The point of contact of 9x+8y–11=0 to the hyperbola $3{\mathrm{x}}^2 –4{\mathrm{y}}^2 = 11$ is

$$\begin{gathered} Given hyperbola 3x^2-4y^2=11 \\ \Rightarrow \frac{{\mathrm{x}}^2}{{\displaystyle \frac{11}{3}}}-\frac{{\mathrm{y}}^2}{{\displaystyle \frac{11}{4}}}=1 \\ \Rightarrow {\mathrm{a}}^2=\frac{11}{3}, {\mathrm{b}}^2=\frac{11}{4} \\ \mathrm{Given} \mathrm{line} 9\mathrm{x}+8\mathrm{y}-11=0 \\ \mathrm{Compare} \mathrm{with} \mathrm{lx}+\mathrm{my}+\mathrm{n}=0 \\ \Rightarrow \mathrm{l}=9, \mathrm{m}=8, \mathrm{n}=-11 \\ \therefore \mathrm{point} \mathrm{of} \mathrm{contact}=\left(\frac{-{\mathrm{a}}^2\mathrm{l}}{\mathrm{n}},\frac{{\mathrm{b}}^2\mathrm{m}}{\mathrm{n}}\right) \\ =\left(\frac{-{\displaystyle \frac{11}{3}}\left(9\right)}{-11}, \frac{{\displaystyle \frac{11}{4}}\left(8\right)}{-11}\right) \\ =(3, -2) \end{gathered}$$