如图,在等腰梯形ABCD中,AB∥CD,且AB=2AD,设∠DAB=θ,θ∈（0,π
如图,在等腰梯形ABCD中,AB∥CD,且AB=2AD,设∠DAB=θ,θ∈（0,π
如图,在等腰梯形ABCD中,AB∥CD,且AB=2AD,设∠DAB=θ,θ∈（0,
π
2
）,以A,B为焦点且过点D的双曲线的离心率为e1,以C,D为焦点且过点A的椭圆的离心率为e2,则（ ）
赞 liu_qinghe 幼苗
共回答了14个问题采纳率：64.3% 举报
连接BD,AC设AD=t则BD=
t2+4t2−2•t•2tcosθ
=
5t2−4t2cosθ
∴双曲线中a=
5t2−4t2cosθ
−t
2
e1=
t
5t2−4t2cosθ
−t
2
∵y=cosθ在（0,
π
2
）上单调减,进而可知当θ增大时,y=
t
5t2−4t2cosθ
−t
2
=
2
1−cosθ
−1
减小,即e1减小∵AC=BD∴椭圆中CD=2t（1-cosθ）=2c∴c'=t（1-cosθ）AC+AD=
5t2−4t2cosθ
+t,∴a'=
1
2
（
5t2−4t2cosθ
+t）e2=
c′
a′
=
t(1−cosθ)
1
2
(
5t2−4t2cosθ
+t)
∴e1e2=
t
5t2−4t2cosθ
−t
2
×
t(1−cosθ)
1
2
(
5t2−4t2cosθ
+t)
=1
t2+4t2−2•t•2tcosθ
=
5t2−4t2cosθ
∴双曲线中a=
5t2−4t2cosθ
−t
2
e1=
t
5t2−4t2cosθ
−t
2
∵y=cosθ在（0,
π
2
）上单调减,进而可知当θ增大时,y=
t
5t2−4t2cosθ
−t
2
=
2
1−cosθ
−1
减小,即e1减小∵AC=BD∴椭圆中CD=2t（1-cosθ）=2c∴c'=t（1-cosθ）AC+AD=
5t2−4t2cosθ
+t,∴a'=
1
2
（
5t2−4t2cosθ
+t）e2=
c′
a′
=
t(1−cosθ)
1
2
(
5t2−4t2cosθ
+t)
∴e1e2=
t
5t2−4t2cosθ
−t
2
×
t(1−cosθ)
1
2
(
5t2−4t2cosθ
+t)
=1
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