已知平行于y轴的动直线l1与曲线C:x²/4+y²=1相交于M、N两点,A1、A2分别为C的左右顶

Question

已知平行于y轴的动直线l1与曲线C:x²/4+y²=1相交于M、N两点,A1、A2分别为C的左右顶点，直线P1M与A2N交于点P，求P的轨迹方程C

Answer 1

由题意得到A1坐标是(-2,0),A2(2,0) 设M坐标是 (xo,yo),则有N(xo,-yo) K(A1M)=yo/(xo+2),K(A2N)=-yo/(xo-2) A1M:方程是y=yo/(xo+2)*(x+2) A2N方程是y=-yo/(xo-2)*(x-2) 联立二方程解得yo/(xo+2)*(x+2)=-yo/(xo-2)*(x-2) (x+2)*(xo-2)=-(xo+2)*(x-2) (xo-2)x+2xo-4=(-xo-2)x+2xo+4 2xox=8 x=4/xo y=yo/(xo+2)*(4/xo+2)=yo/(xo+2)*2(2+xo)/xo=2yo/xo 即有xo=4/x, yo=yxo/2=y*2/x=2y/x 又有xo^2/4+yo^2=1 故有16/x^2)/4+(4y^2/x^2)=1 4/x^2+4y^2/x^2=1 x^2-4y^2=4 即是P的轨迹方程.