For the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ length of semi transverse axis $a=\frac{1}{\sqrt{2}},$

For the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1,$ the foci is $\left(\pm \mathrm{1,0}\right)$

i.e. $\pm 1=\pm \sqrt{a^2+b^2}$

$\Rightarrow b^2=1-a^2=1-\frac{1}{2}=\frac{1}{2}$

Hence, the equation of the required hyperbola is $\frac{x^2}{{\displaystyle \frac{1}{2}}}-\frac{y^2}{{\displaystyle \frac{1}{2}}}=1$