Foci of Ellipse lies on Hyperbola and vice-versa

Question

Let the foci of the hyperbola $\frac{{{x^2}}}{{{A^2}}} - \frac{{{y^2}}}{{{B^2}}} = 1$ , (A,B > 0) be vertices of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ , (a, b > 0) and foci of ellipse be vertices of hyperbola. Let eccentricities of the ellipse and hyperbola be $e_1$ & $e_2$ , respectively $L_1$ and $L_2$ are length of Latus rectum of ellipse and hyperbola respectively. Then find the minimum value of $[e_1+e_2]$ (where [.] denotes greatest integer function)

My approach is as follow

Let $ {e_1} = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} ;{e_2} = \sqrt {1 + \frac{{{B^2}}}{{{A^2}}}} $

Foci of the ellipse are the vertices of the hyperbola and vice-versa then we get the following

$\frac{{{A^2} + {B^2}}}{{{a^2}}} = 1;\frac{{{a^2} - {b^2}}}{{{A^2}}} = 1 \Rightarrow {a^2} - {A^2} = {b^2}$ and ${B^2} = {a^2} - {A^2}$ hence $b=B$ ${e_1} = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} ;{e_2} = \sqrt {1 + \frac{{{B^2}}}{{{A^2}}}} \Rightarrow {e_1} = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} ;{e_2} = \sqrt {1 + \frac{{{b^2}}}{{{a^2} - {b^2}}}} $

Not able to proceed further

Answer 1

Further you can write,
$e_1 = \sqrt{\dfrac{a^2-b^2}{a^2}} , e_2 = \sqrt{\dfrac{a^2}{a^2-b^2}}=\dfrac1{e_1} $

Now we have, $e_1+\dfrac1{e_1}$ where $0<e_1<1 $. For the Floor function to achieve minimum value, $e_1$ must be as close as possible to $1$.

For $0<e_1<1$, $~~2<e_1+\dfrac{1}{e_1}<\infty$

So, $\min\left(\Big{\lfloor}e_1+\frac1{e_1}\Big{\rfloor}\right) =2$

Answer 2

In the given notation the focus of ellipse is $S_1(ae_1,0)$ and vertex is $V_1(a,0)$. For the hyperbola $S_2=(Ae_2,0)$ and $V_2=(A,0)$. Given that $S_1=V_2$ and S_2=V_1$.

we get $e_1=A/a$ and $e_2=a/A$ \implies $e_1 e_2=1$, so by AM-GM, we get $$\frac{e_1+e_2}{2}\ge \sqrt{e_1 e_2} \implies [e_1+e_2]\ge 2.$$