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
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 15:47:43
![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](/uploads/image/z/7627683-3-3.jpg?t=%E4%B8%A4%E4%B8%8D%E7%9B%B8%E7%AD%89%E5%AE%9E%E6%95%B0a%2Cb%2C%E6%BB%A1%E8%B6%B3%E4%B8%8B%E5%88%97%E5%85%B3%E7%B3%BB%E5%BC%8F%EF%BC%9Aa%5E2sin%CE%B8%2Bacos%CE%B8-%28%CF%80%2F4%29%3D0%2Cb%5E2sin%CE%B1%2Bbcos%CE%B1-%28%CF%80%2F4%29%3D0%E4%B8%A4%E4%B8%8D%E7%9B%B8%E7%AD%89%E5%AE%9E%E6%95%B0a%2Cb%2C%E6%BB%A1%E8%B6%B3%E4%B8%8B%E5%88%97%E5%85%B3%E7%B3%BB%E5%BC%8F%EF%BC%9Aa%5E2sin%CE%B8%2Bacos%CE%B8-%28%CF%80%2F4%29%3D0%2Cb%5E2sin%CE%B8%2Bbcos%CE%B8-%28%CF%80%2F4%29%3D0%2C%E8%BF%9E%E6%8E%A5A%28a%5E2%2Ca%29%2CB%28b%5E2%2Cb%29%E4%B8%A4%E7%82%B9%E7%9A%84%E7%9B%B4%E7%BA%BF%E4%B8%8E%E5%9C%86%E5%BF%83%E5%9C%A8%E5%8E%9F%E7%82%B9%E7%9A%84)
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
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?
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
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xsinθ+ycosθ-(π/4)=0上
即点A,B在直线xsinθ+ycosθ-(π/4)=0上
因为点(0,0)到直线xsinθ+ycosθ-(π/4)=0的距离是(π/4)/√(sin^2θ+cos^2θ)=(π/4)