Đáp án:
$M\left( {\dfrac{3}{5};\dfrac{1}{5}} \right)$
Giải thích các bước giải:
Ta có:
$\left( d \right):2x - y - 1 = 0$ và $E\left( {1;6} \right),F\left( { - 3; - 4} \right)$
Mà $M \in \left( d \right) \Rightarrow M\left( {x;2x - 1} \right)$
$\begin{array}{l}
\Rightarrow \overrightarrow {ME} = \left( {1 - x;7 - 2x} \right);\overrightarrow {MF} = \left( { - 3 - x; - 3 - 2x} \right)\\
\Rightarrow \overrightarrow {ME} + \overrightarrow {MF} = \left( { - 2 - 2x;4 - 4x} \right)
\end{array}$
Khi đó:
$\begin{array}{l}
\left| {\overrightarrow {ME} + \overrightarrow {MF} } \right| = \sqrt {{{\left( { - 2 - 2x} \right)}^2} + {{\left( {4 - 4x} \right)}^2}} \\
= \sqrt {4{{\left( {x + 1} \right)}^2} + 16{{\left( {x - 1} \right)}^2}} \\
= \sqrt {20{x^2} - 24x + 20} \\
= \sqrt {20\left( {{x^2} - 2.x.\dfrac{3}{5} + \dfrac{9}{{25}} + \dfrac{{16}}{{25}}} \right)} \\
= \sqrt {20{{\left( {x - \dfrac{3}{5}} \right)}^2} + \dfrac{{64}}{5}}
\end{array}$
Mà ${\left( {x - \dfrac{3}{5}} \right)^2} \ge 0,\forall x$
$\begin{array}{l}
\Rightarrow 20{\left( {x - \dfrac{3}{5}} \right)^2} + \dfrac{{64}}{5} \ge \dfrac{{64}}{5},\forall x\\
\Rightarrow \left| {\overrightarrow {ME} + \overrightarrow {MF} } \right| \ge \dfrac{8}{{\sqrt 5 }}
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
\Leftrightarrow x - \dfrac{3}{5} = 0\\
\Leftrightarrow x = \dfrac{3}{5}
\end{array}$
$ \Rightarrow Min\left| {\overrightarrow {ME} + \overrightarrow {MF} } \right| = \dfrac{8}{{\sqrt 5 }} \Leftrightarrow x = \dfrac{3}{5}$
Hay $Min\left| {\overrightarrow {ME} + \overrightarrow {MF} } \right| = \dfrac{8}{{\sqrt 5 }} \Leftrightarrow M\left( {\dfrac{3}{5};\dfrac{1}{5}} \right)$
Vậy $M\left( {\dfrac{3}{5};\dfrac{1}{5}} \right)$ thỏa mãn đề.
