mathematica 发表于 2021-2-3 09:45
假设出直线BC的方程似乎结果简单一些
Clear["Global`*"];(*mathematica11.2,win7(64bit)Clear all variables*)
(*求解出直线BC与椭圆的两个交点*)
ans=Solve//FullSimplify;
(*对这两个点赋值*)
{x1,y1}={x,y}/.ans[];
{x2,y2}={x,y}/.ans[];
(*AB垂直于BC*)
eqn=(x1-2)*(x2-2)+(y1-1)*(y2-1)//FullSimplify
(*用参数k表达出b的值*)
aaa=Solve
(*检查一下两个交点的坐标，舍弃第二个b值{{b -> 1/3 (-1 - 2 k)}, {b -> 1 - 2 k}}*)
(*舍弃一个b值,因为有一点是(2,1)*)
{x1,y1,x2,y2}/.aaa//FullSimplify;
(*求解出交点,就是垂足的轨迹*)
bbb=Solve[(y==k*x+b&&(y-1)/(x-2)==-1/k)/.{b->1/3*(-1-2*k)},{x,y}]//FullSimplify//Expand//Together
{xx,yy}={x,y}/.bbb[]
(*利用结式来消除变量k*)
Resultant-Numerator,y*Denominator-Numerator,k]//Factor
ParametricPlot[{x,y}/.bbb,{k,-100,100},PlotRange->All,AxesOrigin->{0,0}]
两个坐标的交点分别是
\[\left\{\left\{x\to -\frac{\sqrt{-2 b^2+12 k^2+6}+2 b k}{2 k^2+1},y\to \frac{b-k \sqrt{-2 b^2+12 k^2+6}}{2 k^2+1}\right\},\left\{x\to \frac{\sqrt{-2 b^2+12 k^2+6}-2 b k}{2 k^2+1},y\to \frac{k \sqrt{-2 b^2+12 k^2+6}+b}{2 k^2+1}\right\}\right\}\]
AB与AC垂直的关系是
\[\frac{(b+2 k-1) (3 b+2 k+1)}{2 k^2+1}=0\]
截距是
\[\left\{\left\{b\to \frac{1}{3} (-2 k-1)\right\},\{b\to 1-2 k\}\right\}\]
D点的参数方程是
\[\left\{\left\{x\to \frac{2 \left(k^2+2 k+3\right)}{3 \left(k^2+1\right)},y\to \frac{3 k^2+4 k-1}{3 \left(k^2+1\right)}\right\}\right\}\]
消除掉参量k后的方程是
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