求导求过点P处切线的斜率
2xdx/4+2ydy/3=0
adx/2+2bdy/3=0
dy/dx=-3a/4b
那么直线y-b=-3a/4b(x-a)
即3ax+4by-3a²-4b²=0
x=0时,y=(3a²+4b²)/(4b)
y=0时,x=(3a²+4b²)/(3a)
所以围城的三角形面积=1/2×(3a²+4b²)/(4b)×(3a²+4b²)/(3a)
=(3a²+4b²)²/(24ab)