Cho hai đường thẳng \({d_1}:\,11x - 12y + 1 = 0\) và \({d_2}:\,12x + 11y + 9 = 0\). Xét vị trí tương đối giữa \({d_1}\) và \({d_2}\):

\((E):\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\left( {{a^2} = {b^2} + {c^2};a,b,c > 0} \right)\). Gọi \({F_1}( - c;0) \Rightarrow {F_2}(c;0)\).

Hai đỉnh trên trục nhỏ \({B_1}(0; - b),{B_2}(0;b)\). Ta có hệ:

\(\left\{ {\begin{array}{*{20}{l}}{b = 2c\frac{{\sqrt 3 }}{2}}\\{2(2a + 2b) = 12(2 + \sqrt 3 ) \Leftrightarrow }\\{{c^2} = {a^2} - {b^2}}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{a = 6}\\{b = 3\sqrt 3 }\\{c = 3}\end{array} \Rightarrow (E):\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{27}} = 1.} \right.} \right.\)

\(M \in (E)\). Ta có \(M{F_1} = 2 + \frac{{\sqrt 3 }}{2}x,M{F_2} = 2 - \frac{{\sqrt 3 }}{2}x\).

$F_1{F}_2^2=M{F}_1^2+M{F}_2^2-2M{F}_{1}M{F}_2\cos {60}^{°}\iff 12=4+\frac{9}{4}{x}^2\iff x=\pm \frac{\sqrt{32}}{3}.$

\(\begin{array}{l}{\rm{ V\`i  }}M \in (E){\rm{ n\^e n }}x =  \pm \frac{{\sqrt {32} }}{3} \Rightarrow y =  \pm \frac{1}{3}{\rm{. }}\\ \Rightarrow {M_1}\left( {\frac{{\sqrt {32} }}{3};\frac{1}{3}} \right),{M_2}\left( {\frac{{\sqrt {32} }}{3}; - \frac{1}{3}} \right),{M_3}\left( { - \frac{{\sqrt {32} }}{3}; - \frac{1}{3}} \right),{M_4}\left( { - \frac{{\sqrt {32} }}{3}; - \frac{1}{3}} \right){\rm{. }}\end{array}\)