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.