A hyperbola having the transverse axis of length

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

eccentricity $=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$

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

for hyperbola, given $2 \mathrm{a}=\sqrt{2} \Rightarrow \mathrm{a}=\frac{1}{\sqrt{2}}$

$\therefore$ hyperbola will be

$\frac{x^{2}}{1 / 2}-\frac{y^{2}}{b^{2}}=1$

eccentricity $=\sqrt{1+2 b^{2}}$

$\therefore$ foci $=\left(\pm \sqrt{\frac{1+2 b^{2}}{2}}, 0\right)$

$\because$ Ellipse and hyperbola have same foci

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

$\Rightarrow \quad b^{2}=\frac{1}{2}$

$\therefore$ Equation of hyperbola : $\frac{x^{2}}{1 / 2}-\frac{y^{2}}{1 / 2}=1$

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

Clearly $\left(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}\right)$ does not lie on it.