一道高等数学关于定积分求面积的题设Y^2=2PX与X^2+Y^2=2QX(PQ不相等)交于O、A、B三点,求P,使抛物线与弦AB所围的面积最大.

y^2=2px (x-q)^2+y^2=q^2
x^2+2px-2qx=0
x=0或x=2(q-p) q>p
所以O(0,0) A(2(q-p),2根号(pq-p^2)) B(2(q-p),-2根号(pq-p^2))
抛物线与弦AB所围的面积
S=2∫(0,2(q-p)) 根号(2px) dx
=2根号(2p)*2/3*x^(3/2) |(0,2(q-p))
=16/3*p^(1/2)*(q-p)^(3/2)
=(16/9*根号3)*(3p)^(1/2)*(q-p)^(1/2)*(q-p)^(1/2)*(q-p)^(1/2)
=(16/9*根号3)*{[(3p)*(q-p)*(q-p)*(q-p)]^(1/4)}^2
由于几何平均数