$\left( E \right):{x^2} + 4{y^2} = 4 \Leftrightarrow \dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{1} = 1$
Gọi $M(x;y) \in (E)\Rightarrow \dfrac{x^2}{4}+\dfrac{y^2}{1}=1$;
$\begin{array}{l} M{F_1} = a + ex,M{F_2} = a - ex\\ + M{F_1}M{F_2} + O{M^2} = \left( {a + ex} \right)\left( {a - ex} \right) + {x^2} + {y^2}\\ = {a^2} - {e^2}{x^2} + {x^2}\left( {1 - \dfrac{{{c^2}}}{{{a^2}}}} \right)\\ = {a^2} + {y^2} + {b^2}\dfrac{{{x^2}}}{{{a^2}}} = {a^2} + {y^2} + {b^2}\left( {1 - \dfrac{{{y^2}}}{{{b^2}}}} \right)\\ = {a^2} + {b^2} = 4 + 1 = 5 \end{array}$
