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(E) : $ \frac{{{x^2}}}{4} + {y^2} = 1\,\,\,\,\,\,;\,\,\,\,\,{a^2} = 4$
$\Rightarrow a = 2\,\,;\,\,{b^2} = 1\Rightarrow \,b = 1\,\,;\,\,{c^2} = {a^2} - {b^2} = 3\, \Rightarrow c = \sqrt 3 $
+ Áp dụng định lí côsin trong tam giác F1NF2:
$ \begin{array}{l}
{({F_1}{F_2})^2}\,\, = \,\,NF_1^2\, + \,NF_2^2\,\, - \,2N{F_1}N{F_2}.\cos {60^0}\\
\Leftrightarrow \,\,\,{({F_1}{F_2})^2}\,\, = \,\,{(\,N{F_1}\, + \,N{F_2})^2} - 2N{F_1}.N{F_2} - N{F_1}.N{F_2}\\
\Leftrightarrow \,\,N{F_1}.N{F_2}\,\, = \,\,\frac{4}{3}(\,{a^2} - {c^2}) = \frac{4}{3}\\
\Leftrightarrow \,\,{x^2}\,\, = \,\,\frac{{32}}{9}\,\,;\,\,{y^2} = \frac{2}{{18}}
\end{array} $
Vậy có 4 điểm thỏa yêu cầu bài toán : $ {N_1}\left( {\frac{{4\sqrt 2 }}{3}\,,\,\frac{1}{3}} \right)\,;{N_2}\left( {\frac{{4\sqrt 2 }}{3}\,,\, - \frac{1}{3}} \right)\,;{N_3}\left( { - \frac{{4\sqrt 2 }}{3}\,,\,\frac{1}{3}} \right)\,;\,{N_4}\left( { - \frac{{4\sqrt 2 }}{3}\,,\, - \frac{1}{3}} \right) $
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