The eccentricity of the hyperbola $3{\mathrm{x}}^2+ 7\mathrm{xy} + 2{\mathrm{y}}^2 – 11\mathrm{x} – 7\mathrm{y} + 10 = 0$ is

$3{\mathrm{x}}^2+7\mathrm{xy}+2{\mathrm{y}}^2-11\mathrm{x}-7\mathrm{y}+10=0$

Equation of pair of asymptotes is $3{\mathrm{x}}^2+7\mathrm{xy}+2{\mathrm{y}}^2-11\mathrm{x}-7\mathrm{y}+\mathrm{k}=0$

If $\mathrm{\theta }$ is angle between asymptotes then

$$\begin{gathered} \Rightarrow \mathrm{cos\theta }=\frac{|\mathrm{a}+\mathrm{b}|}{\sqrt{{(\mathrm{a}-\mathrm{b})}^2+4h^2}} \\ where a=3,2h=7,b=2 \\ \Rightarrow \cos \theta =\frac{\left|5\right|}{\sqrt{{\left(1\right)}^2+49}} \\ \Rightarrow \cos \theta =\frac{5}{5\sqrt{2}}=\frac{1}{\sqrt{2}} \\ \Rightarrow \mathrm{cos\theta }=\cos 45\Rightarrow \theta ={45}^0 \\ \mathrm{Now} \mathrm{\theta }=2{\sec }^{-1}\left(\mathrm{e}\right) \\ \Rightarrow \mathrm{e}=\sec \left(\frac{\mathrm{\theta }}{2}\right) \\ \Rightarrow \mathrm{e}=\frac{1}{\cos {\displaystyle \frac{\mathrm{\theta }}{2}}} \\ \Rightarrow \mathrm{e}=\frac{1}{\sqrt{{\displaystyle \frac{1+\mathrm{cos\theta }}{2}}}}=\frac{1}{\sqrt{{\displaystyle \frac{1+{\displaystyle \frac{1}{\sqrt{2}}}}{2}}}} \\ =\sqrt{\frac{2\sqrt{2}}{\sqrt{2}+1}} \end{gathered}$$