设椭圆参数方程
x=√2·cosθ;
y=sinθ;
则:|AB|^2=(xA-xB)^2 + (yA-yB)^2=2(cosθA -cosθB)^2 +(sinθA -sinθB)^2
=2+cos^2 θA +cos^2 θB -4·cosθA·cosθB -2·sinθA·sinθB
即 2+cos^2 θA +cos^2 θB -4·cosθA·cosθB -2·sinθA·sinθB=2^2=4;
cos^2 θA +cos^2 θB -4·cosθA·cosθB -2·sinθA·sinθB=2^2=2;
-sin^2 θA -sin^2 θB -2·sinθA·sinθB=4·cosθA·cosθB ;
即 -(sinθA + sinθB)^2 =4·cosθA·cosθB ;
求直线AB的方程,用两点式:
x-√2·cosθB=[(√2·cosθA-√2·cosθB)/(sinθA-sinθB)]·(y-√2·sinθB);
设其与x轴交点为N,则易求得N坐标
N( (sinθA·cosθB-sinθB·cosθA)/(sinθA-sinθB)/(sinθA-sinθB) ,0 )
即|ON|=|(sinθA·cosθB-sinθB·cosθA)/(sinθA-sinθB)|.
则可知,2·S△OAB=|yA + yB|·|ON|
=|sinθA + sinθB|·|(sinθA·cosθB-sinθB·cosθA)/(sinθA-sinθB)|
=|(sinθA + sinθB)/(sinθA-sinθB)|·|sin(θA-θB)|
=|√(sinθA + sinθB)^2 /((sinθA+sinθB)^2 -4·sinθA·sinθB)|·|sin(θA-θB)|
=|√(-4·cosθA·cosθB /(-4·cosθA·cosθB -4·sinθA·sinθB)|·|sin(θA-θB)|
=|√(cosθA·cosθB /(cosθA·cosθB +sinθA·sinθB)|·|sin(θA-θB)|
=|√(cosθA·cosθB /cos(θA-θB) )|·|sin(θA-θB)|
