(1)当过抛物线y^2=2px（p＞0）焦点F(p/2,0)的直线斜率不存在时，即x=p/2
那么，y^2=2p*(p/2)=p^2
则，y1=p，y2=-p
所以，y1*y2=-p^2=-4
则，p=2
抛物线方程就是：y^2=4x
当过焦点的直线斜率存在，设为k
那么，AB所在直线方程为：y=k[x-(p/2)]
联立直线与抛物线方程就有：y=k*[(y^2/2p)-(p/2)]
所以：ky^2-2py-kp^2=0………………………………………（1）
所以，y1*y2=-p^2=-4
所以，p^2=4
则，P=2
所以，抛物线方程为：y=4x
综上：该抛物线的方程为：y^2=4x
(2)
因为点M在抛物线的准线上，设点M(-p/2,m)
那么，MF所在直线的斜率b=(m-0)/[(-p/2)-(p/2)]=m/(-p)=2
所以，m=-2p
即，点M(-p/2,-2p)
已知点A(x1,y1)，点B(x2,y2)
那么：
a=Kma=(y1+2p)/[x1-(-p/2)]=(y1+2p)/[x1+(p/2)]
c=Kmb=(y2+2p)/[x2-(-p/2)]=(y2+2p)/[x2+(p/2)]
因为y^2=2px
所以，x1=y1^2/2p，x2=y2^2/2p
将其分别代入上述两式，则：
a+c=(y1+2p)/[(y1^2/2p)+(p/2)]+(y2+2p)/[(y2^2/2p)+(p/2)]
=2p*(y1+2p)/(y1^2+p^2)+2p*(y2+2p)/(y2^2+p^2)
=2p*[(y1+2p)/(y1^2+p^2)+(y2+2p)/(y2^2+p^2)]
=2p*[(y1+2p)*(y2^2+p^2)+(y2+2p)*(y1^2+p^2)]/[(y1^2+p^2)*(y2^2+p^2)]
=2p*[y1y2^2+p^2y1+2py2^2+2p^3+y1^2y2+p^2y2+2py1^2+2p^3]/[(y1y2)^2+p^2(y1^2+y2^2)+p^4]
=2p*[y1y2(y1+y2)+p^2(y1+y2)+2p(y1^2+y2^2)+4p^3]/[(y1y2)^2+p^2(y1^2+y2^2)+p^4]
=2p*[(y1y2+p^2)(y1+y2)+2p(y2^2+y2^2)+4p^3]/[(y1y2)^2+p^2(y1^2+y2^2)+p^4]
由前面(1)式知：
y1+y2=2p/k，y1y2=-p^2
那么，y1^2+y2^2=(y1+y2)^2-2y1y2=(4p^2/k^2)+2p^2
将它们分别代入上面a+c的式子中，就有：
a+c=2p*[2p*(4p^2/k^2+2p^2)+4p^3]/[p^4+p^2*(4p^2/k^2+2p^2)+p^4]
=2p*[(8p^3/k^2)+8p^3]/[p^4+(4p^4/k^2)+3p^4]
=2p*[8p^3*(1+1/k^2)]/[4p^4*(1+1/k^2)]
=(2p*8p^3)/(4p^4)
=4。