A hyperbola having the transverse axis of length, 2 has the same foci as that of the ellipse, 3 x 2 + 4 y 2…

Equation of ellipse : $\frac{x^2}{4}+\frac{y^2}{3}=1$

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

Foci of the ellipse: $\left(\pm 2\times \frac{1}{2},0\right)$ i.e.,$\left(\pm 1,0\right)$

For hyperbola: Length of the transverse axis$=2a=\sqrt{2}\Rightarrow a=\frac{1}{\sqrt{2}}$

Foci of hyperbola $\left(\pm 1,0\right)\Rightarrow ae=1$

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

Equation of hyperbola :$\frac{x^2}{{\displaystyle \left(1/2\right)}}-\frac{{\displaystyle x^2}}{{\displaystyle \left(1/2\right)}}=1$

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

Only the point, $\left(\sqrt{\frac{3}{2}},\frac{1}{\sqrt{2}}\right)$ does not satisfy the above equation.