Cho \(\left( E \right)\) có hai tiêu điểm \({F_1}\left( { - 4;0} \right)\), \({F_2}\left( {4;0} \right)\) và điểm \(M\) thuộc \(\left( E \right)\). Biết chu vi tam giác \(M{F_1}{F_2}\) bằng $18$. Khi đó tâm sai của \(\left( E \right)\) bằng

$e = \dfrac{2}{3}$.   

$e = \dfrac{4}{3}$.   

$e = \dfrac{2}{{\sqrt 3 }}$.

$e = \dfrac{4}{{\sqrt 3}}$

\(\dfrac{{{x^2}}}{7} + \dfrac{{{y^2}}}{2} = 1\).

\(\dfrac{{{x^2}}}{{20}} + \dfrac{{{y^2}}}{4} = 1\).

\(\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{1} = 1\).

\(\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{4} = 1\).

$\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{7} = 1$.

$\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{{17}} = 1$.

$\dfrac{{{x^2}}}{{16}} - \dfrac{{{y^2}}}{{17}} = 1$.  

$\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{{16}} = 1$.

\(\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{1} = 1\).

\(\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{2} = 1\).

\(\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{4} = 1\).

\(\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{1} = 1\).

$\cos \varphi  = \dfrac{{\left| {{b^2} - {a^2}} \right|}}{{{a^2} + {b^2}}}$.        

$\cos \varphi  = \dfrac{{\left| {{b^2} - {a^2}} \right|}}{{2\sqrt {{a^2} + {b^2}} }}$.     

$\cos \varphi  = \dfrac{{\left| {{b^2} - {a^2}} \right|}}{{4\left( {{a^2} + {b^2}} \right)}}$.      

$\cos \varphi  = \dfrac{{4\left| {{b^2} - {a^2}} \right|}}{{\sqrt {{a^2} + {b^2}} }}$.

\(\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{1} = 1\).

\(\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{4} = 1\).

\(\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{1} = 1\).

\(\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1\).

\({c^2} = {a^2} + {b^2}\).

\({b^2} = {a^2} + {c^2}\).

\({a^2} = {b^2} + {c^2}\).

\(c = a + b\).