动点P(x,y)满足到点F(0,1)的距离比到直线y=-2的距离小1.
即：动点P(x,y)到点F(0,1)的距离等于到直线y=-1的距离.
故P是以F为焦点,y=-1为准线的抛物线
其方程是：x^2=4y
设AB的方程：y=kx+1,切点为A(x1,y1),B(x1,y2)
将y=kx+1代入抛物线得：x^2-4kx-4=0
x1+x2=4k,x1x2=-4
因为切线方程是MA:xx1=2(y+y1)
MB:xx2=2(y+y2)
MA的斜率是：k1=x1/2
MB的斜率是:k2=x2/2
k1k2=x1x2/4=-1
故：MA垂直MB