Trong mặt phẳng $Oxy$, cho parabol $\left( P \right):y={{x}^{2}}$...

Làm tương tự cách trên, ta có \)">d\left( P \right)A\left( a;{{a}^{2}} \right);B\left( -a-\dfrac{1}{2a};{{\left( a+\dfrac{1}{2a} \right)}^{2}} \right)I\left( P \right){{x}_{I}}=\dfrac{{{x}_{A}}+{{x}_{B}}}{2}\Rightarrow I\left( -\dfrac{1}{4a};\dfrac{1}{16{{a}^{2}}} \right)\left\{ \begin{aligned}
& \overrightarrow{AB}=\left( -2a-\dfrac{1}{2a};1+\dfrac{1}{4{{a}^{2}}} \right) \\
& \overrightarrow{AI}=\left( -\dfrac{1}{4a}-a;\dfrac{1}{16{{a}^{2}}}-{{a}^{2}} \right) \\
\end{aligned} \right.\Rightarrow {{S}_{\Delta IAB}}=\dfrac{1}{2}\left| \left( -2a-\dfrac{1}{2a} \right)\left( \dfrac{1}{16{{a}^{2}}}-{{a}^{2}} \right)-\left( 1+\dfrac{1}{4{{a}^{2}}} \right)\left( -\dfrac{1}{4a}-a \right) \right|\Rightarrow {{S}_{\Delta IAB}}={{a}^{3}}+\dfrac{3}{4}a+\dfrac{3}{16a}+\dfrac{1}{64{{a}^{3}}}d\left( P \right)S=\dfrac{4}{3}{{S}_{\Delta IAB}}=\dfrac{4}{3}{{a}^{3}}+a+\dfrac{1}{4a}+\dfrac{1}{48{{a}^{3}}}S=\left( \dfrac{4}{3}{{a}^{3}}+\dfrac{1}{12a}+\dfrac{1}{12a}+\dfrac{1}{12a} \right)+\left( \dfrac{a}{3}+\dfrac{a}{3}+\dfrac{a}{3}+\dfrac{1}{48{{a}^{3}}} \right)S\overset{Cauchy}{\mathop{\ge }} 4.\sqrt[4]{\dfrac{4}{3}{{a}^{3}}.\dfrac{1}{12a}.\dfrac{1}{12a}.\dfrac{1}{12a}}+4.\sqrt[4]{\dfrac{a}{3}.\dfrac{a}{3}.\dfrac{a}{3}.\dfrac{1}{48{{a}^{3}}}}=\dfrac{4}{3}MinS=\dfrac{4}{3}\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{4{{a}^{3}}}{3}=\dfrac{1}{12a} \\
& \dfrac{a}{3}=\dfrac{1}{48{{a}^{3}}} \\
\end{aligned} \right.$$\Leftrightarrow {{a}^{4}}=\dfrac{1}{16}\Leftrightarrow a=\dfrac{1}{2}$.