求重心坐标

lemondian

已知椭圆方程为$\frac{x^2}{4}+y^2=1$，圆的方程为$(x-\frac{1}{2})^2+(y-\frac{1}{4})^2=2$，求椭圆与圆相交所得的四个交点所形成的四边形的重心坐标。
哦，对了，为了避免不同理解，定义四边形的重心坐标为$(\frac{x_1+x_2+x_3+x_4}{4},\frac{y_1+y_2+y_3+y_4}{4})$.

战巡

回复 1# lemondian


\[(x-\frac{1}{2})^2+(y-\frac{1}{4})^2=2\]
\[x^2-x+\frac{1}{4}+y^2-\frac{y}{2}+\frac{1}{16}=2\]
\[x^2-x+\frac{1}{4}+1-\frac{x^2}{4}-\frac{y}{2}+\frac{1}{16}=2\]
\[y=\frac{3}{2}x^2-2x-\frac{11}{8}\]
\[y^2=(\frac{3}{2}x^2-2x-\frac{11}{8})^2=1-\frac{x^2}{4}\]
\[\frac{9}{4}x^4-6x^3+\frac{x^2}{8}+\frac{11}{2}x+\frac{57}{64}=0\]
故
\[x_1+x_2+x_3+x_4=\frac{6}{\frac{9}{4}}=\frac{8}{3}\]
$y$坐标同理

lemondian

回复 2# 战巡


    四次方程韦达定理，谢谢了。
不知有无其它做法呢？

Tesla35

zhuanlan.zhihu.com/p/40400538

青青子衿

Tesla35 发表于 2018-8-26 15:02
https://zhuanlan.zhihu.com/p/40400538

知乎的这篇文章看不到了……

hbghlyj

也与楼主以前的问题有关:
四边形的重心坐标？


Three Centroids created by a Cyclic Quadrilateral
Christopher Bradley
Abstract: The centroid of the quadrilateral considered to be an area of constant density is G,
the centroid considered as having unit masses at its vertices is N,
the intersection of its diagonals is E.
the centroid considered as having unit masses at its vertices and mass of 2 units at E is F,
It is shown that E, F, N, G are collinear.
people.bath.ac.uk/masgcs/Article141.pdf