设椭圆方程为x^2/a^2+y^2/b^2=1,(a>b>0),该点为(m,n),则切线方程为mx/a^2+ny/b^2=1与两轴的交点为(0,b^2n/n),(a^2/m,0)S=a^2b^2/|2mn|因为m^2/a^2+n^2/b^2=1≥2|mn|/(ab)所以2|mn|≤abS≥ab等号成立条件为m^2/a^2=n^...
