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..
2个回答
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
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
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