Equation of hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{A}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{B}}^2}=1$

Vertices = (A, 0)

$\text{ Foci }=\left({\mathrm{Ae}}_{\mathrm{H}},0\right)$;

 Equation of ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$,

Vertex= (a, 0),Foci=($a{e}_E,0)$

Now $From the given data \left({\mathrm{Ae}}_{\mathrm{H}},0\right)=(\mathrm{a},0) and \left({\mathrm{ae}}_{\mathrm{E}},0\right)=(A,0)$                                        

$\begin{array}{l}\Rightarrow {\mathrm{Ae}}_{\mathrm{H}}\cdot {\mathrm{ae}}_{\mathrm{E}}=\mathrm{A}\cdot \mathrm{a}\\ \Rightarrow {\mathrm{e}}_{\mathrm{H}}\cdot {\mathrm{e}}_{\mathrm{E}}=1\end{array}$

$\left(\mathrm{A}\right)\frac{\mathrm{b}}{\mathrm{B}}=\sqrt{\frac{{\mathrm{a}}^2\left(1-{\mathrm{e}}_{\mathrm{E}}^2\right)}{{\mathrm{A}}^2\left({\mathrm{e}}_{\mathrm{H}}^2-1\right)}}=\sqrt{\frac{{\mathrm{a}}^2-{\mathrm{A}}^2}{{\mathrm{a}}^2-{\mathrm{A}}^2}}=1$

$\text{ (B) Now }\mathrm{A}\cdot \mathrm{M}\ge \mathrm{G}\cdot \mathrm{M}\Rightarrow {\mathrm{e}}_{\mathrm{H}}+{\mathrm{e}}_{\mathrm{E}}>2$

 (C) Angle between asymptotes 

$$\begin{gathered} 2{\tan }^{-1}\frac{\mathrm{B}}{\mathrm{A}}=\frac{2\mathrm{\pi }}{3} \\ \Rightarrow \frac{\mathrm{B}}{\mathrm{A}}=\sqrt{3} \\ \Rightarrow e_H=\sqrt{1+\frac{B^2}{A^2}}=2 \\ \Rightarrow e_{\mathrm{E}}=\frac{1}{2} \\ Now 4e_{\mathrm{E}}=2 \end{gathered}$$

$$\begin{gathered} \text{ D) Solve curves }{\mathrm{x}}^2=\frac{2{\mathrm{a}}^2{\mathrm{e}}_{\mathrm{E}}^2}{{\mathrm{b}}^2\left(1-{\mathrm{e}}_{\mathrm{E}}^2\right)}=\frac{9}{2}, \\ \Rightarrow \frac{{\mathrm{x}}^2}{{\mathrm{y}}^2}=4 \end{gathered}$$