Cho elip (E) : \(\dfrac{{{x^2}}}{{64}} + \dfrac{{{y^2}}}{{48}} = 1.\) Tìm tọa độ những điểm M trên (E) sao cho : \(M{F_1} + 2M{F_2} = 26\).

Ta có \(a = 8\,;\,b = 4\sqrt 3 \,;\,\) \(c = 4\)

\(M(x;y) \in (E)\)\( \Leftrightarrow \dfrac{{{x^2}}}{{64}} + \dfrac{{{y^2}}}{{48}} = 1\,\,\,(1)\)

Ta có: \(M \in \left( E \right)\) \( \Rightarrow M{F_1} + M{F_2} = 2a = 16\)

Mà \(M{F_1} + 2M{F_2} = 26\) nên \(\left( {M{F_1} + M{F_2}} \right) + M{F_2} = 26\)

Hay \(16 + M{F_2} = 26 \Leftrightarrow M{F_2} = 10\)

\(\begin{array}{l} \Leftrightarrow \sqrt {{{\left( {4 - x} \right)}^2} + {y^2}}  = 10\\ \Leftrightarrow {\left( {4 - x} \right)^2} + {y^2} = 100\\ \Leftrightarrow {x^2} - 8x + 16 + {y^2} = 100\end{array}\)

Tọa độ điểm \(M\) thỏa mãn hệ phương trình:

\(\left\{ \begin{array}{l}{x^2} - 8x + 16 + {y^2} = 100\\\dfrac{{{x^2}}}{{64}} + \dfrac{{{y^2}}}{{48}} = 1\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 8x + 16 + {y^2} = 100\\\dfrac{{{y^2}}}{{48}} = 1 - \dfrac{{{x^2}}}{{64}}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 8x + 16 + {y^2} = 100\\{y^2} = 48 - \dfrac{{3{x^2}}}{4}\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 8x + 16 + 48 - \dfrac{{3{x^2}}}{4} = 100\\{y^2} = 48 - \dfrac{{3{x^2}}}{4}\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{{x^2}}}{4} - 8x - 36 = 0\\{y^2} = 48 - \dfrac{{3{x^2}}}{4}\end{array} \right.\)  

\( \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x = 36\left( {loai\,vi\,x < 8} \right)\\x =  - 4\left( {TM} \right)\end{array} \right.\\{y^2} = 48 - \dfrac{{3{x^2}}}{4}\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}x =  - 4\\y =  \pm 6\end{array} \right. \Rightarrow \left[ \begin{array}{l}M\left( { - 4; - 6} \right)\\M\left( { - 4;6} \right)\end{array} \right.\)

Vậy \(M( - 4; \pm 6).\)