Question 4

First, find the locus $S$ by eliminating $\alpha$: From the first line, $\alpha = \frac{4x - 3y}{12}$. Substitute $\alpha$ into the second line: $\left(\frac{4x - 3y}{12}\right)(4x + 3y) = 12$ $(4x - 3y)(4x + 3y) = 144 \Rightarrow 16x^2 - 9y^2 = 144$ Dividing by 144: $\frac{x^2}{9} - \frac{y^2}{16} = 1$. This is a hyperbola with $a^2 = 9$ and $b^2 = 16$. The tangent $T$ is parallel to $4x - \frac{3}{\sqrt{2}}y = 0$, so its slope $m = \frac{4}{3/\sqrt{2}} = \frac{4\sqrt{2}}{3}$. The equation of a tangent with slope $m$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $y = mx \pm \sqrt{a^2 m^2 - b^2}$. $y = \frac{4\sqrt{2}}{3}x \pm \sqrt{9 \left(\frac{32}{9}\right) - 16} = \frac{4\sqrt{2}}{3}x \pm \sqrt{32 - 16} = \frac{4\sqrt{2}}{3}x \pm 4$ Since the tangent passes through $(0, q)$ with $q > 0$, we choose the positive intercept: $y = \frac{4\sqrt{2}}{3}x + 4$. Thus, $q = 4$. For the x-intercept $(p, 0)$, set $y = 0$: $0 = \frac{4\sqrt{2}}{3}p + 4 \Rightarrow p = -4 \cdot \frac{3}{4\sqrt{2}} = -\frac{3}{\sqrt{2}}$ The value of $pq = (-\frac{3}{\sqrt{2}})(4) = -\frac{12}{\sqrt{2}} = -6\sqrt{2}$.