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'

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.