Let $H:\dfrac{-x^2}{a^2}+\dfrac{y^2}{b^2}=1$ be the hyperbola whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4\sqrt{3}$. Suppose the point $(\alpha,6)$, $\alpha>0$, lies on $H$. If $\beta$ is the product of the focal distances of the point $(\alpha,6)$, then $\alpha^2+\beta$ is equal to: