A(-2,0)B(2,0)C(m,0)F(1,0)
(1)PF‖l,于是Xp=1,代入橢圓：Yp=±3/2,即P(1,±3/2)
AM為：x±2y+2=0
(2)設(shè)P為(x0,y0)
直線AM為：y=y0(x+2)/(x0+2),M點(diǎn)：(4,6y0/(x0+2))
同理：N(4,2y0/(x0-2))
直線FM的斜率：2y0/(x0+2)
直線FN的斜率：2y0/3(x0-2)
而：2y0/(x0+2)*2y0/3(x0-2)=4y0^2/[3(x0^2-4)]
考慮到：x0^2/4+y0^2/3=1,即：4y0^2=12-3x0^2=3(4-x0^2)
于是：2y0/(x0+2)*2y0/3(x0-2)=-1
∴FM⊥FN
∴以MN為直徑的圓過F點(diǎn)