两不相等实数a,b,满足下列关系式:a^2sinθ+acosθ-(π/4)=0,b^2sinα+bcosα-(π/4)=0
问题描述:
两不相等实数a,b,满足下列关系式:a^2sinθ+acosθ-(π/4)=0,b^2sinα+bcosα-(π/4)=0
两不相等实数a,b,满足下列关系式:a^2sinθ+acosθ-(π/4)=0,b^2sinθ+bcosθ-(π/4)=0,连接A(a^2,a),B(b^2,b)两点的直线与圆心在原点的单位圆的位置关系?
答
设a^2=x1,a=y1,b^2=x2,b=y2则a^2sinθ+acosθ-(π/4)=0,b^2sinθ+bcosθ-(π/4)=0可得x1sinθ+y1cosθ-(π/4)=0,x2sinθ+y2cosθ-(π/4)=0点A(a^2,a),B(b^2,b)等价于(x1,y1),(x2,y2)则点(x1,y1),(x2,y2)都在直线xsin...