焦点坐标是F1(0,3),即有 c=3,a^2=b^2+c^2=b^2+9
S(MOF1)=1/2OF1*|X|=3/2,X=1
即M坐标是(1,4)
即有1/b^2+16/a^2=1
1/b^2+16/(b^2+9)=1
b^2+9+16b^2=b^4+9b^2
b^4-8b^2-9=0
(b^2-9)*(b^2+1)=0
b^2=9,a^2=18
故椭圆方程是y^2/18+x^2/9=1
(2)OM的斜率是4,那么直线L的斜率也是4,设直线L的方程是y=4x+m
代入到椭圆方程中有x^2/9+(16x^2+8mx+m^2)/18=1
18x^2+8mx+m^2-18=0
x1x2=(m^2-18)/18,x1+x2=-4m/9
y1y2=(4x1+m)(4x2+m)=16x1x2+4m(x1+x2)+m^2=16(m^2-18)/18-16m^2/9+m^2=m^2/9-1
以AB为直径的圆过原点.,则有OA垂直于OB,则有x1x2+y1y2=0
即有(m^2-18)/18+m^2/9-1=0
3m^2=36
m^2=12
m=土2根号3
故存在,直线的方程是y=4x土2根号3