Let E _ 1 : x ^ 2 a ^ 2 + y ^ 2 b ^ 2 = 1, a > b . Let E _ 2 be...

Given ellipse is ${E}_{1}: \dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1, a > b$ 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 $\dfrac{{x}^{2}}{{A}^{2}} + \dfrac{{y}^{2}}{{B}^{2}} = 1, A > B$ is $\sqrt{1 - \dfrac{{B}^{2}}{{A}^{2}}} .$

Thus, for the ellipse ${E}_{1},$ we have $e = \sqrt{1 - \dfrac{{b}^{2}}{{a}^{2}}}$ and for the ellipse ${E}_{2},$ we have ${e}_{1} = \sqrt{1 - \dfrac{{a}^{2}}{{c}^{2}}} .$

Given, $e = {e}_{1}$

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

$⇒ 1 - \dfrac{{b}^{2}}{{a}^{2}} = 1 - \dfrac{{a}^{2}}{{c}^{2}}$

$⇒ \dfrac{{b}^{2}}{{a}^{2}} = \dfrac{{a}^{2}}{{c}^{2}}$

$⇒ {c}^{2} = \dfrac{{a}^{4}}{{b}^{2}}$

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

Also, given that the foci of ${E}_{2}$ are the end points of the minor axis of ${E}_{1},$ thus $b = ce$

$⇒ c = \dfrac{b}{e} . . . \left(ii\right)$

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

$⇒ e = \dfrac{{b}^{2}}{{a}^{2}}$

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

$⇒ {e}^{2} + e - 1 = 0$

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 $x = \dfrac{- b \pm \sqrt{{b}^{2} - 4ac}}{2a},$ we get

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

$⇒ e = \dfrac{- 1 \pm \sqrt{1 + 4}}{2}$

$⇒ e = \dfrac{- 1 \pm \sqrt{5}}{2}$

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