设AB:x=my+a,
代入y^2=8x得y^2-8my-8a=0,
设A(x1,y1),B(x2,y2),则y1+y2=8m,y1y2=-8a,
AP^2=(x1-a)^2+y1^2=(my1)^2+y1^2=(m^2+1)y1^2,
同理,BP^2=(m^2+1)y2^2,
∴1/AP^2+1/BP^2=(y1^2+y2^2)/[(m^2+1)(y1y2)^2]
=[(8m)^2+16a]/[64a^2*(m^2+1)]
=(4m^2+a)/[4a^2*(m^2+1)],是与m无关的常数,
∴a=4.