设P(x1,y1)
则Q(3x1,2y1)
O(0,0)
|PQ|=根号((2x1)²+y1²)
PQ方程为2x1x-y1y=2x1²+y1²=0
O到PQ的距离为|-2x1²+y1²|/根号((2x1)²+y1²)
s△=|-2x1²+y1²|/根号((2x1)²+y1²)×根号((2x1)²+y1²)÷2
=|-2x1²+y1²|/2①
P在椭圆上x1²/4+y1²/3=1
整理得y1²=(12-3x²)/4
带入①得
|-2x1²+(12-3x1²)/4|/2=|12-11x1²|/8
x1∈[-2,2]
|12-11x1²|≤|12-44|=32
所以s≤32/8=4
smax=4