JEE Advance - Mathematics (2025 - Paper 2 Online - No. 4)
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 :
Explanation
$\left.\begin{array}{c}4 x-3 y=12 \alpha \\ 4 x+3 y=\frac{12}{\alpha}\end{array}\right\} \rightarrow 16 x^2-9 y^2=144$
$\begin{aligned} & \frac{x^2}{9}-\frac{y^2}{16}=1 \rightarrow \text { Curve :S } \\ & T: y=m x \pm \sqrt{9 m^2-16} \\ & m=\frac{4 \sqrt{2}}{3} \\ & y=\frac{4 \sqrt{2} x}{3} \pm \sqrt{32-16} \\ & 3 y=4 \sqrt{2} x \pm 12 \\ & \text { as } q>0 \\ & 3 y=4 \sqrt{2} x+12\end{aligned}$
$$ \mathrm{p}=-\frac{3}{\sqrt{2}} \quad \& \quad \mathrm{q}=4 $$
$$ \mathrm{pq}=-6 \sqrt{2} $$
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