$\text{ Equation of the given ellipse in the standard form as }\frac{x^2}{4}+\frac{y}{3}=1$

$\begin{array}{l}\text{ It implies }a=2,b=\sqrt{3}\text{ and }e=\sqrt{1-\frac{3}{2}}=\frac{1}{2}\\ \text{ Focus }=(\pm 1,0)\end{array}$

$\text{ For hyperbola }a=\frac{1}{\sqrt{2}}\text{ and foci are same as foci of ellipse, so that }{a}_2{e}_2=1$

$\begin{array}{l}\Rightarrow e_2=\sqrt{2}\Rightarrow b_2=a_2=\frac{1}{\sqrt{2}}\\ \text{ Hyperbola }\frac{x^2}{a_2^2}-\frac{y^2}{b_2^2}=1\\ \Rightarrow x^2-y^2=\frac{1}{2}\end{array}$

Verify the options, option (2) will not satisfy the above equation.