设以P(2,2)为圆心的圆与椭圆x平方加2y的平方等于1交于A、B两点，求AB中点M的轨迹方程。
热心网友
解：设：A点坐标(x1 , y1) B点坐标(x2 , y2)则: M点坐标[(x1+X2)/2 , (y1+Y2)/2]根据题义得知(1)(x1-2)^2+(y1-2)^2=(x2-2)^2+(y2-2)^2(x1)^2-(x2)^2-4(x1-x2)=(y2)^2-(y1)^2+4(y1-y2)(y1-y2)/(x1-x2)=[(x1+x2)-4]/[4-(y1+y2)](y1-y2)/(x1-x2)=(x-2)/(2-y)(2)(x1)^2+2(y1)^2=(x2)^2+2(y2)^2(x1-x2)(x1+x1)=[-2(y1-y2)(y1+y2)](y1-y2)/(x1-x2)=(x1+x2)/[-2(y1+y2)](y1-y2)/(x1-x2)=x/(-2y)所以:(x-2)/(2-y)=x/(-2y)2x+xy-4y=0。
热心网友
(x1-2)^2+(y1-2)^2=(x2-2)^2+(y2-2)^2(x1)^2-(x2)^2-4(x1-x2)=(y2)^2-(y1)^2+4(y1-y2)(y1-y2)/(x1-x2)=[(x1+x2)-4]/[4-(y1+y2)](y1-y2)/(x1-x2)=(x-2)/(2-y)(2)(x1)^2+2(y1)^2=(x2)^2+2(y2)^2(x1-x2)(x1+x1)=[-2(y1-y2)(y1+y2)](y1-y2)/(x1-x2)=(x1+x2)/[-2(y1+y2)](y1-y2)/(x1-x2)=x/(-2y)所以:(x-2)/(2-y)=x/(-2y)2x+xy-4y=0
热心网友
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