右焦点F(c,0),左准线与x轴的交点Bc^2=a^2+b^2,c>0设P(m,n),m>0,n>0,m、n满足m^2/a^2-n^2/b^2=1.......1)左准线方程x=-a^2/c,令Q(-a^2/c,p)OP垂直平分FQ,O到F和Q的距离相等即c^2=a^4/c^2+p^2................................2)设FQ中点A,则:XA=c-a^2/c=b^2/c,YA=p/2,即A(b^2/c,p/2)OP所在直线方程:y=pcx/2b^2由2)得:p=b√(c^2+a^2)/c则有,y=√(c^2+a^2)x/2b...................... |