The equations of the tangents to the hyperbola $2{\mathrm{x}}^2 – 3{\mathrm{y}}^2 = 6$ which are perpendicular to the line x – 2y + 5 = 0 are

$$\begin{gathered} \mathrm{Given} \mathrm{hyperbola} \frac{{\mathrm{x}}^2}{3}-\frac{{\mathrm{y}}^2}{2}=1 \\ where {\mathrm{a}}^2=3, {\mathrm{b}}^2=2 \\ slope of given line x-2y+5=0 is \frac{1}{2} \\ Now \mathrm{slope} \mathrm{of} \mathrm{tangent} \left(\mathrm{m}\right)=-2 \\ \therefore \mathrm{Equations} \mathrm{of} \mathrm{tangents} \mathrm{are} \mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2} \\ \Rightarrow \mathrm{y}=-2\mathrm{x}\pm \sqrt{12-2} \\ \Rightarrow 2\mathrm{x}+\mathrm{y}\pm \sqrt{10}=0 \end{gathered}$$