Let $H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$ 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

<p>$$\begin{aligned} & H: \frac{x^2}{a^2}-\frac{y^2}{b^2}=-1 \\ & e=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{3} \\ & \Rightarrow 1+\frac{a^2}{b^2}=3 \\ & \Rightarrow \frac{a^2}{b^2}=2 \quad \text{.... (1)}\\ & \frac{2 a^2}{b}=4 \sqrt{3} \end{aligned}$$</p> <p>Using equation (1)</p> <p>$$\begin{aligned} & \frac{4 b^2}{b}=4 \sqrt{3} \\ & \Rightarrow b=\sqrt{3} \\ & a=\sqrt{6} \\ & H: \frac{x^2}{6}-\frac{y^2}{3}=-1 \\ & \frac{\alpha^2}{6}-12=-1 \\ & \frac{\alpha^2}{6}=11 \\ & \begin{array}{c} \alpha^2=66 \\ \text { Focus }:(0, b c)(0,-b c) \\ (0,3),(0,-3) \end{array} \\ & \beta=\sqrt{\alpha^2+9} \times \sqrt{\alpha^2+81} \\ & \beta=105 \\ & \alpha^2+\beta=66+105 \\ & =171 \end{aligned}$$</p>