Phương pháp giải:

Lời giải chi tiết:

Ta có: ${S_{ABC}} = {1 \over 2}.d(A;BC).BC = {1 \over 2}d(A,\Delta ).BC$

Nhận xét: ${S_{ABC}}$ đạt GTLN khi và chỉ khi $d(A;\Delta )$ lớn nhất.

$A({x_0};{y_0}) \in (E) \Rightarrow \,\,\,{{{x_0}^2} \over 8} + {{{y_0}^2} \over 4} = 1$.

Khoảng cách từ A đến D:

$\eqalign{  & d(A,\Delta ) = {{\left| {{x_0} - \sqrt 2 {y_0} + 2} \right|} \over {\sqrt {{1^2} + {{\left( {\sqrt 2 } \right)}^2}} }} = {{\left| {{x_0} - \sqrt 2 {y_0} + 2} \right|} \over {\sqrt 3 }} \le {{\left| {{x_0} - \sqrt 2 {y_0}} \right| + 2} \over {\sqrt 3 }} = {{\left| {2\sqrt 2 .{{{x_0}} \over {2\sqrt 2 }} + \left( { - 2\sqrt 2 } \right).{{{y_0}} \over 2}} \right| + 2} \over {\sqrt 3 }}  \cr   &  \le {{\sqrt {\left[ {{{\left( {2\sqrt 2 } \right)}^2} + {{\left( { - 2\sqrt 2 } \right)}^2}} \right]\left( {{{{x_0}^2} \over 8} + {{{y_0}^2} \over 4}} \right)}  + 2} \over {\sqrt 3 }} = {{\sqrt {16.1}  + 2} \over {\sqrt 3 }} = {6 \over {\sqrt 3 }} = 2\sqrt 3  \cr}$

$\eqalign{  & d{(A;\Delta )_{{\rm{Max}}}} = 2\sqrt 3  \Leftrightarrow \left\{ \matrix{  \left( {{x_0} - \sqrt 2 {y_0}} \right).2 \ge 0 \hfill \cr   {{{x_0}^2} \over 8} + {{{y_0}^2} \over 4} = 1 \hfill \cr   {{{{{x_0}} \over {2\sqrt 2 }}} \over {2\sqrt 2 }} = {{{{{y_0}} \over 2}} \over { - 2\sqrt 2 }} \hfill \cr}  \right. \Leftrightarrow \left\{ \matrix{  {x_0} - \sqrt 2 {y_0} \ge 0 \hfill \cr   {{{x_0}^2} \over 8} + {{{y_0}^2} \over 4} = 1 \hfill \cr   {x_0} =  - \sqrt 2 {y_0} \hfill \cr}  \right. \Leftrightarrow \left\{ \matrix{  {x_0} - \sqrt 2 {y_0} \ge 0 \hfill \cr   {{2{y_0}^2} \over 8} + {{{y_0}^2} \over 4} = 1 \hfill \cr   {x_0} =  - \sqrt 2 {y_0} \hfill \cr}  \right. \Leftrightarrow \left\{ \matrix{  {x_0} - \sqrt 2 {y_0} \ge 0 \hfill \cr   {y_0}^2 = 2 \hfill \cr   {x_0} =  - \sqrt 2 {y_0} \hfill \cr}  \right.  \cr   &  \Leftrightarrow \left\{ \matrix{  {x_0} - \sqrt 2 {y_0} \ge 0 \hfill \cr   {y_0} =  \pm \sqrt 2  \hfill \cr   {x_0} =  - \sqrt 2 {y_0} \hfill \cr}  \right. \Leftrightarrow \left[ \matrix{  \left\{ \matrix{  {x_0} = 2 \hfill \cr   {y_0} =  - \sqrt 2  \hfill \cr}  \right.\,(TM) \hfill \cr   \left\{ \matrix{  {x_0} =  - 2 \hfill \cr   {y_0} = \sqrt 2  \hfill \cr}  \right.(L) \hfill \cr}  \right. \cr}$

Khi đó, $A\left( {2; - \sqrt 2 } \right)$

Chọn: A