Equation of one of the tangents passing through (2, 8) to the hyperbola $5{\mathrm{x}}^2-{\mathrm{y}}^2=5$ is

$$\begin{gathered} \frac{{\mathrm{x}}^2}{1}-\frac{{\mathrm{y}}^2}{5}=1 \\ Eq\mathrm{uation} \mathrm{of} \mathrm{tangent} \mathrm{is} y=mx\pm \sqrt{a^2{m}^2-b^2} \\ \Rightarrow \mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{m}}^2-5} \to \left(1\right) \\ (2, 8) \mathrm{lies} \mathrm{on} \mathrm{this} \mathrm{line} \\ then 8=2m\pm \sqrt{{\mathrm{m}}^2-5} \\ So, {(8-2\mathrm{m})}^2=({\mathrm{m}}^2-5) \\ \Rightarrow 64+4m^2-32m-m^2+5=0 \\ \Rightarrow 3m^2-32m+69=0 \\ \Rightarrow (m-3)(3m-23)=0 \\ \Rightarrow \mathrm{m}=3 or m=\frac{23}{3} \\ \mathrm{Now} sub m=3 in \left(1\right) then we get required \tan gent \\ \mathrm{y}=3\mathrm{x}\pm \sqrt{4} \\ \Rightarrow 3\mathrm{x}-\mathrm{y}\pm 2=0 \end{gathered}$$