(x1+x2)(x1-x2)/2+(y1-y2)(y1+y2)=0,是怎么到[(y1-y2)/(x1-x2)]*[(y1+y2)/2]/[(x1+x2)/2]=-1的.已知椭圆学x^2/2+y^2=1,求斜率为2的平行弦的中点轨迹方程.设斜率为2的弦与椭圆交于A（x1,y1),B(x2,y2),x1^2/2+y1^2=1,(1),x2^2/2+y2^

Question

(x1+x2)(x1-x2)/2+(y1-y2)(y1+y2)=0,是怎么到[(y1-y2)/(x1-x2)]*[(y1+y2)/2]/[(x1+x2)/2]=-1的.
已知椭圆学x^2/2+y^2=1,求斜率为2的平行弦的中点轨迹方程.
设斜率为2的弦与椭圆交于A（x1,y1),B(x2,y2),
x1^2/2+y1^2=1,(1),
x2^2/2+y2^2=1,(2),
(1)-(2)式,
（x1^2-x2^2)/2+(y1^2-y^2)=0,
(x1+x2)(x1-x2)/2+(y1-y2)(y1+y2)=0,
[(y1-y2)/(x1-x2)]*[(y1+y2)/2]/[(x1+x2)/2]=-1/2,
其中,(y1-y2)/(x1-x2)=k=2,
设平行弦中点动点坐标为y=(y1+y2)/2,x=(x1+y1)/2,
2y/x=-1/2,
∴斜率为2的平行弦的中点轨迹方程：y=-x/4

Answer 1

[(y1-y2)/(x1-x2)]*[(y1+y2)/2]/[(x1+x2)/2]=-1不对,应该是-1/2才对.
这是因为 (x1+x2)(x1-x2)/2+(y1-y2)(y1+y2)=0,
可化为 (y1-y2)·(y1+y2) = - (x1-x2)·[(x1+x2)/2],
再左右两边同除以 (x1-x2)·(x1+x2),即可得.