設(shè)AB和X軸交點(diǎn)為N（m,0),
A（x1,y1),B(x2,y2),
AB方程為：y=√2（x-m),
x^2+2*2(x-m)^2=1,
5x^2-8mx+4m^2-1=0,
根據(jù)韋達(dá)定理,
x1+x2=8m/5,(1)
x1x2=(4m^2-1)/5,(2)
設(shè)AB中點(diǎn)P（x0,y0),
用點(diǎn)差法求出,分別用x1,y1,x2,y2代入橢圓方程,再相減,
1/2+√2*y0/x0=0,
y0=-√2x0/4,
P(x0,-√2x0/4),
QP直線斜率k1=-√2/2,（與AB斜率互為負(fù)倒數(shù)）,
QP方程：
y+√2x0/4=- （√2/2）（x-x0),
令y=0,
x=x0/2,
∴Q（x0/2,0),
根據(jù)中點(diǎn)公式,x0=(x1+x2)/2,
x0=4m/5,
Q(2m/5,0),√2x--y-√2m=0,
Q至AB距離h=|√2*2m/5+0-√2m|/√（2+1）=√6|m|/5,
|AB|=√（1+2）[x1-x2)^2
=√3√[（x1+x2)^2-4x1x2]
=（2√3/5）√(5-4m^2),
S△QAB=|AB|*h/2=（3√2/25）√（5m^2-4m^4)
=（3√2/25）√[-4（m^4-5m^2/4+25/64)+25/16]
=（3√2/25）√[-4（m^2-5/8)^2+25/16]
當(dāng)m^2=5/8時(shí),有最大值,
S（max)=(3√2/25）√(25/16)
=3√2/20,
∴S△QAB最大值為3√2/20.