双曲线的焦点为F1（-C,0）,F2（C,0）,过F2且斜率为√3/5的直线交双曲线于P、Q两点,若op垂直oq,pq的
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由已知可设P(x1,x2),Q(x2,y2)及双曲线方程:bx-ay=abx0d把直线y=m(x-c)(注:m=√15/5)代入bx-ay=ab中x0d得：(am-b)x-2acmx+(acm+ab)=0x0dx1+x2=2acm/(am-b),x1x2=(acm+ab)/(am-b)x0d∵OP⊥OQ ∴x1x2+y1y2=0,x1x2+m(x1-c)(x2-c)=0,(注:m=3/5)x0d5x1x2+3(x1-c)(x2-c)=0 即 8x1x2-3c(x1+x2)+3c=0x0d8(acm+ab)/(am-b)-6acm/(am-b)+3c=0x0d3a^4+8ab-3b4=0,(3a-b)(a+3b)=0x0d3a-b=0,b=3a,c=4ax0dx1+x2=2acm/(am-b)=-c/2x0dx1x2=(acm+ab)/(am-b)=-9a/4x0d|PQ|=4,∴PQ的中点到的距离O为2x0d[(x1+x2)/2]+[(y1+y2)/2]=4x0dc/16+[m(-5c/4]=4x0d双曲线方程:3x-y=3 即 x-y/3=1
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