Question

过抛物线y^2=2px(p>0)的焦点F作倾斜角为30度的直线与抛物线交予P,Q两点，

分别过P,Q两点作抛物线准线的垂线于P1，Q1.若PQ的长为2，则四边形PP1QQ1的面积是？

Answer 1

F(p/2, 0)
准线: x = -p/2
直线斜率 = tan30˚ = 1/√3
直线方程: y = (x - p/2)/√3, x = √3y + p/2

与抛物线联立得y² = 2p(√3y + p/2)
y² - 2√3py - p² = 0
y₁ + y₂ = 2√3p, y₁y₂ = -p²
|PQ|² = 4 = (x₁ - x₂)² + (y₁ - y₂)² = (√3y₁ + p/2 - √3y₂ - p/2)² + (y₁ - y₂)²
=4(y₁ - y₂)² => (y₁ - y₂)² = 1

= 4[(y₁ + y₂)² - 4y₁y₂]
= 4(12p² + 4p²) = 48p²
p = √3/6
PP1 = x₁ + p/2
QQ1 = x₂ + p/2
P1Q1 = |y₁ - y₂| = 1
四边形PP1QQ1为梯形, 面积是S = (1/2)(PP1 + QQ1)*P1Q1
= (1/2)*(x₁ + p/2 + x₂ + p/2)*1
= (1/2)(p + x₁ + x₂)
= (1/2)(p + √3y₁ + p/2 + √3y₂ + p/2)
= (1/2)(2p + √3y₁ + √3y₂)
= (1/2)(2p + √3*2√3p)
= 4p
= 2√3/3