Let E 1 : x 2 a 2 + y 2   b 2 = 1 ,   a > b . Let E 2 be another ellipse such that it touches…

Given ellipse is $E_1:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ and $E_2$ is another ellipse which touches the end points of the major axis of $E_1$ and foci of $E_2$ are at the minor axis of $E_1,$ thus the two ellipses are given by the following diagram.

Since, the ellipse $E_2$ touches the major axis of the ellipse $E_1$, hence, the minor axis of the ellipse $E_2$ is $a_1=a.$

Now, let the major axis of the ellipse $E_2$ is $b_1=c$ and it is obvious from the diagram and the given conditions that $b_1>a_1.$

We know that the eccentricity of an ellipse $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1, A>B$ is $\sqrt{1-\frac{B^2}{A^2}}.$

Thus, for the ellipse $E_1,$ we have $e=\sqrt{1-\frac{b^2}{a^2}}$ and for the ellipse $E_2,$ we have $e_1=\sqrt{1-\frac{a^2}{c^2}}.$

Given, $e=e_1$

$\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{a^2}{c^2}}$

$\Rightarrow 1-\frac{b^2}{a^2}=1-\frac{a^2}{c^2}$

$\Rightarrow \frac{b^2}{a^2}=\frac{a^2}{c^2}$

$\Rightarrow c^2=\frac{a^4}{b^2}$

$\Rightarrow c=\frac{a^2}{b} ...\left(i\right)$

Also, given that the foci of $E_2$ are the end points of the minor axis of $E_1,$ thus $b=ce$

$\Rightarrow c=\frac{b}{e} ...\left(ii\right)$

From the above two equations, we get $\frac{b}{e}=\frac{a^2}{ b}$

$\Rightarrow e=\frac{b^2}{a^2}$

Now, using the definition of eccentricity, we get $e=\frac{b^2}{a^2}=1-e^2$

$\Rightarrow e^2+e-1=0$

Now, applying the Sridharacharya's formula for the roots of a quadratic equation, i.e. if $a{x}^2+bx+c=0, a\ne 0,$ then $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a},$ we get

$e=\frac{-1\pm \sqrt{1^2-4\times 1\times \left(-1\right)}}{2\times 1}$

$\Rightarrow e=\frac{-1\pm \sqrt{1+4}}{2}$

$\Rightarrow e=\frac{-1\pm \sqrt{5}}{2}$

But, eccentricity can never be negative, hence $e=\frac{-1+\sqrt{5}}{2}.$