向量PS=λ向量SQ,则△PF1Q的外心S必在直线l上,也即P、Q、S三点共线.根据外心定义,有F1S=QS=PS,故点S为PQ中点,且有PF1⊥QF1,也即△PF1Q为RT△.
c=√(a^2+b^2)
直线l的方程为y=k1*(x-c),代入双曲线方程得
(b^2-k1^2*a^2)x^2+2k1^2*a^2*cx-a^2*(k1^2*c^2+b^2)=0
设P(x1,y1),Q(x2,y2),则有
x1+x2=2k1^2*a^2*c/(k1^2*a^2-b^2) ①
x1x2=a^2*(k1^2*c^2+b^2)/(k1^2*a^2-b^2 ) ②
K2=(y1+y2)/(x1+x2)=[k1*(x1+x2)-2k1*c]/(x1+x2)=k1-2k1*c/(x1+x2)
=k1-2k1*c/[2k1^2*a^2*c/(k1^2*a^2-b^2)]=k1-[k1-b^2/(k1*a^2)]=b^2/(k1*a^2)]
于是k1k2=b^2/a^2＞0
F1(-c,0),P(x1,y1),Q(x2,y2),由F1P^2+F1Q^2=PQ^2得
(x1+c)^2+y1^2+(x2+c)^2+y2^2=(x2-x1)^2+(y2-y1)^2
也即
c(x1+x2)+c^2+x1x2+y1y2=0
c(x1+x2)+c^2+x1x2+k1*(x1-c)*k1*(x2-c)=0
c(1-k1^2)(x1+x2)+(1+k1^2)x1x2+(1+k1^2)c^2=0
代入①、②得
2(1-k1^2)*k1^2*a^2c^2+(1+k1^2)*a^2*(k1^2*c^2+b^2)+(1+k1^2)*c^2*(k1^2*a^2-b^2)=0
4k1^2a^2c^2+b^2(a^2-c^2)+k1^2b^2(a^2-c^2)=0
4k1^2a^4+4k1^2a^2b^2-(1+k1^2)b^4=0
将k1=b^2/(k2*a^2)代入上式化简得
k2^2=(4a^4+4a^2b^2-b^4)/a^4=4+4(b/a)^2-(b/a)^4=-[(b/a)^2-2]^2+8≤8
前已证k1k2=b^2/a^2＞0,故k2≠0.
于是k2的范围为[-2√2,0)∪(0,2√2].
