Since, the vertex of the hyperbola are $A^{\prime }\left(-2, 0\right)$ and $A\left(2, 0\right)$.

Hence, the transverse axis of the hyperbola is $x-$ axis and centre is $O\left(0,0\right)$.

Equation of hyperbola can be taken as

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

$\Rightarrow a=2$ and $ae=3$

$\Rightarrow e=\frac{3}{2}$

Since, $b^2=a^2\left(e^2-1\right)\Rightarrow b^2=a^2{e}^2-a^2=9-4=5$

Hence, equation of hyperbola is $\frac{x^2}{4}-\frac{y^2}{5}=1$

Clearly, point $\left(6, 5\sqrt{2}\right)$ doesn't lies on the hyperbola.