Question

已知双曲线x^2/a^2-y^2/b^2=1.M,N是双曲线上关于原点对称的两点,P是双曲线上任意一点,当直线PM,PN

Answer 1

由题设,可设点M(p,q),N(-p,-q),P(s,t).∴(p²/a²)-(q²/b²)=1,且(s²/a²)-(t²/b²)=1.两式相减得：(s²-p²)/a²-(t²-q²)/b²=0.===>(t²-q²)/(s²-p²)=b²/a².再由斜率公式得：Kpm×Kpn=[(t-q)/(s-p)]×[(t+q)/(s+p)]=(t²-q²)/(s²-p²)=b²/a².∴Kpm×Kpn=b²/a²,该积与点P的位置无关.