Cho hypebol \((H):3{x^2} - 12{y^2} = 12\), có hai...

D

- Hướng dẫn giải

Phương pháp giải:

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

Giải chi tiết:

\((H):3{x^2} - 12{y^2} = 12 \Leftrightarrow \frac{{{x^2}}}{4} - \frac{{{y^2}}}{1} = 1 \Rightarrow \left\{ \begin{array}{l}a = 2\\b = 1\\c = \sqrt 5 \end{array} \right.\)

\(M\left( {{x_0};{y_0}} \right) \in \left( H \right) \Rightarrow M{F_1} = \left| {a + \frac{c}{a}{x_0}} \right| = \left| {2 + \frac{{\sqrt 5 }}{2}{x_0}} \right| = 2 \Rightarrow \left[ \begin{array}{l}2 + \frac{{\sqrt 5 }}{2}{x_0} = 2\\2 + \frac{{\sqrt 5 }}{2}{x_0} =  - 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}{x_0} = 0\\{x_0} =  - \frac{8}{{\sqrt 5 }}\end{array} \right.\)

+) \({x_0} = 0 \Rightarrow {y_0}^2 =  - 1\) (vô lí)

+) \({x_0} =  - \frac{8}{{\sqrt 5 }} \Rightarrow {y_0}^2 = \frac{{11}}{5} \Leftrightarrow {y_0} =  \pm \sqrt {\frac{{11}}{5}}  \Rightarrow {M_1}\left( { - \frac{8}{{\sqrt 5 }};\sqrt {\frac{{11}}{5}} } \right),\,{M_2}\left( { - \frac{8}{{\sqrt 5 }}; - \sqrt {\frac{{11}}{5}} } \right)\,\).

Chọn: D