Cho elip \((E):\,\,{{{x^2}} \over {25}} + {{{y^2}}...

C

- Hướng dẫn giải

Phương pháp giải:

Gọi \(M\left( {{x_0};{y_0}} \right) \in \left( E \right)\), tính \(d\left( {M;\left( D \right)} \right) = {{\left| {{x_0} - 2{y_0} + 12} \right|} \over {\sqrt {{1^2} + {2^2}} }}\)

Sử dụng các bất đẳng thức \(\left| {a + b} \right| \le \left| a \right| + \left| b \right|\), dấu bằng xảy ra \( \Leftrightarrow a.b \ge 0\) và BĐT Bunhia-copxki: \(\left( {{a^2} + {b^2}} \right)\left( {{x^2} + {y^2}} \right) \ge {\left( {ax + by} \right)^2}\), dấu bằng xảy ra \( \Leftrightarrow {a \over x} = {b \over y}\)

Giải chi tiết:

Gọi \(M({x_0};{y_0}) \in (E) \Rightarrow \,\,{{{x_0}^2} \over {25}} + {{{y_0}^2} \over 9} = 1\).

Khoảng cách từ M đến D:

\(\eqalign{  & d(M,D) = {{\left| {{x_0} - 2{y_0} + 12} \right|} \over {\sqrt {{1^2} + {2^2}} }} = {{\left| {{x_0} - 2{y_0} + 12} \right|} \over {\sqrt 5 }} \le {{\left| {{x_0} - 2{y_0}} \right| + 12} \over {\sqrt 5 }} = {{\left| {5.{{{x_0}} \over 5} + \left( { - 6} \right){{{y_0}} \over 3}} \right| + 12} \over {\sqrt 5 }} \le {{\sqrt {({5^2} + {6^2})\left( {{{{x_0}^2} \over {25}} + {{{y_0}^2} \over 9}} \right)}  + 12} \over {\sqrt 5 }}  \cr   &  = {{\sqrt {(25 + 36).1}  + 12} \over {\sqrt 5 }} = {{\sqrt {61}  + 12} \over {\sqrt 5 }} \cr} \).

Vậy,

\(d{(M,D)_{\max }} = {{\sqrt {61}  + 12} \over {\sqrt 5 }} \Leftrightarrow \left\{ \matrix{ \left( {{x_0} - 2{y_0}} \right).12 \ge 0 \hfill \cr   {{{x_0}^2} \over {25}} + {{{y_0}^2} \over 9} = 1 \hfill \cr   {{{{{x_0}} \over 5}} \over 5} = {{{{{y_0}} \over 3}} \over { - 6}} \hfill \cr}  \right. \Leftrightarrow \left\{ \matrix{  {x_0} - 2{y_0} \ge 0 \hfill \cr   {{{x_0}^2} \over {25}} + {{{y_0}^2} \over 9} = 1 \hfill \cr   18{x_0} + 25{y_0} = 0 \hfill \cr}  \right. \Leftrightarrow \left\{ \matrix{  {x_0} - 2{y_0} \ge 0 \hfill \cr   {{{x_0}^2} \over {25}} + {{{y_0}^2} \over 9} = 1 \hfill \cr   {x_0} =  - {{25} \over {18}}{y_0} \hfill \cr}  \right. \Leftrightarrow \left\{ \matrix{  {x_0} - 2{y_0} \ge 0 \hfill \cr   {{{{\left( { - {{25} \over {18}}{y_0}} \right)}^2}} \over {25}} + {{{y_0}^2} \over 9} = 1 \hfill \cr   {x_0} =  - {{25} \over {18}}{y_0} \hfill \cr}  \right.\)

\( \Leftrightarrow \left\{ \matrix{  {x_0} - 2{y_0} \ge 0 \hfill \cr   {y_0}^2 = {{324} \over {661}} \hfill \cr   {x_0} =  - {{25} \over {18}}{y_0} \hfill \cr}  \right. \Leftrightarrow \left\{ \matrix{  {x_0} - 2{y_0} \ge 0 \hfill \cr   {y_0} =  \pm \sqrt {{{324} \over {661}}}  \hfill \cr   {x_0} =  - {{25} \over {18}}{y_0} \hfill \cr}  \right. \Leftrightarrow \left[ \matrix{  \left\{ \matrix{  {x_0} =  - {{25} \over {18}}\sqrt {{{324} \over {661}}}  \hfill \cr   {y_0} = \sqrt {{{324} \over {661}}}  \hfill \cr}  \right.\,\,(L) \hfill \cr   \left\{ \matrix{  {x_0} = {{25} \over {18}}\sqrt {{{324} \over {661}}}  \hfill \cr   {y_0} =  - \sqrt {{{324} \over {661}}}  \hfill \cr}  \right.\,\,(TM) \hfill \cr}  \right.\)

Khi đó, \(M\left( {{{25} \over {18}}\sqrt {{{324} \over {661}}} ; - \sqrt {{{324} \over {661}}} } \right)\)

Chọn: C