焦點(p/2,0)
設AB：y=k(x-p/2)
那么CD：y=(-1/k)(x-p/2)
A、B坐標滿足方程k^2x^2-(pk^2+2p)x+k^2p^2/4=0
C、D坐標滿足方程x^2-(p+2pk^2)x+p^2/4=0
AB=√[(x1-x2)^2+(y1-y2)^2]
=√[(k^2+1)(x1-x2)^2]
=√{(k^2+1)[(x1+x2)^2-4x1x2]}
=√{(k^2+1)[(p+2p/k^2)^2-p^2]}
=|2p(k^2+1)/k^2|
所以1/AB=k^2/[2p(k^2+1)]
CD=√[(x1-x2)^2+(y1-y2)^2]
=√[(1/k^2+1)(x1-x2)^2]
=√{(1/k^2+1)[(x1+x2)^2-4x1x2]}
=√{(1/k^2+1)[(p+2pk^2)^2-p^2]}
=|2p(k^2+1)|
則AB+CD=|2p(k^2+1)|*|1/k^2+1|=|2p*(k+1/k)^2|>=|2p*4|=|8p|