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

Given hyperbola is

$3x^2-2y^2=6or\frac{x^2}{2}-\frac{y^2}{3}=1$

Tangents from the point $\left(\alpha ,\beta \right)$

$y=mx\pm \sqrt{a^2{m}^2-b^2}$

OR ${\left(y-mx\right)}^2=a^2{m}^2-b^2$

OR ${\left(\beta -m\alpha \right)}^2=2m^2-3(\therefore a^2=2and{b}^2=3)$

OR  $m^2{\alpha }^2+{\beta }^2-2m\alpha \beta -2m^2+3=0$

$m^2\left({\alpha }^2-2\right)-2\alpha \beta m+{\beta }^2+3=0$

$m_1.m_2=\frac{{\beta }^2+3}{{\alpha }^2-2}=2=\tan \theta .\tan \varphi \therefore {\beta }^2+3=2\left({\alpha }^2-2\right)$

OR  $2{\alpha }^2-{\beta }^2=7$