Question

数学双曲线求解!F1,F2分别为双曲线X2/A2-Y2/B2=1的左右焦点,过点F2做此双曲线的一条渐近线的垂线,垂足为

Answer 1

过F2(c,0)作渐近线x/a-y/b=0,即bx-ay=0①的垂线：ax+by-ac=0,②

两线交于M(a^2c/(a^2+b^2),abc/(a^2+b^2)),F1(-c,0),

由|MF1|=3|MF2|得

[a^2c/(a^2+b^2)+c]^2+[abc/(a^2+b^2)]^2=9{[a^2c/(a^2+b^2)-c]^2+[abc/(a^2+b^2)]^2},

∴（2a^2+b^2)^2+a^2b^2=9(b^4+a^2b^2),

∴4a^4+5a^2b^2+b^4=9b^4+9a^2b^2,

∴a^4-a^2b^2-2b^4=0,

(a^2-2b^2)(a^2+b^2)=0,a^2+b^2>0,

∴a^2=2b^2,

b^2/a^2=1/2,

b/a=土√2/2,

∴所求渐近线方程是y=(土√2/2)x.