问题补充：
已知抛物线C：x2=4y的焦点为F,过点F作直线l交抛物线C于A、B两点；椭圆E的中心在原点,焦点在x轴上,点F是它的一个顶点,且其离心率e=√3/2（1）求椭圆E的方程（2）经过点AB两边分别做抛物线C的切线L1 L2 L1与L2交与一个点M,证明：AB垂直MF
答案：
(1) x² = 4y = 2py,p = 2
F(0,1)
b = 1,e² = c²/a² = (a² - b²)/a² = 1 - b²/a² = 1 - 1/a² = 3/4
a² = 4
x²/4 + y² = 1
(2)设A(a,a²/4),B(b,b²/4),AB的方程:(y - b²/4)/(a²/4 - b²/4) = (x - b)/a - b)
4y = (a + b)x - ab
过F(0,1):ab = -4 (i)
y = x²/4
y = x/2
L1:y - a²/4 = (a/2)(x - a) (ii)
L2:y - b²/4 = (b/2)(x - b) (iii)
联立(ii)(iii):x = (a + b)/2,y = ab/4 = -4/4 = -1
M((a+b)/2,-1)
AB的斜率p = (a²/4 - b²/4)/(a - b) = (a + b)/4
FM的斜率q = (-1 - 1)/[(a + b)/2 - 0) = -4/(a + b)
pq = -1,AB垂直MF
