\(\overrightarrow{AB}=\left(-a;b\right)\) ; \(\overrightarrow{MA}=\left(a-2;-1\right)\)
ABM thẳng hàng \(\Rightarrow b\left(a-2\right)=a\Rightarrow b=\frac{a}{a-2}\)
Do \(b>0\Rightarrow a>2\)
a/ \(S_{OAB}=\frac{1}{2}OA.OB=\frac{1}{2}ab=\frac{1}{2}.\frac{a^2}{a-2}=\frac{1}{2}\left(a-2+\frac{4}{a-2}+4\right)\ge\frac{1}{2}\left(2\sqrt{\frac{4\left(a-2\right)}{a-2}}+2\right)=3\)
Dấu "=" xảy ra khi \(\left(a-2\right)^2=4\Rightarrow a=4\Rightarrow b=2\)
\(\Rightarrow A\left(4;0\right);B\left(0;2\right)\)
b/ \(OA+OB=a+b=a+\frac{a}{a-2}=a+1+\frac{2}{a-2}\)
\(=a-2+\frac{2}{a-2}+3\ge2\sqrt{\frac{2\left(a-2\right)}{a-2}}+3=3+2\sqrt{2}\)
Dấu "=" xảy ra khi \(\left(a-2\right)^2=2\Leftrightarrow a=2+\sqrt{2}\Rightarrow b=1+\sqrt{2}\)
\(\Rightarrow A\left(2+\sqrt{2};0\right);B\left(0;1+\sqrt{2}\right)\)