A hyperbola having the transverse axis

The given ellipse :

$\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$

$\because c=\sqrt{a^{2}-b^{2}}=\sqrt{4-3}=1$

$\therefore$ Foci $=(\pm 1,0)$

Now for hyperbola :

Given : $2 a=\sqrt{2} \Rightarrow a=\frac{1}{\sqrt{2}}$

$\because c^{2}=a^{2}+b^{2} \Rightarrow 1=\frac{1}{2}+b^{2} \Rightarrow b=\frac{1}{\sqrt{2}}$

So, equation of hyperbola is

$\frac{x^{2}}{\frac{1}{2}}-\frac{y^{2}}{\frac{1}{2}}=1 \quad \Rightarrow 2 x^{2}-2 y^{2}=1$

So, option (d) does not satisfy it.