Cho hypebol \((H):{x^2} - \frac{{{y^2}}}{9} = 1\)....

A

- Hướng dẫn giải

Phương pháp giải:

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

Sử dụng công thức Côsin cho tam giác \(M{F_1}{F_2}\): \({F_1}{F_2}^2 = M{F_1}^2 + M{F_2}^2 - 2.M{F_1}.M{F_2}.\cos \widehat {{F_1}M{F_2}}\)

Giải chi tiết:

\((H):{x^2} - \frac{{{y^2}}}{9} = 1 \Rightarrow \left\{ \begin{array}{l}a = 1\\b = 3\\c = \sqrt {{a^2} + {b^2}}  = \sqrt {10} \end{array} \right.\)

\(M\left( {{x_0};{y_0}} \right) \in \left( H \right) \Rightarrow \left\{ \begin{array}{l}M{F_1} = \left| {a + \frac{c}{a}{x_0}} \right| = \left| {1 + \sqrt {10} {x_0}} \right|\\M{F_2} = \left| {a - \frac{c}{a}{x_0}} \right| = \left| {1 - \sqrt {10} {x_0}} \right|\end{array} \right.\)

Áp dụng công thức Côsin cho tam giác \(M{F_1}{F_2}\): \({F_1}{F_2}^2 = M{F_1}^2 + M{F_2}^2 - 2.M{F_1}.M{F_2}.\cos \widehat {{F_1}M{F_2}}\)

\( \Leftrightarrow {F_1}{F_2}^2 = {\left( {M{F_1} - M{F_2}} \right)^2} + 2M{F_1}.M{F_2} - 2.M{F_1}.M{F_2}.\cos {60^0}\) \( \Leftrightarrow {(2c)^2} = {\left( {2a} \right)^2} + M{F_1}.M{F_2}\)

\( \Leftrightarrow {\left( {2\sqrt {10} } \right)^2} = {\left( {2.1} \right)^2} + \left| {1 + \sqrt {10} {x_0}} \right|.\left| {1 - \sqrt {10} {x_0}} \right| \Leftrightarrow \left| {1 - 10{x_0}^2} \right| = 36\) \( \Leftrightarrow \left[ \begin{array}{l}1 - 10{x_0}^2 = 36\\1 - 10{x_0}^2 =  - 36\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}{x_0}^2 =  - \frac{7}{2}\\{x_0}^2 = \frac{{37}}{{10}}\end{array} \right. \Leftrightarrow {x_0}^2 = \frac{{37}}{{10}}\)

\(M\left( {{x_0};{y_0}} \right) \in \left( H \right) \Rightarrow (H):{x_0}^2 - \frac{{{y_0}^2}}{9} = 1 \Leftrightarrow \frac{{37}}{{10}} - \frac{{{y_0}^2}}{9} = 1 \Leftrightarrow {y_0}^2 = \frac{{243}}{{10}}\) \( \Rightarrow \left\{ \begin{array}{l}{x_0}^2 = \frac{{37}}{{10}}\\{y_0}^2 = \frac{{243}}{{10}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_0} =  \pm \sqrt {\frac{{37}}{{10}}} \\{y_0} =  \pm \sqrt {\frac{{243}}{{10}}} \end{array} \right.\)

Vậy, \({M_1}\left( {\sqrt {\frac{{37}}{{10}}} ;\sqrt {\frac{{243}}{{10}}} } \right);{M_2}\left( {\sqrt {\frac{{37}}{{10}}} ; - \sqrt {\frac{{243}}{{10}}} } \right);{M_3}\left( { - \sqrt {\frac{{37}}{{10}}} ;\sqrt {\frac{{243}}{{10}}} } \right);{M_4}\left( { - \sqrt {\frac{{37}}{{10}}} ; - \sqrt {\frac{{243}}{{10}}} } \right)\).

Chọn: A