The equation of tangent to the hyperbola $4{\mathrm{x}}^2 –5{\mathrm{y}}^2 =20$ parallel to the line x–y=2 is

$$\begin{gathered} \mathrm{Given} \mathrm{hyperbola} 4x^2-5y^2=20 \\ \Rightarrow \frac{{\mathrm{x}}^2}{5}-\frac{{\mathrm{y}}^2}{4}=1 \\ where {\mathrm{a}}^2=5, {\mathrm{b}}^2=4 \\ Given line x-y-2=0 \\ \mathrm{slope} \mathrm{of} \mathrm{given} \mathrm{line}=1 \\ \mathrm{slope} \mathrm{of} \mathrm{tangent} \left(\mathrm{m}\right)=1 \\ \therefore \mathrm{Required} \mathrm{tangent} \mathrm{is} \mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2} \\ \Rightarrow \mathrm{y}=\mathrm{x}\pm \sqrt{5-4} \\ \Rightarrow \mathrm{y}=\mathrm{x}\pm 1 \\ \Rightarrow \mathrm{x}-\mathrm{y}+1=0 \end{gathered}$$