已知E（2,2）是抛物线C：y方=2px上一点,经过点（2,0）的直线l与抛物线C交于A,B两点（不同于E点）,直线EA,EB分别交直线x=-2与点M,N.求抛物线方程及焦点坐标.2.已知O为原点求证角MON为定植.

Question

已知E（2,2）是抛物线C：y方=2px上一点,经过点（2,0）的直线l与抛物线C交于A,B
两点（不同于E点）,直线EA,EB分别交直线x=-2与点M,N.求抛物线方程及焦点坐标.2.已知O为原点求证角MON为定植.

Answer 1

(1)
E(2,2)在C上:4 = 2p*2,p =1,y² = 2x,焦点F(1/2,0)
(2)
设A(a²/2,a),B(b²/2,b)
AB方程:(y - b)/(a - b) = (x - b²/2)/(a²/2 - b²/2)
过(2,0),简化得ab = -4 (i)
EA方程:(y - 2)/(a - 2) = (x - 2)/(a²/2 - a)
令x = -2,y = 2(a - 2)/(a + 2)
M(-2,2(a - 2)/(a + 2))
类似地,N(-2,2(b - 2)/(b + 2))
OM斜率m = -(a - 2)/(a + 2)
ON斜率n = -(b - 2)/(b + 2)
mn = (ab - 2a - 2b + 4)/(ab + 2a + 2b + 4) = (-4 - 2a - 2b + 4)/(-4 + 2a + 2b + 4) = -1
角MON为直角