椭圆x^2+4y^2=16内一点P(1,-1)求过点P的弦的中点的轨迹方程

设弦所在直线的方程是
y+1=k(x-1)
代入椭圆
(4k^2+1)x^2-8(k+1)x+4(k+1)^2-16=0
x1+x2=8(k+1)/(4k^2+1)
y1+y2=[k(x1-1)-1]+[k(x2-1)-1]
=(6k-8k^3-2)/(4k^2+1)
=-(k+1)(8k^2-8k+2)/(4k^2+1)
所以中点是
x=(x1+x2)/2,y=(y1+y2)/2
x/y=-4/(4k^2-4k+1)=-4/(2k-1)^2
k=[√(-4y/x)+1]/2
代入x=(x1+x2)/2=4(k+1)/(4k^2+1)即可

过点P的弦的端点（x1,y1）（x2,y2）中点(x,y)代入
x^2+4y^2=16
相减得
(x1+x2)/(y1+y2)=-4(y1-y2)/(x1-x2)
即x/y=-4*(y+1)/(x-1)

设过P直线交椭圆A（x1,y1) B(x2,y2)AB中点(x0,y0) 设x0不为1设过P直线为y=k(x-1)-1则y0=k(x0-1)-1.k=(y0+1)/(x0-1)将AB代入椭圆x1^2+4y1^2=16x2^2+4y2^2=16两式相减得(x1^2-x2^2)+4(y1^2-y2^2)=0(x1+x2)(x1-x2)+4(y1...