Let $$m_1$$ and $$m_2$$ be the slopes of the tangents drawn from the point $$P(4, 1)$$ to the hyperbola $$H : \frac{y^2}{25} - \frac{x^2}{16} = 1$$. If $$Q$$ is the point from which the tangents drawn to $$H$$ have slopes $$|m_1|$$ and $$|m_2|$$ and they make positive intercepts $$\alpha$$ and $$\beta$$ on the $$x-$$axis, then $$\frac{(PQ)^2}{\alpha\beta}$$ is equal to _____.

We are given the hyperbola $$H: \frac{y^2}{25} - \frac{x^2}{16} = 1$$ and the point $$P(4, 1)$$.

Here $$a^2 = 25$$ and $$b^2 = 16$$.

For a hyperbola $$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$, the tangent line $$y = mx + c$$ satisfies the condition $$c^2 = a^2 - b^2 m^2$$. Let us verify this.

Substituting $$y = mx + c$$ into $$\frac{y^2}{25} - \frac{x^2}{16} = 1$$:

$$\frac{(mx+c)^2}{25} - \frac{x^2}{16} = 1$$

$$16(mx+c)^2 - 25x^2 = 400$$

$$(16m^2 - 25)x^2 + 32mcx + 16c^2 - 400 = 0$$

For tangency, discriminant = 0: $$(32mc)^2 = 4(16m^2 - 25)(16c^2 - 400)$$

Expanding: $$1024m^2c^2 = 1024m^2c^2 - 25600m^2 - 1600c^2 + 40000$$

$$25600m^2 + 1600c^2 = 40000$$

$$16m^2 + c^2 = 25$$

So the tangent condition is $$c^2 = 25 - 16m^2$$.

The tangent through $$P(4,1)$$: since $$1 = 4m + c$$, we get $$c = 1 - 4m$$.

$$(1-4m)^2 = 25 - 16m^2$$

$$1 - 8m + 16m^2 = 25 - 16m^2$$

$$32m^2 - 8m - 24 = 0$$

$$4m^2 - m - 3 = 0$$

$$(4m + 3)(m - 1) = 0$$

So $$m_1 = 1$$ and $$m_2 = -\frac{3}{4}$$.

Therefore $$|m_1| = 1$$ and $$|m_2| = \frac{3}{4}$$.

Now we find the tangent lines with slopes $$|m_1| = 1$$ and $$|m_2| = \frac{3}{4}$$ that make positive x-intercepts.

For slope 1: $$c^2 = 25 - 16 = 9$$, so $$c = \pm 3$$. The x-intercept is $$-\frac{c}{m} = -c$$. For a positive intercept, we need $$c = -3$$. Tangent: $$y = x - 3$$, with $$\alpha = 3$$.

For slope $$\frac{3}{4}$$: $$c^2 = 25 - 16 \cdot \frac{9}{16} = 25 - 9 = 16$$, so $$c = \pm 4$$. The x-intercept is $$-\frac{c}{3/4} = -\frac{4c}{3}$$. For a positive intercept, we need $$c = -4$$. Tangent: $$y = \frac{3}{4}x - 4$$, with $$\beta = \frac{16}{3}$$.

Point $$Q$$ is the intersection of these two tangent lines:

$$x - 3 = \frac{3}{4}x - 4$$

$$\frac{x}{4} = -1 \Rightarrow x = -4, \quad y = -7$$

So $$Q = (-4, -7)$$.

$$(PQ)^2 = (4-(-4))^2 + (1-(-7))^2 = 64 + 64 = 128$$

$$\alpha\beta = 3 \times \frac{16}{3} = 16$$

$$\frac{(PQ)^2}{\alpha\beta} = \frac{128}{16} = 8$$

The answer is $$8$$.