If the vertices of a hyperbola be at

Let the points are,

$A(2,0), A^{\prime}(-2,0)$ and $S(-3,0)$

$\Rightarrow$ Centre of hyperbola is $O(0,0)$

$A A^{\prime}=2 a \Rightarrow 4=2 a \Rightarrow a=2$

$\because \quad$ Distance between the centre and foci is $a e$.

$\therefore \quad O S=a e \Rightarrow 3=2 e \Rightarrow e=\frac{3}{2}$

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

$\Rightarrow \quad$ Equation of hyperbola is $\frac{x^{2}}{4}-\frac{y^{2}}{5}=1$....(1)

$\because \quad(6,552)$ does not satisfy eq (1).

$\therefore \quad(6,5 \sqrt{2})$ does not lie on this hyperbola.