Question

设F是抛物线C:y^2=4x.的焦点，过F的直线l与曲线C相交于A、B两点，O为原点。 1：设直线l的斜率为2，求|A...

设F是抛物线C:y^2=4x.的焦点，过F的直线l与曲线C相交于A、B两点，O为原点。
1：设直线l的斜率为2，求|AB|的大小
2：求证:OA→*OB→为定值

Answer 1

1. y² = 4x = 2px, p = 2, F(1, 0)
   

   AB的方程: y = 2(x - 1), x= y/2 + 1
   代入抛物线: y² = 2y + 4, y - 2y - 4 = 0
   y₁ + y₂ = 2, y₁y₂ = -4
   |AB|² = (x₁ -x₂)|² + (y₁ -y₂)|² = (y₁/2 + 1 - y₂/2 - 1)|² + (y₁ -y₂)² = (5/4)(y₁ -y₂)²
   = (5/4)[(y₁ +y₂)² - 4y₁y₂]
   = (5/4)(4 + 16) = 25

   |AB| = 5
   
   

2. A(a²/4, a), B(b²/4, b)
   AB的方程: (y - b)/(a - b) = (x - b²/4)(a²/4 - b²/4)
   过F(1, 0): -b/(a - b) = (1 - b²/4)(a²/4 - b²/4), ab = -4
   OA•OB = (a²/4)(b²/4) + ab = (a²b²)/16 + ab = (-4)²/16 - 4 = -3
   
   

Answer 2

(1) F(1,0)
设y=k(x-i)代进y^2=4x k=2
得k^2x^2+(4-2k^2)x+K^2=0
x1+x2=（2k^2-4）/k^2=1
(AB)=x1+x2+p=2