Tangents are drawn from the point $(\alpha ,\beta )$  to the hyperbola $3x^2-2y^2=6$  and are inclined at angle $\theta$  and  $\varphi$   to the x-axis. if  $\left(\tan \theta \right)\left(\tan \varphi \right)=2$, then the value of  $\frac{2{\alpha }^2-{\beta }^2}{7}$  is

Given hyperbola is  $3x^2-2y^2=6 or\frac{x^2}{2}-\frac{y^2}{3}=1$
Slope form of tangent is  $y=mx\pm \sqrt{a^2{m}^2-b^2}or {(y-mx)}^2=a^2{m}^2-b^2$
Tangent from the point $(\alpha ,\beta )$ is given by, ${(\beta -m\alpha )}^2=2m^2-3$
 $$\begin{gathered} \Rightarrow m^2({\alpha }^2-2)-2\alpha \beta m+{\beta }^2+3=0 \\ {m}_1{m}_2=\frac{{\beta }^2+3}{{\alpha }^2-2}=2=\tan \theta \tan \varphi \\ \therefore {\beta }^2+3=2{\alpha }^2-4\Rightarrow 7=2{\alpha }^2-{\beta }^2 \end{gathered}$$