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Từ phương trình: $ {\textstyle{{{x^2}} \over 9}} + {\textstyle{{{y^2}} \over 5}} = 1\Rightarrow \left\{ \begin{array}{l}
{a^2} = 9\\
{b^2} = 5
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = 3\\
{c^2} = {a^2} - {b^2} = 4
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = 3\\
c = 2
\end{array} \right. $ $ {F_1}\left( { - 2;0} \right),{F_2}\left( {2;0} \right) $
Ta có M(x, y) thuộc (E) nên $ {F_1}M = a + {\textstyle{c \over a}}x = 3 + {\textstyle{2 \over 3}}x\,;{F_2}M = a - {\textstyle{c \over a}}x = 3 - {\textstyle{2 \over 3}}x $
a. Yêu cầu bài toán $\Leftrightarrow \left[ \begin{array}{l}
{F_1}M = 2{F_2}M\\
{F_2}M = 2{F_1}M
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
3 + {\textstyle{2 \over 3}}x = 2\left( {3 - {\textstyle{2 \over 3}}x} \right)\\
3 - {\textstyle{2 \over 3}}x = 2\left( {3 + {\textstyle{2 \over 3}}x} \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = {\textstyle{3 \over 2}}\\
x = - {\textstyle{3 \over 2}}
\end{array} \right. $
$\Rightarrow M_1\left( {{\textstyle{3 \over 2}};{\textstyle{{\sqrt {15} } \over 4}}} \right)\,\, \vee \,\,M_2\left( {{\textstyle{3 \over 2}}; - {\textstyle{{\sqrt {15} } \over 4}}} \right)\,\, \vee \,\,M_3\left( { - {\textstyle{3 \over 2}};{\textstyle{{\sqrt {15} } \over 4}}} \right)\;\, \vee \,\,M_4\left( { - {\textstyle{3 \over 2}}; - {\textstyle{{\sqrt {15} } \over 4}}} \right) $
b. Ta có $\widehat{F_1MF_2}=60^0$. Xét tam giác $MF_1F_2$ ta có: $ {F_1}F_2^2 = MF_1^2 + MF_2^2 - 2M{F_1}.M{F_2}\cos 60^\circ $
$ \Leftrightarrow {F_1}F_2^2 = {\left( {M{F_1} + M{F_2}} \right)^2} - 3M{F_1}.MF $ $ \Leftrightarrow {\left( {2c} \right)^2} = {\left( {2a} \right)^2} - 3M{F_1}.M{F_2} $
$ \Leftrightarrow M{F_1}.M{F_2} = {\textstyle{{4{a^2} - 4{c^2}} \over 3}} \Leftrightarrow \left( {3 + {\textstyle{2 \over 3}}x} \right)\left( {3 - {\textstyle{2 \over 3}}x} \right) = {\textstyle{{20} \over 3}} \Leftrightarrow {x^2} = {\textstyle{{21} \over 4}} \Leftrightarrow {y^2} = {\textstyle{{25} \over {12}}} $
$ \Leftrightarrow M_1\left( {{\textstyle{{\sqrt {21} } \over 2}};{\textstyle{{5\sqrt 3 } \over 6}}} \right)\,\, \vee \,\,M_2\left( { - {\textstyle{{\sqrt {21} } \over 2}};{\textstyle{{5\sqrt 3 } \over 6}}} \right)\,\, $
$\vee \,\,M_3\left( { - {\textstyle{{\sqrt {21} } \over 2}}; - {\textstyle{{5\sqrt 3 } \over 6}}} \right)\,\, \vee \,\,M_4\left( {{\textstyle{{\sqrt {21} } \over 2}}; - {\textstyle{{5\sqrt 3 } \over 6}}} \right) $
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