A hyperbola passes through the point P( 2 , 3 ) and has foci at (±...

From the standard equation of hyperbola

$⇒ \pm ae = \pm 2$

As we know, ${b}^{2} = {a}^{2}\left({e}^{2} - 1\right)$.

$⇒ {b}^{2} = 4 - {a}^{2}$

$∴$Equation of hyperbola is $\dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{4 - {a}^{2}} = 1$

$∵$It passes through \left(\sqrt{2}, \sqrt{3}\right)$</p><p>$⇒ \dfrac{2}{{a}^{2}} - \dfrac{3}{4 - {a}^{2}} = 1$</p><p>$⇒ {a}^{2} = t$</p><p>$⇒ 2\left(4 - t\right) - 3t - t\left(4 - t\right) = 0$</p><p>$⇒ 8 - 2t - 3t - 4t + {t}^{2} = 0$</p><p>$⇒ {t}^{2} - 8t - t + 8 = 0$</p><p>$⇒ t = 8, 1$</p><p>So,${a}^{2} = 8, 1$.</p><p>$⇒ a = 2\sqrt{2}, 1$</p><p>But for$a = 2\sqrt{2}, b$becomes imaginary. So this case is rejected.</p><p>$∴$Hyperbola is$\dfrac{{x}^{2}}{1} - \dfrac{{y}^{2}}{3} = 1$.</p><p>The equation of the tangent at$P$is$\sqrt{2}x - \dfrac{y}{\sqrt{3}} = 1.

Now checking each points.