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

Xét $A\left( a;{{a}^{2}} \right),B\left( b;{{b}^{2}} \right)$ với $a<b$.
Ta có $\overrightarrow{AB}=\left( b-a;{{b}^{2}}-{{a}^{2}} \right)\Rightarrow \overrightarrow{{{n}_{AB}}}=\left( a+b;-1 \right)$.
$\begin{aligned}
& \Rightarrow AB:\left( a+b \right)\left( x-a \right)-\left( y-{{a}^{2}} \right)=0 \\
& \Rightarrow AB:y=\left( a+b \right)x-ab. \\
\end{aligned}$
Lại có $AB=2\Leftrightarrow {{\left( b-a \right)}^{2}}+{{\left( {{b}^{2}}-{{a}^{2}} \right)}^{2}}=4$.
Phương trình hoành độ giao điểm ${{x}^{2}}=\left( a+b \right)x-ab\Leftrightarrow x\left( x-a \right)-b\left( x-a \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=a \\
& x=b \\
\end{aligned} \right.$.
Diện tích hình phẳng giới hạn bởi $\left( P \right):y={{x}^{2}}$ và đường thẳng AB là
$\begin{aligned}
& S=\int\limits_{a}^{b}{\left| \left( x-a \right)\left( x-b \right) \right|dx}=-\int\limits_{a}^{b}{\left( x-a \right)\left( x-b \right)dx}=\int\limits_{a}^{b}{\left[ \left( a+b \right)x-ab-{{x}^{2}} \right]dx} \\
& =\left[ \left( a+b \right).\dfrac{{{x}^{2}}}{2}-abx-\dfrac{{{x}^{3}}}{3} \right]\left| \begin{aligned}
& ^{b} \\
& _{a} \\
\end{aligned} \right.=\dfrac{1}{2}\left( a+b \right)\left( {{b}^{2}}-{{a}^{2}} \right)-ab\left( b-a \right)-\dfrac{1}{3}\left( {{b}^{3}}-{{a}^{3}} \right) \\
& =\left( b-a \right)\left[ \dfrac{1}{2}{{\left( a+b \right)}^{2}}-ab-\dfrac{1}{3}\left( {{a}^{2}}+ab+{{b}^{2}} \right) \right]=\left( b-a \right).\dfrac{3{{\left( a+b \right)}^{2}}-6ab-2\left( {{a}^{2}}+ab+{{b}^{2}} \right)}{6} \\
& =\dfrac{1}{6}\left( b-a \right)\left( {{a}^{2}}+{{b}^{2}}-2ab \right)=\dfrac{1}{6}{{\left( b-a \right)}^{3}}. \\
\end{aligned}$
Từ ${{\left( b-a \right)}^{2}}+{{\left( {{b}^{2}}-{{a}^{2}} \right)}^{2}}=4\Rightarrow {{\left( b-a \right)}^{2}}\left( 1+{{\left( b+a \right)}^{2}} \right)=4\Rightarrow {{\left( b-a \right)}^{2}}=\dfrac{4}{1+{{\left( b+a \right)}^{2}}}\le 4$
$\Rightarrow b-a\le 2\Rightarrow S=\dfrac{{{\left( b-a \right)}^{3}}}{6}\le \dfrac{{{2}^{3}}}{6}=\dfrac{4}{3}$.
Dấu "=" xảy ra $\Leftrightarrow \left\{ \begin{aligned}
& a+b=0 \\
& b-a=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=1 \\
& a=-1 \\
\end{aligned} \right.\Rightarrow A\left( -1;1 \right),B\left( 1;1 \right)$.