Phương pháp giải:

\(M{F_1} = a + {c \over a}{x_0};\,\,M{F_2} = a - {c \over a}{x_0}\,\,,\,\,\,\left( {M\left( {{x_0};{y_0}} \right) \in \left( E \right)} \right)\)

Lời giải chi tiết:

Gọi \(M({x_0};{y_0}) \in (E) \Rightarrow \,\,{{{x_0}^2} \over {100}} + {{{y_0}^2} \over {36}} = 1\).

\((E):\,\,\,{{{x^2}} \over {100}} + {{{y^2}} \over {36}} = 1 \Rightarrow a = 10,\,\,b = 6\)

Mà \({a^2} - {b^2} = {c^2} \Rightarrow {c^2} = {10^2} - {6^2} = 64 \Rightarrow c = 8\)

\(M{F_1} = a + {c \over a}{x_0} = 10 + {8 \over {10}}{x_0} = 10 + {4 \over 5}{x_0};\,\,\,\,M{F_2} = a - {c \over a}{x_0}\, = 10 - {8 \over {10}}{x_0} = 10 - {4 \over 5}{x_0}\)

\(\eqalign{  & M{F_2} = 4M{F_1} \Leftrightarrow 10 - {4 \over 5}{x_0} = 4\left( {10 + {4 \over 5}{x_0}} \right)\, \Leftrightarrow 10 - {4 \over 5}{x_0} = 40 + {{16} \over 5}{x_0} \Leftrightarrow 4{x_0} =  - 30 \Leftrightarrow {x_0} =  - {{15} \over 2}  \cr   & {{{x_0}^2} \over {100}} + {{{y_0}^2} \over {36}} = 1 \Leftrightarrow {{{{\left( { - {{15} \over 2}} \right)}^2}} \over {100}} + {{{y_0}^2} \over {36}} = 1 \Leftrightarrow {y_0}^2 = {{63} \over 4} \Leftrightarrow {y_0} =  \pm {{3\sqrt 7 } \over 2}  \cr   &  \Rightarrow {M_1}\left( { - {{15} \over 2};{{3\sqrt 7 } \over 2}} \right)\,\,\,,\,\,\,{M_2}\left( { - {{15} \over 2}; - {{3\sqrt 7 } \over 2}} \right)\,\,\, \cr} \)

Chọn: A