Let S denote the locus of the point of inters | vedclass

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(A) Given the lines: $4x - 3y = 12\alpha$ and $4\alpha x + 3\alpha y = 12$.Multiplying the two equations: $(4x - 3y)(4\alpha x + 3\alpha y) = (12\alph

Vedclass · Accepted Answer

$-6$

Vedclass · Answer

(A) Given the lines: $4x - 3y = 12\alpha$ and $4\alpha x + 3\alpha y = 12$.Multiplying the two equations: $(4x - 3y)(4\alpha x + 3\alpha y) = (12\alpha)(12) \implies 16\alpha x^2 + 12\alpha xy - 12\alpha xy - 9\alpha y^2 = 144\alpha$.Dividing by $\alpha$ (since $\alpha 
eq 0$): $16x^2 - 9y^2 = 144 \implies \frac{x^2}{9} - \frac{y^2}{16} = 1$. This is a hyperbola $S$.The slope of the line $4x - \frac{3}{\sqrt{2}}y = 0$ is $m = \frac{4}{3/\sqrt{2}} = \frac{4\sqrt{2}}{3}$.The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.Here $a^2 = 9, b^2 = 16$, so $y = \frac{4\sqrt{2}}{3}x \pm \sqrt{9(\frac{32}{9}) - 16} = \frac{4\sqrt{2}}{3}x \pm \sqrt{32 - 16} = \frac{4\sqrt{2}}{3}x \pm 4$.The tangent passes through $(p, 0)$ and $(0, q)$. The intercept form is $\frac{x}{p} + \frac{y}{q} = 1 \implies y = -\frac{q}{p}x + q$.Comparing $y = \frac{4\sqrt{2}}{3}x + 4$ with $y = -\frac{q}{p}x + q$, we get $q = 4$ and $-\frac{q}{p} = \frac{4\sqrt{2}}{3}$.$-\frac{4}{p} = \frac{4\sqrt{2}}{3} \implies p = -\frac{3}{\sqrt{2}}$.Thus, $pq = (-\frac{3}{\sqrt{2}})(4) = -\frac{12}{\sqrt{2}} = -6\sqrt{2}$.

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