由拋物線的焦點(diǎn)恰好是橢圓的右焦點(diǎn)F知,c=p/2；再由兩條曲線的公共點(diǎn)的連線過F得x=cx^2/a^2+y^2/b^2=1
解得y^2=b^2(1-c^2/a^2)
(1)再由x=cy^2=2px=4cx解得y^2=2pc=4c^2
(2)綜合(1)和(2)得b^2(1-c^2/a^2)=4c^2
于是得(1-e^2)(1-e^2)=4e^2由于0由拋物線的焦點(diǎn)恰好是橢圓的右焦點(diǎn)F知，c=p/2；再由兩條曲線的公共點(diǎn)的連線過F得x=cx^2/a^2+y^2/b^2=1
解得y^2=b^2(1-c^2/a^2)
(1)再由x=cy^2=2px=4cx解得y^2=2pc=4c^2
(2)綜合(1)和(2)得b^2(1-c^2/a^2)=4c^2
于是得(1-e^2)(1-e^2)=4e^2由于0