Let the eccentricity of the ellipse x^2+ a^2 y^2=25 a^2 be b times the eccentricity of the hyperbola x^2- a^2 y^2=5, where a is the minimum distance…

Let the eccentricity of the ellipse $\displaystyle x^2+\mathrm{a}^2 y^2=25 \mathrm{a}^2$ be b times the eccentricity of the hyperbola $\displaystyle x^2-\mathrm{a}^2 y^2=5$, where a is the minimum distance between the curves $\displaystyle y=\mathrm{e}^x$ and $\displaystyle y=\log _{\mathrm{e}} x$. Then $\displaystyle \mathrm{a}^2+\frac{1}{\mathrm{~b}^2}$ is equal to :