Question

设双曲线x^2/a^2-y^2/b^2=1的两条渐近线与直线x=a^2/c分别交于A,B两点,F

Answer 1

把x=a^2/c代入渐近线方程:x^2/a^2-y^2/b^2=0得

y^2=a^2b^2/c^2,

y=土ab/c,

∴A(a^2/c,ab/c),B(a^2/c,-ab/c),F(c,0),

向量FA=(b^2/c,ab/c),FB=(b^2/c,-ab/c),

FA*FB=b^4/c^2-a^2b^2/c^2,

|FA|=|FB|=√(b^4/c^2+a^2b^2/c^2)=b,

依题意,cosAFB=(b^2-a^2)/c^2∈(0,1/2),

∴0