Question

已知抛物线方程为x2=4y，过点M(0,2)作直线与抛物线交于两点A,B，过A,B分别作抛物线的切线，两切线交于P...

已知抛物线方程为x2=4y，过点M(0,2)作直线与抛物线交于两点A,B，过A,B分别作抛物线的切线，两切线交于P，求三角形PAB面积的最小值

Answer 1

设A(a, a²/4), B(b, b²/4), 不影响结果, 不妨设a > 0

AB方程: (y - b²/4)/(a²/4 - b²/4) = (x - b)/(a - b)
M(0, 2)在直线上, 整理得ab = -8, -b = 8/a (1)
y = x²/4, y' = x/2
过A的切线: y - a²/4 = (a/2)(x - a)
过B的切线: y - b²/4 = (b/2)(x - b)
解得P((a + b)/2, ab/4)

过P作轴的平行线l; 过A,B分别作l的垂线, 垂足分别为A'(a, ab/4), B'(b, ab/4)
三角形PAB面积 = 梯形AA'B'B面积 - 三角形PAA'面积 - 三角形PBB'面积
= (1/2)(B'B + A'A)*B'A' - (1/2)PA'*A'A - (1/2)B'P*B'B
= (1/2)(b²/4 - ab/4 + a²/4 - ab/4)*(a - b) - (1/2)[a - (a + b)/2]*(a²/4 - ab/4) - (1/2)[(a + b)/2 - b]*(b²/4 - ab/4)
= (1/2)(b²/4 - ab/4 + a²/4 - ab/4)*(a - b) - (1/4)(a - b)(a²/4 - ab/4) - (1/4)(a - b)(b²/4 - ab/4)
= (1/2)(a - b)(b²/4 - ab/4 + a²/4 - ab/4) - (1/4)(a - b)(b²/4 - ab/4 + a²/4 - ab/4)
= (1/4)(a - b)(b²/4 - ab/4 + a²/4 - ab/4)
= (1/16)(a - b)(a² - 2ab + b²)
= (1/16)(a - b)³
= (1/16)(a + 8/a)³ (利用(1))
≥ (1/16)[2√(a*8/a)³
= (1/16)(4√2)³
= 8√2