记 |PF1| = m,|PF2| = n ,
由已知得 PF1丄PF2,
因此由勾股定理得 m^2+n^2 = |F1F2|^2 =4c^2=4(a^2+b^2) =4(4+1) = 20 ,---------(1)
又由双曲线定义,|m-n| = 2a = 2*2 = 4 ,--------------(2)
(2)两边平方后,减去(1)得 -2mn = -4 ,
所以 mn = 2 ,
即 |PF1|*|PF2|= 2 .
记 |PF1| = m,|PF2| = n ,
由已知得 PF1丄PF2,
因此由勾股定理得 m^2+n^2 = |F1F2|^2 =4c^2=4(a^2+b^2) =4(4+1) = 20 ,---------(1)
又由双曲线定义,|m-n| = 2a = 2*2 = 4 ,--------------(2)
(2)两边平方后,减去(1)得 -2mn = -4 ,
所以 mn = 2 ,
即 |PF1|*|PF2|= 2 .