$\text{ The eccentricity of the hyperbola }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\text{ is reciprocal to that of the ellipse }$

$x^2+4y^2=4\text{ . If the hyperbola passes through a foci of the ellipse, then }$

$\text{ Given ellipse is }\frac{x^2}{4}+\frac{y^2}{1}=1\text{ . }$

$\text{ Eccentricity of the ellipse }=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$

$\text{ Focus of the ellipse }=(-\sqrt{3},0)$

$\begin{array}{l}\text{ Eccentricity of the hyperbola }=\sqrt{1+\frac{b^2}{a^2}}\\ \text{ So, }\sqrt{1+\frac{b^2}{a^2}}=\frac{2}{\sqrt{3}}\Rightarrow \frac{b}{a}=\frac{1}{\sqrt{3}}\end{array}$

$\text{ Since the hyperbola passes through the focus of the ellipse, }\frac{3}{a^2}-0=1$

$\begin{array}{l}\Rightarrow {\mathrm{a}}^2=3\text{ and }{\mathrm{b}}^2=1\\ \text{ Thus, the equation of the hyperbola is }\end{array}$

$\begin{array}{l}\frac{x^2}{3}-\frac{y^2}{1}=1\text{ or }{x}^2-3y^2=3\\ \text{ Focus of hyperbola is }(\pm 2,0)\text{ . }\end{array}$