Cho hypebol \((H):9{x^2} - 16{y^2} = 144\). Tìm đi...

C

- Hướng dẫn giải

Phương pháp giải:

\((H):\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1,\,\,M\left( {{x_0};{y_0}} \right) \in \left( H \right) \Rightarrow \left\{ \begin{array}{l}M{F_1} = \left| {a + \frac{c}{a}{x_0}} \right|\\M{F_2} = \left| {a - \frac{c}{a}{x_0}} \right|\end{array} \right.\)

Giải chi tiết:

\((H):9{x^2} - 16{y^2} = 144 \Leftrightarrow \frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1 \Rightarrow \left\{ \begin{array}{l}a = 4\\b = 3\\c = \sqrt {{a^2} + {b^2}}  = 5\end{array} \right.\)

\(M\left( {{x_0};{y_0}} \right) \in \left( H \right) \Rightarrow \left\{ \begin{array}{l}M{F_1} = \left| {a + \frac{c}{a}{x_0}} \right| = \left| {4 + \frac{5}{4}{x_0}} \right|\\M{F_2} = \left| {a - \frac{c}{a}{x_0}} \right| = \left| {4 - \frac{5}{4}{x_0}} \right|\end{array} \right.\)

Theo đề bài ta có: \(M{F_1} = 2M{F_2} \Leftrightarrow \left| {4 + \frac{5}{4}{x_0}} \right| = 2.\left| {4 - \frac{5}{4}{x_0}} \right| \Leftrightarrow \left[ \begin{array}{l}4 + \frac{5}{4}{x_0} = 8 - \frac{5}{2}{x_0}\\4 + \frac{5}{4}{x_0} =  - 8 + \frac{5}{2}{x_0}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}{x_0} = \frac{{16}}{5}\\{x_0} = \frac{{48}}{5}\end{array} \right.\)

Mà \(M\left( {{x_0};{y_0}} \right) \in \left( H \right) \Rightarrow 9{x_0}^2 - 16{y_0}^2 = 144\)

+) \({x_0} = \frac{{16}}{5} \Rightarrow {y_0}^2 =  - \frac{{81}}{{25}}\,\,\)(vô lí)

+) \({x_0} = \frac{{48}}{5} \Rightarrow {y_0}^2 = \frac{{1071}}{{25}} \Rightarrow {y_0} =  \pm \frac{{3\sqrt {119} }}{5} \Rightarrow {M_1}\left( {\frac{{48}}{5};\frac{{3\sqrt {119} }}{5}} \right),\,\,{M_2}\left( {\frac{{48}}{5}; - \frac{{3\sqrt {119} }}{5}} \right)\)

Chọn: C