Do A; B thuộc (P) nên tọa độ có dạng: \(\left\{{}\begin{matrix}A\left(a^2;2a\right)\\B\left(b^2;2b\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{IA}=\left(a^2;2a-1\right)\\\overrightarrow{IB}=\left(b^2;2b-1\right)\end{matrix}\right.\)
\(\overrightarrow{IA}=4\overrightarrow{IB}\Rightarrow\left\{{}\begin{matrix}a^2=4b^2\\2a-1=4\left(2b-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2=4b^2\\a=\dfrac{8b-3}{2}\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{8b-3}{2}\right)^2=4b^2\)
\(\Leftrightarrow48b^2-48b+9=0\Rightarrow\left[{}\begin{matrix}b=\dfrac{1}{4}\Rightarrow a=-\dfrac{1}{2}\\b=\dfrac{3}{4}\Rightarrow a=\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}A\left(1;-1\right);B\left(\dfrac{1}{4};\dfrac{1}{2}\right)\\A\left(9;3\right);B\left(\dfrac{9}{4};\dfrac{3}{2}\right)\end{matrix}\right.\)
