Question

点A是椭圆C:x^2/a^2+y^2/b^2=1(a>b>0)的短轴位于x轴下方的端点,过A作斜率为1的直线交椭圆于B点

Answer 1

(1) 设AP的方程：y=x-b,则B(1+b,1).向量AB*向量AP=(1+b)^2=9,

∴ b=2,B(3,1)在椭圆C上,9/(a^2)+(1/4)=1,a^2=12,椭圆C的方程为x^2/12+y^2/4=1.

(2) F1(-2√2,0),F2(2√2,0),|QF1|=r1,|QF2|=r2,r1+r2=2a

=4√3,由余弦定理cos=[(r1+r2)^2-2r1r2-4c^2]/(2r1r2)=[8/(r1r2)]-1.∵ r1r2≤[(r1+r2)/2]^2=12,∴ cos∠F1QF2≥-1/4.

(3) A1(-2√3,0),A2(2√3,0),设Q(2√3cosθ,2sinθ),则

k=2sinθ/(2√3cosθ+2√3),设A2Q的斜率为m,则m=2sinθ/(2√3cosθ02√3),k·m=4(sinθ)^2/[3cosθ)^2-1]=-1/3,k=-1/3m,

由-1/2