Statement 1 : Equations of tangents to the hyperbola $2{\mathrm{x}}^2 –3{\mathrm{y}}^2 = 6$ which is parallel to the line y = 3x + 4 is y = 3x – 5 and y = 3x + 5
Statement 2 : For given slope two parallel tangents can be drawn to the hyperbola

Then correct statement is

$$\begin{gathered} \underline{\mathrm{Statement}-\mathrm{I}} : \mathrm{Given} \mathrm{hyperbola} \frac{{\mathrm{x}}^2}{3}-\frac{{\mathrm{y}}^2}{2}=1 \\ Slope of given line y=3x+4 is 3 \\ Now \mathrm{slope} \mathrm{of} required \mathrm{tangent}\left(\mathrm{m}\right)=3 \\ \therefore \mathrm{Equation} \mathrm{of} \mathrm{tangents} \mathrm{are} \mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2} \\ \Rightarrow \mathrm{y}=3\mathrm{x}+\sqrt{3\left(9\right)-2} \\ \Rightarrow \mathrm{y}=3\mathrm{x}\pm \sqrt{25} \\ \Rightarrow \mathrm{y}=3\mathrm{x}\pm 5 \\ \therefore \mathrm{Statement}-\mathrm{I} \mathrm{is} \mathrm{true} \\ \mathrm{Statement}-\mathrm{II} : \mathrm{let} \mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2} \mathrm{be} \mathrm{tangent} \mathrm{of} \\ \mathrm{given} \mathrm{hyperbola}. \mathrm{It} \mathrm{represents} \mathrm{two} \mathrm{parallel} \mathrm{lines}. \\ \therefore \mathrm{both} \mathrm{are} \mathrm{true} \mathrm{and} \mathrm{Statement}-\mathrm{II} \mathrm{is} not the \mathrm{correct} \exp l\mathrm{anation} \mathrm{of} \mathrm{Statement}-\mathrm{I}. \end{gathered}$$