Let the product of the focal distances of the point $$P(4, 2\sqrt{3})$$ on the hyperbola $$H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then $$p^2 + q^2$$ is equal to __________.

The standard form of a rectangular-axis hyperbola with centre at the origin and transverse axis along the x-axis is $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1$$.

The foci are $$(\pm c,0)$$ where $$c^{2}=a^{2}+b^{2} \; -(1)$$.

The given point $$P(4,\,2\sqrt{3})$$ lies on the hyperbola. Substituting in the equation of the curve,

$$\frac{4^{2}}{a^{2}}-\frac{(2\sqrt3)^{2}}{b^{2}} = 1 \;\Longrightarrow\; \frac{16}{a^{2}}-\frac{12}{b^{2}} = 1 \; -(2)$$

Let the focal distances of $$P$$ be $$PF_{1}$$ and $$PF_{2}$$. Because the foci are $$(\pm c,0)$$,

$$PF_{1}^{\,2} = (4-c)^{2} + (2\sqrt3)^{2} = (4-c)^{2} + 12 = c^{2}-8c+28 \; -(3)$$
$$PF_{2}^{\,2} = (4+c)^{2} + 12 = c^{2}+8c+28 \; -(4)$$

The product of the focal distances is given to be 32:

$$PF_{1}\,PF_{2}=32 \;\Longrightarrow\; PF_{1}^{\,2}\,PF_{2}^{\,2}=32^{2}=1024$$

Using $$(3)$$ and $$(4)$$:

$$\bigl(c^{2}-8c+28\bigr)\bigl(c^{2}+8c+28\bigr)=1024$$

Employ $$(A-B)(A+B)=A^{2}-B^{2}$$ with $$A=c^{2}+28,\;B=8c$$:

$$\bigl(c^{2}+28\bigr)^{2}-(8c)^{2}=1024$$
$$c^{4}+56c^{2}+784-64c^{2}=1024$$
$$c^{4}-8c^{2}-240=0$$

Put $$z=c^{2}$$ to obtain $$z^{2}-8z-240=0$$. Solving the quadratic,

$$z = \frac{8\pm\sqrt{8^{2}+4\cdot240}}{2} = \frac{8\pm32}{2}$$

The positive root is $$z=20$$, so

$$c^{2}=20,\qquad c = 2\sqrt5 \; -(5)$$

From $$(1)$$, $$a^{2}+b^{2}=20 \; -(6)$$. From $$(2)$$, $$\dfrac{16}{a^{2}}-\dfrac{12}{b^{2}}=1 \; -(7)$$.

Let $$a^{2}=A,\;b^{2}=B$$. Then $$(6)$$ gives $$A+B=20 \; -(8)$$, and $$(7)$$ gives $$\dfrac{16}{A}-\dfrac{12}{B}=1 \; -(9)$$.

Re-write $$(9)$$: $$\dfrac{16}{A} = 1+\dfrac{12}{B} = \dfrac{B+12}{B}$$,
so $$A = \frac{16B}{B+12} \; -(10)$$.

Substitute $$(10)$$ into $$(8)$$:

$$\frac{16B}{B+12}+B = 20$$
$$16B + B(B+12) = 20(B+12)$$
$$B^{2}+28B = 20B + 240$$
$$B^{2}+8B-240 = 0$$

Solving, $$B = \frac{-8\pm\sqrt{8^{2}+4\cdot240}}{2} = \frac{-8\pm32}{2}$$.

Select the positive value: $$B = 12 \;\Longrightarrow\; b^{2}=12$$.

Using $$(8)$$, $$a^{2}=20-12 = 8$$, hence $$a = \sqrt8 = 2\sqrt2$$.

Conjugate axis length
The conjugate axis has length $$p = 2b = 2\sqrt{b^{2}} = 2\sqrt{12}=4\sqrt3$$,
so $$p^{2} = (4\sqrt3)^{2}=48$$.

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

$$q = \frac{2\cdot12}{2\sqrt2} = \frac{24}{2\sqrt2}= \frac{12}{\sqrt2}=6\sqrt2$$,
so $$q^{2} = (6\sqrt2)^{2}=72$$.

Finally,

$$p^{2}+q^{2}=48+72=120$$.

Hence, the required value is $$\boxed{120}$$.