A hyperbola having the transverse axis of length, 2 has the same...

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

$⇒ e = \sqrt{1 - \dfrac{3}{4}} = \dfrac{1}{2}$

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

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

Foci of hyperbola \left(\pm 1, 0\right) ⇒ ae = 1$</p><p>${b}^{2} = {a}^{2}{e}^{2} - {a}^{2} = 1 - \dfrac{1}{2} = \dfrac{1}{2}$</p><p>Equation of hyperbola :$\dfrac{{x}^{2}}{\left(1 / 2\right)} - \dfrac{{x}^{2}}{\left(1 / 2\right)} = 1$</p><p>$⇒ {x}^{2} - {y}^{2} = \dfrac{1}{2} ⇒ 2{x}^{2} - 2{y}^{2} = 1$</p><p>Only the point, \left(\sqrt{\dfrac{3}{2}}, \dfrac{1}{\sqrt{2}}\right)$ does not satisfy the above equation.