Cho parabol $\left( P \right):y={{x}^{2}}$ và hai điểm $A,B$ thuộc...

Gọi $A\left( a;{{a}^{2}} \right),B\left( b;{{b}^{2}} \right)$ với $a<b$. Ta có $AB=2\Leftrightarrow {{\left( b-a \right)}^{2}}+{{\left( {{b}^{2}}-{{a}^{2}} \right)}^{2}}=4$.
$\begin{aligned}
& AB:\dfrac{x-a}{b-a}=\dfrac{y-{{a}^{2}}}{{{b}^{2}}-{{a}^{2}}}\Leftrightarrow \dfrac{x-a}{1}=\dfrac{y-{{a}^{2}}}{b+a}\Leftrightarrow y=\left( a+b \right)\left( x-a \right)+{{a}^{2}}\Leftrightarrow y=\left( a+b \right)x-ab \\
& S=\int\limits_{a}^{b}{\left( \left( a+b \right)x-ab-{{x}^{2}} \right)dx}=\int\limits_{a}^{b}{\left( x-a \right)\left( b-x \right)dx} \\
\end{aligned}$.
Đặt $t=x-a$.
Suy ra $S=\int\limits_{0}^{b-a}{t\left( b-a-t \right)dt}=\int\limits_{0}^{b-a}{\left( \left( b-a \right)t-{{t}^{2}} \right)dt}=\dfrac{\left( b-a \right){{t}^{2}}}{2}\left| _{0}^{b-a} \right.-\dfrac{{{t}^{3}}}{3}\left| _{0}^{b-a} \right.=\dfrac{{{\left( b-a \right)}^{3}}}{6}$
Ta có: ${{\left( b-a \right)}^{2}}+{{\left( {{b}^{2}}-{{a}^{2}} \right)}^{2}}=4\Leftrightarrow {{\left( b-a \right)}^{2}}\left( 1+{{\left( b+a \right)}^{2}} \right)=4\Leftrightarrow {{\left( b-a \right)}^{2}}=\dfrac{4}{1+{{\left( a+b \right)}^{2}}}\le 4$.
Suy ra $b-a\le 2\Rightarrow S=\dfrac{{{\left( b-a \right)}^{3}}}{6}\le \dfrac{{{2}^{3}}}{6}=\dfrac{4}{3}$.
Dấu "=" xảy ra khi và chỉ khi $\begin{aligned}
& \left\{ \begin{aligned}
& a+b=0 \\
& b-a=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=1 \\
& a=-1 \\
\end{aligned} \right.\Leftrightarrow A\left( -1;1 \right),B\left( 1;1 \right) \\
& \\
\end{aligned}$.