Question

大侠们贡献牛b解法~第7题

Answer 1

F1(-c,0),B(0,b),

∴F1B:x/(-c)+y/b=1,即y=b(x/c+1)①

代入x^/a^-y^/b^=0,得x^/a^-(x^/c^+2x/c+1)=0,

b^x^-2a^cx-a^c^=0,

设P(x1,y1),Q(x2,y2),则

x1+x2=2a^c/b^,

PQ的中点N:xN=(x1+x2)/2=a^c/b^,

由①,yN=b(a^/b^+1)=c^/b,

∴PQ的中垂线:y-yN=(-c/b)(x-xN),

令y=0,得xM=(b/c)yN+xN=c+a^c/b^,

由|MF2|=|F1F2|得a^c/b^=2c,

∴a^=2b^=2(c^-a^),

∴3a^=2c^,(a/c)^=2/3,

设两渐近线的夹角为2α,易知cosα=a/c,

∴cos2α=2(cosα)^-1=2*2/3-1=1/3.选D.