(1/ON²)=6+12/(1+K²).(|k|≤1/√3)

直接看出:k=0,(1/OM²)+(1/ON²)=18为最大值.

|k|=1/√3时,(1/OM²)+(1/ON²)=15为最小值.