$$\begin{gathered} Given hyperbola \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 — ① and foci=(\pm 2,0) \\ \Rightarrow \mathrm{ae}=2 \\ ① \Rightarrow \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2({\mathrm{e}}^2-1)}=1 \\ \mathrm{It} \mathrm{passes} \mathrm{through} \left(\sqrt{2}, \sqrt{3}\right) \mathrm{then} \\ \Rightarrow \frac{2}{{\mathrm{a}}^2}-\frac{3}{4-{\mathrm{a}}^2}=1 \\ \Rightarrow {\mathrm{a}}^4-9{\mathrm{a}}^2+8=0 \\ \Rightarrow {\mathrm{a}}^2=1 \left(\mathrm{or}\right) {\mathrm{a}}^2=8 \\ \Rightarrow {\mathrm{a}}^2=1 (\because \mathrm{e}>1) \\ \Rightarrow {\mathrm{b}}^2=3 \\ \mathrm{Equation} \mathrm{of} \mathrm{tangent} \mathrm{at} \mathrm{P} \mathrm{is} \sqrt{2}\mathrm{x}-\frac{\mathrm{y}}{\sqrt{3}}=1 \\ \mathrm{By} \mathrm{verfication} 3^{\mathrm{rd}} \mathrm{option} \left(2\sqrt{2}, 3\sqrt{3}\right) \mathrm{lies} \mathrm{on} \mathrm{tangent} \end{gathered}$$