来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/09/27 19:15:41
由参数方程求一般方程
希望有详解,
x=2k^2/(1+2k^2)
y=-k/(1+2k^2)
x/y=-2k、k=-x/(2y).
y=-k/(1+2k^2)
=[x/(2y)]/{1+2[-x/(2y)]^2}
=[x/(2y)]/[(2y^2+x^2)/(2y^2)]
=[x/(2y)]*[2y^2/(x^2+2y^2)]
=xy/(x^2+2y^2)
x^2+2y^2=x
(x-1/2)^2+2y^2=1/4
(x-1/2)^2/(1/4)+y^2/(1/8)=1.