Đặt \(\left(\right. E \left.\right) : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 , \left(\right. a > b > 0 \left.\right)\).

Ta có \(\left{\right. & a^{2} = 36 \\ & b^{2} = 25 \Rightarrow \left{\right. & a = 6 \\ & b = 5 \Rightarrow c = \sqrt{11}\).

Tiêu điểm: \(F_{1} \left(\right. - \sqrt{11} ; 0 \left.\right) , F_{2} \left(\right. \sqrt{11} ; 0 \left.\right)\).

Tiêu cự: \(F_{1} F_{2} = 2 c = 2 \sqrt{11}\).

Trục lớn: \(A_{1} A_{2} = 2 a = 12\).

Trục bé: \(B_{1} B_{2} = 2 b = 10\).

Tâm sai: \(e = \frac{c}{a} = \frac{\sqrt{11}}{6}\).

Đặt \(\left(\right. E \left.\right) : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 , \left(\right. a > b > 0 \left.\right)\).

Ta có \(\left{\right. & a^{2} = 36 \\ & b^{2} = 25 \Rightarrow \left{\right. & a = 6 \\ & b = 5 \Rightarrow c = \sqrt{11}\).

Tiêu điểm: \(F_{1} \left(\right. - \sqrt{11} ; 0 \left.\right) , F_{2} \left(\right. \sqrt{11} ; 0 \left.\right)\).

Tiêu cự: \(F_{1} F_{2} = 2 c = 2 \sqrt{11}\).

Trục lớn: \(A_{1} A_{2} = 2 a = 12\).

Trục bé: \(B_{1} B_{2} = 2 b = 10\).

Tâm sai: \(e = \frac{c}{a} = \frac{\sqrt{11}}{6}\).

Đặt \(\left(\right. E \left.\right) : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 , \left(\right. a > b > 0 \left.\right)\).

Ta có \(\left{\right. & a^{2} = 36 \\ & b^{2} = 25 \Rightarrow \left{\right. & a = 6 \\ & b = 5 \Rightarrow c = \sqrt{11}\).

Tiêu điểm: \(F_{1} \left(\right. - \sqrt{11} ; 0 \left.\right) , F_{2} \left(\right. \sqrt{11} ; 0 \left.\right)\).

Tiêu cự: \(F_{1} F_{2} = 2 c = 2 \sqrt{11}\).

Trục lớn: \(A_{1} A_{2} = 2 a = 12\).

Trục bé: \(B_{1} B_{2} = 2 b = 10\).

Tâm sai: \(e = \frac{c}{a} = \frac{\sqrt{11}}{6}\).

Đặt \(\left(\right. E \left.\right) : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 , \left(\right. a > b > 0 \left.\right)\).

Ta có \(\left{\right. & a^{2} = 36 \\ & b^{2} = 25 \Rightarrow \left{\right. & a = 6 \\ & b = 5 \Rightarrow c = \sqrt{11}\).

Tiêu điểm: \(F_{1} \left(\right. - \sqrt{11} ; 0 \left.\right) , F_{2} \left(\right. \sqrt{11} ; 0 \left.\right)\).

Tiêu cự: \(F_{1} F_{2} = 2 c = 2 \sqrt{11}\).

Trục lớn: \(A_{1} A_{2} = 2 a = 12\).

Trục bé: \(B_{1} B_{2} = 2 b = 10\).

Tâm sai: \(e = \frac{c}{a} = \frac{\sqrt{11}}{6}\).

Đặt \(\left(\right. E \left.\right) : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 , \left(\right. a > b > 0 \left.\right)\).

Ta có \(\left{\right. & a^{2} = 36 \\ & b^{2} = 25 \Rightarrow \left{\right. & a = 6 \\ & b = 5 \Rightarrow c = \sqrt{11}\).

Tiêu điểm: \(F_{1} \left(\right. - \sqrt{11} ; 0 \left.\right) , F_{2} \left(\right. \sqrt{11} ; 0 \left.\right)\).

Tiêu cự: \(F_{1} F_{2} = 2 c = 2 \sqrt{11}\).

Trục lớn: \(A_{1} A_{2} = 2 a = 12\).

Trục bé: \(B_{1} B_{2} = 2 b = 10\).

Tâm sai: \(e = \frac{c}{a} = \frac{\sqrt{11}}{6}\).

Đặt \(\left(\right. E \left.\right) : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 , \left(\right. a > b > 0 \left.\right)\).

Ta có \(\left{\right. & a^{2} = 36 \\ & b^{2} = 25 \Rightarrow \left{\right. & a = 6 \\ & b = 5 \Rightarrow c = \sqrt{11}\).

Tiêu điểm: \(F_{1} \left(\right. - \sqrt{11} ; 0 \left.\right) , F_{2} \left(\right. \sqrt{11} ; 0 \left.\right)\).

Tiêu cự: \(F_{1} F_{2} = 2 c = 2 \sqrt{11}\).

Trục lớn: \(A_{1} A_{2} = 2 a = 12\).

Trục bé: \(B_{1} B_{2} = 2 b = 10\).

Tâm sai: \(e = \frac{c}{a} = \frac{\sqrt{11}}{6}\).

Đặt \(\left(\right. E \left.\right) : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 , \left(\right. a > b > 0 \left.\right)\).

Ta có \(\left{\right. & a^{2} = 36 \\ & b^{2} = 25 \Rightarrow \left{\right. & a = 6 \\ & b = 5 \Rightarrow c = \sqrt{11}\).

Tiêu điểm: \(F_{1} \left(\right. - \sqrt{11} ; 0 \left.\right) , F_{2} \left(\right. \sqrt{11} ; 0 \left.\right)\).

Tiêu cự: \(F_{1} F_{2} = 2 c = 2 \sqrt{11}\).

Trục lớn: \(A_{1} A_{2} = 2 a = 12\).

Trục bé: \(B_{1} B_{2} = 2 b = 10\).

Tâm sai: \(e = \frac{c}{a} = \frac{\sqrt{11}}{6}\).

Ta có \(F_{1} \left(\right. - \sqrt{3} ; 0 \left.\right)\), \(F_{2} \left(\right. \sqrt{3} ; 0 \left.\right)\).

Gọi \(M \left(\right. x ; y \left.\right)\), ta có \(M \in \left(\right. E \left.\right) \Leftrightarrow \frac{x^{2}}{4} + y^{2} = 1\) \(\left(\right. 1 \left.\right)\).

Mặt khác ta có \(\overset{\rightarrow}{M F_{1}} \left(\right. - \sqrt{3} - x ; - y \left.\right) ; \overset{\rightarrow}{M F_{2}} \left(\right. \sqrt{3} - x ; - y \left.\right)\).

Do \(M F_{1} \bot M F_{2}\) nên \(\overset{\rightarrow}{M F_{1}} . \overset{\rightarrow}{M F_{2}} = 0 \Leftrightarrow \left(\right. x - \sqrt{3} \left.\right) \left(\right. x + \sqrt{3} \left.\right) + y^{2} = 0 \Leftrightarrow x^{2} + y^{2} = 3\) \(\left(\right. 2 \left.\right)\).

Từ \(\left(\right. 1 \left.\right)\) và \(\left(\right. 2 \left.\right)\) ta có \(\left{\right. & \frac{x^{2}}{4} + y^{2} = 1 \\ & x^{2} + y^{2} = 3 .\)

Suy ra \(M \left(\right. \frac{2 \sqrt{6}}{3} ; \frac{\sqrt{3}}{3} \left.\right)\) hoặc \(M \left(\right. \frac{2 \sqrt{6}}{3} ; - \frac{\sqrt{3}}{3} \left.\right)\) hoặc \(M \left(\right. - \frac{2 \sqrt{6}}{3} ; \frac{\sqrt{3}}{3} \left.\right)\) hoặc \(M \left(\right. - \frac{2 \sqrt{6}}{3} ; - \frac{\sqrt{3}}{3} \left.\right) .\)

Vậy \(M \left(F_{1}\right)^{2} + M \left(F_{2}\right)^{2} = 2 \left(\right. x^{2} + y^{2} \left.\right) + 6 = 12.\)

\(S_{\Delta M F_{1} F_{2}} = \frac{1}{2} M F_{1} . M F_{2} = \frac{1}{2} . \sqrt{\left(\left(\right. x + \sqrt{3} \left.\right)\right)^{2} + y^{2}} . \sqrt{\left(\left(\right. x - \sqrt{3} \left.\right)\right)^{2} + y^{2}} = \frac{1}{2} \sqrt{\left(\left(\right. x^{2} - 3 \left.\right)\right)^{2} + y^{2} \left(\right. x^{2} + 6 \left.\right) + y^{4}} = 1.\)