Write the hyperbola in standard form

Given $\frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$ which is $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1.$

So this is a hyperbola with transverse axis along the $y$-axis.

Use eccentricity

For the hyperbola $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1,$ we have $c^2=a^2+b^2, \qquad e=\frac{c}{b}.$

Given $e=\sqrt{3},$ so $\frac{c}{b}=\sqrt{3} \implies c=\sqrt{3}\,b.$

Squaring: $c^2=3b^2.$

But also $c^2=a^2+b^2.$ Hence $a^2+b^2=3b^2 \implies a^2=2b^2.$

Use length of latus rectum

For the hyperbola $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1,$ the length of the latus rectum is $\frac{2a^2}{b}.$

Given $\frac{2a^2}{b}=4\sqrt{3}.$

Using $a^2=2b^2$: $\frac{2(2b^2)}{b}=4\sqrt{3}$ $4b=4\sqrt{3}$ $b=\sqrt{3}.$

Therefore $a^2=2b^2=2\cdot 3=6,$ and $c=\sqrt{3}b=\sqrt{3}\cdot \sqrt{3}=3.$

Equation of the hyperbola

Thus the hyperbola is $\frac{y^2}{3}-\frac{x^2}{6}=1.$

Find $\alpha$ using the point $(\alpha,6)$

Since $(\alpha,6)$ lies on the hyperbola, $\frac{6^2}{3}-\frac{\alpha^2}{6}=1$ $\frac{36}{3}-\frac{\alpha^2}{6}=1$ $12-\frac{\alpha^2}{6}=1$ $\frac{\alpha^2}{6}=11$ $\alpha^2=66.$

Since $\alpha>0$, $\alpha=\sqrt{66}.$

Find product of focal distances

The foci are $F_1=(0,3), \qquad F_2=(0,-3).$

Let $P=(\alpha,6)=(\sqrt{66},6).$

Then $PF_1=\sqrt{(\sqrt{66})^2+(6-3)^2}=\sqrt{66+9}=\sqrt{75}=5\sqrt{3},$ $PF_2=\sqrt{(\sqrt{66})^2+(6+3)^2}=\sqrt{66+81}=\sqrt{147}=7\sqrt{3}.$

Hence $\beta=PF_1\cdot PF_2=(5\sqrt{3})(7\sqrt{3})=35\cdot 3=105.$

Compute $\alpha^2+\beta$

$\alpha^2+\beta=66+105=171.$

Compare with the stored answer

Derived answer is 171, which matches option B.