Question

求过程和答案！

Answer 1

半圆x^2+y^2=1只有一条直径AB：A(-1,0)、B(1,0)
双曲线以AB为焦点：c=1
设双曲线方程为：x^2/a^2-y^2/b^2=1
a^2+b^2=c^2=1
b^2=1-a^2
x^2/a^2-y^2/(1-a^2)=1
(1-a^2)x^2-a^2y^2=a^2(1-a^2)
(1-a^2)x^2-a^2y^2=a^2-a^4
P(p,√(1-p^2))
∠PAB=π/3

kPA=tan(π/3)=√3
即：[√(1-p^2)-0]/(p+1)=√3
1-p^2=3(p+1)^2
1-p^2=3p^2+6p+3
4p^2+6p+2=0
2(2p^2+3p+1)=0
2(p+1)(2p+1)=0

p=-1或者p=-1/2
∵p=-1时，P(-1,0)是A点，不能使∠PAB=π/3
∴p=-1/2
P(-1/2,√3/2)是双曲线上的点
(1-a^2)(-1/2)^2-a^2(√3/2)^2=a^2-a^4

1-a^2-3a^2=4(a^2-a^4)
4a^4-8a^2+1=0
4(a^4-2a^2+1)=-1+4
(a^2-1)^2=3/4
a^2-1=±√3/2
a^2=1±√3/2
b^2=1-a^2
=±√3/2
∵b^2>0
∴b^2=√3/2
a^2=1-√3/2=(2-√3)/2
双曲线方程：x^2/[(2-√3)/2]-y^2/(√3/2)=1
2x^2/(2-√3)-2y^2/√3=1