Question

Trong mặt phẳng toạ độ Oxy cho A(1;2); B(3;4). Điểm \(M\left(\frac{a}{b};0\right)\) thuộc Ox( với \(a,b\in Z;\left(a,b\right)=1;b>0\)) sao cho \(P=\left|2\overrightarrow{AB}+\overrightarrow{MA}+\overrightarrow{MB}\right|+\left|\overrightarrow{MA}+\overrightarrow{MB}\right|\) nhỏ nhất. Khi đó a+ b=?

Answer 1

Đặt \(\frac{a}{b}=x\)

\(\left\{{}\begin{matrix}\overrightarrow{AB}=\left(2;2\right)\\\overrightarrow{MA}=\left(1-x;2\right)\\\overrightarrow{MB}=\left(3-x;4\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2\overrightarrow{AB}+\overrightarrow{MA}+\overrightarrow{MB}=\left(8-2x;10\right)\\\overrightarrow{MA}+\overrightarrow{MB}=\left(4-2x;6\right)\end{matrix}\right.\)

\(P=\sqrt{\left(8-2x\right)^2+10^2}+\sqrt{\left(4-2x\right)^2+6^2}\)

\(P=\sqrt{\left(8-2x\right)^2+10^2}+\sqrt{\left(2x-4\right)^2+6^2}\)

\(P\ge\sqrt{\left(8-2x+2x-4\right)^2+\left(10+6\right)^2}=4\sqrt{17}\)

Dấu "=" xảy ra khi \(6\left(8-2x\right)=10\left(2x-4\right)\Leftrightarrow x=\frac{11}{4}\)

\(\Rightarrow a+b=15\)