Lời giải chi tiết:

* \({c^2} = {a^2} - {b^2} = 25 - 2 = 23 \Rightarrow c = \sqrt {23} \)

\( \Rightarrow {F_1}\left( { - \sqrt {23} ;0} \right);\,\,{F_2}\left( {\sqrt {23} ;0} \right)\)

* Giả sử \(M\left( {{x_M};{y_M}} \right) \in \left( E \right) \Rightarrow \dfrac{{x_M^2}}{{25}} + \dfrac{{y_M^2}}{2} = 1\,\,\,\left( 1 \right)\)

\(*\,\,\left\{ \begin{array}{l}\overrightarrow {{F_1}M}  = \left( {{x_M} + \sqrt {23} ;{y_M}} \right)\\\overrightarrow {{F_2}M}  = \left( {{x_M} - \sqrt {23} ;{y_M}} \right)\end{array} \right.;\,\,\overrightarrow {{F_1}M} .\overrightarrow {{F_2}M}  = 0 \Leftrightarrow x_M^2 + y_M^2 = 23\,\,\left( 2 \right)\)

* Giải hệ \(\left\{ \begin{array}{l}\left( 1 \right)\\\left( 2 \right)\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x_M^2 = \dfrac{{525}}{{23}}\\y_M^2 = \dfrac{4}{{23}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x = \sqrt {\dfrac{{523}}{{23}}} \\y = \sqrt {\dfrac{4}{{23}}} \end{array} \right. \Rightarrow M\left( {\sqrt {\dfrac{{523}}{{23}}} ;\sqrt {\dfrac{4}{{23}}} } \right)\).

Chọn A.