Question

已知过点M(2p,0)的直线与抛物线y²=2px（p＞0）相交与AB两点， 求证OA⊥OB

当直线AB在什么位置时△AOB的面积最小，并证明。

Answer 1

设A(a²/(2p), a), B(b²/(2p), b)
OA的斜率u = 2p/a; OB的斜率v = 2p/b

AB的方程: (y - b)/(a - b) = [x - b²/(2p)]/[a²/(2p) - b²/(2p)]
y = 2px + ab
过点M(2p,0): 4p² + ab = 0, ab = -4p² (i)
uv = (2p/a)(2p/b) = 4p²/(ab) = 4p²/(-4p²) = -1
OA⊥OB

△AOB的面积S = (1/2)OA*OB
4S² = OA²*OB²
= [a⁴/(4p²) + a²][b⁴/(4p²) + b²]
= (ab)⁴/(16p⁴) + a²b⁴/(4p²) + a⁴b²/(4p²) + a²b²
= (-4p²)⁴/(16p⁴) + (-4p²)²b²/(4p²) + a²(-4p²)²/(4p²) + (-4p²)²
= 16p⁴ + 4p²b² + 4p²a² + 16p⁴
= 32p⁴ + 4p²(a² + b²)
≥ 32p⁴ + 4p²*2|ab|
= 32p⁴ + 4p²*4p²
= 48p⁴
此时|a| = |b|, 即AB与x轴垂直