Tangents are drawn from the point $(\alpha,\beta)$ to the hyperbola $3x^2-2y^2=6$ and are inclined at angles $\theta$ and $\phi$ to the $x-$axis. Tangents are drawn from the point $(\alpha,\beta)$ to the hyperbola $3x^2-2y^2=6$ and are inclined at angles $\theta$ and $\phi$ to the $x-$axis.If $\tan\theta.\tan\phi=2,$ prove that $\beta^2=2\alpha^2-7$

The equation of the pair of tangents from the point $(\alpha,\beta)$ to the hyperbola $3x^2-2y^2=6$ is given by $(3x\alpha-2y\beta-6)^2=(3x^2-2y^2-6)(3\alpha^2-2\beta^2-6)$
When i simplify this expression,i get
$(18+6\beta^2)x^2+(6\alpha^2-12)y^2-12xy\alpha\beta+24y\beta-36x\alpha+18\alpha^2-12\beta^2=0$
This is an equation of pair of lines and the angle between pair of lines $ax^2+2hxy+by^2=0$ is given by the formula $\frac{2\sqrt{h^2-ab}}{a+b}$
So $\tan(\theta-\phi)=\frac{2\sqrt{36\alpha^2\beta^2-(18+6\beta^2)(6\alpha^2-12)}}{18+6\beta^2+6\alpha^2-12}$
$\frac{\tan\theta-\tan\phi}{1+\tan\theta\tan\phi}=\frac{2\sqrt{\alpha^2\beta^2-(3+\beta^2)(\alpha^2-2)}}{1+\beta^2+\alpha^2}$
$\frac{\tan\theta-\tan\phi}{3}=\frac{2\sqrt{\alpha^2\beta^2-(3+\beta^2)(\alpha^2-2)}}{1+\beta^2+\alpha^2}$
I am stuck here.I do not know,how to prove it further.

From this,
$$y=x\tan\psi\pm\sqrt{2\tan^2\psi-3}$$ are always tangents of $$\dfrac{x^2}2-\dfrac{y^2}3=1$$
Now if these tangents pass through $(\alpha,\beta)$
$$\beta=\alpha\tan\psi\pm\sqrt{2\tan^2\psi-3}$$
$$\iff\beta-\alpha\tan\psi=\pm\sqrt{2\tan^2\psi-3}$$
On squaring and rearrangement, 
$$(\alpha^2-2)\tan^2\psi-2\alpha\tan\psi+\beta^2-3=0$$
Now if the two values of $\psi$ are $\theta,\phi$
$$\tan\theta\tan\phi=\dfrac{\beta^2-3}{\alpha^2-2}$$  and we are done!
