Cho hypebol \((H):4{x^2} - {y^2} = 4\). Tìm điểm \...

B

- 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:

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

\(M\left( {{x_0};{y_0}} \right) \in \left( H \right) \Rightarrow \left\{ \begin{array}{l}M{F_1} = \left| {a + \dfrac{c}{a}{x_0}} \right| = \left| {1 + \sqrt 5 {x_0}} \right|\\M{F_2} = \left| {a - \dfrac{c}{a}{x_0}} \right| = \left| {1 - \sqrt 5 {x_0}} \right|\end{array} \right.\)

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

\( \Leftrightarrow {F_1}{F_2}^2 = {\left( {M{F_1} - M{F_2}} \right)^2} + 2M{F_1}.M{F_2} - 2.M{F_1}.M{F_2}.\cos {120^0}\) \( \Leftrightarrow {\left( {2c} \right)^2} = {\left( {2a} \right)^2} + 3M{F_1}.M{F_2}\)

\(\begin{array}{l} \Leftrightarrow {\left( {2.\sqrt 5 } \right)^2} = {\left( {2.1} \right)^2} + 3\left| {1 + \sqrt 5 {x_0}} \right|.\left| {1 - \sqrt 5 {x_0}} \right|\\ \Leftrightarrow 3\left| {1 - 5{x_0}^2} \right| = 16\\ \Leftrightarrow \left[ \begin{array}{l}1 - 5{x_0}^2 = \dfrac{{16}}{3}\\1 - 5{x_0}^2 =  - \dfrac{{16}}{3}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}{x_0}^2 =  - \dfrac{{ - 13}}{3}\\{x_0}^2 = \dfrac{{19}}{{15}}\end{array} \right. \Leftrightarrow {x_0}^2 = \dfrac{{19}}{{15}}\end{array}\)

\(\begin{array}{l}(H):4{x^2} - {y^2} = 4 \Rightarrow 4{x_0}^2 - {y_0}^2 = 4\\ \Leftrightarrow 4.\dfrac{{19}}{{15}} - {y_0}^2 = 4 \Leftrightarrow {y_0}^2 = \dfrac{{16}}{{15}}\end{array}\)

\( \Rightarrow \left\{ \begin{array}{l}{x_0}^2 = \dfrac{{19}}{{15}}\\{y_0}^2 = \dfrac{{16}}{{15}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_0} =  \pm \sqrt {\dfrac{{19}}{{15}}} \\{y_0} =  \pm \dfrac{4}{{\sqrt {15} }}\end{array} \right.\)

Vậy, \({M_1}\left( {\sqrt {\dfrac{{19}}{{15}}} ;\dfrac{4}{{\sqrt {15} }}} \right);{M_2}\left( { - \sqrt {\dfrac{{19}}{{15}}} ;\dfrac{4}{{\sqrt {15} }}} \right);{M_3}\left( {\sqrt {\dfrac{{19}}{{15}}} ; - \dfrac{4}{{\sqrt {15} }}} \right);{M_4}\left( { - \sqrt {\dfrac{{19}}{{15}}} ; - \dfrac{4}{{\sqrt {15} }}} \right)\).

Chọn: B