Question

设动直线L垂直X轴,且与椭圆x^2/4+y^2/2=1交于A、B两点,P是L上满足|PA|×|PB|=1的点,求P点的轨

Answer 1

假设P(x0,y),则直线L：x=x0,A(x0,y0),B(x0,-y0)PA=(0,y-y0),PB=(0,y+y0)椭圆：x^2+2y^2=4,y^2=(4-x^2)/2(y-y0)(y+y0)=1=y^2-y0^2=y^2-(4-x0^2)/2整理后,得到：x0^2+2y^2=6所以,P点轨迹：x^2/6+y^2/3=1