已知抛物线y^2=4x,过原点做两条互相垂直的弦OA、OB(O为坐标原点),求当△ABC面积最小时,直线AB的方程

A(a²/4,a),B(b²/4,b)
OA斜率p = a/(a²/4) = 4/a
OB斜率q = b/(b²/4) = 4/b
pq = -1 = 16/(ab)
ab = -16 (1)
S = (1/2)*OA*OB
4S² = OA²*OB²
= [(a²/4)² + a²][(b²/4)² + b²]
= (a²b²/16)² + a²b² + a²b²(a² + b²)/16
ab为常数,只须考虑a² + b²,a² + b² ≥ 2|ab|
|a| = |b|时,S最小,ab = -16,|a| = |b| = 4
二者横坐标相同(=4),直线AB的方程:x = 4