已知t²+(2+i)t+2xy+(x-y)i=0(x,y为实数),当方程有实数根时,求点(x,y)轨迹方程
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/24 06:07:45
![已知t²+(2+i)t+2xy+(x-y)i=0(x,y为实数),当方程有实数根时,求点(x,y)轨迹方程](/uploads/image/z/2768152-40-2.jpg?t=%E5%B7%B2%E7%9F%A5t%26%23178%3B%2B%EF%BC%882%2Bi%EF%BC%89t%2B2xy%2B%EF%BC%88x-y%EF%BC%89i%3D0%EF%BC%88x%2Cy%E4%B8%BA%E5%AE%9E%E6%95%B0%EF%BC%89%2C%E5%BD%93%E6%96%B9%E7%A8%8B%E6%9C%89%E5%AE%9E%E6%95%B0%E6%A0%B9%E6%97%B6%2C%E6%B1%82%E7%82%B9%EF%BC%88x%2Cy%EF%BC%89%E8%BD%A8%E8%BF%B9%E6%96%B9%E7%A8%8B)
已知t²+(2+i)t+2xy+(x-y)i=0(x,y为实数),当方程有实数根时,求点(x,y)轨迹方程
已知t²+(2+i)t+2xy+(x-y)i=0(x,y为实数),当方程有实数根时,求点(x,y)轨迹方程
已知t²+(2+i)t+2xy+(x-y)i=0(x,y为实数),当方程有实数根时,求点(x,y)轨迹方程
当方程有实数根时,b²-4ac为实数.
(2+i)²-4[2xy+(x-y)i]=4+4i-1-8xy-4i(x-y)=3-8xy+4i--4i(x-y);
由于上式为实数.所以有x-y=1,3-8xy≥0;
x(x-1)≤3/8;(x-1/2)²≤5/8;1/2-√(5/8)≤x≤1/2+√(5/8);
所以点(x,y)轨迹方程为x-y=1,1/2-√(5/8)≤x≤1/2+√(5/8).
祝学习愉快,不懂追问哦.