Let E_1: x^2 a^2+ y^2 b^2=1,a>b. Let E_2 be another ellipse such that it touches the end points of major axis of E_1 and the foci of E_2 are the end…

Given ellipse is ${E}_{1}:\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,a>b$E1​:a2x2​+b2y2​=1,a>b and ${E}_{2}$E2​ is another ellipse which touches the end points of the major axis of ${E}_{1}$E1​ and foci of ${E}_{2}$E2​ are at the minor axis of ${E}_{1},$E1​, thus the two ellipses are given by the following diagram.

Since, the ellipse ${E}_{2}$E2​ touches the major axis of the ellipse ${E}_{1}$E1​, hence, the minor axis of the ellipse ${E}_{2}$E2​ is ${a}_{1}=a.$a1​=a.

Now, let the major axis of the ellipse ${E}_{2}$E2​ is ${b}_{1}=c$b1​=c and it is obvious from the diagram and the given conditions that ${b}_{1}>{a}_{1}.$b1​>a1​.

We know that the eccentricity of an ellipse $\frac{{x}^{2}}{{A}^{2}}+\frac{{y}^{2}}{{B}^{2}}=1,A>B$A2x2​+B2y2​=1,A>B is $\sqrt{1-\frac{{B}^{2}}{{A}^{2}}}.$1−A2B2​​.

Thus, for the ellipse ${E}_{1},$E1​, we have $e=\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}$e=1−a2b2​​ and for the ellipse ${E}_{2},$E2​, we have ${e}_{1}=\sqrt{1-\frac{{a}^{2}}{{c}^{2}}}.$e1​=1−c2a2​​.

Given, $e={e}_{1}$e=e1​

$$\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}=\sqrt{1-\frac{{a}^{2}}{{c}^{2}}}$$1−a2b2​​=1−c2a2​​

$$\Rightarrow 1-\frac{{b}^{2}}{{a}^{2}}=1-\frac{{a}^{2}}{{c}^{2}}$$⇒1−a2b2​=1−c2a2​

$$\Rightarrow \frac{{b}^{2}}{{a}^{2}}=\frac{{a}^{2}}{{c}^{2}}$$⇒a2b2​=c2a2​

$$\Rightarrow {c}^{2}=\frac{{a}^{4}}{{b}^{2}}$$⇒c2=b2a4​

$$\Rightarrow c=\frac{{a}^{2}}{b}...(i)$$⇒c=ba2​...(i)

Also, given that the foci of ${E}_{2}$E2​ are the end points of the minor axis of ${E}_{1},$E1​, thus $b=ce$b=ce

$$\Rightarrow c=\frac{b}{e}...(ii)$$⇒c=eb​...(ii)

From the above two equations, we get $\frac{b}{e}=\frac{{a}^{2}}{b}$eb​=ba2​

$$\Rightarrow e=\frac{{b}^{2}}{{a}^{2}}$$⇒e=a2b2​

Now, using the definition of eccentricity, we get $e=\frac{{b}^{2}}{{a}^{2}}=1-{e}^{2}$e=a2b2​=1−e2

$$\Rightarrow {e}^{2}+e-1=0$$⇒e2+e−1=0

Now, applying the Sridharacharya's formula for the roots of a quadratic equation, i.e. if $a{x}^{2}+bx+c=0,a\neq 0,$ax2+bx+c=0,a=0, then $x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a},$x=2a−b±b2−4ac​​, we get

$$e=\frac{-1\pm \sqrt{{1}^{2}-4\times 1\times (-1)}}{2\times 1}$$e=2×1−1±12−4×1×(−1)​​

$$\Rightarrow e=\frac{-1\pm \sqrt{1+4}}{2}$$⇒e=2−1±1+4​​

$$\Rightarrow e=\frac{-1\pm \sqrt{5}}{2}$$⇒e=2−1±5​​

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