此题有点难度：
1、
设A(x1,y1)、B(x2,y2),
AB：y=k(x-p),即x=(1/k)y + p (为了方便打字,我下面用m代替1/k,即x=my +p)
联立直线、抛物线,消去x,得
y²=2p(my+p),即y²-2pmy-2p²=0
y1y2= -2p² …… ①
原点O到直线AB距离
d=p/√(1+m²)
|AB|=√[(x1-x2)²+(y1-y2)²] (注：x1-x2=(my1+p)-(my2+p)=m(y1-y2) )
=√[ m²(y1-y2)²+(y1-y2)²]
=√[( m²+1)(y1-y2)²]
= |y1-y2|√(m²+1)
∵|MA|=2|MB|,∴|y1|=2|y2|,
又∵y1、y2一定异号,即y1y2＜0,即y1=-2y2
∴|y1-y2|=3|y2|
∴|AB|=3|y2|√(m²+1)
∴S=(1/2)|AB|×d=(1/2)×3|y2|√(m²+1) × p/√(1+m²) = (3/2)·|y2|·p=6
|y2|=4/p
|y1|=8/p
∵y1y2＜0
∴y1y2= -32/p² …… ②
由①②,得p=2
2、设A(x1,y1),则圆心D( (x1+p)/2,y1/2 )
设存在符合条件的直线L：x=n
圆心到L的距离为d=|(x1+p)/2 - n|
|AM|=√[(x1-p)²+y1²]=√[(x1-p)²+2px1] = √(x1²+p²)
∴L被以圆截得的弦长为：(根据那个直角三角形用勾股定理)
2√(R²-d²)=2√[ ( |AM|/2 )² - | (x1+p)/2 - n |² ]
=2√[ (x1²+p²)/4 - | (x1+p)/2 - n | ² ] 全拆开,得
=2√[(-px1/2) + n(x1+ p)-n²]
=2√[(n - p/2)x1 + np-n²]
当n=p/2时,弦长有定值√2 p
直线L：x=p/2