发布网友 发布时间:2024-10-11 07:06
共1个回答
热心网友 时间:2024-10-11 07:25
(1)C1:x^2+y^2=1,故曲线C1是以原点为中心半径为1的单位圆
C2:(x/a)^2+(y/b)^2=1,且a>b>0,故C2是中心在原点、a为长轴、b为短轴的椭圆
当α=0时,射线l:θ=α交C1于点(1,0),交C2于(a,0),
这两个交点间的距离为2即a=3(a=-1与a>0不符,舍去)
当α=π/2时,射线l:θ=α交C1于点(0,1),交C2于(0,b)两个交点重合即b=1
(2)当α=π/4时,交C1于点A1:((1/2)*2^(1/2),(1/2)*2^(1/2))
交C2于点B1:((3/2)*2^(1/2),(1/2)*2^(1/2))
根据对称性:A2:((1/2)*2^(1/2),-(1/2)*2^(1/2))
B2:((3/2)*2^(1/2),-(1/2)*2^(1/2))
显然,A1A2平行于B1B2 ,四边形面积为:
1/2×( ((1/2)*2^(1/2)-(-(1/2)*2^(1/2)))+((1/2)*2^(1/2)-(-(1/2)*2^(1/2))))×((3/2)*2^(1/2)-(1/2)*2^(1/2))=1/2×2×(2^(1/2))×(2^(1/2))=2