yerenju001
幼苗
共回答了19个问题采纳率:89.5% 举报
1.显然直线y=0与双曲线没有交点.
设l:x=my+1,
代入y^2-x^2/3=1,得
(3-m^2)y^2-2my-4=0,
设P(x1,y1),Q(x2,y2),则
y1+y2=2m/(3-m^2),y1y2=-4/(3-m^2),
x1x2=(my1+1)(my2+1)
=m^2y1y2+m(y1+y2)+1
=(-4m^2+2m^2+3-m^2)/(3-m^2)
=(3-3m^2)/(3-m^2),
向量OP*OQ=x1x2+y1y2
=(-1-3m^2)/(3-m^2)≠0.
∴不存在满足题设的直线l.
2.Sn=1/4[(an+1)^2].①
n=1时a1=s1=(1/4)(a1+1)^2,
(a1-1)^2=0,a1=1.
n>1时S=(1/4)[a+1]^2,②
①-②,an=(1/4)(an-a)(an+a+2),
an^2-a^2-2(an+a)=0,
∴(an+a)(an-a-2)=0,
{an}是正项数列,
∴an=a+2,
∴an=1+2(n-1)=2n-1.
1年前
7