Given ellipse $E_1$

$E_1:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b$

So:

• End points of major axis are $(\pm a,0)$
• End points of minor axis are $(0,\pm b)$

Construct ellipse $E_2$

We are told:

• $E_2$ touches the end points of major axis of $E_1$, so $E_2$ passes through $(\pm a,0)$.
• The foci of $E_2$ are the end points of minor axis of $E_1$, i.e. $(0,\pm b)$.

Hence $E_2$ must have center at origin and major axis along the $y$-axis.

Let its equation be $E_2:\frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}=1,\qquad \beta>\alpha$

Since its foci are $(0,\pm b)$, $\sqrt{\beta^2-\alpha^2}=b$ so $\beta^2-\alpha^2=b^2 \qquad ...(1)$

Use the fact that $(\pm a,0)$ lie on $E_2$

Substituting $(a,0)$ in $E_2$: $\frac{a^2}{\alpha^2}=1 \Rightarrow \alpha^2=a^2$ Thus $\alpha=a$

From (1), $\beta^2=a^2+b^2$

Eccentricity of $E_1$

For $E_1$, $e_1=\frac{\sqrt{a^2-b^2}}{a}$

Eccentricity of $E_2$

For $E_2$, semi-major axis is $\beta$, focal distance is $b$, so $e_2=\frac{b}{\beta}=\frac{b}{\sqrt{a^2+b^2}}$

Given $E_1$ and $E_2$ have same eccentricity

$\frac{\sqrt{a^2-b^2}}{a}=\frac{b}{\sqrt{a^2+b^2}}$

Squaring both sides: $\frac{a^2-b^2}{a^2}=\frac{b^2}{a^2+b^2}$

Cross-multiplying: $(a^2-b^2)(a^2+b^2)=a^2b^2$

$a^4-b^4=a^2b^2$

Let $x=\dfrac{b^2}{a^2}$

Then $1-x^2=x$ $x^2+x-1=0$

Solving: $x=\frac{-1\pm\sqrt{5}}{2}$

Since $x=\dfrac{b^2}{a^2}>0$, we take $x=\frac{-1+\sqrt{5}}{2}$

Interpret the required value

The value obtained is $\boxed{\frac{\sqrt5-1}{2}}$

This matches option A.