A hyperbola whose transverse axis is along the major axis of the...

Given conic is& $\dfrac{x^{2}}{12} + \dfrac{y^{2}}{16} = 1$

$e = \sqrt{1 - \dfrac{12}{16}} = \dfrac{1}{2}$

Foci \left(0, 2\right)$and \left(0, - 2\right)$.

So, transverse axis of hyperbola$= 2b = 4$.

$⇒b = 2$

Given $e = \dfrac{3}{2}$

And $a^{2} = b^{2} \left(e^{2} - 1\right)</p><p>$⇒a^{2} = 4 \left(\dfrac{9}{4} -1\right)$</p><p>$⇒a^{2} = 5$</p><p>$∴$The equation of hyperbola is$\dfrac{x2}{5} - \dfrac{y2}{4} = -1.</p><p>Now, Check options.</p><p>Clearly, the point& \left(5, 2\sqrt{3}\right)& does not satisfy above equation.