F(1,0),M(√2,0)
c=1,a=√2
b^2=a^2-c^2=1
(1)
x^2/2+y^2=1
(2),P(2,0)
A(m,n),C(m,-n)
m^2+2n^2=2,2n^2=2-m^2
PA:y=n*(x-2)/(m-2)......(1)
CF:y=-n*(x-1)/(m-1)......(2)
(1),(2):
x=(3m-4)/(2m-3)
y=n/(3-2m)
x^2+2y^2
=[(3m-4)/(2m-3)]^2+2[n/(3-2m)]^2
=(9m^2-24m+16+2-m^2)/(2m-3)^2
=2(2m-3)^2/(2m-3)^2
=2
点B[(3m-4)/(2m-3),n/(3-2m)]在椭圆上，所以BC经过焦点F
我初中都没毕业，高中的俺哪会，骚瑞！
给你思路要不
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