Cho hypebol \((H):\frac{{{x^2}}}{{25}} - \frac{{{y...

A

- Hướng dẫn giải

Phương pháp giải:

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

Giải chi tiết:

\((H):\frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{16}} = 1 \Rightarrow \left\{ \begin{array}{l}a = 5\\b = 4\\c = \sqrt {41} \end{array} \right.\)

\(M\left( {{x_0};{y_0}} \right) \in \left( H \right) \Rightarrow \left\{ \begin{array}{l}M{F_1} = \left| {a + \frac{c}{a}{x_0}} \right| = \left| {5 + \frac{{\sqrt {41} }}{5}{x_0}} \right|\\M{F_2} = \left| {a - \frac{c}{a}{x_0}} \right| = \left| {5 - \frac{{\sqrt {41} }}{5}{x_0}} \right|\end{array} \right.\)

Khi đó, \(O{M^2} - M{F_1}.M{F_2} = {x_0}^2 + {y_0}^2 - \left| {5 + \frac{{\sqrt {41} }}{5}{x_0}} \right|.\left| {5 - \frac{{\sqrt {41} }}{5}{x_0}} \right| = {x_0}^2 + {y_0}^2 - \left| {25 - \frac{{41}}{{25}}{x_0}^2} \right|\)

Mà \(\frac{{{x_0}^2}}{{25}} - \frac{{{y_0}^2}}{{16}} = 1 \Leftrightarrow \frac{{{x_0}^2}}{{25}} = \frac{{{y_0}^2}}{{16}} + 1 \ge 1 \Rightarrow {x_0}^2 \ge 25 \Rightarrow 25 - \frac{{41}}{{25}}{x_0}^2 < 0\)

\( \Rightarrow O{M^2} - M{F_1}.M{F_2} = {x_0}^2 + {y_0}^2 + \left( {25 - \frac{{41}}{{25}}{x_0}^2} \right) = {y_0}^2 - \frac{{16}}{{25}}{x_0}^2 + 25 =  - 16\left( {\frac{{{x_0}^2}}{{25}} - \frac{{{y_0}^2}}{{16}}} \right) + 25 =  - 16 + 25 = 9\)

Chọn: A