Ellipse

Given ellipse $x^2+a^2 y^2=25 a^2 \Rightarrow \;\frac{x^2}{25 a^2}+\;\frac{y^2}{25}=1$$\;\text{eccentricity}\;(e_1)=\sqrt{1-\;\frac{b^2}{a^2}}$$=\sqrt{1-\;\frac{25}{25 a^2}}$$=\sqrt{1-\;\frac{1}{a^2}}$$\Rightarrow e_1^2=1-\;\frac{1}{a^2}$Also, given hyperbola,$x^2-a^2 y^2=5$$\Rightarrow \;\frac{x^2}{5}-\;\frac{y^2}{\frac{5}{a^2}}=1$$\;\text{eccentricity}\;(e_2)=\sqrt{1+\;\frac{b^2}{a^2}}$$=\sqrt{1+\;\frac{5}{5 a^2}}$$=\sqrt{1+\;\frac{1}{a^2}}$ $\Rightarrow e_2^2=1+\;\frac{1}{a^2}$Also given,$e_1=b \times e_2$$\Rightarrow e_1^2=b^2 \times e_2^2$$\Rightarrow 1-\;\frac{1}{a^2}=b^2 \left(1+\;\frac{1}{a^2}\right)$$\Rightarrow \;\frac{a^2-1}{a^2}=\;\frac{b^2 (a^2+1)}{a^2}$$\Rightarrow b^2=\;\frac{a^2-1}{a^2+1}$$y=\log_e^x$ is inverse of $y=e^x$ so it is mirror image of each other with respect to $y=x$ line.Slope of tangent to $y=e^x$ curve$\;\frac{d y}{d x}=e^x$Slope of tangent to $y=\log_e^x$ curve,$\;\frac{d y}{d x}=\;\frac{1}{x}$Both tangents are parallel to $y=x$ line for minimum distance condition.$\therefore$ Slope of $y=x$ line $=$ Slope of both the tangent.$\therefore \;\frac{d y}{d x}=e^x=1 \Rightarrow e^x=e^0=x=0$$\therefore y=e^x=e^0=1$and $\;\frac{d y}{d x}=\;\frac{1}{x}=1 \Rightarrow x=1$$\therefore y=\log_e^1=0$$\therefore$ tangent at $(0,1)$ point of $y=e^x$ curve and tangent at $(1,0)$ point of $y=\log_e^x$ curve are parallel.$\therefore$ Minimum distance between point $(0,1)$ and $(1,0)$ is $=\sqrt{1^2+1^2}=\sqrt{2}$$\therefore a=\sqrt{2}$So, $b^2=\;\frac{a^2-1}{a^2+1}=\;\frac{2-1}{2+1}=\;\frac{1}{3}$$\therefore a^2+\;\frac{1}{b^2}=2+\;\frac{1}{1/3}=5$