Product of Tangent Intercepts on Locus of Intersecting Lines

To find the locus $S$, we consider the given system of lines for $\alpha \neq 0$: $4x - 3y = 12\alpha \quad \text{--- (1)}$ $4\alpha x + 3\alpha y = 12 \implies 4x + 3y = \frac{12}{\alpha} \quad \text{--- (2)}$

Multiplying equation (1) and equation (2), we eliminate the parameter $\alpha$: $(4x - 3y)(4x + 3y) = (12\alpha) \left(\frac{12}{\alpha}\right)$ $16x^2 - 9y^2 = 144$

Dividing both sides by $144$, we obtain the standard equation of a hyperbola: $\frac{x^2}{9} - \frac{y^2}{16} = 1$ Here, $a^2 = 9$ and $b^2 = 16$.

The tangent line $T$ is parallel to the line $4x - \frac{3}{\sqrt{2}}y = 0$, so the slope $m$ of the tangent is: $m = \frac{4}{\frac{3}{\sqrt{2}}} = \frac{4\sqrt{2}}{3}$

The equation of a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ in slope form is: $y = mx \pm \sqrt{a^2m^2 - b^2}$

Substituting $a^2 = 9$, $b^2 = 16$, and $m = \frac{4\sqrt{2}}{3}$: $a^2m^2 - b^2 = 9\left(\frac{4\sqrt{2}}{3}\right)^2 - 16 = 9\left(\frac{32}{9}\right) - 16 = 32 - 16 = 16$

Thus, the tangent lines are: $y = \frac{4\sqrt{2}}{3}x \pm 4$

Since the tangent line $T$ passes through $(0, q)$ with $q > 0$, the $y$-intercept must be positive: $y = \frac{4\sqrt{2}}{3}x + 4$

From this equation:

1. The $y$-intercept $(0, q)$ gives: $q = 4$
2. The $x$-intercept $(p, 0)$ gives: $0 = \frac{4\sqrt{2}}{3}p + 4 \implies p = -\frac{3}{\sqrt{2}}$

Therefore, the value of the product $pq$ is: $pq = \left(-\frac{3}{\sqrt{2}}\right)(4) = -\frac{12}{\sqrt{2}} = -6\sqrt{2}$

Hence, the correct option is (A).