求过定点(0,1)的直线被双曲线x^2-(y^2)/4=1 截得的弦中点轨迹方程_数学解答

求过定点(0,1)的直线被双曲线x^2-(y^2)/4=1 截得的弦中点轨迹方程

人气：194 ℃　时间：2019-09-29 02:49:47

解答

弦A(x1,y1)B(x2,y2)弦中点P(x,y)x1+x2=2xy1+y2=2y(y1-y2)/(x1-x2)=(y-1)/(x-0)x1^2-(y1^2)/4=1x2^2-(y2^2)/4=1两个式子相减(x1-x2)(x1+x2)-(y1-y2)(y2+y1)/4=0(x1+x2)-(y1+y2)(y1-y2)/4(x1-x2)=02x-2y(y-1)/4x=04x^2...