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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Mathematics
Chapter
Pair of Lines

JEE Advanced 2025 · Paper 2 · Mathematics · Pair of Lines

Let S denote the locus of the point of intersection of the pair of...

Let SS denote the locus of the point of intersection of the pair of lines 4x−3y=12α4 x-3 y=12 \alpha 4αx+3αy=124 \alpha x+3 \alpha y=12 where α\alpha varies over the set of non-zero real numbers. Let TT be the tangent to SS passing through the points (p,0)(p, 0) and (0,q),q>0(0, q), q>0, and parallel to the line 4x−32y=04 x-\frac{3}{\sqrt{2}} y=0.
Then the value of pqp q is

Options

  1. A−62-6 \sqrt{2}
  2. B−32-3 \sqrt{2}
  3. C−92-9 \sqrt{2}
  4. D−122-12 \sqrt{2}

Correct answer

A. −62-6 \sqrt{2}

Step-by-step solution

4x−3y=12α4x+3y=12α\begin{array}{l}4 x-3 y=12 \alpha \\ 4 x+3 y=\frac{12}{\alpha}\end{array}\right\} \rightarrow 16 x^2-9 y^2=144x29−y216=1→ Curve :S T:y=mx±9m2−16m=423y=42x3±32−163y=42x±12 as q>03y=42x+12p=−32q=4pq=−62\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 \\ & p=-\frac{3}{\sqrt{2}} & q=4 \\ & p q=-6 \sqrt{2}\end{aligned}$$
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