Let H: -x^2/a^2+y^2/b^2=1 be the hyperbola, whose eccentricity is √ 3 and the length of the latus rectum is 4 √ 3. Suppose the point (α, 6), α>0 lies…

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