Let S denote the locus of the point of intersection of the pair of lines $4x - 3y = 12\alpha$, $4\alpha x + 3\alpha y = 12$, where $\alpha$ varies over the set of non-zero real numbers. Let T be the tangent to S passing through the points $(p, 0)$ and $(0, q)$, $q > 0$, and parallel to the line $4x - \frac{3}{\sqrt{2}}y = 0$. Then the value of $pq$ is (1) $-6\sqrt{2}$ (2) $-3\sqrt{2}$ (3) $-9\sqrt{2}$ (4) $-12\sqrt{2}$

$L_1: \ 4x - 3y = 12\alpha$

$L_2: \ 4\alpha x + 3\alpha y = 12, \qquad \alpha \neq 0$

$S = \text{locus of } L_1 \cap L_2 \text{ as } \alpha \text{ varies}$

$T \parallel 4x - \dfrac{3}{\sqrt2}y = 0, \text{ through } (p,0) \text{ and } (0,q)$

$\text{Find } pq$

$L_1: \ \alpha = \dfrac{4x - 3y}{12}$

$\text{put in } L_2: \ 4\alpha x + 3\alpha y = 12$

$\dfrac{4x-3y}{3}\,x + \dfrac{4x-3y}{4}\,y = 12$

$4x(4x-3y) + 3y(4x-3y) = 144$

$(4x - 3y)(4x + 3y) = 144$

$16x^2 - 9y^2 = 144$

$\Rightarrow \dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$

$1.\ \text{locus } S: \dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$

$2.\ \text{slope of } T = \dfrac{4\sqrt2}{3}$

$3.\ y = mx \pm \sqrt{a^2 m^2 - b^2} \Rightarrow pq$

$4x - \dfrac{3}{\sqrt2}y = 0 \Rightarrow y = \dfrac{4\sqrt2}{3}x \Rightarrow m = \dfrac{4\sqrt2}{3}$

$y = mx \pm \sqrt{a^2 m^2 - b^2}$

$a^2 m^2 = 9 \cdot \dfrac{32}{9} = 32$

$\sqrt{a^2 m^2 - b^2} = \sqrt{32 - 16} = 4$

$y = \dfrac{4\sqrt2}{3}x \pm 4$

$q > 0 \Rightarrow T: \ y = \dfrac{4\sqrt2}{3}x + 4$

$y = 0 \Rightarrow p = -\dfrac{3}{\sqrt2}$

$x = 0 \Rightarrow q = 4$

$pq = \left(-\dfrac{3}{\sqrt2}\right)(4) = -\dfrac{12}{\sqrt2} = -6\sqrt2$

$\Rightarrow \text{option (1)}$