Ellipse 1 : E$_1 = \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ( a > b)$ another ellipse 2 : $E_2$ which passes thru... Problem : 
Ellipse 1 : E$_1 = \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ( a > b)$ another ellipse 2 : $E_2$ which passes thru extremities of major axis of $E_1$ and has its foci at ends of its minor axis.  If eccentricity of both the ellipses are same, then find their eccentricity. 
Since $E_2$ passes through extremities of major axis of $E_1$ therefore its length of minor axis is 2a; and coordinates of focus of $E_2$ = $2b$ but I am not getting further idea to solve this please help. 
 A: Let $e_1,e_2$ is the eccentricity of $E_1,E_2$ respectively. 
First, we have
$$e_1=\frac{\sqrt{a^2-b^2}}{a}.\tag1$$
Let $$E_2\ :\ \frac{x^2}{A^2}+\frac{y^2}{B^2}=1.$$
Here, note that $B\gt A.$
Since we have
$$A=a,\ \ \ \sqrt{B^2-A^2}=b\Rightarrow B=\sqrt{a^2+b^2},$$
we have
$$e_2=\frac{\sqrt{B^2-A^2}}{B}=\frac{b}{\sqrt{a^2+b^2}}.\tag2$$
Hence, from $(1),(2)$, we have
$$\frac{\sqrt{a^2-b^2}}{a}=\frac{b}{\sqrt{a^2+b^2}}\Rightarrow a^2b^2=a^4-b^4$$$$\Rightarrow \left(\frac ba\right)^4+\left(\frac ba\right)^2-1=0\Rightarrow \left(\frac ba\right)^2=\frac{\sqrt 5-1}{2}.$$
Hence, the answer is
$$\frac{\sqrt{a^2-b^2}}{a}=\sqrt{1-\left(\frac ba\right)^2}=\sqrt{1-\frac{\sqrt 5-1}{2}}=\sqrt{\frac{3-\sqrt 5}{2}}=\frac{\sqrt 5-1}{2}.$$
A: So, the equation of $E_2$  $$\frac{x^2}{p^2}+\frac{y^2}{a^2}=1$$ where $p$ is the major axis
So we have $b^2=a^2(1-e^2)\ \ \ \  (0)$ and  $a^2=p^2(1-e^2)\ \ \ \  (1)$
As the coordinates of the foci are  $(0\pm pe),b=pe$ where $e$ is the common eccentricity 
So,from $(0), a^2(1-e^2)=(pe)^2\ \ \ \ (2)$
Divide $(1)$ by $(2)$
