Official solution
Ellipse : \[{{{x^2}} \over 4} + {{{y^2}} \over 3} = 1\]
eccentricity = \[\sqrt {1 - {3 \over 4}} = {1 \over 2}\]
\[ \therefore \] foci = (\[ \pm \] 1, 0)
for hyperbola, given \[2a = \sqrt 2 \Rightarrow a = {1 \over {\sqrt 2 }}\]
\[ \therefore \] hyperbola will be
\[{{{x^2}} \over {1/2}} - {{{y^2}} \over {{b^2}}} = 1\]
eccentricity = \[\sqrt {1 + 2{b^2}} \]
\[ \therefore \] foci \[ = \left( { \pm \sqrt {{{1 + 2{b^2}} \over 2}} ,0} \right)\]
\[ \because \] Ellipse and hyperbola have same foci
\[ \Rightarrow \sqrt {{{1 + 2{b^2}} \over 2}} = 1\]
\[ \Rightarrow {b^2} = {1 \over 2}\]
\[ \therefore \] Equation of hyperbola : \[{{{x^2}} \over {1/2}} - {{{y^2}} \over {1/2}} = 1\]
\[ \Rightarrow \] \[{x^2} - {y^2} = {1 \over 2}\]
Clearly \[\left( {\sqrt {{3 \over 2}} ,{1 \over {\sqrt 2 }}} \right)\] does not lie on it.